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Marine Biological Laboratory Library Woods Hole, Mass. Presented Ly John Wiley and Sons, Inc. July 27, 1956 3B3^^^^^^^^^^^^^BE Elements of Statistics WILEY PUBLICATIONS IN STATISTICS Walter A. Shcwkart, Editor Mathematical Statistics HANSEN, HURWITZ, and MADOW • Sample Survey Methods and Theory, Volume 11 DOOB • Stochastic Processes RAO • Advanced Statistical Methods in Biometric Research KEMPTHORNE • The Design and Analysis of Experiments DWYER • Linear Computations FISHER • Contributions to Mathematical Statistics WALD • Statistical Decision Functions FELLER • An Introduction to Probability Theory and Its Applications, Volume I WALD • Sequential Analysis HOEL • Introduction to Mathematical Statistics, Second Edition (in press) Applied Statistics BENNETT and FRANKLIN • Statistical Analysis in Chemistry and the Chemical Industry (in press) FRYER • Elements of Statistics COCHRAN • Sampling Techniques WOLD and JUREEN • Demand Analysis HANSEN, HURWITZ, and MADOW • Sample Survey Methods and Theory, Volume I CLARK • An Introduction to Statistics TIPPETT • The Methods of Statistics, Fourth EJition ROMIG • 50-100 Binomial Tables GOULDEN • Methods of Statistical Analysis, Second Edition HALD • Statistical Theory with Engineering Applications HALD • Statistical Tables and Formulas YOUDEN • Statistical Methods for Chemists MUDGETT • Index Numbers TIPPETT • Technological Applications of Statistics DEMING • Some Theory of Sampling COCHRAN and COX • Experimental Designs RICE • Control Charts DODGE and ROMIG • Sampling Inspection Tables Relatcil Books of Interest to Statxsticxans ALLEN and ELY • International Trade Statistics HAUSER and LEONARD • Government Statistics for Business Use Elements of Statistics H. C. FRYER Professor of Matlicmatics Statistician, Agricultural Experiment Station Kansas State College //A V. New York • John Wiley & Sons, Inc. London • Chapman & Hall, Limited Copyright, 1954 BY John Wiley & Sons, Inc. All Rights Reserved This book or any part thereof must not be reproduced in any form u'ithout the written permission of the publisher. Copyright, Canada, 1954, International Copyright, 1954 John Wiley & Sons, Inc., Proprietor All Foreign Rights Reserved Reproduction in whole or in pari forbidden. Library of Congress Catalog Card Number: 54-5517 PRINTED IN the UNITED STATES OF AMERICA Prefi ace THIS BOOK HAS BEEN DEVELOPED TO PROVIDE THE BASIS OF AN introductory course in probability and statistics for the college and university student. It includes material which has been tried out in many classes and by several instructors for almost a decade. The instructor using the book as a text or the student interested in the subject will find that college algebra is a necessary and suffi- cient prerequisite for this course, which aims to teach modern but ele- mentary ideas, methods of reasoning, and methods of analysis funda- mental but not peculiar to any particular specialized field. Once the student has acquired a background of elementary methods, prob- ability, and frequency distributions, he can be taught some of the simpler sampling statistics in common use today. Thus, he may learn their importance as well as their application. The serious stu- dent will find included in this book problems to provoke thought and provide practice in statistical methods and reasoning. Some colleges and universities offer statistics courses — often with graduate credit — in which the elementary concepts and methods are not assumed to be known and hence are taught during the first part of the course. It seems to me that one general course in probability and statistics, with emphasis on statistical reasoning and modern methods, helps to avoid useless duplication of mstruction. It also leaves time in subsequent courses to do more advanced work in spe- cialized fields. Such an introductory course also is rapidly becoming a necessary part of a student's education even if he does not use sta- tistics directly in his specialized field. It is helpful to the students during the studies of sampling to pro- vide them with some mathematical models of populations so that they can obtain sampling experiences which — for a whole class — empirically verify for them the sampling distributions given in some of the tables which they will be using. It has been my experience that most stu- dents need this sort of empirical evidence before they really under- stand the nature and the use of sampling distributions. Numbers, and other symbols, written on plastic discs can be made to correspond closely to normal, non-normal, and binomial populations which are vi PREFACE met in actual practice. These populations, if properly employed, will enable the student to understand the more common sampling distri- butions rather well despite a lack of familiarity with mathematical statistics. Some of the problems in this book assume that such popu- lations are available to the students. It is desirable to have calculating machines available so that the students can learn what operations can be performed on them and can solve some of their problems more efficiently. However, I do feel that the acquisition of routine computational skills is not worthy of much college credit; hence, whenever there are heavy computations in a problem in this book, the necessary computations are usually given with the problem. For example, w(X) and -(X^) are given for most problems with even a moderate amount of computation and asking for the mean and the standard deviation. It has been my experience that it takes most of the equivalent of a three-semester-hour course to equip the student with the ideas and methods he needs before he can solve even the most elementary sam- pling problems in any particular field. For that reason we offer at Kansas State College a two-hour course in which the rest of the work on sampling contained herein can be given, and some applications to the students' fields of interest can be considered. It is a pleasure to acknowledge the assistance given me by my col- league J. I. Northam, who pointed out errors in previous lithoprinted versions and made suggestions regarding the way the material should be presented. This book also has derived considerable benefit from the reviews made available to me during the last revisions of the manuscript. Obviously, the responsibility for all remaining short- comings of the book is solely mine. H. C. Fryer Kansas State College Manhattan, Kansas December, 1953 Contents 1. History and Introduction 1 1.1 History 1 1.2 Some of the purposes of statistical reasoning 5 2. The Summarization of Sets of Data Involving One Type of Measurement 9 2.1 The arithmetic mean and the standard deviation 12 2.2 The average (or mean) deviation 18 2.3 Other averages 19 2.4 Frequency distributions 23 2.5 Calculation of the arithmetic mean and the standard devia- tion from frequency distribution tables 28 2.6 Percentiles, deciles, and quartiles 35 2.7 Coefficient of variation 40 2.8 Some of the problems created when only a sample of a popu- lation is available for statistical study 43 Review problems 45 3. Elementary Probability 49 3.1 The determination of probabilities 50 3.2 Permutations and combinations 63 3.3 Repeated trials under specified conditions 67 3.4 Mathematical expectation 70 Review problems 73 4. The Binomial and Normal Frequency Distributions .... 76 4.1 The binomial frequency distribution 78 4.2 The normal frequency distribution 85 4.3 Determination of the proportion of a normal population of measurements included between any specified limits ... 95 4.4 Use of the normal distribution to approximate probabilities for a binomial frequency distribution 101 4.5 Studying the normality of a frequency distribution by recti- fying the r.c.j. curve 102 Review problems 109 5. Sampling from Binomial Populations 113 5.1 Obtaining the sample 117 5.2 Calculation of point and interval estimates of p for a binomial population 121 5.3 Testing predetermined hypotheses regarding p 130 vii 71075 viii CONTENTS 5.4 Testing the hypothesis that two random samples came from the same binomial population 136 5.5 The x"-test when more than one degree of freedom is re- quired 140 5.6 Control charts 146 Review problems 150 6. Introductory Sampling Theory for a Normal Population In- volving Only One Variable 153 6.1 Obtaining the sample 153 6.2 The statistical distribution of sample means, Xi, drawn from a normal population 154 6.3 Estimation of the unknown mean and variance of a popula- tion from the information contained in a sample 159 6.4 A statistical test of a hypothesis that a given sample came from a normal population \\dth a specified mean 170 6.5 A statistical test of the hypothesis that two samples of ob- servations have been drawn from the same normal popula- tion of numerical measurements 176 6.6 Use of the sample range instead of the standard deviation in certain tests of statistical hypotheses 182 6.7 The central limit theorem and non-normal populations . . . 185 Review problems 190 7. Linear Regression and Correlation 192 7.1 Scatter diagrams and types of trend hues 194 7.2 A method for determining linear regression (or trend) lines . 198 7.3 Measurement of the variation about a linear trend hne deter- mined by the method of least squares 208 7.4 Coefficients of linear correlation 216 7.5 Rank correlation 226 Reiiew problems 230 Tables 243 Index 259 CHAPTER 1 History and Introduction 1.1 HISTORY The word "statistics," the associated mathematical analyses, and the general process of statistical reasoning appear to have begun their evolution around the time of Aristotle. This evolution can be de- scribed in terms of the following four phases, some of which occurred simultaneously among different groups of persons: (1.11) An early, highly philosophical, study of "matters of state" which did little more for the statistical science used today than help to suggest its name. (1.12) A semi-numerical and strongly sociological stage typified by the mathematical and philosophical study of large groups of numerical measurements bearing on health, insurance, foreign and domestic trade, and political matters. (1.13) The development of the mathematical theory of probabil- ity starting in the sixteenth century with mathematical attacks on the various problems associated with games of chance. (1.14) The current phase, starting late in the nineteenth century, during which phases (1.12) and (1.13) were combined, improved, and extended to produce a branch of mathematics which can handle a wide variety of problems pertaining to the drawing of valid and use- ful inferences from relatively small groups of numerical measure- ments. During Aristotle's time interest developed in comparative descrip- tions of states. Aristotle is reported * to have written at least one hundred and fifty-eight descriptions of states, covering their histories, public administrations, arts, sciences, and religious practices. It was customary to refer to such compositions as treatises on "matters of state." That apparently is an important part of the origin of the * Harald L. Westergaard (1942), Contributions to the History of Statistics, King. 2 HISTORY AND INTRODUCTION Ch. 1 term "statistics," although the name itself was coined many years after Aristotle's death. For quite a long time after Aristotle a weak interest was main- tained in descriptions of states partly by the intellectuals who enjoyed that pastime and partly by the rulers of the various states through their natural desire to know how many subjects they ruled, and to ascertain the wealth within their realms. Hence it is probable that some sort of crude census taking was attempted. During the seventeenth and eighteenth centuries sufficient interest was generated in the study of the political, sociological, and economic features of states that societies developed for that purpose. In Ger- many this line of intellectual effort caused the development of the Staatenkunde, a name which appears to have led rather directly to the actual coining of the term "statistics." However, the Germans remained content to pursue the philosophical aspects of "matters of state"; hence the Staatenkunde never did become either very mathe- matical in character, or very useful. It merely typifies the last stages of the purely philosophical phase of the development of the science of statistical analysis, and points out its socio-political ancestry. Another, and more fruitful, step in the evolution of the present-day type of statistical reasoning originated in England under the leader- ship of John Graunt. This was a semi-mathematical study of vital statistics, insurance, and economic statistics which came to be known as "Political Arithmetic." Epidemic diseases periodically decimated the populations of European nations; problems of agricultural pro- duction, foreign trade, and public administration became too com- plex to be handled without some form of numerical measurement and an objective means of interpretation of such measurements. Hence there was a natural interest in numbers of births and deaths, in esti- mates of the populations in various areas, in figures on agricultural production and foreign trade, and in methods for administering in- surance against the economic situations created by death and dis- ability. Public interest in specific measurements of populations and of re- sources was heightened by the constant danger of war with a neigh- boring state, and by the advent of an industrial revolution during the eighteenth century. It was the objective of the political arith- meticians to help with the collection and interpretation of data perti- nent to the economic, sociological, and political problems which were becoming increasingly important and numerous. They devised meth- ods for estimating the numbers of persons residing in certain political Sec. 1.1 HISTORY 3 units, and methods for summarizing groups of data. Their efforts to apply mathematical analysis to such problems helped to lay the foundation for the statistical methods now in use. A third step in the evolution of statistical analysis and reasoning came in the development of the mathematical theory of probability, without which statistical reasoning could never have attained its present reliability and usefulness. Games of chance were especially popular among the well-to-do of the sixteenth and seventeenth cen- turies; and many problems involving probability were presented to the mathematicians of the day for solution. For example, an Italian nobleman asked Galileo to explain the following facts: If three dice are thrown, the numbers 9 and 10 can each be obtained from six different combinations of the numbers on the faces of the dice; but it has been found from experience that a sum of 10 appears more frequently than a sum of 9. Why so? By an enumeration of all the physically different ways that three dice can produce sums of 9 or 10, Galileo was able to answer this question clearly and convinc- ingly. His answer appears to be the first published application of the theory of probability.* Other prominent mathematicians such as Pascal, Fermat, James and Daniel Bernoulli, de Moivre, Laplace, Gauss, Simpson, Lagrange, Hermite, and Legendre developed many important theorems and methods of attacking problems involving chance events, and they passed this information on for later use by mathematical statisticians. During the last quarter of the nineteenth century, Sir Francis Galton took the lead in the development of the ideas of regression and correlation when two (or more) measurements are made simul- taneously on each member of a group of objects. He appears to have built his ideas around problems in genetics. Karl Pearson and C. Spearman extended this theory and applied it to studies in the social sciences, especially psychology. Karl Pearson and others also had * The nature of Galileo's solution is as follows. A sum of 9 can be obtained from any of the following combinations of numbers on three dice: 1, 2, 6; 1, 3, 5; 1, 4, 4; 2, 2, 5; 2, 3, 4; or 3, 3, 3. A sum of 10 is obtained from any of these: 1, 3, 6; 1, 4, 5; 2, 2, 6; 2, 3, 5; 2, 4, 4; or 3, 3, 4. There are six different com- binations giving each of the sums 9 and 10; b^it the different combinations do not occur equally frequently. For example, the combination 3, 3, 3 can be thrown but one way whereas the combination 3, 3, 4 can occur on any of three different throws, and hence would tend to appear three times as often as 3, 3, 3. As a matter of fact, a 9 can be thrown in twenty-five different ways, a 10 in twenty-seven different ways, which is the reason that the 10 appears more frequently in games than the 9. ■^'^'^^W A i 4 HISTORY AND INTRODUCTION Ch. 1 begun to study the effects of sampling errors on conclusions drawn from samples. By the end of the nineteenth century the Staatenkunde had ceased to exist, and "Political Arithmetic" had died in name but had devel- oped into a science of statistical analysis, with emphasis on socio- logical and economic applications. The theory of mathematical probability had grown extensively as a branch of pure mathematics, and also was beginning to be associated with applied statistics. Thus the groundwork was laid for the present phase in the evolution of statistical theory and methods. In 1908 William Seely Gosset, who wrote under the pseudonym "Student," published an article in the journal Biometrika which was later to typify the opening of a new era in the statistical analysis and interpretation of sampling data. From 1899 until his death in 1937, Gosset worked for the brewing firm, Messrs. Guiness. His associa- tions with this firm led him into a variety of experiences and sug- gested uses for statistical methods which are typical of several of the present-day applications of statistics. Messrs. Guiness were interested in barley, not just any barley, but in those varieties, growing conditions, and practices which would produce the best barley for breweries to use. These circumstances brought Gosset into contact with agricultural experimentation aimed at the improvement of crops and of agricultural practices. Moreover, Messrs. Guiness did not wish to subsidize the raising of large crops purely for the sake of scientific experimentation; they were a com- mercial firm which wanted to show a profit from their enterprises. That fact, plus the shortage of tillable land in Ireland and England, made Gosset well aware of the importance of small samples and of methods for deriving reliable information from such samples. Furthermore, a large brewery conducts many chemical analyses, and hence needs to take proper account of errors of measurement. And, finally, the firm with which Student was associated was con- fronted with problems concerning industrial statistics: production and marketing analyses, price analyses, and methods for controlling the quality of the products which it manufactures for market. Thus Stu- dent came into contact with a wide variety of agricultural, economic, and industrial problems which would require some form of statistical study. Moreover, those problems had to be solved for a commercial firm, a situation which demanded efficiency and reliability with a minimum cost consistent with these qualities. The twentieth cen- tury renaissance of statistical theory and methods appears to be Sec. 1.2 PURPOSES OF STATISTICAL REASONING 5 based on that attitude toward the purposes of statistical analysis. Unfortunately, the statistical ideas and procedures which Student introduced in 1908 did not become familiar to persons outside his own firm for nearly a decade, at which time R. A. Fisher and his col- leagues in England began to extend and to popularize the theory of small samples and its applications. The theory of statistics was developed extensively by Jerzy Neyman and Karl Pearson's son, Egon S. Pearson. They placed special emphasis on rigor in statisti- cal reasoning and led the way by publishing many papers in this field. Many others have followed their lead since their papers began to appear. The results of this research are being applied in many fields, such as biology, the physical sciences, industry, economics, sociology, medicine, education, and psychology. 1.2 SOME OF THE PURPOSES OF STATISTICAL REASONING Early in his history man displayed a desire to take numerical measurements of the various phenomena involving himself and his environment. At first, those measures probably consisted of simple counts, or of crude measures of weight, volume, length, and area. At present many instruments are available for the precise measurement of those features of man's self and environment which interest him. He constantly is taking groups of numerical measurements because such a procedure can furnish a relatively precise and standard means of obtaining the information desired, of using it efficiently, and of transmitting that information to others. The general purpose of statistical analysis is to assist in the collection and the interpretation of sets of numerical measurements which supposedly have been taken for some useful purpose. Once it is decided that a particular phenomenon should be meas- ured numerically, one of two general classes of data is then ob- tained. It may be that it was both possible and practicable to secure every measurement of that particular kind which exists or could be obtained under the particular circumstances. Such a complete record is one type of statistical population of numerical measurements. An example is the record of the ages of all the legal residents of the state of Kansas on April 1, 1950, as contained in the official United States Census for that date. Another example is a list of the I.Q.'s of all the students entering a particular university in a given year. 6 HISTORY AND INTRODUCTION Ch. 1 However, it is more commonly true that it is impossible, or un- wise, to collect a whole population of numerical measurements. In that event we obtain but a portion of a population for actual analysis, and attempt to draw from it useful conclusions about the popula- tion which was merely sampled. If the sample is to be useful it must be adequately representative of the population; that is, it should faithfully reflect the important features of the population. In the event that the whole population of data is available for analysis, the purpose of statistical analysis is to reduce what is a relatively large bulk of numbers to a comprehensible form by means of graphs and tables and/or by calculating a few figures which con- tain most of the important information theoretically available in the original mass of data. For example, the ACE scores at the beginning of Chapter 2 are numerical measurements which the college took in the belief that they would be of value to the student and to the school, perhaps by helping to determine what profession the student should prepare to enter. Obviously those data are so bulky that they demand some sort of condensation. It is worth noting at this point that even though the necessity to analyze whole populations of data is a rare circumstance, it is not logical to study the statistical analysis of samples without some ade- quate knowledge of the statistical features of the populations from which the samples are taken. Fortunately a considerable amount of useful statistical analysis can be learned and appreciated without studying more than two general types of populations. Whenever we attempt to base conclusions concerning a statistical population of numerical measurements upon relatively few observa- tions (a sample) from that population, we face two important gen- eral questions, (a) How shall the sample be taken so as to maxi- mize its chance of being representative of that population? (6) Having obtained some numerical observations from the population with question a in mind, how do we draw valid conclusions from the sample? As an illustration, consider the following sampling prob- lem. Suppose that a highway commission is considering the pur- chase of some cement for highway construction, and that two com- panies are offering their products for purchase. The commission wishes to compare the seven-day tensile strengths of the two cements before letting the contract. Obviously they must resort to sampling because they can test only a tiny portion of each company's total output of a particular sort of cement. It will be supposed, for pur- poses of illustration, that it has been decided that ten of the stand- Sec. 1.2 PURPOSES OF STATISTICAL REASONING 7 ard laboratory specimens will be tested from each company's prod- uct. The test of each specimen will produce an "observation" from the population of all such tensile strengths possible from that com- pany's cement. All told, there will be twenty samples taken, ten from each company. Would it be satisfactory to inform each company of the plans for testing the cements and ask each company to provide ten specimens of concrete for testing? Or, would it be better to go into the open market and purchase a sack of each company's cement from each of ten stores and have a laboratory uniformly make up the ten testing specimens? Rather clearly the latter method would be much more likely to produce specimens which were representative of the respective strengths of those concretes at seven days of age. One of the purposes of statistical theory is to devise methods for taking samples in such a way that they do yield essentially the same information as is contained in the population which was sampled. For the most part, that phase of statistics lies beyond the scope of this book; hence no attempt will be made to do more than to remind the reader of a few commonsense considerations from time to time. Suppose now that the ten sample specimens of concrete have been tested for tensile strength at seven days of age, with the following results : Cement Seven-Day Tensile Strength (Ib./sq. in.) of Concrete No. 1 425, 410, 425, 460, 430, 445, 445, 415, 450, and 440 No. 2 420, 450, 405, 400, 400, 415, 435, 425, 400, and 430 How do we decide from such evidence whether one concrete will, as a rule, excel the other in tensile strength ; or if either or both conform to pre-assigned standards for such building materials? Casual ob- servation indicates that cement No. 1 tends to produce greater ten- sile strength in its concrete than No. 2; but there are several speci- mens from cement No. 2 that produced greater strength than certain of the specimens from No. 1 cement. For example, five of the No. 1 specimens had tensile strengths at or below 430 pounds per square inch, and two of the specimens of cement No. 2 had strengths greater than 430 pounds per square inch. Without doubt, then, some batches of cement No. 2 are better than some batches of cement No. 1. Such a situation is met quite frequently in sampling studies; only rarely do progressive improvements in methods or materials come on such a large scale that all previous methods or materials are excelled with- out exception. What is needed — and now available to a highly use- 8 HISTORY AND INTRODUCTION Ch. 1 ful degree — is a method of reasoning which enables us to induce from relatively few samples useful information regarding the population sampled. Inductive reasoning based on evidence obtained from samples necessarily runs some risk of error; but as long as the extent of this: risk can be measured, the process offers real hope for useful appli- cation. Much of the recent research in mathematical statistics has been devoted to the development of methods of reasoning based on sampling observations. The reader should not feel from the preceding remarks that sam- pling is useful only in scientific research, because everyone is con- stantly being confronted with sampling studies of one sort or an- other. Radio advertising is quite full of alleged sampling investiga- tions during which various products presumably have been tested and shown to be superior. Life insurance premiums are based on samples of mortality rates among insurable persons. Public opinion polls, economic polls, and the like, often reported in the newspapers, are attempts to reason from a sample to conclusions about a whole population of possible responses to one or more questions. Persons who have visited other parts of the world return and, upon the basis of relatively small samples, attempt to say how whole nations or societies are reacting to certain world events. The reader un- doubtedly can think of many other examples of sampling followed by more or less valid applications of either inductive or deductive reasoning, or what might be better described in this instance as statistical inference. In closing these introductory remarks, it seems fair to warn the student that, as in many other lines of thought, he cannot immedi- ately jump into interesting applications of statistical methods and reasoning. He must first learn some fundamental principles and some statistical tools with which to work, a process which necessarily occupies most of the time in a first course in statistics. There is, however, nothing to prevent him from reading the rest of the book for himself, and from taking other courses in statistics. CHAPTER 2 The Summarization of Sets of Data Involving One Type of Measurement Whenever a statistical investigation is to be made, two initial steps must be taken: (a) A group of objects (persons, plants, bolts, or any- thing capable of being measured) is specified as the subject to be studied. (6) A decision is made regarding the feature of these objects that is to be measured numerically or by some qualitative designation. Such a set of measurements is called a population if it includes every member of the group to be defined in a. For example, suppose that an economist proposes to study the net cash incomes of beef-cattle ranchers in Kansas during the decade from January 1, 1944, to January 1, 1954. It would be necessary first to define the group of ranchers to be included in this study. How many beef animals must he raise? Must the raising of beef cattle be his major source of income according to some standard? Are absentee owners included? There are many other matters which would have to be considered. When a specific group of Kansas ranchers has been defined, part a above has been completed. Next it is necessary to decide upon the specific meaning of the term, net cash income. Is the measurement to be on a per-animal basis, or the total for the ranch regardless of its size? Is any ad- justment to be made for inflation, cost-of-living indexes, and the like? When net cash income has been defined specifically, part b listed above has been completed, and the population is defined. In some situations it is feasible to obtain every possible one of the measurements in a population, as would be the case if every beef-cattle rancher in the group discussed above were to be inter- viewed and his net cash income determined according to the defini- tion adopted by the investigator. Under these circumstances, the purpose of statistical analysis is to summarize the information in the 9 10 SUMMARIZATION OF DATA Ch. 2 data as clearly and as concisely as is possible. A statistical descrip- tion of a population will be found to be important also when the population is to be studied by means of samples rather than in its entirety. There are various widely used methods of accomplishing such a purpose. The choice of a method depends upon what is to be learned from the data, and upon the statistical characteristics of the population which is being summarized. The need for statistical descriptions and summaries is pointed out rather specifically by means of the data in Table 2.01. It contains the 1290 ACE scores made by students entering Kansas State College for the first time in 1947. An ACE score is intended to measure certain features of a student's intellect and aptitudes which are thought to be related to his success in pursuing one of the various possible college curricula. If so, ACE scores should help the student and the staff to do a better job of fitting the students' abilities and interests to the facilities which the college has to offer. The reader is already generally familiar with the term average as some kind of usefully typical number which partially replaces a whole group of numbers; but he may be less familiar with the fact that there are several averages in use. It should be intuitively obvious that no single number, like an average, can be expected to summarize adequately the set of data in Table 2.01. Some measure of the variability exhibited by these ACE scores is needed; that is, an adequate description of the way these scores are dispersed, or distributed, between the lowest and highest scores is needed in addi- tion to a description of the general level of performance. More specifically, we need a standard method of describing any particular student's score relative to the whole group of scores. With such in- formation at hand, a trained adviser may be able to give a student considerable assistance in the choice of a vocation or a profession, or in the solution of personal problems. The statistical procedures described and illustrated in this chapter will make it possible to replace the 1290 ACE scores by relatively few statistical constants, graphs, and tables which still contain all the really pertinent information embodied in the original popula- tion of numerical measurements. Some of these possible procedures will be introduced by means of small sets of data for the sake of convenience. Thereafter, reference again will be made to Table 2.01. TABLE 2.01 1290 ACE Test Scores Made by Students Entering Kansas State College during 1947 (Data furnished by the Counseling Bureau of Kansas State College.) 70 108 84 104 103 92 103 128 96 82 145 139 128 136 94 67 78 49 84 121 86 108 65 47 75 95 67 53 95 104 95 89 79 82 119 106 52 134 131 108 145 110 54 68 123 112 82 123 141 93 75 108 95 82 59 120 105 79 115 118 112 109 104 70 112 125 126 37 1.37 103 80 130 80 93 97 116 48 98 93 102 36 114 119 130 149 95 138 85 104 79 63 63 92 60 92 76 118 125 102 68 117 130 84 118 122 104 123 127 72 99 44 112 78 72 66 94 118 79 129 134 89 90 80 146 105 83 88 134 106 127 77 43 90 95 87 94 57 46 68 124 134 120 136 111 92 128 88 110 65 157 113 145 118 116 72 120 62 52 90 91 115 128 61 58 106 100 106 99 82 121 71 103 75 112 107 88 142 112 71 135 124 99 98 95 119 93 101 99 118 150 100 91 109 83 87 93 120 92 77 80 101 140 112 60 116 67 97 114 60 111 102 107 117 77 129 80 56 105 88 113 93 73 127 111 107 76 55 102 90 78 105 87 113 138 108 68 134 85 97 76 111 85 98 97 78 79 110 143 62 84 110 94 135 156 71 112 70 123 69 58 79 123 121 117 84 64 72 97 84 115 133 90 128 120 120 148 68 46 116 103 124 61 43 98 99 111 110 66 40 75 100 63 72 71 96 72 87 52 102 73 65 80 71 91 100 64 52 78 130 85 87 66 103 94 111 159 88 85 115 108 111 41 120 111 129 89 183 78 99 108 55 91 80 103 86 58 99 132 109 118 92 105 109 95 88 153 70 76 109 121 102 119 82 103 45 50 69 69 44 100 93 71 99 45 78 99 124 99 108 72 87 122 98 96 83 52 94 121 63 108 94 102 103 69 114 95 41 28 65 107 66 76 99 33 115 121 146 83 89 124 118 98 131 95 100 97 89 74 102 103 144 76 91 53 103 92 158 65 99 107 114 90 104 54 136 86 76 114 118 114 84 112 98 139 61 23 127 115 94 68 108 93 89 75 90 113 97 64 100 101 109 100 91 37 128 118 64 72 92 76 113 166 118 125 82 104 87 113 95 112 105 73 101 121 70 69 98 72 140 115 106 114 83 113 95 91 75 128 50 68 92 127 92 91 57 120 106 117 84 74 121 134 92 91 102 78 81 106 119 106 82 114 146 110 46 112 136 41 87 23 120 119 131 68 99 123 87 113 69 102 76 127 88 99 96 65 113 103 77 104 123 105 79 107 108 61 93 81 143 93 112 139 109 123 93 134 110 65 116 75 147 136 139 125 103 108 47 107 115 56 101 80 56 95 82 66 112 76 54 62 49 51 99 131 101 134 94 68 68 53 120 124 73 120 70 129 78 151 55 134 69 80 109 125 49 36 134 97 118 76 78 137 61 73 72 111 95 162 97 128 91 58 93 54 90 93 80 94 101 96 69 99 43 111 108 133 101 121 66 77 71 105 156 85 89 150 104 97 94 128 100 138 105 49 85 116 98 89 123 153 129 93 78 98 82 84 65 93 54 109 57 96 139 65 126 129 110 86 72 100 61 105 131 101 108 96 138 87 69 112 154 116 88 134 79 92 41 93 121 99 74 117 100 93 127 99 108 95 95 76 120 130 111 111 131 104 145 93 97 96 109 98 99 60 83 60 94 69 97 120 88 108 74 110 62 68 77 86 94 93 94 70 119 120 120 120 112 38 49 128 91 84 101 65 87 72 128 88 126 122 53 29 100 111 117 127 107 83 138 80 88 54 104 105 72 115 119 113 80 90 64 128 89 109 61 135 127 100 96 81 76 148 69 55 136 98 77 44 105 125 118 109 92 119 94 108 33 101 108 122 88 102 56 92 117 114 117 130 65 99 60 88 102 130 128 109 105 115 64 120 127 87 81 110 81 120 115 99 73 105 122 109 95 85 83 99 60 48 123 127 43 104 96 103 84 131 104 115 69 108 113 37 63 106 78 52 57 102 121 135 72 158 89 106 94 105 99 86 75 94 114 109 91 114 151 138 135 64 80 80 91 99 107 100 94 110 91 100 47 113 94 89 106 64 84 99 90 67 145 74 87 92 67 81 113 116 89 99 69 98 109 84 77 106 123 74 107 92 98 87 80 74 61 125 101 72 77 92 50 108 91 118 159 145 94 110 95 107 81 112 73 82 85 116 98 44 105 140 109 41 105 43 119 148 112 83 73 81 117 135 63 91 102 117 74 66 90 124 64 117 58 126 100 108 65 88 24 88 80 34 107 101 44 78 97 133 77 137 90 114 72 110 163 56 102 55 61 103 80 50 71 110 88 44 123 99 160 65 87 113 71 92 107 124 134 96 126 98 62 89 54 115 131 81 101 91 106 44 102 90 110 106 84 70 70 78 104 73 69 125 109 81 76 103 86 106 67 109 103 120 117 91 76 122 69 105 154 68 107 127 58 125 122 55 102 65 83 149 58 45 119 119 129 151 78 81 50 78 103 117 70 79 111 124 57 93 134 120 117 75 90 77 130 84 102 97 76 135 139 49 97 136 84 89 104 52 71 87 43 96 131 80 86 107 139 127 61 113 71 109 67 126 96 113 82 55 84 71 62 70 97 107 115 95 64 111 88 118 118 95 78 178 88 88 109 88 98 32 39 71 68 60 95 139 123 91 99 61 104 78 178 98 85 95 89 97 110 80 94 87 64 100 42 105 99 39 124 109 67 108 85 101 130 39 114 127 115 62 149 137 64 113 97 93 104 113 154 93 47 79 44 84 102 124 132 131 43 102 140 42 75 123 63 122 73 99 107 111 98 80 92 120 60 100 74 99 78 87 127 61 142 131 106 116 98 86 77 130 67 107 106 82 120 72 70 63 47 82 108 78 112 101 113 107 50 106 106 117 100 53 107 79 115 96 78 70 100 44 80 71 82 114 107 80 62 88 92 89 88 129 11 12 SUMMARIZATION OF DATA Ch. 2 2.1 THE ARITHMETIC MEAN AND THE STANDARD DEVIATION When it is necessary to analyze a population of statistical meas- urements it often is desirable to calculate a single number which will be typical of the general level of magnitude of the measurements in the population. Logically, the first question is: What features should averages have in order to be typical of the data in some useful sense? Therefore, the following properties of averages are suggested as being either required of averages, or desirable: (2.11) An average should be close, on the scale of measurement, to the point (or interval) of greatest concentration of the measure- ments in the population. (2.12) It should be as centrally located among the numbers as is compatible with property (2.11). (2.13) An average should be simple to compute if that is achiev- able under the circumstances. (2.14) It should be tractable to mathematical operations so that useful theoretical information can be derived by means of mathe- matical methods. (2.15) The average should be such that measures of the scatter of the data about the average can be obtained and also have prop- erties (2.13) and (2.14). A simple but crude average which sometimes is quite useful is the midrange, MR. It is defined as that number (not necessarily one of those being studied) which is halfway between the extreme numbers in the set being summarized. For example, the extreme ACE scores in Table 2.01 are 23 and 183. The difference is 160; hence 23 -4- 183 MR = 23 + 160/2 = 103, also = . because this is the number which is halfway between 23 and 183. Among the desirable properties of averages listed above, the midrange is centrally located (in the sense that it is midway between the extremes) , and it often is in the region of the greatest concentration of the data. It also is easily calculated, but it does not possess the other properties listed. In addition, the midrange does not appear to be a very reliable average because its size depends on only two of the numbers. Sec. 2.1 ARITHMETIC MEAN AND STANDARD DEVIATION 13 An average called the arithmetic mean has been found to possess all the properties (2.11) to (2.15) to a rather high degree for a broad class of statistical populations. In addition, it is extremely useful in the analysis of sampling data, as will be shown later. Hence the arithmetic mean is a highly recommended average. The arithmetic mean, /a (Greek letter mu), of A^ measurements: Xi, . . . , Xx is calculated by dividing the sum of the N measurements by N. Symbolically, (2.11) M = Zi+X2 + -Y3+---+Z N N E (Xi) i=l N or for brevity, n = N N To illustrate, suppose that Xi = 2, X2 = 5, X3 = 1, X4 = 3, and X5 = 4; then the arithmetic mean is M = (2 + 5+1+3 + 4)/5 = 3.* Problem 2.11. Suppose that eight players are on the traveling squad of a basketball team, and their weights are 152, 170, 165, 185, 201, 174, 191, and 210 pounds. What is the arithmetic mean of these weights? The first question which may occur to the student is: What are the A^t in this instance? It is a well-known assumption in arithmetic and algebra that the same sum is obtained for a given set of numbers no matter what the order of addition; that is, 3 + 6 + 15 = 24 = 6 + 15 + 3 = 15 + 3 + 6, or any other possible order of addition. Likewise, in the present problem, it makes no difference which weight is symbolized as Xi, which as X2, etc. It is convenient just to let A"i = the first weight listed, A% = the second weight on the list, and so on. If that is done in this problem, A'l = 152, X2 = 170, A'3 = 165, * Although the discussion in this chapter is chiefly devoted to methods and ideas appropriate to populations of data — which usually contain a large number of measurements — small groups of numbers will be used in examples and prob- lems for the purpose of facilitating and shortening discussions. Obviously, most of these problems and examples resemble samples far more than populations. However, the methods introduced will apply to populations and will not neces- sarily be correct or efficient for sampling studies, as will be noted later in the book. 14 SUMMARIZATION OF DATA Ch. 2 Z4 = 185, X5 = 201, Ze = 174, X^ = 191, and Xg = 210, all in pounds. Therefore, 8 X^Zi = Xi + X2+---+X8 = 152+ 170+-.-+210 = 1448 1 so that /x, = 1448/8 = 181 pounds, which is the arithmetic mean of the weights. Although the number 181 pounds gives a useful impression of the general weight size of the players of problem 2.11, it is obvious that the same mean weight could have been obtained for many other groups of eight weights, some of which might be considered to be quite different from those above. For example, each of the following sets of eight weights (in pounds) has /x = 181: Set 1. 185, 180, 181, 184, 182, 179, 177, and 180. Set 2. 190, 190, 190, 182, 184, 183, 190, and 139. Set 3. 172, 180, 165, 160, 175, 168, 180, and 248. In Set 1, the extreme weights are but 8 pounds apart; three weights are higher than the mean, four are lower than the mean, and one is the same as the mean weight. This set differs from that of problem 2.11 chiefly by being more uniform. In Set 2, the difference between the extremes is 51 pounds, and every weight but one is higher than the mean. This set differs from that in problem 2.11 chiefly in the fact that the mean actually is not very representative of the weights of the squad members. Similar remarks hold for Set 3 except that seven of the eight weights are below the mean. To summarize: Sets 1, 2, and 3 differed from the example of Problem 2.11 in the amount of dispersion, or non-uniformity, among the numbers and in the manner in which it occurred. All sets of data had the same arith- metic mean. A measure of the dispersion, or variation, of the measurements, Xi, about their arithmetic mean, (i, logically would be based upon the amounts by which the Xi are greater than or less than that mean. It is customary to symbolize those amounts by Xi — Xi — fx, and to call the Xi the deviations from the mean. It is observed that when an X is smaller than the mean, the x is negative ; when the X is larger than the mean, the corresponding deviation, x, is positive. In the first numerical example of this chapter, Xi = ~1, X2 = +2, Xz = —2, Xi = 0, and ^5 = +1. For problem 2.11, X]_ = —29, X2 — — 11, xs = —16, Xi = +4, X5 = +20, Xq — —7, Xt = +10, and Xs = +29. It is observed that, at least in these instances, Sx = 0. Sec. 2.1 ARITHMETIC MEAN AND STANDARD DEVIATION 15 The truth of the general theorem that the sum of the algebraic values of the Xi always is zero is established as follows. By definition and simple algebra, ^x = "^{X - /x) = ^X - ^iii) ; but 2X = iV/*, and SC/x) also = NiJi because this symbol requires that we add N terms obtained by letting i have values from 1 to A^, inclusive. The ju, stays constant for each i; therefore, SC/x) = Nfx. Since S.'C = Nfi — N^., it is always equal to zero, as was to be shown. As a consequence of the truth of the above theorem, a measure of the variation about the arithmetic mean cannot be based upon the algebraic sum of the Xi. Therefore, one of two actions should be taken: (a) Ignore the signs of the Xi and obtain their mean there- after. Or (6), find some other relatively simple function of the Xi which has more of the desirable properties (2.11) to (2.15) than are obtained by method a. The latter procedure has proved to be the more successful and therefore will be considered first. As a matter of fact, it involves a function of the squared deviations, x^. The quantity (2.12) a = Vx{xi^)/N , where a, the Greek letter sigma, has been found by statisticians to be a good measure of the variability of a set of numerical measurements about their arithmetic mean. Just why it should be so useful cannot be shown to the student at this time, but it does have more of the desirable properties of measures of variation than any other such measure which has yet been devised. The quantity defined by formula 2.12 is called the standard deviation of the X^ about their mean, /*. It would be zero if all the Xi were equal; the more dis- persed they are about the mean, the larger the standard deviation tends to be. For example, consider the weights of problem 2.11 and of Set 1. The former obviously are more dispersed and generally more variable than the latter. The two standard deviations are 18.0 and 2.4, respectively, which certainly is a concise way to point out that, although the mean weights of the two squads are the same, their dispersions about that mean are far from the same. The square of the standard deviation, o-^, is called the variance of the Xi about fi. There are some relatively advanced statistical procedures in which it is preferable to work with the variance in- stead of the standard deviation, but the latter will be used most of the time in this book. From the definition of a contained in formula 2.12 it appears that each Xi must be calculated and squared, but such is not the case. If 16 SUMMARIZATION OF DATA Ch. 2 the x's are difficult to compute, the following results are useful. In view of the fact that x- = {X - fx)- = {X^ - 2/.X + fi^) , it should be clear that 2x- = 2(X- - 2fxX + fxr) = 2X- - 2/i2X + 2^. But 2/i- = N/x~, as explained earlier, and jn = SAV^V; therefore, —2fi%X = -2(2X)VA' and V = +1(2X)VA'- It follows that Sa;^ = -^X- - (S-X") VA". If this substitution is made into formula 2.12, the follow- ing alternative method for computing the standard deviation, o-, is obtained : (2.13) . = ^ ^^. For a numerical example considered earlier in this chapter, formula /55 - (15)75 /- 2.13 becomes <t = ^ = V 2 . The variance is o- = 2. \ 5 To further illustrate the uses to which a^ and o- can be put, con- sider again the ACE scores of Table 2.01. The arithmetic mean is 95.7. The standard deviation is calculated to be 26.1 (see problem 11 at the end of this section), with the extreme scores being 23 and 183. It is noted that 95.7 is a bit less than midway between the lowest and highest scores, but in general it is quite centrally located in that respect. To obtain a clearer picture of the dispersion of the scores between the extreme scores, and about the mean, the standard deviation will be found to be very useful in subsequent discussions. It is entirely possible for different sets of data to have essentially the same extremes but very different distributions with respect to the mean. The o-^ and a will help to describe these differences. This use of the variance and standard deviation is illustrated, in part, by the following discussion. As the student can verify, approximately 67.1 per cent of the ACE scores in Table 2.01 lie within a distance (on the ACE scale) of la- below or above the mean, ju,; that is, approximately 67.1 per cent of the 1290 scores are among the numbers from 70 to 121, inclusive. This fact can be put in the following brief form: /x ± lo- includes 67.1 per cent of the scores. In a strictly normal population the cor- responding percentage is 68.3. Such information sometimes is con- sidered useful in the summarization of sets of numbers like Table 2.01. Likewise the interval /u, ± 2o- (which includes scores from 44 to 147^ inclusive) contains 95.2 per cent of the 1290 scores in the table. If Sec. 2.1 ARITHMETIC MEAN AND STANDARD DEVIATION 17 this population were perfectly normal, that percentage would be 95.4. Also the interval /^ ± So- includes 99.8 per cent of the ACE scores, whereas a normal population would have 99.7 per cent of its members in that interval. The reader can determine how closely the population of Table 2.01 conforms to the normal requirement that 38.3 per cent of the measurements shall lie not more than one-half a standard deviation above or below the mean, fx. More discussion of normality and of population distributions will come later; the point of the above discussion is that knowledge about the mean and the standard deviation is useful in the study of one of the most important types of populations of data. PROBLEMS 1. Calculate the arithmetic mean and the standard deviation of the following numbers: 2, 3, 9, 7, 5, 4, 10, 6, 3, 1, and 5. 2. Make up three sets of numbers, each of which has fi = 7. 3. Compute the x^ for problem 1 and verify that Sx^ = 0. 4. Given the numbers 0, 8, 0, 1, 1, 1, 10, 2, 1, 1, 2, 3, 0, and 1, compute m- Does /I seem to you to be a good average for these numbers? Why? Ans. 2.21. 5. Suppose that the mathematics grades for a certain class were 54, 95, 68. 71, 87, 75, 84, 63, 76, 81, 70, 90, 73. 77, and 61. Calculate m and <r, using the indi- vidual Xj first and then using formula 2.13 for a-. 6. The following percentages of protein in samples of pasture grasses were made available by Dr. George Wise, formerly of the Department of Dairy Husbandry at Kansas State College. Compute /jl, cr^, and <r, given that 2X = 1423.33 and that 2Z2 = 21,924.2025. 22.59 15.26 15.63 13.52 10.82 9.29 12.25 21.07 18.83 16.52 15.54 16.17 10.03 15.71 12.79 13.11 14.85 11.45 11.97 11.07 15.26 9.31 12.30 13.04 14.19 12.94 14.36 15.02 11.15 12.08 15.41 8.56 9.09 13.07 12.51 18.91 14.28 14.54 13.68 11.78 14.22 13.07 14.27 10.27 11.01 11.66 8.19 6.75 14.48 15.98 14.36 15.24 14.48 14.05 15.02 15.41 10.02 9.96 12.34 16.26 10.19 14.20 12.56 9.74 14.34 13.07 12.33 11.57 15.48 11.74 9.39 6.47 25.09 23.23 16.75 10.62 16.30 17.29 20.63 13.76 11.88 10.57 8.37 26.20 26.74 22.02 20.60 20.36 15.83 14.11 22.42 19.47 17.98 20.32 14.83 13.03 10.31 8.95 11.57 16.03 Ans. 14.23, 16.66, 4.08. 7. The following are bearings taken with a radio direction finder on a signal sent repeatedly from a fixed location. Compute their arithmetic mean and standard deviation as though these data constituted a population. X: 9, 8, 6, 4, -10, 6, 7, 10, 8, 9, 7, 6, 8, 8, 10, 10, 8, 8, 10, 9, 10, 7, 7, 3, and 8 (degrees from north). 18 SUMMARIZATION OF DATA Ch. 2 8. The following data are like those of problem 7, but taken on a different direction finder. Obtain the variance and the a for these data given that SX2 = 116,830, ^X = 1708. X: 66, 68, 69, 68, 71, 70, 66, 70, 68, 67, 68, 73, 68, 65, 72, 73, 68, 67, 69, 65, 64, 66, 67, 70, and 70 (degrees from north). Ans. 5.58, 2.36. 9. Work problem 8, after subtracting 60 degrees from each bearing. How much were yu, and a changed? How much were the x^ changed? What if only 50 degrees had been subtracted? 10. Use one-sixth of the range in problem 6 as an estimate of the standard deviation, and compare this estimate with the true standard deviation. Ans. 3.38. 11. Given that for Table 2.01 SX = 123,445, and SX2 =. 12,693,988. Calculate the arithmetic mean and the variance about the mean, /x. 12. Given the following six yields of Ponca wheat at Manhattan, Kansas, compute their mean after first subtracting 27 from each number. (Data provided by Department of Agronomy, Kansas State College.) Yield (bushels/acre) : 27.2, 40.9, 46.0, 38.1, 43.8, 46.3. Ans. 13.2 bushels per acre; therefore, true mean = 40.2. 13. The test weights corresponding to the bushel yields of problem 12 were as follows (data from same source) : 59.3, 60.7, 60.6, 60.2, 61.9, 58.1. Calculate the midrange, the arithmetic mean, and the variance. 14. In problems 12 and 13, which of the types of measurement, yield or test weight, gives the more consistent results according to this evidence? Give reasons. 15. Write down every fiftieth score in Table 2.01, starting in the upper left- hand corner of the table and working from left to right. Complite the arith- metic mean of the sample thus obtained and compute the percentage error relative to the true mean, 95.7. 2.2 THE AVERAGE (OR MEAN) DEVIATION A measure of the variation about the arithmetic mean based on the numerical values (signs are ignored) of the Xi was mentioned in the preceding section. If we were to set out to devise a simple and logical way to measure the dispersion of a group of numbers about some point, such as the arithmetic mean, we might well decide to use what is called the average {or mean) deviation. It is the arithmetic mean of the Xi each taken as a positive number regardless of its actual sign. For example, consider the weights of problem 2.11. The numerical deviations from the mean, jx, were found to be: 29, 16, 11, 7, 4, 10, 20, and 29 pounds. On the average — that is, con- sidering the arithmetic mean as the average — the weights of those Sec. 2.3 OTHER AVERAGES 19 basketball players differed from the mean weight of the group by (29 + 16 4- 11 + 7 + 4 + 10 + 20 + 29)/8 = 126/8 = 15.75 pounds. Then the average deviation for these weights is 15.75 pounds. Symbolically the average deviation is defined by SLY - /x I si rr I (2.21) AD = — -^ or AD = N N where | a: | = a deviation from ju, taken as a positive number whether the corresponding X was larger than [x or smaller than ju,. For the weights just used for illustration, a = 18.11 pounds. The standard deviation is larger than the mean deviation, as is usual. The standard deviation is much more widely used than the mean deviation partly because it has many useful applications in sampling studies, which after all is by far the more fruitful and interesting field of statistical analysis. 2.3 OTHER AVERAGES Another average which is simple to compute and of rather wide application for descriptive purposes is the median, symbolized as md. The median of a set of numerical measurements is intended to be a number such that one-half the numbers are less than or equal to the median, and the other half are greater than or equal to the median; that is, the median is exactly in the middle of the set of numbers in order of size, if such is possible. It is necessary — either actually or effectively — to list the numbers in order of size before the median can be determined accurately. Such an ordered group of numbers is called an ordered array. Thus the numbers 1, 5, 2, 3, 0, 1, 8, and 10 do not form an ordered array, whereas these same numbers listed as 0, 1, 1, 2, 3, 5, 8, and 10 do constitute an ordered array. With the definition of an ordered array established, it is con- venient to define the median of N numbers: Xi, X2, . . . , X}f as the [ (A^ -f l)/2]th number in the array, starting with the lowest num- ber in the array. It is noted that only if N is odd does such an ordinal number exist; but it is sufficient herein to define an "ordinal" number like 4.5 to be a number which is just midway in magnitude between the fourth and fifth numbers in the array. For example, for the 20 SUMMARIZATION OF DATA Ch. 2 array used above, N = 8, so that {N -\-l)/2 = 4.5. Hence the median is )nd = 2.5, a number midway in size between the 2 (which is the fourth number in the array) and the 3 (which is the fifth num- ber in the array) . It can be seen, with a little study, that the median is an average which will be nearer the region of concentration of the numerical measurements in a population than the arithmetic mean if there are a few "stray" numbers at one end of the scale of measurement. For example, consider the following simulated annual salaries (in thou- sands of dollars) of college instructors in one department: 3.1, 3.5, 3.5, 3.6, 3.6, 3.6, 3.8, 3.8, 3.8, 3.9, 3.9, 4.0, 4.0, 4.0, 4.4, 4.8, 5.0, 6.5, 8.4, 8.7, and 8.8. For these data N = 21, 2X = 96.1, /x = 4.7, and md = 3.9. It is seen that eleven of that staff are receiving within $300 of the median salary whereas only three are that close to the arithmetic mean. The arithmetic mean exaggerates the typical salary in a very real sense for all but the fortunate six at the top. In situa- tions of this sort — which will be described later as skewed distribu- tions when more data are involved — the median is a better average than the arithmetic mean when its purpose is to describe the typical measurement in the population. If a fairly large group of numbers is to be summarized and the median is a desirable average to use, the midrange can be helpful in reducing the necessary labor. For example, the MR for Table 2.01 is 103; hence we can hope that the median has about the same size. On this assumption we can count the scores greater than or equal to 100 and thereafter determine exactly the 645.5th number in the array without excessive labor. It thus is found that md = 97. There are three other averages which will be considered, and which will find occasional application to numerical measurements. One is the mode (MO). The mode is defined to be that measure- ment which occurs in a given set of numbers with the greatest fre- quency, if such a number exists. For example, the mode of the set 5, 8, 9, 10, 10, 10, 11, 13, and 15 is MO = 10. If some number in a group of data decisively occurs with the greatest frequency, the mode may well be the average to use; but such is rather rarely the case. The geometric mean of A^i, . . . , X^^ is defined as the A'^th root of the product of these numbers. Symbolically, (2.31) GM = VZ1Z2X3 -■■ Xn Sec. 2.3 OTHER AVERAGES 21 Under most circumstances it is easier to compute the geometric mean from the relation (2.32) log(GM) =-S(logZ). N As an illustration consider the numbers 2, 5, 8, and 15. By definition GM = the fourth root of the product (2) • (5) • (8) • (15) ; but, using logarithms to the base 10, one has log GM = (log 2 + • • • + log 15) /4 = 0.7698. The antilog 0.7698 is approximately 5.9, which is the geometric mean of the given set of numbers. The geometric mean is useful in the calculation of certain index numbers, in studies of biological growth, and, in general, whenever the statistical array indicates that a geometrical series is involved. Obviously the geo- metric mean is not used if any X = or if the product under the radical is negative. The last average to be considered herein is called the harmonic mean. It also is used only in specialized circumstances, but the pos- session of some information about it will help to round out the reader's knowledge regarding statistical averages. The harmonic mean is defined as the reciprocal of the arithmetic mean of the reciprocals of a given group of numbers; or 1 N (2.33) HM = [S(i/z,)]/iv 2;(i/z,) For example, if the Z's are 3, 8, 2, 5, and 2 the denominator is 2(lAYi) = 1/3 + 1/8 + 1/2 + 1/5 -f 1/2 = 1.6583, approximately; hence HM = 5/1.6583 = 3.02. One use of the harmonic mean comes when rates of some sort are involved. Consider this problem. A man drives the first 50 miles of a trip at 50 mph, and the second 50 miles at a rate of 60 mph. What is his average rate for the trip? By the usual definition, the average rate is obtained by dividing the total distance traveled by the total time taken to go that distance. The distance traveled was 100 miles. The first 50 miles took one hour, and the second 50 miles took five-sixths of an hour; hence the total time was 11/6 hours. Therefore, the average rate of speed was 100/(11/6) = 600/11 = 54 and 6/11 mph. The harmonic mean of 50 and 60 also is 54 and 6/11 mph; that is, the required average rate is just the harmonic mean of the two rates in this instance. It is noted that the distance traveled was the same for each rate of speed. Now suppose that a person drives for one hour at 50 mph and then the second hour at 60 mph. What is the average rate of speed during 22 SUMMARIZATION OF DATA Ch. 2 this trip? The total distance traveled is 110 miles, and it took 2 hours; therefore the average rate is 55 mph. But that is just the arithmetic mean of the two rates. It is seen that when time (hours) was fixed in the problem, the appropriate average was the arith- metic mean; when the distance (miles) was fixed and time was variable according to the speed of travel, the appropriate average was the harmonic mean. In general, the proper average to use in any particular situation either will be determined at the outset by previous practices in the particular sort of work, or it can be determined by a bit of prelim- inary study of the matter. Hence no attempt will be made to lay down rules. However, it should be apparent to the reader that when a body of data is to be summarized statistically there may be several possible choices of averages and also of measures of variation. We should be fully conscious of this fact when we compute averages, or when we interpret those averages computed by someone else. PROBLEMS 1. The following numbers are salaries (in dollars) in a public school system before World War II : 1300, 1500, 1300, 1350, 1600, 1250, 1400, 1350, 1800, 4500, 1450, 3000, 2200, 1250, 1300, 1550, 1700, 1600, 1350, 1400, 1450, 1750, 1500, 1600, and 1400. Calculate the arithmetic mean and the median, and state which aver- age you consider the more typical of these salaries. 2. Suppose, in problem 1, that the following raises in salary were given: $2200 to $3000, $3000 to $3500, $4500 to $5000, $1800 to $2200, $1750 to $2200; and all others are given a $100 raise. The salaries of problem 1 add to $41,850, and the total of the raises is $4650, approximately 11 per cent of $41,850. Is it then fair to state that those teachers received an 11 per cent increase in salary, on the average? 3. Compute the geometric mean of 76.3 and 85.1. 4. Sometimes the median can be used as an average when numerical measure- ments are not employed. For example, some radio direction finder networks rate their bearings as to their quality, ranging from A, (best) through B, C, F, and P. If the median does not turn out to be indeterminate (as by falling be- tween two different letters) it may be useful in describing the average quality of the readings. Obtain the median quality of the following quality ratings: A, C, C, B, A, B, C, B, P, F, C, C, F, B, C, A, and B. Ans. C. 5. What is the modal quality rating for problem 4? 6. Compute the geometric mean of the salaries in problem 1 to the nearest dollar. Am. $1591. 7. Suppose that peaches were bought in three different areas for $3 per bushel, $2 per bushel, and $4 per bushel, respectively. Suppose also that $24 was spent for peaches at each price level. What was the average price paid per bushel? 8. Do as in problem 7 except to consider that 10 bushels of peaches were bought at each price. Ans. $3 per bushel. Sec. 2.4 FREQUENCY DISTRIBUTIONS 23 9. Suppose that 5 bushels of the $3 peaches, 10 of the $2 peaches, and 15 bushels of the $4 peaches were purchased. What was the average price per bushel? 2.4 FREQUENCY DISTRIBUTIONS To introduce the method of constructing frequency distributions, and to show what sort of information can be derived from them, reference is made to the numbers of Table 2.01. It is possible by means of problem 11, section 2.1, to calculate that /a = 95.7 and o- = 26.1. These statistical constants furnish some useful information about the population of scores, but they fail quite badly to sum- marize them adequately. For example, a person wdio made a score of 120 could not be told accurately how he compared with the others taking this same test, and that information usually is important in the use of such tests. One way to obtain this sort of information is to construct frequency distributions and graphs which display the out- standing features of the population. Two types of summaries of distributions will be considered both numerically and graphically: a frequency and a relative cumulative frequency (or r.c.j) distribution. Both distributions will be de- scribed by means of a grouping of the individual scores into con- venient score classes, even though such frequency distributions could be made without grouping the members of the population into classes. The scores then lose their individual identities and become members of ten to twenty groups. The data become more manageable, and little accuracy is lost in the process. To illustrate, consider Table 2.01 again. The extreme scores have been noted previously to be 23 and 183 so that the range is 160. If the range is divided by 10 a quotient of 16 is obtained. Classes of that length would give the minimum acceptable number of classes; hence for convenience in tallying (as shown below) the class interval will be taken as 15. Table 2.41 was constructed starting with the lowest score at 10 purely because it was convenient and the lowest class included the lowest ACE score in Table 2.01. The actual tallying of the data is shown, as is the summarization of the tallies into a frequency (/) for each class. In Table 2.42 a more concise form of the frequency distribu- tion is shown along with the r.c.j. distribution. The latter distribu- tion gives the decimal fraction of the ACE scores which were less than or equal to the upper limit of the corresponding score class at the left. For example, practically one-third (actually .332) of the scores were at or below a score of 84, according to Table 2.42. 24 SUMMARIZATION OF DATA Ch. 2 TABLE 2.41 Frequency Distribution Table for the Data of Table 2.01, Showing Tallying Score Class / 175-189 /// 3 160-174 //// 4 145-159 mi m m m m m //I 33 130-144 mimimimiimrwmimmmmmmm mm/ 81 115-129 mmmmimmimimimmimmimm mmmimimimimmimmmmmm mmimimimmimmmi ise 100-114 numimimimmimimmmimmimm mmmmimmimmimmimimimm mimmmimrwrwmiiHjniimmmm mmmimmimmmmmimimim/// 278 85-99 mmmmmmimmmimimimimimi mimmimmmimmmimmmimim mjmmmimjmmimimimmjfwmm immmmimimiiHimimimimirHim// 277 70-84 mjmmimimmmjmjmmimimmm mimmmimimmjmjmimmmimimi mimimmimimimimimmimimim iiii 209 55-69 mmmmmmmmmmmmmm mmmmmmmmmmmmii 132 40-54 mmmmmmmmmmmmm 11 11 69 25- ^9 m m m 15 10- 24 /// 3 Total 1290 To assist in the drawing of conclusions from Tables 2.41 and 2.42, the information they contain is presented in graphic form in Figure 2.41. Conclusions such as the following can be drawn from that figure and the tables pertaining to Table 2.01 : (2.41) Only about 7 per cent of the students made a score less than 55. This information can be read directly from Table 2.42 and verified approximately from the r.c.f. curve of Figure 2.41. Also, approximately 50 per cent of the students made a score of 98 or more, a fact which corresponds closely with the fact that the exact median is 97. Information of this sort can be obtained from Figure 2.41 by reading horizontally from r.c.f. = .50 over to the r.c.f. curve and then vertically downward to the scale of ACE scores. Sec. 2.4 FREQUENCY DISTRIBUTIONS 25 TABLE 2.42 The Relative Cumulative Frequency Distribution op the Data IN Table 2.01 Score Class / c.f. r.c.f. 175-189 3 1290 1.000 160-174 4 1287 .998 145-159 33 1283 .995 130-144 81 1250 .969 115-129 186 1169 .906 100-114 278 983 .762 85- 99 277 705 .546 70- 84 209 428 .332 55- 69 132 219 .170 40- 54 69 87 .067 25- 39 15 18 .014 10- 24 3 3 .002 Total 1290 (2.42) With specific reference to the student mentioned above who made a score of 120, it is learned that about 82 per cent of the students did no better; or only about 18 per cent beat him. Hence he should be considered to be quite high in aptitude and intelligence relative to those who took that same test, and would be expected to do rather well in college. 85 100 115 Test score Figure 2.41. Frequency distributions of the ACE test scores listed in Table 2.01. Data furnished by the Counseling Bureau of Kansas State College. 26 SUMMARIZATION OF DATA Ch. 2 Other similar information can be read from the above tables and graphs. Moreover, it will be shown later that the r.c.f. graph can be employed to determine quartile, decile, and even percentile limits if the graph is drawn with sufficient care. In addition it will be shown in the next section that in some situations good approximations to fi and a- can be computed from the frequency distribution table. In brief, the frequency and r.c.f. tables and graphs serve as visual guides and as sources of good approximations. If more precise information is needed or desired (and it seldom is) , one can analyze the individual observations. The graph of the r.c.f. distribution sometimes is called an ogive. PROBLEMS 1. Following are some numbers of house flies counted on individual dairy cows which had been sprayed with a 3 per cent solution of Thanite in 40 oil: 35, 37, 41, 103, 174, 7, 11, 32, 23, 7, 6, 3, 14, 23, 23, 36, 25, 27, 3, 3, 13, 14, 14, 6, 15, 9, 11, 21, 9, 12, 3, 15, 19, 29, 26, 1, 8, 4, 9, 7, 12, 5, 1, 3, 5, 60, 11, 6, 4, 7, 22, 28, 5, 3, 6, 15, 1, 2, 11, 4. 27, 1, 0, 0, 19, 6, 2, 3, 4, 13, 5, 12, 11, 14, 45, 4, 38, 5, 17, 27, 39, 33, 13, 9, 8, 33, 19, 6, 12, 32, 11, 35, 18, 11, 25, 23, 45, 30, 4, 4, 15, 15, 16, 11, 16, 18, 32, 49, 129, 7, 21, 26, 76, 40, 5, 7, 5, 7, 4, 62, 91, 133, 61, 59, 20, 26, 10, 12, 6, 7, 8, 8, 2, 24, 21, 51, 110. 11, 6, 4, 4. 5. 5, 5, 13, 3, 6, and 7. Construct frequency and relative cumulative frequency distributions for these data, esti- mate the median, and decide whether fi or vtd is the preferable average for these data. Is this a skewed distribution? Use class intervals: 0-8, 9-17, etc. 2. Graph the distributions asked for in problem 1. 3. Compute the aiithmetic mean and the median of the counts in problem 1. Compare them, and draw appropriate conclusions. SX = 3031, SX^ = 163,439. 4. Estimate from the ogive (r.c.f. curve) for problem 1 what percentage of the fly counts lie between 5 and 25, inclusive. Check your calculation by actually counting in problem 1. Ans. Graph, 46; by count, 45. 5. Use the following frequency distribution table of 8-week weights (in grams) of male White Rock chickens raised at the Kansas State College Poultry Farm and the accompanying graphs to: (a) estimate /x, (b) determine what percentage of the weights exceeded 800 grams, (c) determine the range covered by the "middle" 50 per cent of the weights, that is, excluding the upper and lower 25 per cent of the weights. 6. Construct the frequency and the relative cumulative frequency distribu- tions for the following counts, which are similar to those numbers in problem 1 : 6, 8, 13, 36, 48, 65, 34, 24, 49, 24, 40, 18, 20, 34, 87, 28, 14, 30, 24, 53, 57, 93, 36, 80, 33, 48, 57, 98, 73, 135, 21, 40, 32, 58, 4, 20, 30, 33, 20, 22, 28, 11, 23, 46, 41, 41, 44, 23, 18, 41, 48, 81, 80, 70, 5, 2, 13, 21, 21, 171, 1, 7, 10, 5, 2, 17, 9, 35, 6, 8, 10, 23, 19, 3, 25, 16, 131, 19, 19, 24, 12, 10, 4, 5, 2, 14, 17, 18, 10, 8, 4, 0, 4, 12, 14, 111, 17, 33, 3, 2, 7, 10, 17, 4, 5, 2, 48, 4, 11, 31, 18, 32, 26, 3, 18, 19, 101, 10, 10, 3, 27, 14, 29, 24, 13, 26, 31, 5, 20, 16, 13, 6, 7, 32, 17, 25, 6, 8, 5, 24, 13, 8, 7, 2, 1, 3, 26, 38, 44, 3, 5, 6, 22, 28, 16, 22, 8, 19, 12, 3, 24, 10, 8, 33, 20, 29, 3, 15, Sec. 2.4 FREQUENCY DISTRIBUTIONS 27 Relative Cumulative Cumulative Frequency, Frequency, Frequency, Veight Class / c./. T.C.f. 1025-1074 1 1217 1.000 975-1024 16 1216 .999 925- 974 29 1200 .986 875- 924 66 1171 .962 825- 874 148 1105 .908 775- 824 169 957 .786 725- 774 265 788 .647 675- 724 210 523 .430 625- 674 155 313 .257 575- 624 85 158 .130 525- 574 51 73 .060 475- 524 17 22 .018 425- 474 5 Total 1217 5 .004 250 - 200 - 2? 150 - it 100 - 50 - 425 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 /\ <*"" — 1 1 1 — / \ - — / \ ^ / \/ — — / ''v — / ' \ — / / \ / ' \ Ms \ — / / \ ' ^ \ — / \ — — y ^ ^^>.,^^ — r tn4-rT 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 r>^ 1.00 .90 |. .80 i cr <u .70 4= <u .60 -^ 2. .50 I .40 \ .30 ;| .20 ^ .10 525 625 725 825 925 1025 Weight 18, 20, 8, 13, 17, 27, 23, 23, 10, 25, 13, 10, 12, 10, 6, 6, 14, 24, 61, 25, 26, 21, 12. 15, 18, 19, 26, 21, 11, 0, 0, 8, 34, 66, 32, 7, 8, 23, 20, 24, 62, 8, 15, 19, 33, 20, 51, 11, 20, 13, 27, 15, 10, 16, 16, 5, 4, 24, 30, 37, 26, 17, 14, 15, 6, 3, 22, 53, 54, 74, 1, 10, 12, 22, 49, 52, 31, 7, 20, 23, 28, 56, 2, 6, 6, 30, 30, 38, 1, 2, 4, 21, 51, 14, 5, 17, 21, 28, 9, and 7. 7. Construct the relative cumulative frequency distribution for the data of the preceding problem, and read from it the value of the median. Check that result with the value obtained from an ordered array of those data. 8. Use the graph of the relative cumulative frequency distribution for the counts in problem 6 to determine what percentage is less than or equal to ten flies per cow. Ans. 29 per cent. 28 SUMMARIZATION OF DATA Ch. 2 9. Within what extremes did the lowest one-fourth of the counts listed in problem 6 lie? The middle one-fourth? The highest one-fourth? 10. Take any newspaper with at least one hundred bond or stock quotations and make frequency and relative cumulative frequency distributions of those prices. 2.5 CALCULATION OF THE ARITHMETIC MEAN AND THE STANDARD DEVIATION FROM FREQUENCY DISTRIBUTION TABLES If the frequency distribution table has class intervals of equal lengths, approximate values can be computed for /x and o- with a considerable saving in labor as compared to their computation from the individual measurements. The method of computation involves the sole assumption that the numbers grouped into each class actu- ally were at the midpoint of their class. Although that assumption is not strictly correct, the individual discrepancies usually balance out so well that the net error is unimportant in practice. If it should be decided that some additional accuracy is needed, Sheppard's cor- rections for grouping can be employed. (See, for example, Kenney, Mathematics of Statistics, Part One, D. Van Nostrand.) Table 2.51 presents methods for computing /a and o- which follow directly from the definitions of these quantities if all the data in a class are considered to be at the midpoint of the class. For example, the data in Table 2.51 would be considered to be 22.5, 22.5, 17.5, 17.5, 17.5, 17.5, 17.5, 17.5, 12.5, 12.5, 12.5, 12.5, 12.5, 12.5, 12.5, 12.5, 12.5, 12.5, 7.5, 7.5, 7.5, 7.5, 7.5, 2.5, and 2.5 each midpoint appearing pre- cisely that number of times indicated by the class frequency, /. The student should check the fact that the sum of these products is 317.5, which is shown in Table 2.51 as the total of the column headed "f-z." The symbol z is employed to denote the midpoint of the class interval. For convenience and for uniformity of procedure, the midpoint of a class of data measured on a continuous scale is defined to be the lower limit of the class (as recorded in the table) plus one-half the length of the class interval. Also, the length of the class interval, for such data, is defined as the numerical difference between any two successive left- or right-hand end points of classes. Thus the mid- point of the class "20-24.999 . . ." is 20 + (1/2) (5) = the z for this class interval. The reader should note that there will be circumstances in practice which will justify a different determination of the midpoint, z. For Sec. 2.5 CALCULATION OF m AND a FROM TABLES 29 example, if some objects have been weighed to the nearest pound it is reasonable to suppose that an interval written as 20-24 actually means 19.5 to 24.5. If the class intervals are so written, the above- stated rules apply. The length of the class interval will be 5 as be- fore, but the midpoint will be computed as 19.5 + (1/2) (5) = 22.0 instead of 22.5, as it would be if computed on the assumption that the interval started at 20. If we are summarizing data which can only be integers, a class interval such as 20-24 should include only the numbers 20, 21, 22, 23, and 24. It then is reasonable to take z = 22. The length of the class interval should be taken as 5 again so that the numerical dis- tance between midpoints will coincide with the length of the class interval, which seems to be a reasonable requirement. The proper procedure for other methods of measurement can be figured out along the lines just outlined. TABLE 2.51 Illustration of a Method of Calculating h and <r from the Data in a Frequency Distribution Table with Class Intervals of Equal Lengths Class Midpoint Fr( squency Interval z / f-z Z - fi /•(2-m)^ 20-24.9 22.5 2 45.0 9.8 192.08 15-19.9 17.5 6 105.0 4.8 138.24 10-14.9 12.5 10 125.0 - 0.2 0.40 5- 9.9 7.5 5 37.5 - 5.2 135.20 0- 4.9 2.5 Totals 2 25 5.0 -10.2 208.08 317.5 674.00 Zf-z 317.5 2/ 25 12.7; 0- = ^f-iz-^r 2/ = 5.19, approximately. Another, and easier, method of computing /x and o- from a frequency distribution table with equal class intervals is illustrated in Table 2.52 along with a partial demonstration of the generality of the method. The procedure involves the same assumption made above and produces exactly the same values for fx and a. However, in this method the class interval is employed as the computational unit, with the result that the sizes of the numbers needed in the process are smaller than those of Table 2.51. This makes the computations simpler. 30 SUMMARIZATION OF DATA Ch. 2 The procedures outlined in Table 2.52 can be justified as follows. It is obvious that the midpoints, Zi, of Table 2.51 can be rewritten as follows: 2.5 = 12.5 - 2(5), 7.5 = 12.5 - 1(5), 12.5 = 12.5 + 0(5), 17.5 = 12.5 + 1(5), 22.5 = 12.5 + 2(5). Then S(/-0) = 2[12.5 + 2(5)] + 6[12.5 + 1(5)] + 10[12.5 + 0(5)] + 5[12.5 - 1(5)] + 2[12.5 - 2(5)]; = (2 + 6 + 10 + 5 + 2)(12.5) + [2(2) + 6(1) + 10(0) + 5(-l) + 2(-2)](5); or, a bit more generally, 2(/-.) = (S/)(12.5)+Z(/-cf)(7), if d = +2 for the top class of Tables 2.51 and 2.52, d = +1 for the next class down, etc., until d — —2 for the bottom class of each of those tables. (The symbol I stands for the length of the class in- terval.) Therefore, S(/-.) 2(/-rf)(7) (2.51) M = = 12.5 -\ ^ \ 2(/) S(/) is the approximation to ju, which can be obtained from the frequency distribution table. It should be clear that some other midpoint be- sides 12.5 could have been used without changing the answer ob- tained. Hence if there is any best choice of a midpoint to use as a base point (with d = 0) , that choice must rest on its leading to simpler computations. Generally, the d should be taken as zero for the class with the greatest frequency. If the distribution is quite non-symmetrical, it is advisable to shift the choice one way or the other so that the positive and negative Jd's will be more nearly bal- anced. This rarely will be more than two classes from the one with the greatest frequency, /. Sec. 2.5 CALCULATION OF ^l AND <r FROM TABLES 31 It can be seen in Tables 2.51 and 2.52 that when d = for the class with the largest frequency (/) the resulting arithmetic involves smaller numbers than for the other methods. It should be noted, again, that all three of the methods illustrated give exactly the same answers; the only differences lie in the ease of computation. TABLE 2.52 Illustration of a Simplified Method for Computing ij. and a from a Frequency Distribution Table with Equal Class Intervals Method A (d taken Method B {d taken Mid- Fre- for interval with for interval with Class point quency greatest frequency) lowest frequency) Interval z / d f-d f-d' d J-d f-d-^ 20-24.9 22.5 2 +2 4 8 15-19.9 17.5 6 + 1 6 6 -1 - 6 6 10-14.9 12.5 10 -2 -20 40 5- 9.9 7.5 5 -1 -5 5 -3 -15 45 0- 4.9 2.5 2 -2 -4 8 -4 - 8 32 Totals 25 + 1 27 -49 123 M = (z for class with d = 0) + ^^ (/) = 12.5 + (l/25)(5) = 12.7. ^^(^) j zjf.^P) - [2(/-d)]V _2Cf) ^ (5) /27-^a)V25 ^ ^^^ V S(/) V 25 The derivation of the formula shown for o- is more difficult than that for fx, as might be expected, but it can be obtained by elementary algebra, formula 2.13, and by expressing each z in terms of the one for which d = 0. This derivation will be left as an exercise for the ambitious student. The methods just described can be applied to obtain satisfactory approximations to the arithmetic mean and the standard deviation of the ACE scores in Table 2.01, a task which clearly would be quite laborious if formulas 2.13 and 2.11 were to be employed directly on the 1290 numbers in that table. The d are taken as zero for the class with a frequency of 277 (Table 2.42) instead of the class with / = 278 because they are essentially the same size and the distribution is a bit non-symmetrical (or skewed) in the direction of the lower ACE scores. The general result is to have smaller d's with the larger fre- 32 SUMMARIZATION OF DATA Ch. 2 quencies, and hence to make the computations somewhat easier. Fol- lowing are the required calculations based on Table 2.42; it is assumed that the scores are necessarily integers. 2 f d f-d f-d'~ 92 3 6 4 5 33 4 81 3 186 2 278 1 277 209 -1 132 -2 69 -3 15 -4 3 -5 18 108 20 100 132 528 243 729 372 744 278 278 -209 209 -264 528 -207 621 - 60 240 - 15 75 2(/) = 1290 S(/-d) = +308 4160 = ^if-d^) By the formulas previously used, M = 92 + (308/1290) (15) = 95.6 compared to the true mean of 95.7. /4160 - (308)Vl290 0" = (15) .. / = 26.7 compared to the true value of \ 1290 26.1. In view of the fact that the scores were integers, these approxima- tions certainly would be considered satisfactory, and the time and labor saved by these methods are considerable. The distribution of the population of ACE scores is rather sym- metrical, that is, there is a region of high frequency about halfway between the extremes, and the frequency of occurrence of scores away from this region diminishes at about the same rate as scores are considered equally far above and below the region of highest fre- quency. This distribution is shown in Figure 2.41. With this type of distribution the arithmetic mean is an excellent average to use as a part of the description of the population. Other distributions may be non-symmetrical, or skewed. For such populations the median often serves as the more descriptive average. As a matter of fact, the difference between the sizes of the arithmetic mean and the median is an indication of the degree of skewness or lack of symmetry, in the frequency distribution. If the distribution Sec. 2.5 CALCULATION OF ^ AND <x FROM TABLES 33 is perfectly symmetrical (no skewness) the arithmetic mean and the median are equal. The more skewed the distribution, the farther apart the median and this mean may become. PROBLEMS L Compute the arithmetic mean of the following numbers by Method A of Table 2.52: A': 24, 8, 7, 14, 21, 10, 12, 14, 17, 9. 11, 5, 15, 16, 8, 2, 13, 18, 12, 3, 15, 4, 16, 19, and 11. Use class intervals 2-5.9 ... , etc., to 22-25.9 2. Compute the arithmetic mean and the standard deviation for problem 1 exactly, and compare with the values obtained by the methods of Table 2.52. Ans. fjL = 12.2, o- = 5.5; they are 12.8 and 5.3 by table. 3. Put the numerical measurements of problem 1, section 2.4, into a fre- quency distribution table with class intervals of equal lengths, and compute the standard deviation of those counts. 4. Do as in problem 3 for the data of problem 6, section 2.4. Ans. 23.7. 5. Calculate the mean and standard deviation for the hypothetical data in the following table. Also, compare six times the standard deviation with the range as nearly as it can be derived from the table. Class Interval Frequency 28-29.9... 5 26-27.9.. 16 24-25.9.. 29 22-23.9.. 41 20-21.9.. 50 18-19.9.. 45 16-17.9.. 32 14-15.9.. 20 12-13.9.. 9 10-11.9.. 3 Total 250 6. Graph the relative cumulative frequency distribution for problem 5 and read from it the percentage of the measurements which exceed 23. Which ex- ceed the arithmetic mean. Which lie between the mean and 23. Ans. 28%, 50%, 22%. 7. What is the median for problem 5 as read from the r.c.f. cui've? Which is the modal class? Would you expect the mode and the median to differ by as much as two units ; or less than two units? Give reasons. 8. Graph the following actual or estimated age distributions of the United States population and draw appropriate conclusions regarding apparent trends during the decades covered. Consider top class as 0-4 and bottom one as 75-79. 34 SUMMARIZATION OF DATA Ch. 2 (Data from Current Population Re-port, Population Estimates, August 10, 1950, Bureau of the Census.) Numbers are thousands. Age Class 1940 1950 1960 Under 5 years 10,542 16,580 13,121 5-9 10,685 13,959 15,693 10-14 11,746 11,349 17,439 15-19 12,334 10,561 13,860 20-24 11,588 11,585 11,274 25-29 11,097 12,161 10,725 30-34 10,242 11,439 11,771 35-39 9,545 10,960 12,211 40-44 8,788 10,061 11,377 45^9 8,255 9,231 10,713 60-54 7,257 8,254 9,583 55-59 5,868 7,440 8,469 60-64 4,760 6,210 7,205 65-69 3,748 4,611 5,980 70-74 2,561 3,282 4,428 75 years and over 2,655 3,716 5,083 9. Change the top and bottom classes in problem 8 to 0-4 and 75-79, and then compute IX and md. Which do you consider the better average here? 10. Given the following frequency distribution of the minimum annual tem- peratures in 116 cities of Kansas, compute the arithmetic mean and the stand- ard deviation: Temperature Temperature Interval / Interval / -17.4 tb -15.0 3 -32.4 to -30.0 17 -19.9 -17.5 7 -34.9 -32.5 1 -22.4 -20.0 17 -37.4 -35.0 2 -24.9 -22.5 24 -39.9 -37.5 -27.4 -25.0 24 -42.4 -40.0 1 -29.9 -27.5 20 2(/) = 116 Ans. -25.7°, 4.4°. 11. Given the following table of average lengths of growing season for ninety- five Kansas counties, compute the mean length of growing season, and also the median length. Which would be the most descriptive average here? Class Interval / 198-202 days 3 193-197 8 188-192 13 183-187 18 178-182 8 173-177 12 Class Interval / 168-172 days 163-167 158-162 153-157 16 9 7 2 2(/) = 95 Sec. 2.6 PERCENTILES, DECILES, AND QUARTILES 35 2.6 PERCENTILES, DECILES, AND QUARTILES The standard deviation about the arithmetic mean, the range, the average deviation, and the comparative magnitudes of the median and the mean (all presented earlier) provide useful information re- garding the dispersion of the numerical measurements in a group of data which is being analyzed. However, there are some circum- stances in which it is desirable to divide the ordered array into seg- ments each containing a stated percentage of all the numbers in the set. More specifically, it may be convenient to partition a large body of data into four, ten, or one hundred subgroups, each contain- ing approximately the same number of measurements from the set, and with the subgroups corresponding to successive segments of the array. The subgroups will be called quartiles if four divisions are employed, deciles if there are ten subgroups, and percentiles if there are one hundred subgroups.* The aim in stating the upper limit of the first quartile, for example, is to designate a number such that one-fourth the numbers in the array are less than or equal to that upper limit. Although the upper and the lower limits of the quartiles, deciles, and percentiles could be read from a carefully drawn r.c.f. curve if the data are sufficiently numerous, it is desirable to have precise definitions for them. This could be done in a variety of ways, not essentially different, so that certain convenient and reasonably stand- ard definitions will be adopted rather arbitrarily. Before general rules and methods for determining the limits on the quartiles, deciles, and percentiles are considered attention is called to the following two arrays and to some general problems inherent in the determination of such subgroups as quartiles : Setl. 1,2,3,4,6,8,8,9,10,10,11,12,15,18,18. N ^ 15. Set 2. 1, 2, 4, 7, 9, 9, 11, 11, 12, 15, 15, 18, 20, 24, 25, 27. N = 16. Suppose that we wish to divide these sets of numbers into four sub- groups, each containing equally many numbers, if possible, and com- ing as close as possible to equality in other instances. Two facts are * It seems to the author that the term percentile should refer to an interval which includes approximately one per cent of all the measurements. However, most textbooks use this term to designate only one end point of what is called a percentile herein. Similar remarks apply to the terms decile and quartile. Since we usually speak of a score being in a percentile rather than at it, usage seems to support the point of view taken herein. 36 SUMMARIZATION OF DATA Ch. 2 immediately clear, (a) When A^ is not a multiple of four, we cannot define four groups each containing one-fourth of A^" measurements; and [b] repetitions of numbers will pose a problem in some instances because numbers of equal size logically must be in the same subgroup, and yet to put them there sometimes will cause one subgroup to con- tain more than its stated proportion of all the measurements. It will be convenient first to describe the method to be used to determine percentile limits because deciles and quartiles can be de- fined in terms of percentiles. The general aim in defining percentiles is to divide the ordered array into 100 subgroups, each of which con- tains one per cent of the numbers in the set, as nearly as this is pos- sible. This result will be accomplished by defining the upper limit P of the pth. percentile to be the th number in that (iV+ 1) 100 array. For example, if A^ = 1290, as in Table 2.01, the upper limit of the ninetieth percentile is the [^\ (1291) ]th, or the 1161.9th, number in the array or along its scale of measurement. Such an "ordinal" number as 1161.9 will be defined to be the number which is nine-tenths of the way between the 1161st and the 1162nd numbers from the bottom of the array. It is seen from Table 2.42 that there are 1169 numbers less than or equal to 129. With this information it is found that the 1161st and the 1162nd scores in order of size are 128 and 129, respectively. Hence, the 1161.9th number along the scale of the ACE scores is 128.9, which, then, is the upper limit of the ninetieth percentile. The lower limit of this percentile is just the upper limit of the eighty-ninth percentile. By definition, this is the [89(1291)/100]th number along the array of the ACE scores. Since 89(1291)/100 = 1148.99, the lower limit of the ninetieth percentile is a number which is .99 of the way between the 1148th and 1149th scores from the bottom of the array. The 1148th score is 127, whereas the 1149th score is 128; hence the lower limit of the ninetieth per- centile is 127.99. It follows then that the ninetieth percentile con- tains scores of 128 only. Actual enumeration discloses that there are 13 scores of 128, which is as close to one per cent of 1290 as is possible with integers. Such close agreement with the ideal will not be attained with most of the percentiles, especially in the neighbor- hood of the mean and the median, because there will be repetitions of scores which will cover more than one percentile. It may be better when much of this occurs to be content with the coarser subgroups given by deciles or even quartiles. Sec. 2.6 PERCENTILES, DECILES, AND QUARTILES 37 The upper limit of the first decile is the same as the upper limit of the tenth percentile, and similarly for the other deciles. The upper limits of the first, second, and third quartiles are the same as the upper limits of the twenty-fifth, fiftieth, and seventy-fifth percentiles. It should be clear that the median is the upper limit of the second quartile. It is traditional to designate the upper limits of the first and third quartiles as Qi and Q3, respectively, even though the term quartile may be used differently from the way they are used in this book, as was mentioned earlier in a footnote. PROBLEMS 1. Following are the average temperatures for July in Topeka, Kansas, from 1901 to 1930, inclusive, in degrees Fahrenheit: 86.6, 77.0, 77.6, 75.0, 74.1, 74.8, 78.7, 76.0, 78.0, 79.4, 78.8, 79.9, 81.8, 80.2, 74.0, 81.9, 80.4, 78.0, 81.6, 76.8, 79.8, 76.4, 79.0, 75.2, 78.6, 79.0, 76.6, 78.3, 79.0, and 82.4. Obtain Q^, md, and Q3. 2. Determine and interpret the limits of the second decile for the data of problem 1. Also compute the median. Atis. 74.88 to 76.08; md = 78.65. 3. What are the limits of the third quartile of the data of problem 6 of section 2.1? 4. What are the limits of the first quartile for the fiy counts given in prob- lem 6, section 2.4? Ans. to 8 inclusive. 5. Calculate the limits on the ninth decile for the counts of problem 1, section 2.4. What information can you derive from these limits? 6. Use Figure 2.61 on page 40 to determine the approximate sizes of Q^, md. and Q3 for the birth weights recorded in Table 2.61. What information about the birth weights do these numbers gi\-e? Ans. 66, 81, 95 grams. 7. Construct a frequency distribution table and a graph for the 4-day gains of Table 2.62 and compute the mean gain. 8. Determine the limits on the 10th percentile of the 4-day gains of Table 2.62 and interpret these numbers statistically. Ans. —30 to —1.7, inclusive. 9. Construct a relative cumulative frequency distribution table for the birth weights listed in Table 2.61, using the class limits indicated in Figure 2.61. 10. Suppose that a student entering college takes the following tests: a general psychological test, a reading test, a mathematics aptitude test, a social science aptitude test, and a physical science aptitude test. If his respective percentile ratings are 90, 87, 50, 92, and 63, what advice would you give him regarding a choice of a curriculum, assuming that you have faith in these tests? Explain your reasoning. 11. Determine the lower limit of the upper (tenth) decile for the ACE scoi'es of Table 2.01. 38 SUMMARIZATION OF DATA Ch. 2 TABLE 2.61 Birth Weights of Female (F) and Male (M) Guinea Pigs Born DURING A Particular Experiment (Data obtained from H. L. Ibsen, Kansas State College.) Jan., F 65.3 106.0 100.7 52.0 81.6 83.9 89.6 Nov., M 77.6 73.3 106.2 74.4 60.3 93.6 90.0 87.2 118.9 100.2 80.0 84.7 63.0 53.9 43.9 92.2 83.7 108.4 66.7 67.2 69.4 65.0 73.1 84.6 79.2 98.2 75.9 62.7 64.3 79.5 70.0 81.5 87.0 105.5 98.6 72.7 77.1 May, F 85.5 77.3 63.7 115.3 82.2 122.3 82.9 63.4 93.5 72.4 67.4 64.3 68.0 123.6 99.6 59.4 57.7 84.5 90.7 58.2 54.6 62.6 86.8 89.4 99.0 85.2 75.6 67.5 80.6 79.3 Jan., M March, M 77.5 97.3 62.2 59.3 Sept., F 90.0 106.7 87.4 94.0 77.8 57.0 62.2 81.8 69.6 95.5 97.3 80.5 76.6 June, M 41.1 74.2 67.4 109.8 105.1 97.6 84.3 91.1 112.0 66.6 61.7 70.2 119.1 107.3 73.7 57.3 44.5 41.5 64.7 63.0 Oct., F 130.4 86.7 79.1 96.7 36.1 50.7 63.2 80.7 60.9 102.7 58.9 92.6 112.4 60.3 48.6 61.6 77.4 92.9 100.6 75.3 82.1 79.8 63.0 48.0 81.2 91.2 93.5 46.1 90.0 100.3 69.1 43.3 Aug., M 80.9 96.4 134.0 56.2 100.8 66.1 51.6 102.7 76.8 50.4 136.3 Feb., F 63.2 91.0 53.8 65.8 117.2 64.0 76.2 113.5 77.3 61.8 84.6 82.1 45.8 63.7 96.6 116.6 75.0 109.6 88.0 63.0 65.9 56.0 87.3 Oct., M 74.9 72.3 76.5 69.0 62.5 63.4 75.2 89.2 124.9 80.0 77.6 68.9 75.6 79.4 67.3 85.2 109.0 107.0 98.2 105.5 67.9 83.9 94.6 46.6 76.4 91.2 91.7 89.7 98.8 57.8 72.1 74.0 49.0 47.4 121.1 119.2 82.2 88.7 78.8 97.4 82.0 49.9 66.6 91.5 109.6 106.7 76.5 73.9 104.3 87.6 55.3 82.0 87.6 107.8 56.7 75.5 67.0 80.5 79.6 78.8 99.6 Dec, F 90.1 68.2 64.1 53.6 Aug., F 66.5 91.0 101.9 110.6 138.4 65.0 94.6 78.0 53.3 62.9 112.2 68.1 97.9 51.4 94.2 70.3 124.5 88.8 73.6 94.7 72.0 76.8 133.4 59.8 83.2 74.3 56.9 87.8 94.2 44.4 56.2 July, F 45.0 88.3 76.0 Nov., F 89.3 Feb., M 50.6 37.0 85.9 92.2 67.7 94.0 87.9 96.5 Apr., F 54.7 85.8 68.4 91.2 90.8 81.5 105.0 89.8 May, M 51.6 89.0 45.6 93.1 117.7 79.6 80.8 95.5 110.0 65.7 71.2 98.0 84.8 103.5 67.0 76.5 82.2 112.7 53.7 121.9 92.0 99.9 88.2 52.5 104.9 69.2 109.4 76.1 120.4 94.7 85.7 45.6 104.2 67.2 113.4 54.3 78.3 95.7 Sept., M 96.7 87.3 88.2 79.3 115.6 47.3 76.1 98.3 68.6 95.6 70.5 104.0 68.9 102.7 43.3 110.8 99.3 72.7 78.3 105.0 75.5 110.6 42.1 91.3 72.5 63.0 105.3 Dec, M 83.1 75.2 80.3 40.8 99.9 98.4 84.0 108.3 72.3 84.2 68.1 102.6 65.8 83.8 50.1 67.1 103.7 111.6 94.5 84.5 77.8 38.7 72.4 70.2 116.2 108.3 94.7 98.0 102.6 102.5 47.3 55.9 33.5 70.3 88.0 114.0 80.0 77.6 82.6 54.6 56.5 30.0 73.2 110.3 93.6 67.2 63.9 59.7 68.9 73.8 68.4 63.2 96.9 92.2 70.0 59.9 67.8 74.0 68.9 105.9 88.0 March, F Apr., M 80.5 56.5 84.8 75.4 65.9 97.8 136.9 88.0 110.3 77.5 43.6 67.0 77.2 90.8 99.5 94.6 90.0 96.8 61.5 78.3 75.2 98.5 95.1 87.5 99.3 51.8 113.5 July, M 76.6 75.8 86.6 102.0 94.7 50.2 117.8 127.2 97.8 82.0 79.2 84.1 92.4 84.6 84.5 64.4 54.6 80.1 77.7 90.8 101.7 66.5 82.6 71.2 June, F 38.6 88.9 86.7 87.1 109.0 105.2 69.8 34.9 70.8 77.8 81.2 107.5 67.8 67.2 57.1 49.5 88.2 34.2 102.1 • Sec. 2.6 PERCENTILES, DECILES, AND QUARTILES 39 TABLE 2.62 Four-Day Gains of Female (F) and of Male (M) Guinea Pigs Described in Table 2.61 (Gains in grams. Negative sign denotes loss of weight.) Jan., F 4.8 5.7 7.3 15.9 17.3 19.1 Oct., M 0.6 3.7 3.8 0.6 11.3 31.5 21.1 15.2 22.0 18.9 3.8 1.1 - 0.7 8.1 26.0 25.8 25.6 20.4 2.9 1.7 4.6 0.5 19.2 14.4 12.0 17.6 6.8 2.5 Apr, ., F 20.4 -12.8 22.5 18.1 1.5 5.5 31.2 _ 4.2 24.8 6.7 5.5 22.0 12.1 12.5 31.2 Jan., M 2.8 13.0 21.3 23.5 12.9 21.2 28.7 4.1 3.1 June, F 7.5 8.0 7.8 10.9 15.8 28.1 6.9 0.9 -13.6 20.0 4.8 9.7 -30.0 9.3 3.9 7.3 6.1 July, M 21.4 21.2 0.3 -22.1 14.7 5.8 21.7 21.2 19.2 17.6 - 5.3 -23.3 21.4 5.3 Apr. ,, M 8.3 - 6.0 11.4 6.7 21.1 0.9 9.8 14.5 - 0.6 11.8 12.6 Sept., M Nov., F 2.8 1.3 12.1 17.3 12.1 30.9 -26.3 - 0.8 11.4 Feb.. F 1.9 15.2 8.5 12.7 13.6 4.5 - 0.4 4.4 6.1 20.4 7.3 16.6 21.7 3.7 -21.7 Dec, F 0.2 13.5 18.6 10.8 19.9 14.8 25.6 13.6 - 0.2 May, F 15.7 15.8 12.0 21.1 26.4 8.4 13.6 4.5 3.3 13.9 8.7 18.7 10.8 -10.3 10.4 6.3 3.6 June, M 14.9 13.6 4.8 9.5 2.6 12.0 17.3 5.4 8.9 10.7 16.0 5.8 13.7 4.2 -10.7 4.8 19.9 ii.e 12.4 12.8 6.0 14.7 4.5 - 7.9 10.0 10.1 15.4 12.1 13.8 17.7 7.2 24.6 - 5.4 - 1.6 7.9 7.7 15.4 20.0 13.5 5.7 23.3 - 8.0 9.0 20.9 12.3 16.8 16.1 18.7 8.6 -14.3 Feb., M 5.0 20.6 15.2 - 1.7 20.9 18.3 7.3 - 5.4 -29.2 0.7 15.4 13.7 - 3.1 27.5 - 2.3 7.1 1.3 1.8 17.8 13.4 - 2.6 16.3 22.8 15.0 Dec, M 4.4 10.1 7.4 14.1 - 4.8 -21.5 8.9 -16.5 13.8 5.1 9.4 20.3 4.9 - 2.3 28.5 26.1 15.2 12.1 8.9 12.5 13.2 Sept., F 23.4 8.6 20.7 March, F 7.2 25.4 10.6 10.4 - 8.1 13.0 11.0 18.1 5.1 3.8 25.1 - 1.6 17.4 10.6 14.8 13.7 15.6 13.6 3.9 20.5 8.9 19.8 -18.9 - 0.6 - 0.8 9.8 3.3 11.2 17.8 19.5 9.2 10.1 5.7 15.8 4.5 0.8 16.8 9.5 Aug., M 12.4 14.1 13.2 36.1 7.7 3.2 7.5 20.6 - 9.7 16.4 16.8 3.9 -12.3 July, F 9.4 2.8 10.8 29.4 2.1 - 1.9 May, M 8.4 9.1 14.7 11.1 Nov., M 2.4 0.1 20.7 15.9 Aug., F 10.1 10.4 5.2 29.5 5.8 20.4 14.6 11.0 11.5 12.5 6.9 26.0 11.4 18.5 8.0 2.4 6.5 9.2 10.7 - 3.3 -12.7 18.7 8.5 28.4 28.2 20.8 20.0 0.7 12.5 16.0 21.9 21.6 12.5 20.3 March, M 5.3 16.8 28.5 24.3 12.5 Oct., F 9.1 3.5 2.5 15.0 21.5 14.5 20.9 17.6 5.8 13.1 7.3 4.3 16.9 19.4 16.1 25.8 0.6 9.7 9.7 3.8 13.0 18.4 8.9 19.7 6.2 40 SUMMARIZATION OF DATA Ch. 2 1.00 Figure 2.61. 60 70 80 90 100 110 120 130 140 Birth weight in grams Graph of the relative cumulative frequency distribution for the data of Table 2.61. 2.7 THE COEFFICIENT OF VARIATION There is considerable need for a measure of relative variability in one set of numbers as compared with another when the units of meas- ure, or the levels of magnitude of measurement, are quite different. The standard deviation, the mean deviation, and the range are ex- pressed in the same units as those in which the data were taken; so they obviously reflect the general size of those units. Suppose that one bushel of a particular sort of crop weighs 60 pounds, on the average. Then the frequently used unit "pounds per 1/1000 acre" is but 0.06, or 6 per cent, of the size of the unit "bushels per acre." Hence to convert yields in pounds per 1/1000 acre into bushels per acre would require multiplication by 163^. To see what effect such a procedure has on the standard deviation, consider the general case of two measurements X and kX, where the k is any con- stant. For example, X could be the number of pounds per 1/1000 acre so that k = 16%. By formula 2.13, ax = vy2 {^Xf/N N is the standard deviation of X. For the measurement, kX, (TkX = N 2(fcX)2 - (XkX)yN k^i:X^ - A;2.(2X)ViV N = k- o-x- Sec. 2.7 THE COEFFICIENT OF VARIATION 41 In other words, the standard deviation of the yields in bushels per acre is 16% times that of the same yields expressed as pounds per 1/1000 acre. Hence, even though o- is an excellent and widely used measure of the variability exhibited by a group of numerical meas- urements, its size docs depend directly upon the units of measure involved, and also upon the level of magnitude of those measurements. To illustrate the point regarding the level of magnitude of measure- ment, suppose that one were interested in knowing if the weights of thirty-year-old males in Manhattan, Kansas, were more (or less) variable than the weights of twelve-year-old boys in that city. Sup- pose also that the average weight of the men is known to be approxi- mately twice that of the youths. The analysis just presented shows that if the boys' weights were each to be doubled so they would be on a level comparable to that of the men, their standard deviation automatically would be doubled too. It does not seem reasonable that doubling all of the A''s in a set of measurements should change their fundamental variability relative to another set of measure- ments; hence there is need for a measure of variability which would not be so affected. The coefficient of variation is that sort of measure of relative variability. It is easy to see that the mean of kX is k times the mean of .Y be- cause ix„x = ^(kX)/N = k-${X)/N = k^x- Therefore, the ratio of the standard deviation to the arithmetic mean will be a measure of relative variability in a useful sense because <^kx _ kax _ (^x f^kX kfJLx M.Y regardless of the size of kij^O) . It is customary to express this ratio of the standard deviation to the arithmetic mean as a percentage, and to define the coefficient of variation (CV) by (2.71) CV = 100(t/m. To illustrate formula 2.71 from previously discussed data, the stu- dent can verify that, for ACE scores, CV = 27.3; for the birth weights of guinea pigs, CV = 25.2; and for problem 5, section 2.5, CV = 18.9, each as a per cent. A person acquainted with ACE scores might then observe that the scores at Kansas State College during 1947 were relatively more variable than the national scores, which (it is sup- posed for illustration) had CV = 20 per cent. Concerning the birth weights, we might learn that some other group of these animals has a standard deviation of only 15 grams, and hastily (and erroneously) 42 SUMMARIZATION OF DATA Ch. 2 conclude that they were more uniform in weight than those whose weights are reported in Table 2.42. However, if the second group has a mean weight of /x = 50 grams, it then is apparent that CV = 100(15/50) = 30 per cent. Hence Professor Ibsen's guinea pigs had less variability in weight at birth than the other group of guinea pigs when account is taken of the fact that they were generally heavier. PROBLEMS 1. Compute the coefficient of variation for eacli of the following and draw appropriate conclusions : X (N. Y. Curb Issues): 4, 3, 88, 1, 108, 42, 1, 25, 18, 5, 3, 6, 2, 22, and 70; Y (Bond Quotations): 88, 115, 104, 113, 119, 80, 66, 40, 31, 101, 48, 43, 100, 84, and 15. 2. Using the X^ as -2, 5, 8, 3, 1, 0, -2, 4, 3, and 6, and using k = 2, demon- strate the (Tj^-^ = k-ff^. 3. Suppose that a group of measurements of the yield of corn in a certain area of Iowa had yu = 70 bushels per acre, with o- = 10 bushels, whereas an area in Kansas, growing the same variety of com and employing the same agronomic method of culture, gave yields with /a = 40 bushels per acre and a = 8 bushels per acre. Are the yields in that part of Iowa relatively more variable than those in Kansas, less variable, or about the same, according to these data? 4. Suppose that during a certain period of years the prices of a certain com- modity averaged $1.25, with standard deviation of 25 cents. The prices of this same commodity during another period averaged but 80 cents, with a standard deviation of 10 cents. During which period were the prices of this commodity relatively more stable? Give reasoning. 5. Following are simulated breaking strengths of samples of concrete (in hun- dreds of pounds per square inch) : 40, 65, 50, 33, 48, 57, 60, 52, 50, 46, 70, 55, 51, 41, 49, 53, 56, 44, 47. 50, 46, 53, and 55. Compute the coefficient of variation. 6. In problems 7 and 8, section 2.1, for which direction finder were the read- ings relatively less variable? Would that result bear on the choice of one instrument over the other if such a choice were to be made? 7. Use the data of Table 2.61 to determine if the birth weights of female guinea pigs born during January and February were relatively more or less variable than those of males born during the same period. 8. Solve as in problem 7 for June and July considered together. 9. Use Table 2.62 to determine if the four-day gains of the males born during January, Februaiy, and March were relatively more or less uniform than those of females born during the same period. 10. Solve as in problem 9 for animals born during October, November, and December. Sec. 2.8 PROBLEMS CREATED BY DRAWING SAMPLES 43 2.8 SOME OF THE PROBLEMS CREATED WHEN ONLY A SAMPLE OF A POPULATION IS AVAILABLE FOR STATISTICAL STUDY Suppose that we wished to study ACE scores of college students but could not afford the time or the expense required to analyze all their scores, and hence took only a portion of them, say 50. Although an economy of time and money will be obtained, several new prob- lems will be created. First, how should the 50 students be chosen for the sample? Ideally, they should be representative, in all important respects, of the whole group which is being sampled. But this cannot be ascer- tained without studying the ACE scores of the whole group — and then no sampling would be needed. If the first 50 on an alphabetical list were to be taken, the Macintoshes, McTaverishes, Swensons, and Swansons never would be chosen ; and they might differ fundamentally from those who would be chosen. If the first 50 who came into the counseling bureau were taken as a sample, they might differ as re- gards ACE scores from those who came in later, or who never came into the bureau at all. In view of these and similar dangers of acquiring a biased sample from such procedures, it is necessary to devise a sampling method such that every eligible student has an equal and independent opportunity to be chosen in the sample. The net result of these requirements is to make it true that every pos- sible sample of the chosen size (50 in the example above) will be equally likely to be drawn. This is the fundamental requirement of random sampling. There are various ways to draw a random sample of 50 from among 1290 members of a population. One would be to assign each person who took the ACE test a different number, place these numbers on pieces of cardboard, and draw 50 of them at random (in the popular sense) from a bowl containing all of the pieces of cardboard. If the scores in the population are recorded in rows and columns, as in Table 2.01, we can assign numbers to the rows and to the columns, and then draw a row number and a column number at random as before. These two numbers together will uniquely designate a score for the sample. If this is done 50 times — ignoring any repeats of exactly the same row-column combination — this sample also will be a random sample because every possible set of 50 scores among the 1290 in the population will have had an equal opportunity to have been drawn. 44 SUMMARIZATION OF DATA Ch. 2 The following random sample of 50 scores from Table 2.01 was obtained by the second method described above: 131, 66, 117, 117, 145, 71, 118, 99, 128, 111, 95, 78, 88, 55, 86, 89, 97, 98, 87, 80, 100, 76, 124, 89, 79, 101, 89, 156, 111, 98, 103, 68, 110, 76, 99, 100, 102, 61, 50, 125, 92, 106, 63, 117, 124, 87, 95, 100, 58, and 99. These particular measurements were obtained by chance from among many possible different sets of 50. This fact suggests that the theory of probability is needed in the analysis of sampling data. It is found in the usual manner that the mean and the standard deviation for the sample above are 96.28 and 22.55, respectively. The range of scores in this sample is 106, the median is 98, and the coefficient of variation is 23.4 per cent. It is known that these sta- tistical measures are not likely to be exactly the same as the popula- tion parameters, but it is to be hoped that they are not far from those values. Another sample was drawn in the same manner as the sample just described. The following were calculated for this second sample: mean = 99.22, standard deviation = 27.30, range = 144, median = 100.5, and the coefficient of variation is 27.5 per cent. It is noted that each of these statistical measures is different for the two samples, yet only the ranges differ by a large percentage. It is typical of random samples that they usually differ from each other in several respects because the particular members of such samples are in the sample by chance. It also is true that the sizes of such statistical measures as the sampling mean will follow some predictable pattern over considerable sampling experience. If this were not true, nothing much could be learned from sampling. It will be seen in later chap- ters that probability theory is needed to study these matters. To illustrate the effect of the type of population on the results obtained from random sampling, consider two samples drawn from the data in problem 1 at the end of section 2.4. For convenience, samples of 10 numbers each were drawn even though this is a some- what larger fraction of the population than was taken from Table 2.01. These samples were obtained by considering that the fly counts were numbered serially from left to right, starting with the top row. There are 148 fly counts in this population; hence a sample of 10 was drawn by effectively drawing 10 numbers at random from among the numbers 1, 2, 3, ... , 148. When these 10 ordinal numbers were drawn, the corresponding fly counts were obtained by counting in Ch. 2 REVIEW PROBLEMS 45 the manner indicated above. Following are the summaries of the two samples: Sample 1 Sample 2 Mean = 23.2, median = 19 Mean = 13.4, median = 10 Standard deviation = 16.2, range = 45 Standard deviation = 9.7, range = 28 CV = 69.8 per cent CV = 72.4 per cent For these two samples, each of the five statistical measures is dif- ferent again. Moreover, considering the fact that these fly counts are generally smaller numbers than the ACE scores, the relative differences are much larger between the two samples than was true for the ACE scores. For example the mean for one sample of the counts is almost twice the size of the mean of the other sample. Much the same is true of each of the other measures except the coefficient of variation. This is an illustration of the fact that a statistical study of samples requires some information about the frequency distribu- tions of the populations sampled. Hence this matter, and probability, must be studied before more can be done about the analysis of sam- pling data. These are the aims of chapters 3 and 4. REVIEW PROBLEMS 1. The effectiveness of penicillin in controlling bacterial growth can be meas- ured by the "inhibition zone" produced when a standard amount of penicillin is properly added to a plate of agar containing the type of bacterial growth one wishes to study. Following are 54 such determinations arranged in 9 groups of 6 tests each. (From an article by Jackson W. Foster and H. Boyd Woodruff, Journal of Bacteriology, August, 1943.) Calculate the arithmetic mean of each set of tests, and then compute the standard deviation of these nine means. Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 Test 7 Test 8 Test 9 28.1 28.5 28.0 27.5 29.0 28.0 29.0 28.5 28.0 28.0 28.0 28.2 28.0 28.0 28.0 29.0 28.2 29.0 27.5 28.0 27.5 28.1 28.3 28.0 27.3 28.2 28.6 27.8 28.0 28.0 28.5 27.3 28.0 27.0 27.5 27.7 28.0 27.5 28.0 28.0 29.0 28.1 29.0 28.0 27.5 27.5 28.0 28.0 28.2 28.0 27.5 29.0 28.0 29.0 2. Determine the range for each set of data in problem 1, and compute the standard deviation for the set with the greatest range. Ans. 0.6, 1.0, 0.7, 1.0, 1.7, 0.6, 2.0, 1.0, 1.5; 0.88. 3. If a list of the farm acreages in a certain county in Kansas forms a statis- tical array of numbers from 35 to 4000; and if n =600 and md = 350 acres: (a) Which average would you think might be more typical of the size of farm in that county? (b) Would you expect the high point of the frequency distribution to be about over the mean, to its right, or to its left? Give reasons. 46 SUMMARIZATION OF DATA Ch. 2 4. Determine the median and also the upper hmits of the first and third quartiles for the data of problem 1 when all results are considered as one group of data. Alls. 28.0, 27.95, 28.22. 5. Take any newspaper which gives quotations from New York bond prices and make frequency and r.c.f. distributions for all the closing prices as listed for that particular day. 6. Graph the distributions for problem 5. Then determine graphically the proportion of the prices which exceed 100. Check this result by actual count. You are given the following information as a basis for working problems 7 to 11, inclusive. These data are from the Ohio Psychological Tests given to 602 students at Kansas State College during 1945. The scores are represented by the symbol X in the following summary, and are given only as integers: Score Class / Score Class / At least 111 3 55-61 70 104-110 9 48-54 74 97-103 16 41^7 72 For ungrouped data 90- 96 21 34^0 65 SX = 36,000 83- 89 32 27-33 49 SZ2 = 2,400,000 76- 82 49 20-26 12 69- 75 59 Less than 20 5 62- 68 66 Total 602 7. Make the r.c.f. distribution and graph it. 8. Compute approximations to fi and o-2 after changing the top and bottom classes to 111-117, and 13-19, respectively. Ans. 58.6, 423.0. 9. What percentage of the students had scores above 100? Between 50 and 75? What range of scores is included between the 80th and 90th percentiles? 10. Graph the frequency distribution and state which average you would employ to convey the best impression of the level of performance by these students on this standard test. 11. If 30 scores were to be selected at random from these 602, how many of them would you expect to fall at or below 40? 12. Given that N = 25, 2F = 110, and 272 = gOO for any particular set of measurements, calculate the coefiicient of variation. Ans. 48.9 per cent. 13. Given the following frequency distributions for a certain group of prices of bonds, construct gi-aphs of these distributions. Briefly describe the sorts of information which can be obtained from such figures, and give several illustrations. Price Class / r.c.f. Price Class / r.c.f. At least 121 5 1.00 85-90.99 70 .25 115-120.99 10 .99 79-84.99 30 .11 109-114.99 40 .97 73-78.99 15 .05 103-108.99 90 .89 67-72.99 7 .02 97-102.99 130 .71 Less than 67 3 .01 91- 96.99 100 .45 Total 500 Ch. 2 REVIEW PROBLEMS 47 14. Calculate the median and the lower limits of the second quartile and of the 85th percentile of the data in problem 13. Ans. 98.14, 90.99, 107.32. 15. What percentage of the bonds of problem 13 had prices below 100? 16. Within what limits do the prices of the "middle" 60 per cent of the bonds lie (that is, excluding the lower and also the upper 20 per cent of the prices in the array)? Ans. 88.85 to 105.99. 17. The figures recorded below are the batting averages of all American Asso- ciation players who were at bat at least 100 times, as reported by the Kansas City Star on July 29, 1951. The averages are arrayed from highest to lowest. Make a frequency distribution table, and compute /j, and a. Aver- Aver- Aver- Player age Player age Player age Walker .389 Federoff .295 Lerchen .266 Cerv .354 Thorpe .292 Daugherty .265 Crowe .335 Segrist .291 Markland .262 Thompson .335 Mangan .290 Marchio .262 Sullivan .332 Beard .288 Montag .262 Wright .323 Milne .287 Scherbarth .262 Richter .322 Bollweg .286 Atkins .260 Katt .322 Lyons .284 Henley .260 Clarkson .322 Unser .283 Lund .259 Dandridge .317 Barnacle .282 Basso .256 Whitman .315 Pendleton .280 Antonello .256 Benson .314 Brancato .280 Hoak .254 Reed .312 Deal .279 Marshall .254 Mordarski .311 Chapman .279 Fernandez .252 Hoderlein .310 Marquis .278 Ivropf .250 Mavis .310 Tipton .278 Turner .250 Carey .309 Zauchin .277 Ruver .248 Campbell .309 McQuillen .277 Conway .247 Courtney .309 Ozark .276 Lu Cadello .243 O'Brien .307 Dallasandro .276 Aliperto .237 Broome (L) .307 Stevens .274 Natisin .235 Saffell .305 Thomas .274 McAlister .234 Kalin .304 Cole .274 House .231 Mozzali .304 Wright .274 Teed .230 Cassini .301 Olmo .270 Thomson .223 Merson .300 Klaus .269 Rocco .212 Repulski .300 Gilbert .268 Morgan .179 Broome (C) .295 De La Garza .267 Okrie .165 18. Referring to problem 17, write numbers from 1 to 84, inclusive, on pieces of pasteboard, mix them well, and then draw 5 at random. Consider each number drawn as the rank of a person's batting average in the above list, starting at the top. Do this 10 times, and record the range of batting averages in each set of 5 so drawn. Compare the average range with the standard deviation obtained in problem 17. If another such set of batting averages had twice as large a standard deviation, what general effect do you think this would have on the range? 48 SUMMARIZATION OF DATA Ch. 2 REFERENCES Dixon, Wilfrid J., and Frank J. Massey, Jr., Introduction to Statistical Analysis, McGraw-Hill Book Company, New York, 1951. Freund, John E., Modern Elementary Statistics, Prentice-Hall, New York, 1952. Kenney, John F., Mathematics of Statistics, Part I, Second Edition, D. Van Nostrand Company, New York, 1947. Neiswanger, W. A., Elementary Statistical Methods, Macmillan Company, New York, 1943. Waugh, Albert E., Elements of Statistical Method, Second Edition, McGraw- Hill Book Company, New York, 1943. CHAPTER 3 Elementary Probability Several relatively small populations of data have been studied be- cause it is not feasible to use large groups of data in the classroom. Quite commonly, populations actually involve a very large number of numerical measurements; so large, in fact, that their number can be considered as infinite without doing appreciable violence to the subsequent analyses. Obviously, no more than a portion (sample) of the measurements in an infinite population can be obtained for study. Sampling theory requires certain probability considerations and some definite assumptions regarding the distribution of the meas- urements in the population (as noted in section 2.8). Hence it is appropriate to consider some of the more basic and widely used fre- quency distributions before attacking the problems of sampling. That is done in this and the following chapter. Probability is involved whenever the occurrence, or non-occurrence, of any anticipated event is dependent to some degree upon chance. An "event" can be any sort of occurrence or non-occurrence which has been specified in advance. In the classroom, red and green marbles might be placed in a sack, thoroughly mixed, and one drawn out without looking into the sack. The drawing of a green marble could be considered as the event E in this instance. Likewise, if a bridge deck is thoroughly shuffled and one card drawn at random from it, the appearance of the ace of spades then might be the event E. Another wide application of probability in everyday life lies in the determination of the premiums for life insurance policies and annuities. If a man aged 35 years purchases an annuity which will pay him $100 per month starting at age 60 if he is alive, there are three major matters to be considered: (a) interest on the money involved, (6) the probabilities that the man will live to receive each successive payment, and (c) operating expenses and a fair profit for 49 50 ELEMENTARY PROBABILITY Ch. 3 the company. Whether or not such a person does live to receive a particular payment must be regarded as a chance event and, there- fore, requires some use of the theory of probability. Public opinion polls regarding political matters, buyers' preferences, and foreign affairs involve chance in the selection of the persons who are to be interviewed. The reader should be able to think of many other everyday events in which the theory of probability is involved. 3.1 THE DETERMINATION OF PROBABILITIES Before a method is presented for determining the probability that an event E will occur under specified conditions it is useful to dis- tinguish between what will be called single events and classes of events. For the purposes of this book this distinction can be made by means of examples. Suppose that two dice are placed in a can, shaken vigorously, and rolled out upon a flat, hard surface. Many "events" can occur with each die, but just six usually are of interest: a 1, 2, 3, 4, 5, or a 6 appears on the upper face of each die when it stops rolling. How the dice were turned when they were thrown, where on the surface they came to rest, or how many turns they made while in motion are ordinarily of no interest. Moreover, it would be at least impracticable, if not impossible, to relate those phenomena to the number of dots on the upper face of a die. Hence the six possible events which will be considered herein are the ap- pearance of a 1, 2, 3, 4, 5, or a 6 on the upper face of each die. Since these events cannot be further decomposed, we shall refer to them as single events. If, with each die, these faces tend to appear with equal relative frequencies over many trials, the dice are each said to be unbiased. It is with single events occurring with equal relative fre- quencies that we shall be primarily concerned in the subsequent dis- cussion. If both dice are considered simultaneously and an event is considered to consist of a number on one die and a number on the other die, thirty-six single events are possible because any one face on the first die can appear with any of the six faces on the other die. Each possible pair of faces defines an observable event. If attention is turned to the sum of the numbers of dots appearing on the upper faces of two dice w^hich have been thrown simulta- neously, any one of eleven different sums is possible. The different possible sums define eleven classes of single events (occurring with equal relative frequencies). For example, the class of events (com- Sec. 3.1 THE DETERMINATION OF PROBABILITIES 51 posed of such single events), "sum = 7," contains the following single events : 1 on die 1, 6 on die 2 2 on die 1, 5 on die 2 3 on die 1, 4 on die 2 6 on die 1, 1 on die 2; 5 on die 1, 2 on die 2; 4 on die 1, 3 on die 2. The class, "sum = 2," includes but one single event because there is but one way that it is possible to get a sum of 2. The class, "sum = 3," includes two single events: a 1 on die 1, a 2 on die 2; or a 2 on die 1, a 1 on die 2. The class, "sum = 4," includes three single events, etc., until all thirty-six of the possible single events have been put into one of the eleven classes of events. We could define other classes of events among the thirty-six single events possible when two unbiased dice are tossed. For example, we could have class 1 = "sum = 7" and class 2 = "sum is not = 7." There are six single events in class 1 and thirty single events in class 2. The preceding discussion has brought out the fact that single events and classes of events differ in one important respect. The single events are expected to occur with equal relative frequencies over many trials under the specified conditions, whereas the classes of events consist of groupings of single events, and hence would be ex- pected to occur with relative frequencies which depend upon the numbers of single events in the classes. Upon the basis of the preceding discussion, a useful method for determining probabilities can be devised for instances in which the single events occur with unequal relative frequencies. Suppose that under certain specified conditions any one of A^^ possible single events can occur and that they form an exhaustive set; that is, some one of these single events must occur on any trial under the specified condi- tions. Assume also that the single events are grouped into s non- overlapping classes of events, with Ui in class 1, W2 in class 2, . . . , and with Wg in class s. Then the probability that the single event which actually does occur on one future trial will belong to class i {i varies from 1 to s) is given by (3.11) PiEi) = ni/N. As an illustration of the use of formula 3.11 consider the dice prob- lem discussed above in which thirty-six single events are possible. Certain classes of events, the single events which each class includes, and the probabilities associated with each class of events are given in Table 3.11. 52 ELEMENTARY PROBABILITY Ch. 3 TABLE 3.11 Single Events and Classes of Events Involved When Two Unbiased Dice Are Thrown, and the Probabilities Associated with Those Classes of Events Classes of Events Single Events Die 1 Die 2 Number of Single Events Probabilities for Classes Sum = 2 1 1 1 (= m) 1/36 Sum = 3 1 2 2 1 2(=n2) 2/36 Sum = 4 1 3 2 3 1 2 3 (= 713) 3/36 Sum = 5 1 4 2 3 4 1 3 2 4(=n4) 4/36 Sum = 6 1 5 2 4 3 5 1 4 2 3 5(=n5) 5/36 Sum = 7 1 6 2 5 3 4 6 1 5 2 4 3 6(=n6) 6/36 Sum = 8 2 6 3 5 4 6 2 5 3 4 5(=n^) 5/36 Sum = 9 3 6 4 5 6 3 5 4 4(-n8) 4/36 Sum = 10 4 6 5 6 4 5 3(=ri9) 3/36 Sum = 11 5 6 6 5 2 (= nio) 2/36 Sum = 12 6 6 Xm = N 1 (= nil) = 36 1/36 Sec. 3.1 THE DETERMINATION OF PROBABILITIES 53 Other classes of single events could be defined, of course, such as the two classes: "sum ^ 5" and "sum > 5." From Table 3.11 it is apparent that 10 of the 36 possible single events produce sums which are less than or equal to 5, whereas the remaining twenty-six single events yield sums which are greater than 5. Therefore, in this case, N = 36, s = 2, rii = 10, and ^2 = 26; so that F(sum = 5) = n2/N = 26/36 - 13/18, or = .72. Two useful facts are derivable from formula 3.11: (3.12) < P(-E') < 1 because no n^ can be larger than A^; and, (3.13) P{E) + P(not E) = 1 because rii/N -\- {N - n^)/N = N/N = 1. Other laws follow from formula 3.11. Two of the more important theorems will be proved and illustrated. Suppose that Ei and E^ denote two mutually exclusive classes of events; that is, single events in classes E^ and E'o cannot occur simultaneously on any one trial. Suppose also that there are Wi and no single events in classes Ei and £Jo, respectively. If a total of A^ single events is possible, the prob- ability that an event in either class Ei or class E2 will occur on one random trial is, by definition, (3.14) P{Ex or E2) = (ni + n2)/N = Ui/N + ng/iV = P(E,) + PiE2). The same reasoning and algebra are sufficient to show that for r classes of events: Ei, E2, . . . , Er with Ui single events in class Ei (i = 1 to r) , the probability that some one of the mutually exclusive events Ei, E2, . . . , Er will occur on one random trial is given by (3.15) P{E,,E2,---,ovEr) = P{E0+P(E2)-h----\-PiEr). This result is known as the Law of Total Probability for Mutually Exclusive Events. To illustrate formula 3.15, suppose that a sack contains 10 green, 15 red, 5 white, and 20 purple marbles, all identical save for color. What is the probability that a colored marble will be drawn on one future random trial? Let Ei stand for the drawing of any one of the green marbles. There are 10 green marbles, and each is equally likely to be drawn; hence there are 10 single events in the class Ei. Also, let Eo represent the drawing of any red marble, E-i stand for the drawing of a white marble, and E4 equal the drawing of any purple 54 ELEMENTARY PROBABILITY Ch. 3 marble. Then Wi = 10, ^2 = 15, ns — 5, and n^ = 20; hence P{Ei, E2, or ^4) = (10 + 15 + 20) /50 = 10/50 + 15/50 + 20/50 = .90. In the discussion leading to the Law of Total Probability the events El, E2, . . . , Er were assumed to be mutually exclusive, that is, only one of those events could occur on any one random trial. Suppose now that Ei and E2 are independent events in the sense that each can occur simultaneously on one trial without interfering or helping with the occurrence of the other in any way. For example, the obtaining of a 6 on one die and a 5 on another die on a single throw of the pair of dice is an illustration of independent events. If Ei and E2 are independent events, they can occur together in ^1-772 combinations of single events because each of the rii single events in Ex can occur with each of the n2 single events in E2. The classes of events, Ei and E2, will each belong to a general class of events, which will be supposed to contain Ni and A'^2 single events, respectively. There- fore, the total number of combinations of single events possible on random trials now is Nx-N2. Of those possible single events, ni-n2 will belong to both Ex and E2. Therefore, the probability that an even in Ex will occur simultaneously with an event in E2 is given by P{Ex and ^2) = (.ni-n2)/{Nx-N2) = (ni/7Vi) • (ris/iVa) = P{Ex)-PiE2). As an illustration of the above discussion and results, suppose that a game consists in throwing a penny and an unbiased die simul- taneously, with the thrower winning if he throws a head on the penny along with a 5 or a 6 on the die. Let Ex represent throwing a head on the coin, and E2 throwing a 5 or a 6 on the die. A single event now consists of a particular result on the coin plus a particular result of throwing the die. The coin can turn up either of two ways, the die any of six ways; hence there are 2(6) = 12 combinations of events, each equally likely to occur on any one trial. In these cir- cumstances, nx = 1, 712 = 2, Nx = 2, and A''2 = 6; hence (1)(2) P(H and a 5 or a 6) = — — - = (1/2) (2/6) = 1/6. (2)(6) The reasoning and algebra above can be extended easily to prove that, if the occurrence of events in classes Ex, E2, ■ . ■ , and E,- are independent and can occur in Wj out of Ni ways, respectively {i = 1 to r) , the probability of the simultaneous occurrence of these r events Sec. 3.1 THE DETERMINATION OF PROBABILITIES 55 on one future random trial can be obtained from the following- formula : (3.16) P{E,, E2, ■■■, and Er) = P{E,)-P{E2) • ■ ■ P{Er). This result is known as the Law of Co7npound Probability for Inde- pendent Events. A similar and more general law than (3.16) can be established for situations involving dependent, rather than independent, events. Suppose that the occurrence of event E2 depends on another event El having occurred previously, or that it is useful to regard Eo's oc- currence as depending on the prior happening of Ei, perhaps for the sake of convenient computation. For example, suppose that a bridge deck is to be well shuffled, and then two cards drawn successively and at random without replacing the first card drawn. What is the probability that the second card to be drawn will be an ace? It should be apparent that the answer depends somehow on the out- come of the first draw from the deck so that the second event is dependent upon the first event. To attack the problem rather generally, suppose that n single events are possible under a given set of circumstances and that an event Ei is associated with ni of these n single events. Assume also that an event E2 occurs on ni2 of the events on which Ei also occurs. Then the probability that both Ei and E2 will occur on one trial is PiEiEi) = ni2/n, which can be rewritten in the following way: (3.17) PiEiE^) = — = (-—) = P{Ei) -PiE^/E,), n \n / \ni / where P(E'o/£'i) is the probability that E2 will occur after it is known that El has occurred. The probability law expressed in (3.17) actually includes the law of (3.16) as a special case. If E2 is independent of Ei, the number of single events on which E2 can occur will be the same regardless of the prior occurrence or the non-occurrence of E^; hence the prob- ability P{E2/Ei) will be just P{E2), and the formula 3.17 becomes (3.16). Problem 3.11. What is the probabihty that two successive aces will be drawn from a well-shufBed bridge deck if the first card drawn is not replaced before the second draw is made? The number of aces available for the first draw is 4, and any of 52 cards might be drawn; hence P{Ei) = 4/52. On the second draw 56 ELEMENTARY PROBABILITY Ch. 3 — assuming that an ace was drawn on the first draw — there are 3 aces among 51 cards remaining in the deck. Hence, PiEo/Ei) = 3/51; then, by formula 3.17, P{A, A) = (4/52) • (3/51) = 1/221. There are many situations in which those chance occurrences which would be considered as the single events do not occur with equal relative frequencies. For example, a coin may be biased so that the heads side turns up more frequently than the tails side. Un- der such conditions, we cannot assign a probability of 1/2 to the occurrence of heads (and likewise for tails) and employ the simple arguments used above. However, we can think of determining the appropriate probabilities for these single events by empirical means, that is, by many actual trials under the specified conditions. For example, we could toss the coin in question many times and then use the observed proportion of heads as an approximation to the true probability, p. Thereafter, formulas 3.15, 3.16, and 3.17 can be used. An interesting and instructive application of the probability meth- ods introduced above can be made to the study of human blood groups. If the red blood corpuscles of one individual's blood are mixed with the blood serum of another person (as in transfusions), one of two general results will be observed to follow: the red cor- puscles will disperse evenly through the recipient's blood as though in their own serum, or they will form clumps of cells. The latter reaction is called agglutination, and it is so undesirable that there is considerable interest in preventing its occurrence. To that end, bloods are classified according to certain systems. One such system is based on the known existence of factors A and B each (or both) of which may be either present or absent from any person's blood. The following four blood groups are based on the A and B factors: (1) type 0: neither A nor B present in the blood; (2) type A: factor A present but not factor B; (3) type B: factor B present but not factor A; (4) type AB: both of the factors A and B are present in the blood. There are several interesting features about the A and B factors in blood. (1) They are inherited essentially in accord with simple Mendelian laws of inheritance, a circumstance which requires meas- ures of probability. (2) Various racial or geographic groups tend to differ from each other in the proportions carrying the A and/or B factors, thus providing a source of some additional evidence about racial origins. (3) The mode of inheritance of the A-B groups can be used in genetic studies and in some legal problems. Sec. 3.1 THE DETERMINATION OF PROBABILITIES 57 Each person is considered to have in the cells of his body twenty- four pairs of chromosomes, one member of each pair having come from each parent. On these chromosomes are carried genes which are believed to govern the inheritance of various human character- istics. The genes which determine the presence or the absence of the A and B factors in the blood are carried on one of the twenty-four pairs of chromosomes. Attention here is centered solely on the chro- mosomes of that pair, one of which came from the father, the other from the mother. Moreover, attention is to be fixed upon one spe- cific gene position on each such chromosome, namely, that position occupied by the genes which cause the presence or the absence of the A and B factors. If the gene at this position produces neither the A nor the B factor in the blood, it is marked (diagrammatically) as an gene. Similarly there are A and B genes so that the blood type can be indicated by showing what genes the two chromosomes carry. Symbolically, there are the following four blood types: 0/0 = type 0; A/0 or A/A = type A; B/0 or B/B = type B; and A/B = type AB. The information presented in the preceding paragraphs makes it possible to predict the proportions of the various blood groups among the progeny of any particular combination of parents, provided that a large number of such parents and children are involved. Suppose that one parent has blood of type AB and the other has type blood. Then the possible blood types which can occur among their offspring are as follows: Father Mother Parents A/B X 0/0 Genes passed on A or B or B/O Possible offspring B/O A/0 [A/0 Of the four possible pairings of chromosomes from the father and the mother, two produce type B blood in the child because only the gene for the factor B is carried on the chromosomes. Similarly, the other two possible pairings of chromosomes produce type A blood in the children. There is no reason to doubt the usual hypothesis that each of these four possible pairings occurs the same percentage of the 58 ELEMENTARY PR0B7\£ILITY Ch. 3 time in the long run of many such matings ; hence the individual pair- ings are considered to be single events. It follows that the probability that any specified future child of these parents will have type B blood is P(B) = 2/4 = 1/2. Similarly, P(A) = 1/2. There is no possibility that these parents will produce a child of either type or type AB; hence P(AB) = P(0) = 0/4 = 0. In view of the fact that the inheritance which a child receives from its father is an independent event with respect to its inheritance from its mother, formula 3.16 can be applied. This gives the follow- ing simple solution for the probability that a child with blood type B will be produced: P(B from father) = 1/2; P(0 from mother) = 1; therefore, P(B from father and from mother) = (1/2) (1) = 1/2. Since that is the only way a child with B-type blood can be produced by these parents, that is the solution to the problem. If one parent is type and the other type A, something must be known or assumed regarding the specific type A, that is, A/0 or A/A. If one parent is A/0 and the other is type 0/0, P( A) = P(0) = 1/2. No other type is possible. But if the parent with type A blood is actually A/A, all children will be A/0. If it is known only that the parents are of types A and 0, respec- tively, and if it assumed that type A is equally frequently A/0 and A/A, the probability that any particular future child will be type A is P(type A child) = P(A/0 parent) -PIA child/ (A/0 parent)] + P(A/A parent) •P[A child/ (A/A parent)] = (l/2)-(l/2) + (l/2)-(l) =3/4 by the probability laws of (3.15) and (3.17). Table 3.12 was derived by the above methods under the assump- tion that type A is equally frequently A/A and A/0; and similarly for type B. (Actually this assumption is unrealistic, but it is con- venient here.) The reader should verify several of the probabilities in this table, noting particularly where the assumption regarding the relative frequency of A/A and A/0 among type A parents (or likewise for type B) affects the calculations. Sec. 3.1 THE DETERMINATION OF PROBABILITIES 59 TABLE 3.12 .ND PrOPOETIONS OP OFFSPRING FROM THE ! Indicated M Mother's Father's Type Type Types of Progeny Probability of That Type of Progeny 1 A A 1/4 3/4 B B 1/4 3/4 AB A B 1/2 1/2 A A 1/4 3/4 A A 1/16 15/16 B A B AB 1/16 3/16 3/16 9/16 AB A B AB 1/2 1/8 3/8 B B 1/4 3/4 A A B AB 1/16 3/16 3/16 9/16 B B 1/16 15/16 AB A B AB 1/8 1/2 3/8 AB A B 1/2 1/2 A A B AB 1/2 1/8 3/8 B A B AB 1/8 1/2 3/8 AB A B AB 1/4 1/4 1/2 60 ELEMENTARY PROBABILITY Ch. 3 Another method of classifying human bloods is based on the M and N factors, which are inherited independently of the A and B factors; that is, the genes for M and N are on a different pair of chromosomes from that which carries the gene for A and B. Appar- ently, both M and N are never both absent. There are, therefore, just three types: M, N, and MN if we ignore the subtypes which have been discovered recently. The following symbolism will be employed in the discussion of the M-N blood types: M/M = type M, N/N = type N, and M/N = type MN. The inheritance of these types can be studied in the manner already established for the A-B blood groups. In view of the fact that the three M-N types are classifications which are independent of a person's A-B type, the two blood group- ings considered simultaneously make it possible to distinguish 3 X 4 = 12 different blood types even without bothering with the subdivisions of the A-B and M-N groups, which are serologically determinable. Problem 3.12. If a woman's blood belongs to types and MN, and her hus- band's blood is AB and N, what are the possible blood types for their first child, and what is the probabiUty associated with each type? The mother can pass on to her child one of the pairs of genes 0, M OT 0, N because her genetic constitution as regards blood types is (0/0) (M/N). Likewise, the father can transmit either A, N or B, N to his offspring. Therefore, the possible gene combinations in the child are: 0, M with A,N;0,M with B,N;0,N with A, N; and 0, N with B, N. This makes the following four classifications of bloods possible for their first child: A-MN, A-N, B-MN, and B-N, each being expected to occur equally frequently. Hence, each of these four classifications has a probability of one-fourth occurrence in their first child. No other type can occur. Two other general blood groups will be mentioned: one involves the Rh factor, the other is based on the P factor. Each is inherited independently of the other and of the A-B and M-N types. Recent discoveries of subdivisions of these groups will be ignored. There- fore, P+ and P— groups will be recognized, and also Rh-|- and Rh — types. Genetically, PP and Pp are P + , and RhRh and Rhrh are Rh-|-, leaving pp as P— and rhrh as Rh — . Sec. 3.1 THE DETERMINATION OF PROBABILITIES 61 The information given previously on blood groups can be sum- marized as follows: Blood Typing System I(A-B) II(M-N) III(P) IV(Rh) 0/0 = group O M/M = type M P/P Rh/Rh or or A/O N/N = type N P/p = type P+ Rh/rh = type Rh+ or A/A = group A M/N = type MN p/p = type P- rh/rh = type Rh- B/0 or B/B = group B A/B = group AB It is seen that there are 4 X 3 X 2 X 2 == 48 mutually exclusive and serologically distinguishable blood classifications, even with the simplified groups discussed herein. By using all the known sub- groups of blood types, there are many more distinguishable and mutu- ally exclusive classifications of human bloods. The availability of these classifications has been helpful in legal cases involving dis- puted parentage, heirship claims when alleged maternity is doubted, identification of blood stains, genetic studies, anthropological inves- tigations, and the identification of corpses when other methods have failed. Apparently the chief use of blood types in legal cases occurs when one can prove the impossibility of an allegation. For example, an 0-type father and an AB-type mother cannot (under the informa- tion set forth above) have an 0-type child. Or, as another case, if an accused person has blood stains on his clothing and claims that they resulted from his having had a nosebleed, the finding that his blood is A, M whereas the stains are A, N would disprove his claim. PROBLEMS 1. What is the probability that a wife with type A blood and a husband with type B blood will have three children whose blood types all are O? 2. Suppose that a husband has type A blood and that his wife's blood group is AB. What is the most likely type of blood for their first child under the assumptions made in Table 3.12? W^hat type is impossible? Ans. A, O. 3. If the parents of problem 2 claim five children all of blood type B, would you doubt the blood types or, perhaps, the alleged parentage? Give probability argument. 62 ELEMENTARY PROBABILITY Ch. 3 4. If a name is to be selected at random from among those persons who were residents of the United States in 1950 and who then were between the ages of 35 and 74, inchisive, estimate the probabiHtj^ that the person so chosen will be aged 50 to 59, inclusive. (See problem 8, section 2.5 for the appropriate table.) Ans. .26. 5. Compute the probability of throwing either a sum of seven or of eleven on one throw of two unbiased dice by enumerating the single events in these two classes of events. Verify your answer by applying the Law of Total Probability. 6. If five unbiased coins are to be flipped simultaneously, calculate the proba- bility that there will be a 3:2 division of heads and tails, either way. Ans. 5/8. 7. Verify the probabilities given in the second and tenth lines of Table 3.12 by listing all the possible combinations of chromosomes. Where does the matter of single events come into these calculations? 8. Use the laws of Total and Compound Probabilities to solve problem 7. 9. Suppose that two bags— identical in appearance — contain, respectively, 20 red and 30 blue marbles; and 40 red and 10 blue marbles. If one bag is to be selected at random and then one marble withdrawn from that bag, what is the probability that it will be red? That it will not be red? 10. If three unbiased dice are to be thrown once, what is the probability that a sum of 4 will be thrown? A sum of at least 4? Ans. 1/72, 215/216. 11. If the throw described in problem 10 is to be made twice, what is the probability that a sum of 4 will be thrown both times? What is the probability that exactly one sum of 4 will be thrown on the two throws? 12. Suppose that two babies have been born almost simultaneously in a cer- tain hospital, and that one of the families subsequently claims that the babies were interchanged either willfully or accidentally. The blood classes of the babies and of the parents are as follows: Mr. Timofeef is A, MN, P+, and Rh + ; Mrs. Timofeef is B, N, P+, and Rh-; Mr. Brown is B, M, P+, andRh-; Mrs. Brown is O, N, P - , and Rh - . The child the Timofeefs now have is O, MN, P-, and Rh + . The child the Browns now have is 0, MN, P+, and Rh — . Have the babies been inter- changed? Or is it impossible to tell from this information? Give reasons. Ans. No interchange has occurred. 13. Suppose that a few days after a wealthy man has died a woman claims that a certain girl is her daughter and that the deceased was the father. Also suppose that the following facts about blood classes have been established: (1) The deceased's blood was B, M, Rh^-, and P+. (2) The deceased had a son whose blood was in group O and also was Rh — . (3) The alleged mother's blood is A, MN, Rh+, and P+. (4) The girl's blood is O, M, Rh-, and P-. What conclusions can you draw about the paternity of the girl? Justify your statements with probability evidence based on the following assumptions: (1) For a person whose blood is B it is assumed that the chances are two out of three that the specific type is B/O if no other pertinent information is available Sec. 3.2 PERMUTATIONS AND COMBINATIONS 63 to change th\s assumption. (2) Similarly, for type A. (3) The probability that a person who tests Rh+ is specifically Rh/rh is 2/3, and similarly for P+ if there is no other information available which would change the probability. 14. Suppose that a tortoise (land turtle) is wandering at random on a 50 by 50-foot lawn enclosed by a fence. He is equally likely to be on any particular square foot of lawn one could designate in advance. What is the probability that at any specified future time he will be within 10 feet of the fence? If it is known that he is not within 10 feet of the south fence, what is the probability that he is not within 10 feet of any of the four fences? Ans. 16/25, 9/20. 15. Ignoring the refinements, the Rh factor is inherited as described above. The discovery of this factor in 1940 led to an explanation of one type of infant mortality, erythroblastosis. In a large majority of the cases, the father is Rh+, the mother is Rh — , and the child is Rh+. Only a fraction of the cases wherein the child is Rh+ and the mother is Rh — , which are potentially erythroblastotic, actually result in trouble; but why some do and others do not is not presently known. Obviously, the father could be either Rh/Rh or Rh/rh, but the mother must be rh/rh. Assume that the population of potential parents is divided for each sex as follows: 30 per cent RhRh, 60 per cent Rhrh, and 10 per cent rhrh What is the expected proportion of potential erythroblastotics among their children? 3.2 PERMUTATIONS AND COMBINATIONS Probability has been calculated in such a way that two numbers need to be determined: (a) the number of single events in the class of events whose probability of occurrence is being determined, and (b) the total number of single events possible under the prescribed conditions or in the mathematical model. For example, the prob- ability of throwing a sum of seven with two unbiased dice is the ratio of the number of single events which give a seven to the total number of ways a sum can be produced. In this instance it is easy to determine those two numbers, but it is not usually easy. The determination of the necessary two numbers often is greatly facili- tated by the use of the mathematical concepts, permutations and combinations. In the process of introducing these concepts, it is con- venient to develop certain useful formulas in terms of abstract let- ters. Thereafter, these symbols will be employed to represent per- sons, heads and tails on a coin, physical objects, etc. A set of letters, such as ABC, can differ from another set of the same number of similar marks in one, or both, of two ways: the same letters may appear in a different order, or exactly the same letters may not be present in both sets. For example, ABC and ACB are different orderings of the same three letters, whereas ABC and BCD 64 ELEMENTARY PROBABILITY Ch. 3 involve different letters. Two sets of three (or of n) letters are said to be different permutations of letters if they differ in either the order in which the letters are arranged, or if some different letters are in- volved in the two sets. Two groups of n letters are considered to be different combinations of letters only if some different letters are included in the two sets. A reordering of the same letters forms a new permutation but not a new combination. Hence, just one com- bination can be formed from n given letters if all the letters are used at once. This question then arises: How many different permuta- tions can one form from n letters, using all n letters each time? The process of constructing a permutation consists in determining a first, a second, . . . , and finally an nth letter. The first letter is chosen from among n, the second from among the remaining (n — 1) letters, the third from among the {n — 2) then remaining unchosen, etc., until finally only one letter is left for the nth choice. These n choices can be made in n(n — 1) (n — 2) (n — 3) ... (2) (1) ways, which is then the number of different permutations possible with n letters if all n of them are used in each permutation. To illustrate, suppose that there are three letters: A, B, and C. The following out- line shows how the choices can be made: For first letter, A, or B, or C. For second letter, B or C A or C A or B, For third letter, C or B C or A B or A, Therefore the permutations are: ABC, ACB, BAG, BCA, CAB, CBA, and there are 3(2) (1) = 6 of them. It is convenient to denote the product n{n — 1) (n — 2) ... (2) (1) by the symbol n !, and to call it n factorial. Hence, if P„, „ is adopted as the symbol for the number of permutations of n marks arranged in sets of n, we have (3.21) P„. n = nl as the formula for computing the number of such permutations. If n = 3, as above, P3, 3 = 3! = 3-2-1 = 6, as before. More often, it is necessary to make up permutations of marks in which only r of the n marks are used at any one time. For example, we might be choosing a batting order of 9 men from a squad of more than 9 men. To see how the process goes, suppose that it is required to make up all the possible two-letter permutations from the letters A, B, C, and D. There are 4 choices for the first letter and 3 choices for the second; therefore, there are 4(3) = 12 possible choices of two Sec. 3.2 PERMUTATIONS AND COMBINATIONS 65 from among four. In general, the symbol Pn, ,■ stands for the num- ber of r-lctter permutations which can be formed from n letters. Then, for n = 4 and r — 2, P4, 2 = 4(3) = 12, or, in general, (3.22) Pn. r = n{n - l)(n - 2) • • • (n - r + 1). It is possible and useful to express F„, ,■ in terms of factorials. To do so, we deliberately create the factorials from formula 3.22 by multiplying and dividing by {n — r) (n — r — 1) ... (2) (1) to make the numerator into n! and the denominator into {n — r) !. The final result is n\ (3.23) Pn, r = (n — r) ! From the definitions of permutations and combinations it follows that every set of r letters can be formed into but one combination, using all r of the letters; whereas, r letters can be formed into r! permutations. Hence, it is concluded that there are r\ times as many permutations of n marks taken r at a time as there are different combinations of n letters taken r at a time. Therefore, if the symbol Cn, r is adopted to indicate the number of possible combinations of n letters taken in groups of r letters each, the formula for that num- ber is whichever of n(n — l)(n — 2) • • • (n — r -{- 1) (3.24) Cn. r = ~ -^ ^ ^ or i^O.Zb) ^n, r we wish to employ, r\{;n — r)\ Problem 3.21. In how many different orders can 4 cars be parked among 6 consecutive parking places along a curb? It should be clear in this situation that the order in which the cars are parked makes a difference because the different orders are dis- tinguishable, and would be considered as different by a policeman checking parking. Therefore, this is a problem in numbers of permu- tations and can be worked by either formula 3.22 or 3.23. By formula 3.23 6! 6-5-4-3-2-1 Pe. 4 = = = 360. (6-4)! 2-1 66 ELEMENTARY PROBABILITY Ch. 3 Problem 3.22, How many different (as to cards held) 5-card poker hands are possible from the usual 52-card deck? In view of the fact that the order in which the cards were dealt does not affect the actual cards held, this is a problem in numbers of combinations of 52 objects taken 5 at a time; hence the number of poker hands is C52. 5 = (52!)/(5!47!) = 53,040 after common fac- tors in numerator and denominator are divided out and the remain- ing factors multiplied together. Problem 3.23. What is the probability that 5 cards dealt from a well-shuffled poker deck will all be spades? Two numbers need to be determined before formula 3.11 can be applied: (1) the total number of 5-card hands which are all spades, and (2) the total number of 5-card hands of any sort which possibly could be dealt from the deck. In view of the fact that the order in which the cards were dealt is unimportant, this is a matter of finding numbers of combinations, namely, C13, 5 and C52, 5- Therefore, the required probability is P(all spades) = C13, 5/<^52, 5 = -0005, or 1 chance in 2000. Problem 3.24. What is the probability that 5 cards dealt from a well-shuffled poker deck will include exactly 3 aces? Three aces can be chosen from among the 4 available in C4, 3 or 4 ways. Likewise, C48, 2 = 1128 is the number of different pairs of cards which do not include any aces. All possible 5-card hands with exactly 3 aces must necessarily be the same as all the possible ways to put some 3 aces with one of the 1128 pairs of cards which are not aces; hence there must be 4(1128) = 4512 different 5-card hands which include exactly 3 aces. Therefore, the probability of being dealt such a hand is P(exactly 3 aces, 2 non-aces) = 4512/C52, 5 = .0016, or 1 chance in 625. PROBLEMS 1. In how many ways, which differ as regards the persons in particular chairs, can 4 men and 4 women be seated around a dinner table, with men and women seated alternately? 2. Suppose that there are 10 persons in a room, and that they have the fol- lowing blood types: 1 is AB, 3 are A, 2 are B, and 4 have type O blood. If 2 Sec. 3.3 REPEATED TRIALS 67 are chosen at random what is the probability that they will have the same type of blood? Ans. 2/9. 3. Suppose that a baseball team has 4 men who can bat in any of the first 3 positions, 5 who can bat in any of the fourth, fifth, and sixth positions, and 7 who can bat in any of the last three positions. How many possible batting orders are there? 4. Assume that 7 insecticides are to be tested as to their effectiveness in killing house flies. If each spray is to be tested against every other spray once in a separate test, how many tests will this require? Ans. 21. 5. Suppose that a housewife buys 3 cans of peaches, 6 cans of apricots, and 4 cans of pears; and suppose that her child tears off all the labels on the cans. If the housewife needs 2 cans of fruit for dinner, what is the probability that the first 2 cans chosen will contain the same kind of fruit? 6. How many 13-card bridge hands are there with no card higher than 8? Ans. 37,442,160. 7. If 7 unbiased coins are flipped simultaneously, how many single events are there in the class: 3 heads, 4 tails? 8. Compare the coefficients of {x -\- y)^ with Cg g, C^ ^, Cg 3, Cg 9, Cg j, and Cg Q given that 0! = 1. 9. What is the probability that 5 cards dealt from a well-shufHed poker deck will include 3 queens and 2 aces? Three queens and at least 1 ace? 10. What is the probability that 13 cards dealt from a well-shuffled bridge deck will include exactly 8 honor cards (honor cards are 10, J, Q, K, and Ace)? Ans. .040. 11. In how many ways can 6 boxers be paired off for 3 bouts being held simultaneously? 3.3 REPEATED TRIALS UNDER SPECIFIED CONDITIONS Situations involving the numbers of occurrences and non-occur- rences of an event E on repeated trials under the same original con- ditions are of particular interest in statistical analysis. The prin- ciples involved will be seen to be important to the study of frequency distributions, and to sampling studies. The probability problems created when trials are repeated under fixed conditions can be illustrated by means of mathematical models of these problems. Suppose that a coin is flipped n times and the number of heads noted. On such a set of repeated trials any number of heads is possible from to n, that is, there are (n -f- 1) possible classes of event: heads, n tails; 1 head, (n — 1) tails; 2 heads, (n — 2) tails; . . . ; n heads, tails. Each class of events includes some number of single events (if the coin is unbiased) from 1 to whatever the maximum size of C„, r is for the given n. For example, 68 ELEMENTARY PROBABILITY Ch. 3 if an unbiased coin is tossed 5 times, there are 6 classes of events, and the specific single events in the class 4iH, IT are Toss First Second Third Fourth Fifth T H H H H H T H H H H H T H H H H H T H H H H H T 5! It is seen that C5, 4 = — — - = 5 = the number of single events in the class which includes exactly 4 heads. It also should be observed that the outcome of each toss is an independent event relative to the out- come of any other toss; hence the probability of the first result listed above, THHHH, is (1/2) (1/2) ••• (1/2) = (1/2)^ With this un- biased coin, that also is the probability for any of the other single events in this class of events. Therefore, the probability of an event in the class 4/f, 17" is C5, 4(l/2)^(l/2)\ in which the exponent 4 refers to the number of heads and the exponent 1 refers to the number of tails. The reader can verify the fact that for any specified number of heads from to 5 the probability of exactly r heads is C5, r(l/2)'' X (1/2)^"'', where ?• takes any value from to 5. In general, if n unbiased coins are to be flipped (or one such coin be flipped n times) the probability of the appearance of any specific number of heads, say r, is (3.31) P[r heads, (n - r) tails] = Cn, r(l/2)'"(l/2)"-'". To extend this result a bit, let an event E have a constant prob- ability, p, of occurrence on each of n repeated trials. Then the probability that E will occur on exactly r of the trials [and fail on the other {n — r) trials] is given by the following formula: (3.32) P[rE'8, (n - r) not-^'s] = Cn, ApYil - vT~"- The student can verify that this formula becomes (3.31) if p = 1/2, E = H, and (not-£;) = T. One more generalization can be obtained regarding formulas 3.31 and 3.32 by considering the expansions of the two binomials Sec. 3.3 REPEATED TRIALS 69 (1/2 + 1/2)" and [q + p)", in which q = 1 — p. To see these generalizations, consider the following binomials expansions: (1/2 + 1/2)2 ^ 1(1/2)0(1/2)2 + 2(1/2)1(1/2)1 + l(l/2)2(l/2)«, = C2, 0(1/2)0(1/2)2 + C2, 1(1/2)1(1/2)1 + C2. 2(l/2)2(l/2)«, = P(0//', 2r) + P(1H, IT) + P(2//, OT). That is, the successive terms of the expansion of (1/2 + 1/2)^ are given by formula 3.31 if r = 0, 1, and 2, successively; and those three terms give the probabilities for the three possible classes of events in terms of the number of heads appearing. The generaliza- tions for 3, 4, . . ., or n tosses should be apparent. For the more general situation in which the probability of the occurrence of an event E is constantly p under repeated trials, (q + P? = Up)\q)' + 2(p)i(?)i + l(p)2(g)«, = P(0 ^'s, 2 not-^'s) + P(l E, 1 not-^) + P(2 E's, not-^'s) ; and again it should be apparent that these successive terms corre- spond to formula 3.32 for r = 0, 1, and 2, successively. PROBLEMS 1. What is the probability that if 6 unbiased pennies are tossed simulta- neously, exactly 3 heads will appear? 2. What is the probability that at least 3 heads will appear under the condi- tions of problem 1? Ans. 21/32. 3. If one parent is Rhrh and AO, and the other parent is rhrh and BO, what is the probability that both their first two children will be Rh— and AB? 4. Suppose that a sample of 100 bolts is taken from a very large batch which contains exactly one-half of 1 per cent of unacceptable bolts. What is the probability that at least 2 bolts in the sample will be unacceptable? Ans. .09. 5. If 5 bolts among the 100 in the sample of problem 4 are found to be im- acceptable products, what would you conclude about the hypothesis that only one-half of 1 per cent were faulty in the whole batch? Give reasons. 6. Write out the series for {x -\- y)"^ and show that the coefficients are num- bers of combinations, C^ ,., with r = to 4. 7. Suppose that the teams listed on a football parlay card are so handicapped that you actually have a 50-50 chance on each team you pick. What is the probability that you will pick exactly 9 winners out of 10? Would this proba- bility justify odds of 25 to 1 for this accomplishment? What about odds of 250 to 1 for getting 10 out of 10 correct? 70 ELEMENTARY PROBABILITY Ch. 3 8. If a pair of unbiased dice is to be thrown 6 times in succession, what is the probability that exactly 3 sevens will be thrown? What would you think the most likely number of sevens would be? Ans. .054. 9. If a certain manufacturing process is producing machine parts of which 10 per cent have some serious defect, what is the probability that all of the 10 parts chosen at random will be acceptable (that is, have no serious defect)? How many would you have to take in the sample before the probability of all being acceptable will be no greater than .05? 10. Graph /(p) = (1 — p)'^^, and relate this graph to problems like problem 9. 3.4 MATHEMATICAL EXPECTATION The discussions earlier in this chapter have involved the occur- rences of chance events as a result of what have been termed "trials" under specified conditions. The outcome of a trial is de- scribed in one of two general ways: (a) Something happens a certain number of times on a specified number of trials, or (b) we simply note whether or not an event E has, or has not, occurred and asso- ciate with that occurrence some value, say a financial loss, as in in- surance. With either type of situation it may be important to be able to predict what will be the average outcome of trials under the stated conditions, over the long run of experience. For example, an insurance company needs to know wdiat amounts it should expect to have to pay out in death benefits during a particular period of time, one year, for instance. In case a, the prediction needed is to be presented in the form of an expected number of occurrences of an event E" on a set of n future trials. A formula for this expected number can be justified heuristi- cally as follows. If the probability of £" is p, the p is just the fraction of the time that E should occur over many trials. Hence, if there are to be n trials, it is reasonable to say that the expected number of occurrences oi E on n trials is (3.41) Expected number = E{r) = p-(n). Problem 3.41. If 6 unbiased coins are to be tossed simultaneously, what is the expected number of heads? In this circumstance p = 1/2 and n = 6; hence the expected (or long-run average) number of heads is E (r) = (1/2) (6) = 3. Actu- ally, our intuition would lead to the same conclusion. Problem 3.42. Suppose that an insurance company has insured 50,000 persons who are each 30 years old, and that records from past experience show that 6/1000 of such persons die before reaching the age of 31. What is the expected number of deaths during the first year of the insurance contract? Sec. 3.4 MATHEMATICAL EXPECTATION 71 For this situation p = .006, n = 50,000; therefore the mathemat- ically expected number of death benefits among those thirty-year- olds is E{r) = .006(50,000) = 300. The reader will realize that it would be unsound financially for the company to be prepared to pay only 300 death benefits because this time the number of deaths might be considerably higher. All that is being said is that over a period of years of such calculations the average number of deaths among thirty -year-olds in this same insurance class will be very close to 300. When chance occurrences are of the type b described above there may be associated with the occurrence of E some value, say a finan- cial gain or loss. Then we may wish to predict the loss or gain to be expected on the average under the given conditions. For example, suppose that you are going to roll a pair of unbiased dice and are to be paid 60 cents if you get a sum of 7. How much should you pay to play such a game if you just wish to break even? Obviously you will receive either 60 cents or zero cents after each game; but over many games what will be your average winnings per game? That is the amount you can pay and break even. Because the prob- ability of throwing a sum of 7 is 1/6 you expect, mathematically, to win 60 cents on about one-sixth of your throws and to win zero cents on the other five-sixths of the throws. Hence, the mathematical expectation logically is (1/6) ■ (60 cents) + (5/6) • (0 cents) = 10 cents. Therefore you can expect to break even in the long run if you pay 10 cents to play each game. The game just described can be extended to include a reward of 90 cents if you throw a sum of 11 on the two dice. In this circum- stance you can win in either of two mutually exclusive ways, that is, you can throw a 7 or an 11. Therefore attention is centered on three classes of events and the corresponding rewards: A sum of 7 with a reward of 60 cents, a sum of 11 with a reward of 90 cents, and a sum other than 7 and 11 with a reward of cents. Therefore, over a large number of games you will tend to win 60 cents on one-sixth of the throws, 90 cents on one-eighteenth of the throws, and cents on the other seven-ninths of the throws. Hence your mathematical expectation on this game is (1/6) (60 cents) -|- (1/18) (90 cents) + (7/9) (0 cents) = 15 cents, because that is the 72 ELEMENTARY PROBABILITY Ch. 3 average winnings per game and is, therefore, the amount you could pay to play this game and expect to break even. The preceding ideas and methods can be generalized and sym- bolized in the following manner. Let all the single events possible under a specified set of conditions be grouped into s mutually ex- clusive classes of events. Let Xi be the reward, loss, or in general the "value" of the occurrence of an event in the ith class; and let Pi be the probability that an event in the ith class will occur on any designated future trial under the stipulated conditions. Finally, let Eix) stand for the total mathematical expectation under the given conditions. It follows from the reasoning outlined above that s (3.42) E(X) = p,X, + P2-T2 + • • • + PsXs = E (Pi^i)- i=l PROBLEMS 1. Suppose that you are 20 years of age and that you are to inherit $10,000 at the age of 30 if you are aHve then. What is the expected value of this in- heritance if you have a probability of .92 of living to be 30 years of age? (This probability is derived from the American Experience Mortality Table.) 2. It is approximately true that brown and blue eye colors are inherited in a manner similar to that explained for the A-B blood groups. If b/b = blue eye color and either B/b or B/B = brown eye color, what is the expected num- ber of blue-eyed children among 500 fiom parents who are B/b and b/b, re- spectively? Ans. 250. 3. Answer the same question as in problem 2 for parents who are both B/b. 4. If in each three-month period 1 car in 20 of the type which you drive has an accident costing an average of $75 for repairs, how much insurance against such a loss should you pay each quarter if you allow the company 15 per cent beyond mathematical expectation for handling the business, and if you ignore interest on your money? Ans. $4.31. 5. How much would one be justified mathematically in wagering against one dollar that on 10 throws of two unbiased dice a sum of 7 will appear less than 3 times? 6. Suppose that you have the choice of receiving $10,000 at age 65 if j'ou are alive then, or of taking a cash payment now. From a purely mathematical point of view and ignoring interest on money, what should the size of the pay- ment be if your probability of living to be 65 is .56? An.s. $5600. 7. Suppose that a concession at a fair offers a 50-cent prize if you pay 10 cents for 3 throws and knock down all of a stack of milk bottles on the 3 throws. Suppose also that 3^ou have 1 chance in 10 to knock down the bottles. If the operator of the concession has to pay $75 per day for the privilege of doing business there, how many customers must he have per day in order that he can expect (mathematically) to make some money? 8. Suppose that a person who is 40 years of age is to receive $1000 on each of his sixtieth, sixty-first, and sixty-second birthdays if he is alive to receive them. Also suppose that interest on money is to be ignored. Given that his proba- Ch. 3 REVIEW PROBLEMS 73 bilities of living to those successive ages are .74, .72, and .70, what should this person pay for an annuity of this sort if the company is allowed 15 per cent for overhead? Ans. REVIEW PROBLEMS 1. What is the difference between a population of numerical measurements and a sample of such measurements? 2. What was Political Arithmetic? With what sorts of problems were the political arithmeticians mainly concerned? 3. Who was Student and how was his work connected with present types of statistical problems? 4. Expand (1/3 + 2/3)4 into a series and state specifically what probability is given by each term if p = 2/3 is the probability that a certain loaded penny will turn up heads on any particular future throw. Describe the mathematical procedure needed here to define the single events. 5. What is a frequency distribution? A relative cumulative frequency dis- tribution ? 6. Suppose that a college freshman has earned the following percentile ratings on the indicated tests: (a) general intelligence, 90; (b) achievement in social sciences, 65; (c) achievement in ph3'sical sciences, 92; (d) achievement in mathematics, 95. What can you say about the student's probable future success in courses in chemistry, physics, mathematics, history, and sociology if it is assumed that the tests are trustworthy and if no serious personal problems interfere ? 7. Given that for a set of numerical measurements X^, Xo, ■ ■ . , X^q, "^X = 95 and 2(x-) = 2.06, calculate the coefficient of variation. 8. Calculate the geometric, arithmetic, and harmonic means of 1/2, 2, and S and discuss the choice of the best average for these numbers. Ans. 2, 3.5, 8/7. 9. Suppose that the following probabilities regarding football games have been determined reliably: A to beat B. p^ = 2/3; C to beat D, p^ = 1/2; and E to beat F, p^ = 5/6. What are the odds that A, D, and E all win? 10. Given that the graph below is the r.c.f. curve for a certain group of scores, determine from it the median, the upper limit of the third quartile, and the upper limit of the sixth decile. Also interpret these results statistically, with some indication of the uses to which such information can be put. Arts. 31.5. 41.0, 35.0. .uu .90 _ 1 1 1 1 1 1 1 1 I^^U^.-^— -T^ .80 - ^x^"""^ - .70 - y^ - .60 - y^ - .50 - > < - .40 - ^^ — .30 >^ — .20 - ^,„>^ — .10 n I-j Ti 1 1 1 1 1 1 1 1 1 ~ 5 10 15 20 25 30 35 40 45 50 55 60 65 Score 74 ELEMENTARY PROBABILITY Ch. 3 IL What proportion of the scores summarized in problem 10 lay between 40 and 60 inclusive? What proportion exceeded 60? 12. What is the probability that on 15 flips of an unbiased penny one will get either 7 or 8 heads? Will get neither 7 or 8 heads? Ans. .39, .61. 13. Suppose that for a certain strain of chickens the probability that a late- feathering chick will be hatched from any egg selected at random is 1/16. What is the expected number of such late-feathering chicks among 800 newly hatched chicks? 14. If a pair of true dice is rolled 60 times, what is the mathematically ex- pected number of sevens? Of either sevens or elevens? Of sums greater than 9? Ans. 10, 13%, 10. 15. Assume that the semester grades in a large chemistry class have the ogive graphed below. If the letter grades are to be distributed as follows: 7 per cent A, 20 per cent B, 46 per cent C, 20 per cent D, and 7 per cent E, what are the grade ranges covered by each letter grade? 1.00 aj .80 60 70 Grade 16. What is the median numerical grade for the data of problem 15 above? What are the upper limits of the quartiles? Ans. 69; 59.5, 69, 77, 100. 17. Suppose that 6 unbiased pennies are to be tossed simultaneously. What is the probability that no more than 2 will show heads? That at least 2 will turn up heads? 18. Assume that the true odds on each of 3 horses to win a particular race are determined to be as follows: horse A, 3:2; horse B, 1:3; and horse C, 1:9. What is each horse's probability of winning? What is the probability that some one of these 3 hordes will win? Ans. .60, .25, .10; .95. 19. Given that for three separate statistical populations of data : /Uj = 25, o-j = 4; lUo — ^0, 0-0 = 5; and /u.^ = 100, a.^ = 13. Which group of data would you consider as relatively the more variable? Give specific statistical evi- dence to back your answer. 20. Compute the mean deviation and the standard deviation for the fol- lowing data: 13, 9, 10, 17, 15, 20, 11, 5, 2, 10, 14, 13, 19, 21, 16, 8, 14, 6, 3, 29, 16, 17, 15, 15, 18, and 2. Which measure of variation do you think best describes the dispersion of these data about their arithmetic mean? Give reasons. You are given that SZ = 338, 2X2 :^ 5406. Ans. 4.92, 6.24. Ch. 3 REFERENCES 75 21. According to Figure 2.41, in which quartile would you place a score of 85? In. which decile would this score fall? 22. If all students whose ACE scores (Table 2.01) fell among the lower 15 per cent, approximately, of all scores made were to be advised to consider seriously dropping out of college, what would be the highest score whose re- cipient would receive such advice? Ans. 67. 23. From Figure 2.41 determine approximately the percentage of those scores which were not more than one times the standard deviation either greater than or less than the mean, /x. 24. In the game of "craps" two unbiased dice are thrown successively by the same person. He wins if: (a) he throws a sum of either 7 or 11 on his first throw; or (b) he throws a 4, 5, 6, 8, 9, or 10 on the first throw and repeats his number on a subsequent throw before he throws a 7. What is his probability of winning within 2 throws? Ans. 97/324. 25. In a certain gambling game you are paid 15 for 1 if you throw a 1 and a 2 (either order) on two unbiased dice. On 1800 games on each of which the player pays one dollar, what is the expected percentage profit for the house relative to the amount taken in? REFERENCES Arley, Niels, and K. Rander Buch, Introduction to the Theory of Probability and Statistics, John Wiley and Sons, New York, 1950. Kenney, John F., Mathematics of Statistics, Part II, Second Edition, D. Van Nostrand Company, New York, 1947. Levy, H., and L. Roth, Elements of Probability, Oxford University Press, Lon- don, 1936. CHAPTER 4 The Binomial and Normal Frequency Distributions The discussions and illustrations of Chapter 2 involved situations in which groups of measurements (usually numerous) had been taken under specified conditions, and we had in mind only an efficient summarization of the data. The ACE scores in Table 2.01 were cited as an example. In a sense, we simply took what we got and thereafter applied statistical methods to reduce a bulk of data to a more comprehensible form without losing any essential information. More generally, however, populations of numerical measurements must be studied by means of samples because so many measurements are involved that it is not feasible, efficient, or even possible to obtain and to analyze the whole of the population. Two different types of populations will be considered. In one type, the chance variable will be a qualitative one such as male or female, dead or alive, own an automobile or do not own an automobile. The population will consist of individual members, each falling into one of just two classes according to the qualitative designation adopted. The other type of population to be considered will be based upon a variable which is measured along a continuous scale, such as the weight of an individual, the volume of a gas, or the bushel yield of a variety of wheat. As regards populations in which a qualitative variable is used, attention herein will be confined to what is called a hinomial 'popula- tion because each member of the population falls into one of only two classes. The proportion of a binomial population which belongs to one of the two classes will be measured by the fraction p, leaving the fraction falling into the other class to he 1 — p = q. For example, if all the babies born in New York City during a given year were to be classified as male or female, p might be the fraction who were males, li p = .51, then q- = 1 — .51 = .49. The sex would be the qualitative variable mentioned above, and has but two "values": 76 Ch. 4 BINOMIAL AND NORMAL DISTRIBUTIONS 77 male and female. If a baby were to be chosen at random from among those born in the specified year, its classification as male or female would be a member of this binomial population. Under the above assumptions, the probability that such a selection will turn out to be male is p = .51. If n repeated observations, or trials, are made on a binomial pop- ulation in which the proportion p is staying fixed, and if attention is fixed upon the number of individuals in each of the two classes, these numbers are variable from one set of n trials to another. For exam- ple, it was noted in Chapter 3 that the probability that r males, say, and n — r females would be observed is given by C„, rCp)*"!! — p)"^'". In other words, r is a chance variable. The relative frequencies with which r will have the values 0, 1, 2, ... , and n after a great many sets of n random trials from a binomial population constitutes a binomial frequency distribution. This distribution will be of more direct in- terest to us than the binomial population in itself because the binomial frequency distribution describes results which are obtained in the process of sampling a binomial population. There are many types of populations for which the random vari- able is of the second type discussed at the beginning of this chapter, namely, a measurement referred to a continuous scale, such as weight. Probably, the most important populations of this sort are those called normal populations. It will be convenient to describe this type of population by means of a mathematical formula for its frequency distribution. This will be done in a later section. It seems obvious that we cannot possibly learn much by sampling a population which cannot be clearly and concisely described ; hence there is need for a mathematical description, or classification, of populations. We choose to study types of populations by means of their frequency distributions because that — or something equivalent — constitutes the fullest description we can obtain for a particular population. As noted above, the discussion in this chapter will be devoted to two of the most important types of frequency distribu- tions: one, the binomial, is appropriate to qualitative measurements of a certain kind ; the other, the normal, typifies continuous numerical measurements of types quite frequently met in practice. Between these two theoretical distributions, a great many of the uses of statistical analysis will be introduced. 78 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 4.1 THE BINOMIAL FREQUENCY DISTRIBUTION As was stated above, a binomial population and the corresponding binomial frequency distribution are involved when every single event which can occur under prescribed conditions must belong to one of two classifications. This fact corresponds to the meaning of the prefix bi- in the word binomial. For example, if you take out a term insurance policy for a period of 10 years you and the company are interested in your subsequent classification as "dead" or "alive" before, or at, the end of the 10 years. Of course, the company in- sures many persons and regards them as a group, some of whom will be classifiable as "dead" and the remainder as "alive" at the expiration of the 10-year term. What the insurance company and its clients need to know, then, is this: Given a group of n persons insured for a 10-year term, what are the probabilities associated with each of the possible numbers of "dead" and "alive" insured persons during the 10-year period of the insurance contract? For any spe- cific n the relative frequency — over a great deal of experience — of the occurrence of 0, 1, 2, 3, ... , (n — 1), or all n "dead" after 10 years will be the binomial frequency distribution mentioned above. It is upon the basis of this sort of information that insurance premiums are calculated. Suppose, for simplicity, that a company has insured 10 persons who are 30 years of age for a 10-year term. What can the company expect to pay out in death benefits? It is obvious that at the end of the 10-year period any one of 11 events may have occurred. There can be 0, 1, 2, ... , or all 10 classified as "dead." Also, over the experience of many such groups of contracts for 10-year periods those 11 possible outcomes will occur with unequal relative frequencies which depend both on the number, such as n = 10, and on the prob- ability of death for persons in this age interval. Clearly, this bi- nomial frequency distribution depends on n and on p = probability of death between the ages of 30 and 40 years. No one can state theoretically what the probability of death is for any particular person during the age period of 30 to 40 years; but tables have been compiled from experience which give the best avail- able estimate of the desired probability. For example, the American Expedience Table of Mortality indicates that the average probability is approximately one-tenth that an insurable person (determined by examination before the company will insure) now 30 years of age will die before he is 40 years of age. Sec. 4.1 THE BINOMIAL FREQUENCY DISTRIBUTION 79 If 10 persons are insured under conditions to which the American Experience Mortality Table applies, it follows from the discussion of section 3.3 that the 11 corresponding probabilities of occurrence of these numbers of deaths are given by the successive terms of the following binomial series: (.9 + .1)10 ^ (.9)10+ io(.9)9(.l)i + 45(.9)^(.l)2+..-+(.l)io It is not necessary to devise some game with p = .1 and discover from experience that a fraction (.9)^^ of the trials will show no occurrences of the event E because the only interest is in the relative number of 250 _ 1 1 1 1 1 1 1 _ 200 - — |l50 I'lOO - - 50 - - 1 1 ~ w ] L 2 : Nun iber ofh eads ) i (r) r 8 9 10 Figure 4.11. Graph of the binomial fiequency distribution for p = 1/2 and n = 10. occurrences, and that is what the probability gives. Hence, the above series gives the frequency distribution of the eleven possible classes of events. To re-illustrate the discussion of the preceding paragraphs with an example which the reader can reproduce easily and, in addition, to show how to graph a binomial frequency distribution, attention again is called to a mathematical model. Suppose that an unbiased coin is to be flipped 10 times and the number of heads is to be recorded after each set of 10 throws. In these circumstances, n. = 10, p = 1/2, and q = 1 — p — 1/2; hence the successive terms of the following binomial series give the probabilities for 0, 1, 2, 3, . . . , or 10 heads on any future set of 10 throws: (l/2)io + 10(l/2)«(l/2)i + 45(1/2)8 (1/2)2 + ...+ (1/2)10; or 1/1024 + 10/1024 + 45/1024 +••• + 1/1024. In view of the fact that each of the denominators is 1024, we obtain a useful and simpler expression for the relative frequency of occurrence of 0, 1, 2, ... , or 10 heads on 10 throws, by using only the numerators. From them a graph can be constructed to depict the relative frequency for each possibility, as is done in Figure 4.11. 80 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 This figure also can be described as the graph of the binomial fre- quency distribution when ?i = 10 and p — 1/2. It is apparent that the actual form of a binomial frequency dis- tribution depends upon two numbers, n and p. If p = 1/2 = 1 — p, 2 20 1 1 1 1 1 1 1 1 X g> _ c 1 10 - ■ - n 1 ♦ ; 1 1 . 1 _... Figure 4.12^. 3 4 5 6 7 Number of occurrences (r) (p-2/9) 10 Graph of the binomial frequency distribution with p = 2/9 and n= 10. 30 2 20 - 1 1 1 1 1 1 1 i i i : T ■ - 3 4 5 6 7 Number of occurrences (r) (p = 13/16) 10 Figure 4.12B. Graph of the binomial frequency distribution with p = 13/16 and w- 10. the graph is symmetrical, as in Figure 4.11. If p > 1/2, the event E is more likely to occur than to fail to occur; hence the higher ordinates of the graph will be toward the right-hand side of the graph. If p < 1/2, the reverse situation is expected. These remarks are illus- trated in Figures 4.12.4 and B. For Figure 4.12A, p = 2/9; and for Figure 4.125, p = 13/16. In both cases n — 10. The series for the binomials (7/9 + 2/9) i" and (3/16 + 13/16) i*^ were employed in the Sec. 4.1 THE BINOMIAL FREQUENCY DISTRIBUTION 81 constructions of these figures, using only the numerators of the terms as explained above. The r.c.j. distribution for a binomial situation is discontinuous — as is expected — and involves successive ordinates, each at least as large as the preceding one to its left on the graph. Such a graph is shown in Figure 4.13. If we were to draw a smooth curve through the tops of the ordinates, it would have the same general appearance as the r.c.j. curves drawn in Chapter 2. Fundamentally the frequency and r.c.j. tables corresponding to Figures 4.11 and 4.13 are as shown in Table 4.11. The meaning and TABLE 4.11 Frequency and r.c.f. Distributions for the Binomial Distribution Defined by p = 5 = 1/2, n = 10. Total Frequency Taken = 1024, the Sum of the Numerators of the Series for (1/2 + 1/2)^" Number of Oc- currences of E r / c.f. r.c.j. 10 1 1024 1.000 9 10 1023 .999 8 45 1013 .989 7 120 968 .945 6 210 848 .828 5 252 638 .623 4 210 386 .377 3 120 176 .172 2 45 56 .055 1 10 11 .011 1 1 .001 2(/) = 1024 use of Table 4.11 are fundamentally the same as for similar tables in Chapter 2, but some differences should be noted. The major dif- ference arises from the fact that the class "intervals" now are just isolated points on a scale of measurement appropriate to r. For example, 5^ per cent (0.055) of the observed values of r (over a very large number of observations on r) will be at or below r = 2. How- ever, these observed numbers of occurrences of E will be 2's, I's, and O's only: there is no such r as 1.6, for example. Another difference between Table 4.11 and similar tables in Chapter 2 is that the former is a theoretical table which fits any situation for which p = 1/2 and n = 10. The frequency tables in Chapter 2 were relevant only to the 82 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 particular situation which produced the data summarized in a given table. We might wish to know what the median r is for a binomial dis- tribution. By Table 4.11, 37.7 per cent of the numbers are seen to be O's, I's, 2's, 3's, and 4's. If 5's are included, the percentage runs past 50 (needed for the median) to 62.3; therefore the median r must be 5. It is not some decimal fraction between 4 and 5 because no such numbers even exist on the scale of measurement of r. J i L 8 10 1.00 .90 ^ - .80 § .70 f .60 I .50 1 .40 I .30 ^ H.20i .10 CC .00 12 3 4 5 6 7 Number of occurrences (r) Figure 4.13. The r.c.j. distribution for the binomial distribution with p = 1/2 and n = 10. The median of the binomial distribution just considered also can be obtained from the r.c.j. distribution of Figure 4.13 by reading horizontally from the point where r.c.j. = .50 until we come to the first ordinate on the left, which is high enough to be intersected by the horizontal line from r.c.j. = .50. It is interesting to compare a frequency distribution which was obtained by actual trials with that which would be expected mathe- matically under the specified conditions. This is done approximately in Table 4.12 for a situation in which 5 pennies were flipped 2000 times. It was assumed that the pennies were unbiased, although this is known not to be strictly true for any actual coin. It should be apparent from previous discussions that the mathematically ex- pected proportions of the 6 possible combinations of heads and tails listed in column 1 of Table 4.12 are 1:5:10:10:5:1. The resulting expected numbers of occurrences of each of the possibilities are given to the nearest whole number under the heading "Exp." in columns 3, 5, 7, and 9. Sec. 4.1 THE BINOMIAL FREQUENCY DISTRIBUTION 83 TABLE 4.12 COMPAKISON OF OBSERVED AND ExPECTED FREQUENCIES OF HeADS (H) AND Tails {T) When 5 Pennies (Assumed Unbiased) Are Flipped 100, 500, 1000, AND 2000 Times Combination of H and T r, (n - r) 100 Throws 500 Throws 1000 Throws 2000 Throws Obs. Exp. Obs. Exp. Obs. Exp. Obs. Exp. 5H,0T 4H, IT 3H,2T 2H,ST 1H,4T OH, 5T 4 10 33 31 18 3 3 16 31 31 16 3 16 71 151 163 86 13 16 78 156 156 78 16 33 135 287 342 168 35 31 156 313 313 156 31 63 293 600 641 332 71 62* 312 625 625 312 63* * Actually each of these numbers is 62.5 but was rounded off this way to keep the sum of the observed and expected frequencies equal. After the 2000 trials involving 10,000 tosses the ratio of heads to tails is 0.94 to 1. Hence there apparently is a weak but definite tendency for tails to appear more frequently than heads; that is, p is not exactly equal to 1/2. Methods will be described in Chapter 5 for deciding when a coin, say, is biased, and for estimating the de- gree of bias. If the observed frequencies in any column of Table 4.12 are taken as the / and the r is listed merely as 5, 4, 3, 2, 1, and 0, we have an observed frequency distribution, as in Chapter 2. If the expected frequencies (which follow a mathematical law) are used as the / column and r again is listed as 5, 4, 3, 2, 1, and 0, we have a the- oretical frequency distribution of the sort being discussed in this chapter. In view of the existence of a general mathematical expression for the binomial frequency distribution (as in formula 3.32), we might be curious to know if such statistical measures as the arithmetic mean and the standard deviation can be determined just from the n and p which determine the distribution. This is, in fact, true, as will be shown partially below. The discussion of mathematical expectation given in Chapter 3 included the information that the arithmetic mean of the number of occurrences of an event E over many trials coincides with the ex- 84 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 pected number, E{r), for any designated future trial. Hence it al- ready has been found from experience and intuition that for the binomial situation fi = np. This result can be established for any binomial frequency distribution, but such will not be done herein. It also can be shown by somewhat more difficult mathematics that the standard deviation for a binomial frequency distribution is given by (T = vnpq ; consequently, given n and p, we can compute the mean and standard deviation very easily. This will be found to be helpful later in this chapter. Sometimes when dealing with binomial distributions it is advanta- geous to work with the fractional number of occurrences, r/n, rather than with the actual number, r. In this case, the arithmetic mean is p, rather than np; and the standard deviation is wpq/n , instead of 'Vnpq. To illustrate the use of these formulas both for r and for r/n consider again the insurance example above in which n = 10 and p = .1. Under these circumstances the mean r is np = lO(.l) = 1, and the standard deviation is wnpq = V 10(.1)(.9) = 0.949, approxi- mately. The mean r/n is p = .1, and the standard deviation of the fraction dead is \^pq/n = v (.1)(.9)/10 = 0.095, approximately. If the number n were sufficiently large that it would be practical, such information as that just derived might be useful to an insurance com- pany in anticipating the average number (or fraction) of death benefits it could expect to pay, and in making sufficient allowance for chance deviations from those average numbers so that adequate funds would be available to pay death benefits. PROBLEMS 1. Use the coefficients of (1/2)'" in the series for (1/2 + 1/2)6 to graph the binomial frequency distribution appropriate to sets of 6 trials with an event whose probability of occurrence is constantly p = 1/2. 2. Under the conditions of problem 1, what is the probability that the event will occur at least 4 times on 6 trials. Ans. 11/32. 3. Graph the frequency distributions for the binomial with p = 1/2 and n = 4, 8, and 12, successively. Compute the /j, and a- in each instance and locate on the scale of r: /t ± lo-, /i =h 2o-, and fi ± So-. 4. Graph the binomial frequency disti'ibution for p = 1/4, n ~ 4, and read from it the probability that r will be 2, 3, or 4. 5. Check the result obtained in problem 4 by constructing the r.c.f. graph and reading the answer from this graph. 6. Graph the binomial frequency distribution for p = .7, n = 4, and determine the probability that on one random set of 4 trials E will occur at least fi times, where fi = arithmetic mean. Sec. 4.2 THE NORMAL FREQUENCY DISTRIBUTION 85 7. Flip 3 pennies 80 times, recording the number of heads after each toss of the 3 pennies. Then compare the observed and the mathematically expected numbers of occurrences in each of the four possible classes of events in terms of number of heads. 8. Perform the operations of problem 7, except to compare the observed and the theoretical values of the arithmetic mean. 9. Suppose that a large gi'oup of fruit flies consists of members who are classified as either "normal" or "sooty." Among 10 of these flies selected at random, 3 were found to be "sooty." How frequently would that result be obtained if half the flies in the population are "sooty"? How frequently if 25 per cent are "sooty"? 10. Suppose that under the conditions of problem 9, 100 flies are chosen at random and 30 (same percentage as in problem 9) are "sooty." Answer the same questions as in problem 9. Ans. 23 times in 1,000,000; 12 times in 1 billion. 11. Construct the r.c.f. distributions for problem 3 and then determine the median r in each distribution. 12. Suppose that there are two political parties interested in a certain college election, and that 60 per cent of the eligible voters are Progressives and 40 per cent are Independents. If a random sample of 10 persons is taken, what is the probability that a plurality of them will be Independents, even though they constitute the minority part.v? Ans. .17. 13. Referring to problem 12, how large must the sample be before the proba- bility is less than 1/4 that there will be more Independents than Progressives in the sample? 14. Suppose that 6 persons out of 10 selected at random in a certain city favor a particular flood-control policy. What is the probability of such a result when only 45 per cent of those in that city actually favor that policy? Ans. .16 15. For problem 9 determine the median number of "sooty" flies among 10. What is the probability that the actual number observed will exceed the median? 4.2 THE NORMAL FREQUENCY DISTRIBUTION A frequency distribution for a population of numerical measure- ments is intended to display in some manner the density with which the measurements are distributed along the scale on which they are measured. Such a frequency distribution indicates the region (along the scale of measurement) in which the measurements tend to be most numerous, and also shows the way in which they are dispersed about that region of concentration. The reader should see that these are the same two general matters of concern considered in Chapter 2. Averages were used to measure general level of performance (as on ACE scores), and measures like the standard deviation, mean deviation, range, and quartiles were employed in the description of the dispersion of the data along the scale of measurement. This sort 86 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 of information is essential to any adequate description of a popula- tion, and also is vital when considering sampling problems. In order that we may be able to perform certain useful statistical analyses it usually is necessary to assume (after investigation) that the data conform to some general type of frequency distribution, such as the binomial frequency distribution considered in the preced- ing section. In that section a formula was used to determine the frequency distribution for a binomial population when the basic information {n and p) was available. The formulas and the proce- dures for their use are appropriate for discontinuous measurements which fall into only two categories, such as heads and tails. Likewise, we need a mathematical formula which is appropriate when continuous measurements (such as weights, heights, and ages) can be expected to conform to what is called a normal jrequency dis- tribution. Mathematicians long ago derived the necessary formula, in fact, it has been derived several different ways, all of which — as rigorous derivations — are inappropriate to this book. However, it is possible here, and useful, to show how the normal distribution is re- lated to the binomial frequency distribution. As the number of trials (n) is increased the number of ordinates which graphically represent the binomial frequency distribution also becomes greater. As the n increases the discontinuity of the dis- tribution may become less important and less noticeable for many practical purposes. This matter is illustrated in Figures 4:21 A, B, and C, for which p — 1/2 and n = 5, 20, and 100, respectively. In Figure 4.21 A our eyes have to search a bit for the actual form of the distribution; for n = 20, the points rather definitely follow a certain symmetrical curve quite well ; and for n — 100, the points of the graph dot out a symmetrical bell-shaped curve quite clearly. To put the matter another way, if the instructor were to ask each member of the class to draw a smooth curve which seemed to the student to fit the points of the figures best, there would be considerable hesitation and disagreement about Figure 4.21A, much less trouble with Figure 4.215, and practically unanimous accord concerning the curve needed for Figure 4.21C. The student will realize that the labor involved in the construction of figures 4.21 A, B, and C becomes increasingly great as n varies from 5 to 100. In view of the fact that Figures 4.215 and C are closely approximated by continuous curves, we might hope that a relatively simple formula for a continuous curve might be employed instead of Cn, r-p^q^~^, or instead of a sunmiation involving this formula. For- Sec. 4.2 THE NORMAL FREQUENCY DISTRIBUTION 87 12 3 4 5 Number of heads (r) Figure 4.21^. The binomial frequency distribution with p = 1/2 and n = 5. 20 18 16 14 1^12 a> g-10 i 8 6 4 2 1 1 1 1 1 M 1 1 1 1 1 1 1 1 1 1 1 1 1 • 1 1 - • • - - • • - — • • - - • • - 1 1 1 M M 1 1 1 1 1 1 1 1 1 I n 1 1 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Number of heads (r) Figure 4.21B. The binomial frequency distribution with p = 1/2 and n = 20. 10 1 i 1 1 1 1 1 1 1 1 1 1 M 1 M 1 1 M 1 1 1 1 M 1 • • • 9 — • • 8 • • r? 7 — • • <u 6 • • — Q) 5 - • • — ' 4 — • • ~ 3 — • • 2 — • • — 1 1 t I, • 1 • 1 1 1 1 1 • M 1 1 M 1 1 1 1 1 1 1 M 1 1 T t - 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 Number of heads (r) Figure 4.21C The binomial frequency distribution for p = 1/2 and n = 100. BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 tunately, it can be shown mathematically that if n is fairly large and p is not far from 1/2, the numbers obtained from Cn, r'P^i^ — vY~^ by setting r successively equal to 0, 1, 2, • • • , and n are much the same as those obtained from (1/ V 27r-cr) .g-(^ - **) /2<' ^ wherein X replaces r, np = iJL, cT = vnpq , and e = the base for natural logarithms. In particular, if p = 1/2 so that fi = n/2 and cr = vn/2, it is found that the approximation is very close for n = 20 or more. Table 4.21 shows the approximation when n = 20. TABLE 4.21 Illustration of the Goodness with Which the Normal Frequency Curve Fits the Binomial Frequency Distribution When p = q = 1/2 AND 71 = 20 r or X Binomial Normal Error r or X Binomial Normal Error .000 .000 11 .160 .161 001 1 .000 .000 12 .120 .120 000 2 .000 .000 13 .074 .073 001 3 .001 .001 000 14 .037 .036 001 4 .005 .005 000 15 .015 .015 000 5 .015 .015 000 16 .005 .005 000 6 .037 .036 001 17 .001 .001 000 7 .074 .073 001 18 .000 .000 8 .120 .120 000 19 .000 .000 9 .160 .161 001 20 .000 .000 10 .176 .178 002 If the relative frequencies calculated the two ways shown in Table 4.21 are plotted on a common set of axes, Figure 4.22A is obtained. Graphically, the normal frequency distribution fits this binomial dis- tribution almost perfectly at the points where the binomial distribu- tion exists. The sum of all the relative frequencies (ordinates) for the binomial frequency distribution is 1 because it is the sum of the probabilities for all of the {n -\- 1) mutually exclusive events which are possible under the specified conditions. Likewise the sum of all the ordinates of the normal curve at the points where A" = 0, 1, 2, . . . , 19, and 20 will add to approximately 1. If rectangles of width 1 and heights 1 Vi = Viott ,-(Xi- io)Vio where i = to 20, inclusive, are constructed as in Figure 4.225, their total area also is approximately 1. Moreover, the total area of the Sec. 4.2 THE NORMAL FREQUENCY DISTRIBUTION 89 .lOU .160 1 1 1 1 r\ ' ' ' ' .140 / \ frequency o o : / \ : Relative o o o o : / \ : .040 / \ .020 / \ .000 L^ 1 1 1 1 1 TSU 2 4 6 8 10 12 14 16 18 20 Number of heads (r orx) (• = binomial; x = normal approximation) Figure 4.22^. The normal curve fitted to tlie binomial frequency distribution with w = 20 and p = 1/2. 2 4 6 8 10 12 14 16 18 20 Number of heads (r orx) (• = binomial; x = normal approximation) Figure 4.225. Illu.stration of the relationship between the area under the normal curve and the probabilities which can be derived from a binomial distribution with p = 1/2 and n = 20. 90 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 rectangles is approximately the same as the area under the normal curve, as the reader can verify visually. With the preceding remarks in mind, consider the following two facts: (a) To obtain the exact probability that r will have one of the values from a to 6, inclusive, we need to sum the ordinates, Cn, r(l/2") for n = 20 and r = a, a + 1, a +2, . ..b. (b) The opera- tion described in a is approximately the equivalent of finding the area under the normal curve between the points X — a — 1/2 and X = b -\- 1/2. The operation of a is very laborious; hence if b can be accomplished with much less work and is satisfactorily accurate, it should be the better method. As a matter of fact, this is the case, as will be shown by some of the subsequent discussion of this chapter. If the relative frequency of occurrence of a normally distributed measurement, X, is denoted by yi, we have the following general formula for yi : (4.21) ^i = — L-e-^^-'^^^/^^l 27ro- Hence, if the /t and a are known and the measurement X is known to have a normal distribution, we can graph the frequency distribu- tion by the usual methods of algebra. For example, if /a = 60 and o- = 10, formula 4.21 becomes (4.22) yi = —/= e V27r(10) -(X - 60)V200 Table 4.22 was prepared from this formula, and Figure 4.23 then was constructed from the pairs of values (A", ^i) in that table. It will be left as an exercise for the student to verify the values given for yx by using Table VI (end of book) to obtain e~"', where w = (A" — 60)V200. Thereafter, division by 10 gives the numbers in Table 4.22, under the heading y^. The following information can be obtained easily from Figure 4.23: (a) The normal distribution curve is symmetrical about a vertical line through the point where A^ = /a = 60; (6) the median A', the modal A"^, and the arithmetic mean of the A's are equal and each is equal to 60; and (c) after (X — 60) becomes at least twice the size of the standard deviation, either positive or negative, the corresponding ordinates, ?/i, are very small. In fact, when (A — 60) becomes three times the size of the standard deviation, the corre- sponding y\ is practically zero. Hence, it follows that for a truly Sec. 4.2 THE NORMAL FREQUENCY DISTRIBUTION 91 normal distribution the useful range of the Z's is about six times the size of the standard deviation. The reader should recall that this Figure 4.23. The normal frequency distribution curve for a population with /n = 60 and a = 10. approximate relationship between the range and the standard devia- tion was used in Chapter 2 when the normal frequency distribution was mentioned first, TABLE 4.22 Coordinates of Points Satisfying Formula 4.22 for a Normal Frequency Distribution with fx = 60 and c = 10 X 2/1 X 2/1 30 .000 65 .035 35 .002 70 .024 40 .005 75 .013 45 .013 80 .005 50 .024 85 .002 55 .035 90 .000 60 .040 Most students have asked (or heard someone else ask) an in- structor: "Do you grade on the curve?" The curve which the stu- dent has in mind is the normal frequency distribution curve; but it appears from the discussion above that there is a different normal curve for each combination of fx and a. This is correct ; but the stu- dent who is asking such a question is chiefly interested in his per- formance relative to the other persons who took the same examina- tion. He hopes that a grade of 40 on one test is as good as a grade of 70 on another test if its relative rank among all grades on that 92 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 examination is the same in both instances. In other words, what is of interest is the general form of the frequency distribution of a set of grades and a system for comparing one person's grade with all the other relevant grades. The discussion which follow^s is intended to show how we can reduce all formulas for particular normal fre- quency distributions to one general — and simpler — formula which preserves all the information which we usually desire from such a formula. Multiply through formula 4.21 by a and then make the following substitutions of variables: let y = o-i/i and let A = (A" — ju,)/o-. The result of these substitutions is the following formula for the standard normal frequency distribution: (4.23) y = ^e-''^'. V27r What has been done by means of these substitutions can be described graphically as follows: (a) Both the vertical and the horizontal axes have been marked off in multiples of the standard deviation, o-; and (b) the peak of the curve (which is above the point where X = ju, = md = MO) has been placed above the point where X — 0. Hence the .Y's which are less than fx now correspond to negative values of X, those which are greater than ju, now correspond to positive A's. The first two columns of Table III give the numbers needed to construct the graph of equation 4.23. Figure 4.24 was constructed by means of this table. Figures 4.23 and 4.24 are essentially the same curve; the only difference lies in the way the vertical and horizontal axes are scaled. In Figure 4.23 A would be under the point where A' = 60, would be +1 under the point where X =70 because 70 is one times the standard deviation larger than 60, the mean. The A would be —1 under the point where X = 50 because 50 is one times the standard deviation smaller than the mean, 60. The other corresponding values of A and X can be determined in the same manner. An illustration of the application of standard normal frequency distributions to a generally familiar situation can be obtained from the batting averages of baseball players. The conditions which might affect batting averages may change from season to season or from league to league so that such averages for the different situations are not directly comparable. For example, the ball may be livelier one season than during another; or perhaps the pitching may generally Sec. 4.2 THE NORMAL FREQUENCY DISTRIBUTION 93 be poorer one season than another. Hence an average of .350 during a season with a lively ball and generally mediocre pitching might not represent a better batting performance than an average of .320 at- tained with a less lively ball and more effective pitching. These matters should be reflected in the general level of batting averages and in the consistency with which players' averages grouped about the general average. That is, the mean and the standard deviation of the batting averages should be taken into account. This is precisely what is done when standard normal units are employed. Figure 4.24. Graph of the standard normal frequency distribution whose for- mula is given by equation (4.23). There also seems to be evidence indicating that batting averages can be assumed to be reasonably normal in their distribution. Batting averages for some of the better batters from the National and American leagues (undifferentiated here but kept separate when the standard normal units were computed) are presented in an ordered array in Table 4.23, first as they usually appear and then in terms of standard normal units. ■ Some interesting conclusions can be drawn from Table 4.23, al- though they might be disputed upon the basis of other evidence and other points of view. For example: (a) The best batter listed for 1940 (Deb Garms) ranks fifteenth in standard normal units, con- sidering all 4 years together. (6) The batter with the best average of all (Ty Cobb) — when the level and dispersion of batting averages within a league and year are taken into account — had an actual average of .385, which was bettered by five other batters unless 94 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 TABLE 4.23 Ordered Batting Averages for Indicated Years Before and After Conversion to Standard Normal Units (Includes higher ranking batters who were in at least 75 games during the season.) Actual Average Standard Normal Units 1910 1920 1930 1940 1910 1920 1930 1940 .385 .405 .401 .355 3.43 3.17 2.45 1.96 .364 .388 .393 .352 2.85 2.72 2.24 1.88 .340 .382 .386 .348 2.19 2.57 2.06 1.78 .331 .376 .383 .344 1.94 2.41 1.98 1.68 .325 .370 .381 .342 1.77 2.25 1.93 1.63 .322 .369 .379 .340 1.69 2.23 1.88 1.58 .321 .360 .379 .340 1.66 1.99 1.88 1.58 .320 .355 .374 .337 1.63 1.86 1.75 1.51 .312 .351 .373 .326 1.41 1.75 1.73 1.23 .309 .340 .368 .322 1.33 1.47 1.60 1.13 .308 .339 .367 .320 1.30 1.44 1.57 1.08 .306 .338 .366 .319 1.25 1.41 1.55 1.06 .305 .338 .366 .318 1.22 1.41 1.55 1.03 .304 .334 .359 .317 1.19 1.31 1.37 1.01 .302 .333 .359 .317 1.14 1.28 1.37 1.01 .301 .333 .357 .316 1.11 1.28 1.31 0.98 .300 .332 .356 .316 1.08 1.26 1.29 0.98 .300 .328 .355 .316 1.08 1.15 1.26 0.98 .298 .328 .354 .314 1.02 1.15 1.24 0.93 .298 .328 .350 .313 1.02 1.15 1.13 0.90 standard normal units are employed. In these units, his average is a decided stand-out, being 0.26 unit ahead of the runner-up. (c) In general, there is reason to believe that the batters in the year 1940 were not up to the standards of the other years shown in Table 4.23, especially those of 1920 and 1930. PROBLEMS 1. Graph the normal frequency distribution with /i = 4 and o- = 2 directly from equation 4.21. 2. Graph the normal distribution of problem 1, using formula 4.23. 3. Graph the normal curve which approximates the binomial frequency dis- tribution with n = 8 and p = 1/2. Do likewise for the binomial with n = 8 and p = 1/4, and note the decrease in the goodness of the approximation. Sec. 4.3 FRACTION OF X'S WITHIN GIVEN LIMITS 95 4. Graph as in problem 3, with n = 12 instead of 8, and comment on the effect of increasing the size of n. 5. Graph the frequency curve for a normal population with fi — 10 and (7 == 2, and estimate roughly from the graph the proportion of all the measure- ments in this population which are greater than or equal to 12. 6. Make a frequency distribution table for the birth weights of male guinea pigs as recorded in Table 2.61, compute the fi and <r, and then graph the normal curve with the same fi and o-. How does the graph compare with the frequency distribution curve made directly from your distribution table? 7. Perform the operations of problem 6, using the records for the female guinea pigs in Table 2.61. 8. Perform the operations of problem 6, using the 4-day gains of male guinea pigs as listed in Table 2.62. 9. Perform the operations of problem 6, using the 4-day gains of female guinea pigs as given in Table 2.62. 10. Graph the binomial frequency distribution for n = 16 and p = 1/2 and then plot the corresponding normal distribution on the same axes, adjusting the height to fit the binomial. Also construct for each value of r a rectangle of base r — 1/2 to r + 1/2 and height = C^g j.-p^q'^^^''. Then indicate on your graph the area under the normal curve which is approximately equal to P{8^r ^ 11), the probability that r will have a value from 8 to 11, inclusive. 11. Perform the operations of problem 10, with p = 3/5. 12. Choose any available source and compare the batting averages in the National League for 1940 and 1950, using the leading 25 players in each year and converting the batting averages to standard normal units. 4.3 DETERMINATION OF THE PROPORTION OF A NOR- MAL POPULATION OF MEASUREMENTS INCLUDED BETWEEN ANY SPECIFIED LIMITS In Chapter 2 the student was given the opportunity to learn how to construct an r.c.f. distribution, how to graph it, and how to deter- mine from this graph the limits on A" which would include any specified proportion of the data so summarized. Furthermore, the inverse process also was discussed, namely, the determination of the proportion of the data which lies within specified limits. It is de- sirable to be able to obtain the same sort of information for nor- mally distributed groups of measurements. The basis for such a procedure was given in the preceding section. There is, however, one major difference between the process taught in Chapter 2 and that which is necessary to handle the standard normal frequency distribution. In the latter situation there is no distribution table with class intervals and cumulative frequencies determined by means of certain arithmetic procedures. Instead the r.c.f. distribution must be derived from the formula for the normal 96 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 distribution function. The mathematical procedures needed in this process are beyond the level of this course; but the reader can under- stand that the r.c.j. curve of Figure 4.31 plays the same general role in the analysis of normal data that the r.c.j. curves did in Chapter 2. 1.00 Figure 4.31. Relative cumulative frequency distribution for the standard nor- mal frequency distribution described by formula 4.23. The following problems will illustrate the uses to which Figure 4.31 can be put. Problem 4.31. Determine the limits on X for the third quartile of a standard normal population of measurements. The limits required are obviously the median and Qs, respectively. If we read horizontally from .50 on the vertical scale over to the normal r.c.j. curve and then downward to the horizontal scale, we find that A = 0, as is to be expected. Doing likewise for .75 on the vertical scale, we find that A = 0.68; therefore, the limits on the third quartile are A = to A = 0.68. Since these limits apply to any standard normal distribution, the limits of the third quartile for any particular normal distribution in terms of a measurement, .Y, can be obtained from the relation: A = (X — }*) /a. Problem 4^2. What is the probability that a measurement chosen at random from a normal population with /^ = 50 and o- = 5 will be found to lie between 50 and 52? Between 48 and 50? Between 60 and 65? To reduce this specific normal distribution to the standard normal distribution, substitute /a = 50 and o- = 5 into A = {X — /a)/o- so Sec. 4.3 FRACTION OF X'S WITHIN GIVEN LIMITS 97 that A = {X - 50) /5. If Z = 50, A = 0; and if X = 52, A = 0.40. To answer the specific question asked regarding probabilities it is necessary now to extend somewhat the concept of probability pre- viously employed herein. When the possible events correspond to positions along a con- tinuous scale of measurement, the number of possibilities (previously denoted by N) is infinite. Moreover, the likelihood of occurrence changes along the scale. It no longer is useful to ask for the prob- ability that X will have a specific size along this scale on any future trial. Instead, an event E will consist of X lying within certain limits. The probability that a randomly chosen X will fall between the limits X = a to X = b now will be defined to be the proportion of all the X's in the population which are included in that interval. Graphically, this will be the proportion of the whole area under the frequency distribution curve which lies between X = a and X = b. Therefore, in problem 4.32, we need to know what proportion of this normal population lies between A = and A = 0.40. From Figure 4.31 it is learned that 50 per cent of this population has values less than and about 66 per cent has values less than 0.40; therefore, about 16 per cent of the numbers in a normal population have A's between and 0.40. It follows that P(50 ^ X :^ 52) = .16. It is concluded from the symmetry of the normal curve that P(48 ^ X ^ 50) = .16 also. Furthermore, P(60 ^ X ^ 65) = .025 because 2.5 per cent of the A''s have sizes within the limits 60 to 65. As a final illustration of the use to which Figure 4.31 can be put consider a problem of grading "on the curve." Problem 4.33. Given that a large group of grades in psychology conform to a normal distribution with ^ = 75 and <t = 7, suppose it is required to put letter grades on these scores in the proportions: 7A:20BAQC:20D:7F. What are the numerical limits on each letter grade? It is useful first to translate the proportionality above into a dif- ferent form as follows. Starting with the lowest grade, F (which will be represented at the left-hand end of the scale of A) , we have the following facts: .07 of the grades are to be F ; ,27 of the grades are to be D or F; .73 of the grades are to be C, D, or F; and .93 of the grades are to be B, C, D, or F. 98 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 It is learned from Figure 4.31 that, for a normal population, .07 of the X's correspond to A = —1.48; .27 of the X's correspond to A — —0.65; .73 of the A''s correspond to A ^ +0.41; and .93 of the A''s correspond to A — +1.50. In terms of the X's, we have the following facts obtained from the relation: A = (X — /x)/o-: .07of theX's are^65-; .27 of the X's are ^ 70+ ; .73 of the X's are ^ 78-; and .93 of the X's are ^86-; therefore, the required numerical limits on the letter grades are as follows: A = 86 on; B = 78 to 85; C = 71 to 77; D = 65 to 70; F = below 65. The preceding applications of Figure 4.31 have given approximate answers to the questions asked, and these answers are as accurate as the graph used and our ability to read values from it will allow. It seems rather obvious that a more accurate and, if possible, more convenient method is desirable. A method of this sort is available through the use of statistical tables. They perform essentially the same service as Figure 4.31. Although their derivation is not appro- priate to this book, the reader can simply keep in mind the fact that the information obtained from Table III is the same as that which can be derived directly from Figure 4.31, but is in a more accurate and convenient form. It will be left as an exercise to rework problems 4.31 to 4.33, inclu- sive, using Table III in place of Figure 4.31 as was done above. It is worth while to investigate a set of data from Chapter 2 to see if it seems to be following a normal frequency distribution, at least approximately. Actually it is not feasible at this level of statis- tics to decide this matter rigorously; but some useful information can be obtained nonetheless. Consider first the ACE scores of Table 2.01, their frequency dis- tribution in Table 2.42, and the graph of Figure 2.41. Obviously, some approximation is introduced by using such a summary — espe- cially one with only 12 class intervals — but the approximate distri- bution will serve the purpose here. The graphs of Figure 2.41 would Sec. 4.3 FRACTION OF X'S WITHIN GIVEN LIMITS 99 resemble those of Figures 4.23 and 4.31 rather closely if the former were smoothed curves instead of broken-line graphs. Hence, it appears, superficially, that the population of ACE scores follows a normal distribution fairly well if the more general and important features are the only ones considered. To be more definite, consider the following information: (a) For the ACE scores, /i, = 96, approximately, and the median is 97. These averages are equal in a normal distribution but the discrepancy is not at all large. (b) The following table shows the corresponding proportions within stated, and important, intervals on X: Percentage of the Population Included Interval on X ACE Normal Difference M±0.5<r 37.6 38.3 -0.7 Mil.Ocr 67.1 68.3 -1.2 /J. ± 1.5a 85.3 86.6 -1.3 Mrt2.0cr 95.2 95.4 -0.2 Mrt2.5(r 99.2 98.8 +0.4 M± 3.0(7 99.8 99.7 +0.1 Although the deviations from normal expectancy are somewhat systematic, there being a small deficiency in the middle and a smaller excess in the tails of the distribution, the ACE distribution still seems to be approximated by the normal quite well. If, then, it is assumed that the ACE scores do essentially conform to a normal distribution, the substitution X — {X — 96) /26 would convert the scores of Table 2.01 into standard normal measurements. The graph of their frequency distribution essentially would be Fig- ure 4.23, the r.c.f. curve would be given approximately by Figure 4.31, and Table III would present the distribution in tabular form. The statistical analysis of these data then might be more easily and efficiently accomplished than would otherwise be the case, and little or no important information would be lost in the process. PROBLEMS 1. If a binomial frequency distribution has p = 1/4 and n = 80, calculate Pir > 25) by means of the normal approximation to this binomial distribution. 2. Suppose that all the residents of a certain city definitely have made up their minds about a particular civic issue, and that 55 per cent favor one specific decision. What is the probability that on a random sample of 100 interviews 100 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 less than 50 will favor this decision, that is, it will seem that the residents are against this decision when they actually favor it? Ajis. .13. 3. Suppose that an event ^^ occurs with a relative frequency p = 1/2, and that n random observations are to be made under these conditions. How large must n be before the number of occurrences, r, of E^ will fall within one per cent of its mathematical expectation with probability equal to .10? That is, you must choose n so that P{ti/2 - n/200 ^ r ^ n/2 + ?(/200) = .10. 4. Suppose that a basketball team has established in previous games that it is safe to assume that the probability on each shot by a team member that a goal will be scored is .35. What is the probability that in a game in which they take 60 shots from the field they will hit less than 18 if the idealized assump- tions just stated are good? Ans. .17. 5. Suppose that a pair of unbiased dice are to be rolled 50 times. What is the probability that a 6 or a 7 or an 8 will appear on 20 to 25, inclusive, of these throws? 6. According to certain records the average length of growing season at Man- hattan, Kansas, is 172 days. If the standard deviation about this mean is 13 days, and if lengths of growing seasons in this area are normally distributed, what is the probability that the next growing season will be long enough to mature a crop which requires 190 days to complete its development? Ans. .084. 7. Suppose that when wood blocks of a certain type, 2 by 2 by 8 inches, are tested for strength with the proper engineering equipment, their strengths are normally distributed with mean equal 13,000 pounds and standard deviation equal 3600 pounds. How many blocks out of 100 tested would you expect to have strengths below 6000 pounds? Between 10,000 and 15,000 pounds? 8. If you are told that the heights of 10,000 college men closely follow a normal distribution with yu, = 69 inches and o- = 2.5 inches: (a) How many of these men would you expect to be at least 6 feet in height? Ans. 1150. (b) What range of heights would you expect to include the middle 50 per cent of the men in the group? Ans. 67.3 to 70.7. 9. Assuming that the wages of certain laborers in the building trades are normally distributed about a mean of $1.80 per hour with a standard deviation of 30 cents: (a) What proportion of the laborers receive at least one dollar per hour? (6) What range includes the middle two-thirds of these laborer's wages? 10. Suppose that tests have indicated that certain silk fabrics have breaking strengths which are normally distributed about a mean of 27 pounds, with ff = 8; whereas, materials with a mixture of silk and rayon have fi = 37 pounds and (7 = 9. How likely is it that a piece of silk selected at random will be at least as strong as the average for the silk and rayon mixture? How likely is it that a randomly chosen piece of the silk-rayon mixture will be no stronger than the average for silk? Ans. P = .11, .13. 11. Suppose that the persons whose ACE scores are in Table 2.01 are to be given letter grades on the assumption that these scores are normally distributed with n = 95.7 and a = 26.1. If 10 per cent are to get A's and 22 per cent D's, compute the score limits on each letter grade. Sec. 4.4 APPROXIMATE BINOMIAL PROBABILITIES 101 12. Suppose that 52 per cent of the voters in a certain city are in favor of a particular one of the possible sites for a new high school. If 100 voters are to be selected at random, what is the probability that less than 50 per cent will vote in favor of this site? If the poll is so taken that 60 per cent of those who favor that site will not participate in the poll, what now is the probability that less than 25 per cent of a sample of 100 will vote for the site in question which 52 per cent of the voters actually favor? Ans. .31, .12. 4.4 USE OF THE NORMAL DISTRIBUTION TO APPROXIMATE PROBABILITIES FOR A BINOMIAL FREQUENCY DISTRIBUTION Another important use to which the normal r.c.j. distribution can be put has been suggested previously, namely, the approximation of the summation of C„, rP''q"~^ from r = a to r = b, when n is at all large and p is close to 1/2. It has been indicated that this sum is approximately equal to that area under the normal curve between the points Xi = a — 1/2 and X2 = b -\- 1/2. Moreover, it has been shown that the area under the normal curve between any two points along the X-axis can be obtained quite easily from an r.c.f. curve or from Table III. To illustrate this process and to indicate its accuracy, suppose n = 20 and V — Q — 1/2, and that it is required to determine P(r^ 12). For this binomial distribution, ju, = nip = 10 and a- = yjnpq = V5; hence the normal distribution with these parameters will be employed in the approximation. Also, A^i = 11.5, and X2 = 20.5. In terms of standard normal units, Xi = (11.5 - 10)/2.24 = +0.67, and X2 = (20.5 - 10)/2.24 = +4.69. By means of Table III and some interpolation it is found that ap- proximately 25 per cent of a standard normal population has num- bers between these A-limits; hence P(r^ 12) = .25, approximately. Using the last column of Table VII from r = 12 on down, and using a divisor of 2^° = 1,048,576, the exact probability — to 4 deci- mals — that r will have some size from 12 to 20, inclusive, is found to be .2517. Certainly the normal approximation of .25 is excellent for most purposes, and the labor saved is considerable. 102 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 PROBLEMS 1. Given a large number of college grades which follow a normal distribution with |U = 65 and a = 10, what proportion of the grades would you expect to lie in the interval from 50 to 70, inclusive? 2. Referring to Figure 4.31, how probable is it that 3 random selections from this population will each have \'s ^2? Ans. P = .000027. 3. What proportion of the measurements in a normal population would you expect to lie beyond A' = 1.1 if /x = 0.5 and tr = 0.25? 4. What proportion of the data described in problem 3 lies at least 0.15 unit from the arithmetic mean if the numbers are in an array? .Ans. .55. 5. Certain frost data collected in the neighborhood of Manhattan, Kansas, over a 69-year period indicates that the average date of the last killing frost in the spring is April 24, with a standard deviation of 10 days. Assuming a normal frequency distribution and assuming that the date of the last killing frost cannot be predicted from a current year's weather, what is the probability that the last killing frost next spring will come on or after May 1? 4.5 STUDYING THE NORMALITY OF A FREQUENCY DISTRIBUTION BY RECTIFYING THE r.c.f. CURVE A method was given in a preceding section for detecting gross non- normality by calculating the proportions of a population lying within such intervals as ji^ ± ko- and comparing these with those proportions which are typical of a perfectly normal population of measurements. A graphic method will be presented in this section which will make it quite easy to compare the whole of a population with a standard normal population. The graphic method has these advantages: (1) Like other graphs, it utilizes the eye-mindedness of many persons. (2) It compares all the distribution with a standard normal instead of comparing a few segments such as /x ± 0.5cr, /a i lo-, etc. How- ever, this graphic procedure has the disadvantage that it may encour- age a hasty acceptance of the assumption that the given population is sufficiently near normal for the purposes at hand. More rigorous tests of normality exist in more mathematical textbooks, which can be consulted if the situation demands that additional care. It will be seen when the Central Limit Theorem is discussed in a later chapter that a considerable amount of non-normality can be toler- ated in sampling studies; hence a precise — and laborious — test for normality is not often employed. In such situations a graphic test may be sufficiently reliable. The process of rectifying a curve y — f{x), which is the basic procedure of this section, is one of changing the scale of measure- Sec. 4.5 RECTIFICATION OF THE NORMAL CURVE 103 merit of either, or both, x and y so that the new graph of ?/ = /(.r) becomes a straight line. That is, a curved line is straightened out by a change of scale. Because the reader is assumed to be familiar with logarithms, the description of a method for rectifying a normal curve will be pre- ceded by a similar discussion regarding logarithmic and exponential curves, li Y — logio X, as in Table 4.51 for selected A^'s, the pairs of values (X, Y = logio X) plot on the curve of Figure 4.51. TABLE 4.51 Some Pairs op Numbers Which Satisfy the Equation Y = logio X X Y X Y X Y 1 0.00 100 2.00 500 2.70 3 0.48 150 2.18 600 2.78 8 0.90 200 2.30 700 2.85 10 1.00 250 2.40 800 2.90 40 1.60 300 2.48 900 2.95 50 1.70 350 2.54 1000 3.00 70 1.85 400 2.60 100 200 300 400 500 600 700 800 900 1000 X Figure 4.51. Graph of Y = fog^Q X for X in the interval 1^X^1000. It is obvious that as the size of X increases, the size of Y = logio X increases less and less for equal increases in A". For example, when A changes from 500 to 600, log A changes by 0.08; but when A changes from 900 to 1000 (another increase of 100), log X changes by only 0.05. It is typical of straight-line (linear) mathe- matical relationships that Y changes the same amount for equal increases in A^. In other words, Y changes uniformly with increas- ing X. If log A is put on a uniform scale, as in Figure 4.52, and 104 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 the corresponding A''s matched with their logarithms, the X-scale is what is called a logarithmic scale. Figure 4.53 shows the effect of graphing Y = log X against .Y when X is on a logarithmic scale. Uniform scale, logX: 0.0 Corresponding X: 1 0.1 0.2 I 0.3 2 I 0.4 0.5 3 0.6 4 I 0.7 5 I I I 0.8 0.9 1.0 6 7 8 9 10 Figure 4.52. Matching of the logarithmic and the arithmetic scales of a measm-ement, X. 10 100 X on logarithmic scale 1000 Figure 4.53. Graph of Y = log^Q X when X is scaled according to the log^Q X as derived from Figure 4.52. As can be seen, the graph is a straight line, and, for any equal dis- tance along the horizontal axis, the Y changes by the same amount. It is noted that the A'-axis falls into parts of equal length: one for numbers from 1 to 10, one for numbers from 10 to 100, and another for X's between 100 and 1000. This corresponds to numbers whose logarithms have characteristics of 0, 1, and 2, respectively. Graph paper with one scale logarithmic and the other arithmetic will be called semi-log paper. When it has three repeated sections along one axis (X-axis in Figure 4.53) it is called three-cycle semi-log paper. The three cycles correspond to any three successive char- acteristics of logarithms, that is, to numbers which fall between any three successive powers of 10. Figure 4.54 illustrates the use of semi-log paper to rectify an exponential curve. In this case Y = 2e^-^, but any base for the power could be used. Clearly logio Y — logio 2 + 3X logio e; or Sec. 4.5 RECTIFICATION OF THE NORMAL CURVE 105 logio Y — 1.30A' + 0.30, approximately. This will be a straight line if Y is measured on a logarithmic scale, as in Figure 4.54. Table 4.52 gives the values used in plotting Figure 4.54. TABLE 4.52 Values of 2e^^ for Selected X's X Y = 2e'^ X Y = 2e^^ 2.00 1.15 63.00 0.25 4.23 1.25 85.04 0.50 8.96 1.50 180.03 0.75 18.98 1.75 381.14 0.85 25.61 2.00 806.86 1.00 40.17 1000 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 X Figure 4.54. Graph of F = 2e^^ on semi-log paper. 106 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 With the foregoing introduction to the method of rectifying curves the same general process will be applied to the normal r.c.j. curve. As noted earlier, one of the cjuestions which may arise in prac- tice is whether or not a given type of numerical measurement does fol- low a normal frequency distribution. Although the graphic proce- dure to be illustrated is definitely not a rigorous test for normality, it may be sufficient for practical purposes. The vertical scale to be employed will have wdiat is called a nor- mal r.c.j. scale marked off in whole percentages. The horizontal scale will be an arithmetic (or uniform) scale for the A used previ- ously in discussions of the standard normal frequency distribution. Figure 4.55 illustrates the process of obtaining the vertical scale in a manner which is analogous to that illustrated earlier for semi-log paper. It w^as constructed with the aid of Table III by plotting the normal r.c.j. as a percentage (top scale) directly over the cor- responding A, and then interpolating for the "integral r.c.j. per cent" found on the middle scale of Figure 4.55. This middle scale is the one to be used here in studying the approximate normality of fre- quency distributions. Normal r.c.f. as a % 2.3 1 15.9 50.0 84.1 1 1 1 97.7 1 (Table III) Integral % ' 1 1 1 2 3 5 1 1 1 1 1 1 1 1 1 10 20 30 40 50 60 70 80 90 1 1 95 97 I (Scale for 99 vertical axis) Standard normal units, X 1 1 -2 1 i 1 1 1 1 -1 +1 1 1 + 2 1 (Scale for horizontal axis) Figure 4.55. Determination of the scales for a normal-arithmetic graph. As is to be expected after the discussion of semi-log graph paper, it is not necessary to go through the work back of Figure 4.55 be- cause graph paper already exists on which we can do this graphing. Figures 4.56A and B were constructed on normal-arithmetic paper to illustrate the way the normality or the non-normality of a dis- tribution affects a graph on such paper. Four distributions are employed in these illustrations: (a) Truly normal distribution of Table III; (b) ACE scores of Table 2.01; (c) the data on farm acreages in Table A (below) ; and (d) the definitely non-normal distribution of Table 4.53. Sec. 4.5 RECTIFICATION OF THE NORMAL CURVE 107 TABLE 4.53 A Fictitious Non-normal Frequency Distribution Class Interval / r.c.J. X Class Interval / r.c.j. 55-59.9. . 1 100.0 3.9 10-14.9. 60 80.1 0.61 50-54.9.. . 2 99.8 3.5 5- 9.9. 80 69.5 0.25 45-49.9.. . 5 99.5 3.2 0- 4.9. 100 55.5 -0.11 40-44.9.. . 8 98.6 2.8 - 5 to - 0.0. 90 37.9 -0.48 35-39.9.. . 10 97.2 2.43 -10 to - 5.0. 80 22.0 -0.84 30-34.9.. . 12 95.4 2.07 -15 to -10.0. 30 7.9 -1.21 25-29.9.. . 15 93.3 1.70 -20 to -15.0. 10 2.6 -1.57 20-24.9.. . 20 90.7 1.34 -25 to -20.0. 5 0.9 -1.94 15-19.9.. . 40 87.1 0.98 568 ji = 6.6, approximately; and a = 13.73. 99 95 90 80 c 70 ^60 § 50 ^ 40 "30 20 (A)« Normal (Table III) ® ACE scores I I I I I I + 1 +2 +1 +2 -2 -1 X X Figures 4.56. Some graphs of r.c.j. distributions on normal-arithmetic paper. It should be evident from Figures 4.56 that the following are true: (a) The frequency distribution from Table III yields a perfectly straight line when r.c.j. as a percentage is plotted against A on a normal-arithmetic graph paper. (6) The frequency distribution of the ACE scores apparently is quite near to normal because the points of their r.c.j. graph on 108 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 normal-arithmetic graph paper appear to deviate only slightly from a straight line. (c) The distribution of the farm acreages in Ness County, Kan- sas, is essentially normal except that the lower end of the distri- bution is missing, that is, the distribution is truncated. (d) The fictitious distribution of Table 4.53 clearly is not nor- mal because the points of the r.c.f. graph definitely do not follow a straight line on normal-arithmetic paper. It should be noted with respect to the conclusions above that only gross (and hence certainly serious) non-normality will show up under this sort of scrutiny. A look at the frequency distributions associated with (6) and (c) above shows that there certainly is some lack of normality. Figures 4.56 show this clearly ; but whether or not the relative departure from a straight line is negligible will depend on the particular circumstances. Discussion to be given in Chapter 6 will be helpful in this decision. PROBLEMS 1. Plot the following pairs of values of X and Y as points on a graph, using semi-log paper with Y measured on the logarithmic scale. Then determine the slope of the straight line through the points and relate it to the way Y changes per unit increase in X. X: 1, 2, 3, 4, 5, 6. Y: 2, 6, 18, 54, 162, 486. 2. Plot Yi = logio A' and F2 = 5 logio X on the same sheet of arithmetic graph paper and also on the same sheet of semi-log paper. What effect does the coeffi- cient 5 have on these graphs? J -(X-l)2 3. Plot the r.c.f. curve for y = — -f= e ^° on normal-arithmetic graph paper. 5V27r 4. Plot the following tabular r.c.f. distribution on normal-arithmetic paper and comment on any lack of normality revealed by your graph. Class Class Class Interval r.c.f. Interval r.c.f. Interval r.c.f. 80-82.99.. 1.00 65-67.99... .33 50-52.99... .09 77-79.99.. .90 62-64.99... .26 47-49.99... .07 74-76.99.. .74 59-61.99... .20 44-46.99... .05 71-73.99.. .50 56-58.99... .16 41-43.99... .02 68-70.99.. .40 53-55.99... .13 5. Make an r.c.f. distribution for the fly counts of problem 1, section 2.4. Plot this distribution on normal-arithmetic paper, and discuss any apparent non-normality of this distribution. Ch. 4 REVIEW PROBLEMS 109 6. Perform the operations required in problem 5 for all the birth weights of female guinea pigs listed in Table 2.61. 7. Perform the operations required in problem 5 for all the birth weights of male guinea pigs listed in Table 2.61. 8. Perform the operations required in problem 5 for all the 4-day gains of female guinea pigs listed in Table 2.62. 9. Perform the operations required in problem 5 for all the 4-day gains of male guinea pigs listed in Table 2.62. 10. Perform the operations required in problem 2 for logarithms to the natural base e rather than 10, and comment on the effect of this change on the graph. REVIEW PROBLEMS 1. If you are among 1000 persons, each of whom purchases a one-dollar lot- tery ticket for a prize of $1000, what is the expected value of your ticket in the mathematical sense? 2. Determine the expected frequencies of sums of 3, 4, 5, and 6, respectively, when three unbiased dice are thrown simultaneously 1000 times. Ans. 4.6, 13.9, 27.8, 46.3. 3. If 25 pennies and 15 dimes are placed in individual envelopes, thoroughly mixed, and presented to you for the selection of one envelope, what is the probability that you will get a dime? What is your mathematical expectation on such a draw? 4. How many two-digit numbers can you make up by selecting any number from 1 to 9, inclusive, for each digit? How many numbers could you form if none is to contain the same digit twice? Ans. 81, 72. 5. Suppose that a turtle is hatched at point A and then wanders over a uni- form terrain in search of food. If he never wanders more than 1000 yards radially from spot A, and if he moves over the area in such a way that he is equally likely to be on any preassigned areas of a specified size, what is the probability that he will be within a circular area of 100 square yards whose center is 300 yards from, and northeast of, the spot Al 6. Table A (below) and Figure A present the distributions of the various sizes of farms in Ness County, Kansas. If a stratoliner were to drop a package by parachute so that it will be sure to land on Ness County, but the pilot cannot tell where, what is the probability that it will fall on a farm of more than 1000 acres if 10 per cent of the county is not in farm land and that 10 per cent is uniformly distributed over the county? Ans. .30. 7. If 100 farmers are to be selected from Ness County without knowledge of the areas of their farms, and if one supposes one farmer per farm, what is the mathematically expected number of representatives from farms covering less than 500 acres? What fraction of the county's farm acreage do they represent? 8. Determine graphically the lower limit of the sixtieth percentile and of the third decile for the data of Table A. An&. About 520 acres; 260 acres. 9. Table B presents a summary of the years of schooling had by all legal residents of Kansas who were 25 years of age or older on April 1, 1940. Con- struct what appears to you to be a good graphic presentation of these data. 110 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 10. If a roving reporter were to go all over Kansas, impartially asking persons their opinions on a certain educational matter, what proportion of his inter- views would you expect to be with persons who have had at least two years of college education if he talked only to persons who were at least 25 on April 1, 1940? What percentage would have no college education? Ans. 9 per cent, 88 per cent. 11. If an insecticide is known to be 99 per cent lethal to a certain species of insect, what is the probability that less than 5 will survive if 150 selected at random are sprayed with this spray? 12. Suppose that a particular variety of apple grown under specified condi- tions produces yields (per tree) which are normally distributed with fx = 8 bushels and o- = 2.5 bushels. What is the probability that a randomly chosen tree will be found to yield less than 5 bushels? That two such trees will each be found to yield less than 5 bushels? Ans. .12, .014. 13. Obtain records like those in Table B from the latest census, make an r.c.f. graph for those data, and determine the median years of schooling. Compare with the median for 1940 and draw any appropriate conclusions. Use college years as 13, 14, 15, 16, and 17. 14. Plot the r.c.j. distribution of Table A on normal-arithmetic paper and comment on the apparent normality, or lack of it, for this distribution of farm 1 201 401 601 Figure A. 801 1001 1201 1401 1601 1801 2001 Acres Two types of relative cumulative frequency distributions for the sizes of farms in Ness County, Kansas, in 1940. Ch. 4 REVIEW PROBLEMS 111 TABLE A. Distribution of Farm Acreages in the Whole of Ness County, Kansas (Data furnished by W. H. Pine, Department of Economics and Sociology, Kansas State College.) Acreage Frequency, Frequency Interval / Over 2000 28 1801-2000 7 1601-1800 14 1401-1600 21 1201-1400 28 1001-1200 35 801-1000 77 601- 800 194 401- 600 236 201- 400 320 1- 200 274 Total 1233 IX = 550 acres B. Years of Schooling squency, Percentage of r.c.f. Total Acreage 1.00 1.00 .98 .88 .97 .86 .96 .83 .94 .78 .92 .73 .89 .67 .83 .57 .67 .37 .48 .21 .22 .06 md = = 420 acres 25 Years Old on April 1, 1940 Years Number, Completed / None 11,975 Grade 1 2,136 2 5,507 3 14,833 4 29,745 5 33,628 6 45,722 7 54,326 8 388,512 High School Year 1 62,173 2 61,935 3 31,315 4 173,580 College Year 1 29,113 2 32,374 3 12,973 4 35,347 At least 5 12,580 Total 1,037,774 112 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 REFERENCES Dixon, Wilfrid J., and Frank J. Massey, Jr., Introduction to Statistical Analysis, McGraw-Hill Book Company, New York, 1951. Hald, A., Statistical Theory with Engineering Applications, John Wiley & Sons, New York, 1952. Kenney, John F., Mathematics oj Statistics, Part I, Second Edition, D. Van Nostrand Company, New York, 1947. Waiigh, Albert E., Elements oj Statistical Method, Second Edition, McGraw- Hill Book Company, New York, 1943. CHAPTER 5 Sampling from Bmomial Populations When a population of numerical measurements involves so much data that it is either impossible or unwise to attempt to analyze the whole of it, sampling must be relied upon to furnish the desired information. As a matter of fact, most of the statistical analyses now performed involve sampling data. A multitude of examples could be sited to illustrate the need for sampling, but the following will suffice for the purposes of this discussion. (5.01) Public opinion polls. Only a small percentage of the per- sons eligible for interview actually are questioned about the matter under study. The sole objective of the study is to estimate the pro- portions of the citizens favoring the various points of view. If the question to be asked has only a yes or a no answer the results of the poll will constitute a sample from a binomial population, and we would be attempting to estimate p. (5.02) A study of the toxicities of two insecticides conducted by spraying insects of a certain species with the insecticides and count- ing the dead insects. This is another case of sampling a binomial population; but the purposes of the investigation may be different. The following question is to be answered: Is one of the sprays more toxic to these insects than the other? Statistically, the question becomes: Is it reasonable to suppose that the two sets of data ob- tained with the two sprays are samples from the same binomial population? Of course, such a study also may include the estima- tion of p as mentioned in (5.01). (5.03) Testing the breaking strengths of concrete columns, of wood or of metal beams, and of other engineering materials. Break- ing strengths are measured on a continuous scale of numbers; hence their populations have continuous frequency distributions. Problems 113 114 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 of this sort include the estimation of true average breaking strengths, and comparisons of the strengths of different materials. (5.04) Studies involving two variables such as -prices of selected stocks and the volume of production of finished steel, ACE score and grade average in college, stand counts of wheat and the yield of a plot, etc. In such investigations it would be necessary to estimate from sampling data the relationship between the two variables, ex- press it mathematically, and then use it in accordance with the pur- poses of the investigation. It can be seen from the examples above that two general types of statistical problems must be considered in sampling studies. One is to derive from the sample observations some numbers which can be used satisfactorily in place of one or more unknown population parameters. These numbers which will be derived from the sample are called sampling estimates of the parameters. They are change- able from sample to sample and, being dependent upon chance events, are subject to the laws of probability. The other general problem is to test hypotheses regarding pop- ulations against actual sample evidence. For example, if the popula- tions of the breaking strengths of two types (different shapes, for example) of concrete columns each follow a normal frequency dis- tribution with the same variance, o-^, these populations can differ only in their means fxi, and /xo. That is, it is supposed that the en- gineers in charge are satisfied that the two types of columns have the same uniformity of performance from test to test, but it is yet to be decided whether they have the same average strength. If so (that is, if jxi — 1x2) , the populations of breaking strengths are iden- tical normal populations. It then becomes a problem of deciding from samples taken from each population whether or not /xi is in fact equal to ixo- It usually is convenient statistically to assume that /Ai does equal /xo, and then to see how reasonable this hypothesis is in the light of sample observations. It should be clear — intuitively, at least— that decisions based on samples may be in error, and that we do not know in any particular case if our sample is so unusual that it is misleading us. How, then, can sample evidence become a satisfactory basis for making decisions about populations? The answer lies in the fact that, while no one can say whether a particular decision is right or wrong, it is possible to determine the relative frequency with which correct decisions will be made over the long-run of much experience if we are following Ch. 5 SAMPLING FROM BINOMIAL POPULATIONS 115 certain rules for acting upon the basis of sampling evidence. It follows that the probability of making a correct decision from any specified future sample (say the next one we are going to take) also can be stated. To illustrate some of the preceding discussion, suppose you are about to engage in a coin-tossing game in which "heads" is the event which is of particular interest to you. Assume, also, that you are not satisfied that the coin is unbiased but are not going to woriy about bias unless the probability of heads is as low as 1/3. Before playing the game you are going to flip the coin 15 times and then come to a decision regarding the bias of the coin. What rules for action should you adopt and how effective will they be in detecting bias as bad as p = 1/3? It is being assumed that you are not enter- taining the possibility of bias toward too many heads. As long as the coin has two sides and one is heads, the other tails, any result from to all 15 heads can occur on 15 flips regardless of bias in the coin. However, it should be clear that the relative fre- quencies of occurrence of the 16 possible results are dependent upon the size of p. For p = 1/3, for example, such a result as 15 heads on 15 throws is an extremely rare occurrence. The actual rarity, in terms of probability, can be derived from the binomial series for {q + p)", with p, q, and n given. If p = 1/2 and n = 15, the binomial series is (1/2 -f 1/2)1^ = .000 + .000 + .003 + .014 + .042 -f .092 -\- .153 r: 1 2 3 4 5 6 + .196 + .196 + .153 + .092 + .042 + .014 + .003 r: 7 8 9 10 11 12 13 -f .000 + .000. r: 14 15 When p is unknown and a sample has produced r = 0, 1, 2, or 3 heads on 15 random flips, you probably would be very reluctant to accept the hypothesis, Hq{p = 1/2) because the total probability of the occurrence of one of these 4 mutually exclusive events is but .017, or about 1 chance in 59. Although it is true that one of those 4 results can be obtained when the coin is unbiased — and you knew this before you tossed the coin 15 times — you are now faced with the necessity to decide if the coin is biased or not, and you must do 116 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 SO upon the basis of the sample's evidence. If you decide to reject the hypothesis that p = 1/2 whenever the observed number of heads is one of the 4 cases just listed, you will unjustly reject Hq 1.7 per cent of the time because that is how frequently such cases occur by chance when p does equal 1/2. Nevertheless, some rules for action must be adopted or else nothing can be decided from samples. Hence, it will be supposed that the following rules will be followed after 15 sample tosses of the coin in question: (a) If r = 0, 1, 2, or 3 heads, you will reject Hoip = 1/2) and assert that the coin is biased against heads. (b) If r — 4, you will accept Ha and play the game on the assump- tion that the coin is not biased against heads. These two rules can lead you to correct conclusions and actions, and they also can cause you to make one of two kinds of errors: (1) The hypothesis Hoip = 1/2), which is being tested by sam- pling, may be rejected when it is true. This will be called an error of the first kind. In the above example, the probability that such an error would occur was noted to be .017 under the rules a and b. (2) The Hq may be accepted when it is false. This will be called an error of the second kind. It should be clear that the likelihood of committing an error of this kind depends on what possibilities — or alternative hypotheses — there are. It is customary to set up the hypothesis Ho in such a way that it is considered more serious to make an error of the first kind than it is to accept a false hypothesis. When this is done the probability of committing an error of the first kind (to be designated by a) is kept low — usually a< .10 — and the rules adopted for acting upon the basis of sampling evidence are chosen so that for a given a the prob- ability that an error of the second kind will be made (to be desig- nated by (3) is as small as possible under the circumstances. Referring back to the coin-tossing problem, we see that a = .017. Also, the person who was trying to decide from 15 throws if the coin was seriously biased would not care if p had some size between 1/2 and 1/3, but did wish to detect a p as low as 1/3. Hence the alterna- tive hypothesis whose truth could lead to errors of the second kind includes all p's at or below 1/3. For the sake of simplicity it will be assumed that the only alternative hypothesis to Hoip = 1/2) is i/i(p = l/3). Sec. 5.1 OBTAINING THE SAMPLE 117 The /? can be determined from the following series: (2/3 + 1/3)^5 = .002 + .017 + .060 + .130 + .195 + .214 + .179 r: 1 2 3 4 5 6 + .115 + .057 + .022 + .007 + .002 + .000 + .000 r: 7 8 9 10 11 12 13 + .000 + .000. r: 14 15 Hence if p actually is 1/3 so that the hypothesis of no bias should be rejected, the probability is .002 + .017 + .060 + .130 = .209 that Ho will be rejected. Or the probability that Ho will not be rejected when it should be — an error of the second kind — is ^ = 1 — .209 = .791. Obviously, the rules a and b would not be good ones if it is serious to fail to detect the bias indicated by p = 1/3. However, if the most serious mistake is to accuse someone of employing a biased coin when he is innocent, rules a and 6 may be quite satisfactory. In practice we seldom can compute yS as simply as above. Usually the a is set at an appropriate level and then standard tests are em- ployed without actually knowing the /3. However, it can be said here that the tests to be discussed in this, and the next, chapter have been chosen with the idea of making the /? as small as possible under the circumstances and for the chosen a. As the heading of this chapter indicates, the subsequent discussion will be confined to samples from binomial populations. Later chap- ters will take up the normal and the two-variable situations. 5.1 OBTAINING THE SAMPLE Before a method for obtaining the sample is devised, the popula- tion which is to be sampled must be defined clearly. It is recalled from Chapter 4 that a binomial population is possible only if the units in some definable group have attributes which may be described by just two classes. Moreover, the fractional part of the population falling into each class must stay fixed. For example, all the farmers in Finney County, Kansas, on July 1, 1953, could be classified un- ambiguously into two classes as regards membership in some co- operative association: those who do belong to some cooperative and those who do not belong to any such association. The units would 118 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 be the individual farmers' designations as member or as non-member. The fraction of the total number of farmers in that county who were classed as members could be the parameter p of Chapter 4. Then 1 — V would be the fraction who were classified as non-members on the stated date. It is entirely possible for those same farmers to be the basis for other populations. Their answers for or against a proposed new federal farm policy could constitute another binomial population of interest, their per-acre incomes during a specified period could be another (non-binomial) population, and the sizes of their families on July 1, 1953, could be still another (non-binomial) population which might be of interest to some group of persons. The chief criterion of a good definition of a population which is about to be sampled is that it make entirely clear in all important respects the larger group of units to which the conclusions drawn from the sample will pertain. Given a well-defined population, the sample obviously must be taken in such a manner that the impression it produces through statistical analyses will have the greatest possible chance to be accurate and dependable. Naturally the facilities and economic re- sources available for the sampling may be limiting factors; but it will be assumed in the discussion to follow that those resources and facilities are at least good enough to justify undertaking the sam- pling study at all. For purposes of illustration, suppose that we wish to determine public opinion in a large city regarding a political issue of current interest, and that our resources allow us to interview only one person per each hundred in the population. How should this one per cent sample be taken? If we were to visit the few major business districts we could obtain our allotted number of interviews more quickly and with less cost because more persons are concen- trated in these small areas during business hours; but we have no assurance that the opinions of the persons we would meet there are the same as those we would find in the outlying districts, for example. We might consider using the telephone directory until we thought of the fact that some residents do not have telephones. If we are interested only in the opinions of registered voters — as is easily pos- sible — ^we could use an official listing of those persons. It is possible that for some questions which might be asked we could take a random sample of names from this list and interview them as our sample. Such a random sample could be taken by numbering the entries con- secutively from one to the number of voters on the list and then Sec. 5.1 OBTAINING THE SAMPLE 119 drawing numbers at random until a sample of the desired size was obtained. Such a procedure would make it true that every possible sample of the size n had been equally likely to be drawn at the outset of the sampling, and this is necessary in random sampling. In many circumstances the procedure of sampling just outlined would be unsatisfactory. A city might be made up of racial and economic groups of such diverse opinions on the matter being studied that it would not do to leave their representation in the sample to pure chance, as in the random sampling just described. It would be necessary to sample each group in accordance with its proportionate part of the city's total registration of voters. It is noted that the sampling discussed above has supposed that the sample will be taken by means of personal interviews. Any such systems as calling persons on the telephone or mailing questionnaires, which depend on voluntary and selective responses, or on their being at a certain place at a certain time, are almost certain to produce biased samples. The cause of their not responding, and hence not being in the sample, may be associated with the type of response they would have given. The theory and techniques of sampling in such a way that the conclusions which can be drawn therefrom will be accurate and re- liable are very extensive and cannot be covered here. The remarks above merely point out a few of the more important and general considerations. However, the reader can be warned to be critical of any conclusions drawn from samples until he is satisfied that the samples were taken in such a way that they should be representative of the population about which conclusions have been drawn. If one brand of cigarettes is said to be the favorite of a certain ])rofessional group, we should at least wonder if that group was properly sampled. Or, if someone returns from a foreign country and asserts that the residents of that country hold certain points of view regarding a matter of world-wide interest, we should wonder if he did an ade- quate job of sampling public opinion in that country. Or, as a final example, if someone seeks to obtain a sample by means of a mailed questionnaire, we should wonder if those who do not respond have a different opinion, say, from that generally expressed by those who did return their questionnaires. If so, what population did those who returned their questionnaires represent? 120 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 PROBLEMS 1. Suppose that you were sent out to ascertain the public opinion in a cer- tain community regarding the necessity for flood control of a certain type in that area. How would you obtain your sample so that it would be representa- tive of the whole community? 2. Referring to problem 1, would it make any difference in the manner in which you took your sample if it were taken in July of 1936 during a severe drouth oi' in July of 1951 right after a record flood? Justify your answer. 3. Suppose that a roving reporter goes into a city with the intention of ascertaining public opinion on a matter of foreign policy. He is going to walk about the streets asking persons at random a specific question requiring one of three answers: Yes, No, or No Opinion. Will it make any difference what hours of the day, between 7 a.m. and 6 p.m., he does this? Would the day of the week matter? Would the type of city — industrial, college site, farming community, rich suburb, and the like — have anything to do with the answers to these questions? 4. Suppose that a company which is manufacturing candies develops a new product whose originators believe is especially good. Which of the following possible ways of testing the public's reaction to this new confection would you prefer to use? Why? (a) Sit back and see how the sales go. ib) Have some trained persons take samples out to the public and ask peo- ple to taste the candy, to record their reactions, and to give these records to the field representatives directly. (c) In each of the first 10,000 packages manufactured, place a stamped and addressed card requesting that the purchaser record his opinion of the candy and mail the card to the company. (d) Ask a panel of expert candy tasters to decide the matter. (e) Do as in d, first, then a. (/) Do as in d, first, then b. ig) Do as in d, first, then c. (h) Have all the firm's employees recoid their opinions of the candy and decide from these records if mass production is wise. (i) Combine h and a. (;■) Combine h and b. (k) Combine h and c. (l) Combine others above. Specify. (m) Specify another method if you have one you prefer. 5. Suppose that some engineering concern wishes to test the strength of two types of structural beams, each produced and recommended by a different com- pany. Which of the following sampling procedures would you recommend if the engineering group has two laboratories, each with its operating personnel, available for the tests? (a) Ask each company to send a specified number of beams for testing and have each laboratory test half of each company's product. Sec. 5.2 POINT AND INTERVAL ESTIMATION OF p 121 (6) As in a, but have one laboratory test all one company's beams, the sec- ond laboratoiy testing all the second company's beams. (c) Go into the public market and purchase the necessary number of beams of each type, and then do as in a. (d) As in c, but replace a by b. (e) Specify other ways. 6. An agronomist wishes to run critical yield, protein, and test weight studies on a proposed new variety of wheat before the variety is released to the public. He proposes to use a standard and widely planted variety for comparison with the new one. Plenty of land is available for this study, but it is quite non- uniform in soil qualities, moisture content, and exposure to weather. Which of the following outlines for such a study would you prefer, and why? (a) Plant the new variety on the east half of the available land, the standard variety on the west half (or vice versa, as decided by flipping a coin), harvest and measure wheat from each half, determine test weight and protein content on the yield from each half separately. (6) Divide the available area into 20 equal-sized plots and plant 10 plots to each variety, choosing the variety for a plot by drawing the names from a hat. Then determine yield, protein, and test weight separately from each plot's wheat. (c) Do as in b, except that the plots are grouped into 10 pairs and each pair has both varieties planted side by side. (d) Save the land for some other purpose, send out samples of each wheat to 10 farmers, and ask them to report the yields and test weights and send in samples for protein analysis. 5.2 CALCULATION OF POINT AND INTERVAL ESTI- MATES OF p FOR A BINOMIAL POPULATION It was indicated in Chapter 4 that a binomial frequency distribu- tion can be defined when individuals are identified only as belonging to one of two possible classes of attributes such as male or female, dead or alive, acceptable product or unacceptable product, and the like. Moreover, the proportions falling into the two classes of at- tributes are constantly p: (1 — p). If n members of a binomial population are selected at random, the particular individuals drawn are the result of chance occurrences. Hence, we may find that any number from r = to r = n of those individuals possess the attribute A, say, even though a fixed propor- tion, p, have that attribute in the whole population. The possible outcomes of such a sampling vary from r = to r = n and form a binomial frequency distribution with mean ix = np and with standard deviation a = y/npq, as was shown in Chapter 4. The reader is reminded that the probability that exactly r of the n members of 122 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 the sample will possess an attribute with probability of occurrence = p for any specified future trial from the population is given by the formula: C„, r'P''(l - p)"-*". The point of view in the preceding paragraph is that of Chapter 4 in which the size of p was assumed to be known. More commonly, p is not known and we have only a sample estimate of its size. This estimate is r/n, which varies under repeated sampling from to 1. Even though r/n is a variable quantity, useful and reliable conclu- sions can be drawn from samples taken from a binomial population, as will be shown shortly. Three types of such conclusions will be considered in this chapter: (a) Given a sample, what can we say about the size of p? (6) Given a sample from a binomial distribu- tion, how well does it agree with a predetermined hypothesis con- cerning the magnitude of the p for that population? (c) Given two random samples, did they probably come from the same binomial population? The present section is concerned with question a. When the true proportions of the two types of members of a binomial population are not known, they can be estimated by means of a sample, as suggested above. This estimation can take either of two forms: (a) a point, or specific, estimate of p, which would be used in lieu of the p, or (b) an interval estimate which would have a preassigned probability of bracketing the size of p. This latter process is called placing a confidence interval on p. The confidence we can have that the bracket, or interval, does actually include the unknown parameter is described by the confidence coefficient. Statistical research indicates that the best point estimate of p is ob- tained from p = r/n, the observed fraction of the sample which pos- sess the particular attribute that is being studied. Some of the reasons for this decision are: (5.21) The p has an expected value E{p) = E{r/n) = E(r)/n = np/n = p for any particular sample size, n. That is, the long-run av- erage size of p is exactly equal to the true population parameter p. It is customary to call point estimates unbiased estimates if their mathemat- ical expectation is the parameter which is being estimated. We gen- erally prefer to employ unbiased estimates, like p, unless some more important property is missing. (5.22) The estimate p = r/n has a variance = pq/n because the variance of r is npq — as shown in Chapter 4 — and the effect of dividing the r by n is to divide the variance by n^, as was shown in the section of Chapter 2 which dealt with the coefficient of variation. This vari- Sec. 5.2 POINT AND INTERVAL ESTIMATION OF p 123 ance, pq/n, of the estimate p will be quite small for a sample of almost any useful size because the p and the q are each less than unity. This indicates that p will not vary greatly from sample to sample, especially if the sample size is fairly large. As a matter of fact, the size of the variance of p can be made as small as desired by taking the n suffi- ciently large. Hence, this estimate, p, is considered to be a very efficient estimate of p. In view of the fact that p is almost always in error to some degree in spite of the fact that it is the best point estimate possible, there are many circumstances in which an interval estimate of p is desirable. The interval estimate also is more difficult to compute and to inter- pret; hence it will be considered in some detail. The situation is this: n members of a certain binomial population have been drawn at random so that each member of the population had an equal opportunity to be in the sample, and r of them have been found to have the specified attribute A. Given the proportion r/n observed in the sample, what useful limits can we place on the true proportion, p, of A members in the whole population, and what confidence can we have in those limits? It is customary to call such interval estimates confidence limits, or to say that these limits con- stitute a confidence interval. The degree of confidence which we can place in such limits on p is measured by the probability that the sample has given an interval which actually does include p. As might be expected, this probability is the relative frequency with which the sampling process used will produce an interval which does include p. It will be convenient to use the symbol CI95, for example, to designate the confidence interval which has — at the start of the sampling proc- ess — 95 chances out of 100 of including the parameter which is being estimated. Suppose that a relatively small manufacturing concern is produc- ing roller bearings which are to be shipped to a larger company manufacturing farm machinery. There will be certain specific stand- ards, such as maximum or minimum limits on diameter, which the bearings must meet before they are considered to be acceptable products. Hence, any large batch of bearings could be grouped into two subgroups marked as "acceptable" and "unacceptable," respec- tively, if every bearing were to have its diameter measured with perfect accuracy. It will be assumed here for simplicity of discus- sion that the company which is to receive the bearings requires that each shipment must be 90 per cent "acceptable" or it can be re- jected. The concern which is producing the bearings will have to 124 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 inspect its products by means of samples because it is inconceivable that every bearing should be carefully measured. Assume that a sample of 10 bearings has been inspected and that all 10 were found to be acceptable. Is this sufficient evidence that the shipment probably is up to the standard? In this connection, consider a binomial population with p only .80; that is, it is well below the standards set above. The probability that every member of a sample of 10 will be acceptable is (.80) ^•'j which is .11; hence there is about 1 chance in 9 that this definitely substandard batch of bearings will show none unacceptable on a sample of but 10. Ob- viously, if p were less than .80, p^° would be less than .11; and, con- versely, if p were larger than .80, p'*^ would be greater than .11. Therefore, it should be clear that the result, 10 acceptable bearings out of 10 inspected in a sample, could be obtained from any one of a whole range of possible binomial populations corresponding to values of p ranging from to 1. As a matter of fact, the sample discussed above could be drawn at random from any binomial pop- ulation with as many as 10 acceptable bearings among the individ- uals. Of course, with n. = 10, a sample with r also equal 10 is more likely to come from a population with p near 1 than from a popula- tion with p near 0. The above discussion re-emphasizes the fact that we cannot attain certainty in conclusions drawn from samples: there always must be some risk that the sample has led to a false conclusion. We choose a risk of error which we can afford to take and express it in terais of the confidence coefficient described earlier. If it be supposed that an event which is as unlikely to occur as 1 time in 20 can be ignored, what confidence interval (CI95) can we set on p as a result of the above sample in which r = 10 acceptable bearings out of 10 observed in the sample? We use what will be called a central 95 per cent of all possible r's by determining a range on r which is such that not more than 2^2 per cent of all samples with the same n and p will fall beyond each end (separately) of the range so determined. For example, in the series below for (1/4 + 3/4)^'^ the first five terms — to the left of the brace — add to .0197, which is less than 2% per cent, or .0250. If the sixth term from the left is added, the sum exceeds .0250. There- fore, among all possible samples of 10 observations from a binomial distribution with n = 10 and p = 3/4 the sample number, r, will be below 5 for a bit less than 2^/4 per cent of all such samples. At the other end of the series for (1/4 -f 3/4)^° no term is less than or equal to .0250; hence, the "central 95 per cent" will be occupied by samples Sec. 5.2 POINT AND INTERVAL ESTIMATION OF p 125 for which r = 5, 6, 7, 8, 9, or 10. Consequently, if you have drawn a sample with n = 10 and p is unknown, it is quite unlikely that p was as large as 3/4 if it was found in the sample that r = 0, 1, 2, 3, or 4. As a matter of fact, you could just form the habit of assuming that p never is as large as 3/4 whenever r is 0, 1, 2, 3, or 4 and you would be wrong less than 5 per cent of the time because samples of that sort occur less than 5 per cent of the time when n = 10 and p is as large as 3/4. To answer the question posed earlier, we consider the following reasoning. If n = 10, the probability series for p set successively equal to 2/3, 3/4, .69, and .70 (for reasons which will appear soon) are obtained as in Chapter 4 and lead to the following conclusions: (1/3 + 2/3)^° = .0000 + .0003 + .0031 + .0163{+ .0569 + .1368 (sum = .0197, is <.0250) r: 1 2 3 4 5 + .2276 + .2601 + .1951 + .0867}+ .0173 is <.0250 r: 6 7 8 9 10 (Note that the "central 95 per cent" does not include the observed number of occurrences, r = 10.) (1/4 + 3/4)^^ = .0000 + .0000 + .0004 + .0031 + .0162{+ .0584 (sum = .0197, is <.0250) r: 1 2 3 4 5 + .1460 + .2503 + .2816 + .1877 + .0563} none ex- cluded r: 6 7 8 9 10 (The central 95 per cent does include the sample result, r = 10, but still might do so with a smaller p; hence p = 3/4 may be too large to be the lower end of the 95 per cent confidence interval. Therefore, p = .70 will be tried.) (.3 + .iy^ = .0000 + .0001 + .0014 + .0090{+ .0368 + .1029 (sum = .0105, is <.0250) r: 1 2 3 4 5 + .2001 + .2668 + .2335 + .1211 + .0282} none excluded r: 6 7 8 9 10 126 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 (The central 95 per cent still includes the observed number, r = 10, and it again is possible that p could be smaller and still keep r = 10 in the central 95 per cent. Hence, try p = .69.) (.31 + .69)^^ = .0000 + .0002 + .0018 + .0108 {+ .0422 + .1128 (sum = .0128, is <.0250) r: ; 1 2 3 4 5 + .2093 + .2662 + .2222 + .1100} + .0245 is <.0250 r: 6 7 8 9 10 (The central 95 per cent now just barely excludes the observed num- ber, r = 10; therefore, the smallest value of p which has been con- sidered here and which still keeps r = 10 within the central 95 per cent is .70. However, it is clear that if three decimal places were used, the lower end of the confidence interval would be nearer to .69 than to .70; hence, .69 is taken as the lower end of the 95 per cent confidence interval.) To determine the upper end of the 95 per cent confidence interval, it is necessary to find out by a similar procedure how large p can become and still leave the observation, r = 10, in the central 95 per cent of the binomial population with n = 10. Obviously, p can go all the way to 1.00, or 100 per cent, and still not exclude the case when r = 10; hence, p = 1.00 is the upper limit of the 95 per cent confi- dence interval when r has been found to be 10 when n = 10. There- fore, it is concluded that if with n — 10, r is observed to be 10 also, the 95 per cent confidence interval on the true percentage in the population is .69 ^p^ 1.00. At the same time, the person doing the sampling is aware that there are 5, or less, chances in 100 that his sample has been sufficiently "wild," or unusual, that it has pro- duced a confidence interval which fails to include the true propor- tion, p, of acceptable products in the population which was sampled. The work done above is illustrative of the principles involved but is too laborious to be repeated each time a confidence interval is needed, especially when n > 10. Therefore, advantage is taken of some work done by C. J. Clopper and E. S. Pearson, published in Volume 26 of Biometrika. Table 5.21 was obtained by reading from their graphs the 95 and 99 per cent confidence intervals on p for n = 50, 100, and 250. If n is smaller than 50, the confidence intervals are so wide that they are of doubtful value in practice. However, Sec. 5.2 POINT AND INTERVAL ESTIMATION OF p 127 TABLE 5.21 The 95 and 99 Per Cent Confidence Limits on p for Samples of 50, 100, AND 250 Taken from Binomial Populations (L = lower limit) (Based on graphs by C. J. Clopper and E. S. Pearson, Volume 26 of Biometrika.) n = 50 n = 100 n = 250 95% 99% 95% 99% 95% 99% r/n L U L U L U L [/ L U L U .000 8 10 4 5 2 2 .025 12 15 8 1 9 1 5 1 6 .050 1 16 19 2 11 1 13 3 9 2 10 .075 2 19 1 22 3 14 2 17 4 11 4 13 .100 3 22 2 26 5 18 4 20 6 15 5 16 .125 5 25 3 29 6 21 5 23 8 17 7 19 .150 6 28 4 32 8 24 7 26 11 20 9 22 .175 8 31 6 35 10 26 8 29 13 23 11 25 .200 10 34 7 38 12 29 10 32 15 26 13 27 .225 12 37 9 41 15 32 12 35 18 29 15 30 .250 14 39 11 44 17 35 14 38 20 31 18 33 .275 16 42 12 47 19 38 16 40 22 34 20 35 .300 18 45 15 49 21 40 18 42 24 36 22 38 .325 20 47 17 52 23 42 21 45 27 39 25 40 .350 22 50 18 54 26 45 23 48 29 41 27 43 .375 24 52 21 57 28 48 25 51 31 44 30 46 .400 26 55 23 59 30 50 27 54 34 46 32 49 .425 29 57 25 62 32 53 30 56 36 49 34 51 .450 31 60 27 64 35 55 32 58 39 51 37 54 .475 33 62 29 66 37 58 34 61 41 54 39 56 .500 35 65 31 68 40 60 37 63 44 56 41 58 .525 38 67 34 71 42 62 39 65 46 59 44 60 .550 40 69 36 73 45 65 41 68 48 61 46 63 .575 43 72 39 75 47 67 44 70 51 64 49 65 .600 45 74 41 77 50 69 46 73 54 66 51 68 .625 48 76 43 79 52 72 49 75 56 69 54 70 .650 50 78 46 81 55 74 51 77 59 71 56 73 .675 52 80 48 83 57 77 54 79 61 73 59 75 .700 55 82 51 85 60 79 57 81 64 76 62 77 .725 58 84 54 87 62 81 60 83 66 78 65 80 .750 60 86 56 89 65 83 63 85 69 80 67 82 .775 63 88 59 91 67 85 65 87 71 82 70 84 .800 66 90 62 92 70 87 68 90 74 85 72 86 .825 68 92 65 94 73 90 71 91 77 87 75 89 .850 71 94 68 95 76 91 74 93 80 89 78 91 .875 75 95 71 97 79 93 76 95 82 91 81 93 .900 78 96 74 98 82 95 80 97 85 93 84 95 .925 81 98 77 99 85 97 83 98 89 95 87 96 .950 84 99 81 99 88 98 86 99 91 97 90 98 1.000 92 100 89 100 96 100 95 100 98 100 97 100 128 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 Table 5.21a has been added to show, through the preceding discus- sions, just how the numbers in Table 5.21 could be got. Obviously, if n were as large as 50 the work illustrated above would become tremendously laborious. In Table 5.21 the observed fraction, r/n, was used instead of r because it was convenient to do so. TABLE 5.21a 95 Per Cent Confidence Limits with n = 10 r L U r L U r L U 32 4 11 75 8 44 97 1 46 5 18 82 9 54 100 2 2 57 6 25 89 10 69 100 3 6 66 7 34 94 The use of Table 5.21 will be illustrated by some examples. Problem 5.21. Suppose that a random sample of 250 primers for cartridges has been taken from a large batch and has been tested b3' actual firing. If 6 of the primers fail to fire, place a 99 per cent confidence interval on the true percentage of duds in the whole batch, and interpret these limits. For this sample r/n = 6/250 = .024, which is so near to the value of .025 listed in Table 5.21 that interpolation is unnecessary. There- fore, the required confidence interval is read from the table as 1 to 6 per cent duds in the whole batch. If future action regarding these primers is based on the assumption that at least 1 per cent but not more than 6 per cent of them are duds, a risk of only 1 in 100 is being run that the sample has been misleading. If 6 per cent is more than the allowable proportion of duds, this sample indicates that the batch may be substandard. Whether the primers would be rejected or additional evidence obtained would depend upon the particular circumstances. Problem 5.22. Suppose that a concern which manufactures roller bearings must meet a standard of 95 per cent acceptable according to certain prescribed measurements. If a sample of 250 yields 3 unacceptable bearings, is the ship- ment up to the required standard or not? In this instance r/n = .012, so the 95 and 99 per cent confidence intervals are found to be to 3, and to 4, respectively, by interpola- tion in Table 5.21. Therefore we could conclude that the shipment has less than 5 per cent unacceptable with considerable confidence Sec. 5.2 POINT AND INTERVAL ESTIMATION OF p 129 because even the upper limit of the 99 per cent confidence interval on the true proportion of duds is below 5 per cent. The procedures demonstrated above are not suggested as sufficient quality control measures in themselves, but they do illustrate prin- ciples which are basic to acceptance sampling. PROBLEMS 1. Suppose that 100 bolts have been taken at random from a large group and that 2 have been found to be defective. What is the 99 per cent confidence interval on the true proportion of defectives in the group sampled? 2. Suppose that a sample of 250 Germans showed that 101 had type blood. Place 95 per cent Hmits (to nearest per cent) on the percentage of such Ger- man persons having type O blood. Ans. 34 to 46. 3. The little fruit fly, Drosophila melanogaster, has been used so extensively in genetic research that a great deal is known about the genes which it carries on its chromosomes. Among these genes are some which produce what are called recessive lethals because they kill the potential offspring at an early stage of development if both chromosomes carry the gene for that particular lethal. Mating studies are able to show if only one chromosome of a fly carries a particular lethal-producing gene. Suppose that a sample of 250 flies is found to include 10 which are carrying one particular lethal. What can you say about the true proportion of lethal-carrying flies in this population? 4. Suppose that two different strains of fruit flies have been developed in a laboratory upon the basis of the numbers of eggs that the females laid per day. Suppose also that a particular recessive lethal, l-^, has been discovered in both strains; and that samples of 250 flies from each strain gave these results: strain A had 18 lethals, strain B had 32 flies carrying lethals among the 250 examined. What can you conclude about the true proportion of lethal-carrying flies in each strain? Are these two proportions probably equal? 5. Suppose that 50 apples have been selected at random from a tree which has a very large number of apples. If 5 apples were foimd to suffer from a certain blight, what percentage of blight do you estimate for the whole tree if you wish to run a risk of only 1 in 20 that your answer is wrong as a result of an anomalous sample? Do likewise for a risk of only 1 in 100 being in error. 6. Suppose that 100 eggs are selected at random from a large shipment, and that 5 are found to be stale. What would you set as the upper limit on the percentage stale in the whole shipment if you can afford a risk of sampling error of only 1 in 100? Ans. 13. 7. If a sample of 250 gun barrels in a large shipment has been examined for defects and none found to be defective, place 99 per cent confidence limits on the true proportion of defective barrels in the whole shipment. Would a sample of 250 be large enough if the shipment must contain one per cent, or less, defective? 8. If 250 one-pound cartons of butter are to be selected from a carload at random and examined for mold particles, what is the maximum number which can be found to contain too much mold before you should conclude that 5 per 130 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 cent, or more, of the cartons probably contain too many mold particles? Use a 99 per cent confidence interval as the basis for your answer. Ans. 2L 9. Suppose that 100 cattle selected at random from a very large group have been tested for tuberculosis. If 15 were found to be reactors, place an upper limit on the proportion of reactors in the whole group if 95 per cent confidence in the answer is considered adequate in these circumstances. 10. The United States Department of Agriculture publication, Agricultural Statistics, 1946, indicates that among United States herds of cattle which are infected with Bang's disease at all, an average of 12 per cent of the cows have the disease. Suppose that a large herd which has some incidence of the dis- ease is to be tested by taking a random sample of 50 cattle. How many out of the 50 must be free of the disease before the owner can be assured (at the 99 per cent level of confidence) that his herd is above average in freedom from Bang's disease? Ans. 46. 11. Calculate directly from the binomial series (q-\-p)'^ the 75 per cent con- fidence interval on p if 3 out of 4 items sampled are found to be acceptable in the sense employed earlier. Obtain the answer to the nearest whole per cent. 12. Verify the entry in Table 5.21o for r = 5. 13. Suppose that an entomologist wishes to know what percentage of the com plants in a large field have been infested to some degree by the southwest corn borer. He thinks that the percentage is somewhere between 20 and 80, but he wants to reduce that uncertainty to an interval of not over 15 percent- age points. If he is willing to accept a risk of 5 in 100 of drawing an erroneous conclusion, how large a sample must he take? 14. Suppose that you are helping to administer a farm management associa- tion and wish to learn what percentage of the members use a certain procedure recommended for poultrymen. Suppose, also, that a random sample of 250 interviews reveals that 200 in the sample do use the recommended practices. What can you say about the true percentage using this practice in the whole association? Ans. CI95: 74-85% 1 2-86% J Ql . 72_c««/- I ^^^ practices. 5.3 TESTING PREDETERMINED HYPOTHESES REGARDING p In some fields of investigation the probable magnitude of p can be deduced from what appear to be reasonable theoretical considera- tions, as was illustrated in the discussions of the A-B, and other, blood groups. As another illustration, the theory of sex inheritance might lead geneticists to conclude that male and female offspring of human beings should be produced in equal proportions. If so, p = 1/2 when children are classified merely as male or female. Abundant statistical evidence now exists to show that more than one-half the children born in the United States are male; therefore, the original hypothesis that p = 1/2 is known to be false. However, mankind cannot afford to wait many years until the collection of a great Sec. 5.3 PREDETERMINED HYPOTHESES REGARDING p 131 volume of data makes it possible to determine, virtually without error, the truth or falsity of many of the hypotheses which play im- portant roles in everyday life and in scientific investigations. In these circumstances samples can be taken and made the basis for satisfactory conclusions. The statistical methods needed for a test of a predetermined hy- pothesis regarding some binomial population are intended to decide whether or not it is reasonable (as defined by an accompanying prob- ability statement) to suppose that a given sample actually has been drawn from the binomial population which is specified by the hy- pothesis being tested. In order that such a decision can be made, a basis must be established for comparing a particular sampling result with results to be expected from sampling ij the hypothesis being tested is strictly correct. How should this be accomplished? Actu- ally, the problem is a very complex one whose full solution cannot be attempted at the reader's present stage of statistical development; but some useful and informative rationalizations can be presented. Strange as it may seem, a large part of the complexity of this problem comes from the fact that there are so many possible solu- tions that the more difficult job is to choose the best one. This was indicated in the introductory part of this chapter. In that intro- duction a rather simple example was considered and a hypothesis was judged for reasonableness by means of the binomial expansion {q -{- p)'\ It was possible with the aid of that expansion to say that if p = 1/2 only 1.7 per cent of a large number of random sam- ples would have r as small as 0, 1, 2, or 3. The rarity of such occur- rences was made the basis for rejecting Hoip — 1/2). Although the risk of falsely rejecting Hq when p actually is 1/2 is only .017, the likelihood of falsely accepting Hq when p actually is 1/3 was seen to be high; nearly four chances out of five. It was stated in con- nection with that example that the choice of the best procedure for making decisions from samples depends on this latter probability of an error of the second kind because the probability of an error of the first kind usually is fixed in advance. In the example just reviewed, a sampling frequency distribution was employed, and events which fell in the lower frequency intervals — that is, the extreme sizes of r — constituted what is called the region of rejection. On the scale of measurement of r, the points 0, 1, 2, and 3 comprised the region of rejection. The general problem of choosing best tests of hypotheses regarding population parameters consists of finding functions of the sample observations and of the 132 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 population parameters for which the best regions of rejection can be defined. The best region of rejection, among several choices, is the one which for a given a will make (i the smallest; that is, for a fixed probability of rejecting a true hypothesis it will give the lowest probability of accepting a false hypothesis in consideration of the other possible hypotheses. Statistical research has shown that a good function to use in the solution of the problem set up for this section is one which is called chi-square, and is denoted by the symbol x". Its magnitude depends upon the numbers of individuals, or other units, observed in the sample to fall into each of the possible classes of attributes. It also depends upon the numbers which are expected mathematically to fall in those classes, which in turn depends upon the predetermined hypothesis regarding the population parameter p. For example, sup- pose that we have sufficient reason to believe that one-half the off- spring of guinea pigs should be males. The predetermined hypothesis now is that p = 1/2. If a sample group of progeny selected at random from a whole population of actual or possible progeny is found to have 38 males and 32 females, is it reasonable to believe that this is a sample from a population for which p = 1/2? The mathematically expected number of males out of 70 offspring is E{r) = (1/2) (70) = 35; hence the number of males in the sample is 3 greater than ex- pectation. It follows automatically that there are 3 fewer females than expected mathematically. The function x" will be defined by the following formula: .^^ (observed number in class — expected number)^ expected number in class where the summation includes two terms, one for males and one for females. It is apparent from this formula that if the observed num- bers in the two classes agree well with those numbers which are expected mathematically considering the assumed magnitude of p, X^ will be relatively small; but if the numbers observed to fall in each class notably disagree with those expected from the predeter- mined hypothesis, x" will be relatively large. The decision that x^ is relatively large or small is based upon the proportion of all such sample values of x" which would be at least that large ij the hypoth- esis being tested were, in fact, true. For the illustration above, x' = (38 - 35)735 + (32 - 35)735 = 0.51. The remaining question is: Is it reasonable to suppose that Sec. 5.3 PREDETERMINED HYPOTHESES REGARDING p 133 X^ got so large as this purely as the result of the chance occurrences of sampling? As was done earlier, attention will be called first to some actual sampling experiences, and then a mathematical table will be employed to obtain the required information more quickly and more accurately. Table 5.31 summarizes the results obtained from 652 samples from a population for which p was known to be 1/2. Figure 5.31 is the graph of the r.c.j. distribution presented in Table 5.31. TABLE 5.31 Observed Frequency and r.c.J. Distributions for x^ When the Two Classes, Male and Female, Determine the Population, and p = 1/2 Class Interval / r.c.j. Class Interval / r.c.J, > 5.50 9 1.000 2.00-2.49 50 .885 5.00-5.49 6 .986 1.50-1.99 48 .808 4.50-4.99 4 .977 1.00-1.49 70 .735 4.00-4.49 7 .971 0.50-0.99 101 .627 3.50-3.99 10 .960 0.00-0.49 308 .472 3.00-3.49 12 .945 2.50-2.99 27 .926 Total = 652 1.00 T - / X 1 1 1 1 1 1 1 ^ .90 1 -80 % -70 •I -60 1 .50 3 .40 - — .1 .30 1 .20 " .10 - - 1 1 I 1 1 1 1 1 1 1 1 .50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 5.50 Figure 5.31. Sampling distribution of x" with one degree of freedom, as deter- mined from 652 samples taken from a binomial population with p = 1/2. If we read upward from x" = 0.51 to the graph of Figure 5.31 and then horizontally to the vertical scale, it appears that about 52 or 53 per cent of all such sampling values of x^ would exceed 0.51. Ob- viously, then, 0.51 is not an unusual sampling size for x", provided the hypothesis upon the basis of which the expected numbers were calculated is exactly correct. Hence it is entirely reasonable to sup- pose that this sample of male and female guinea pigs deviated from 134 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 a 50:50 sex ratio purely as a consequence of the chance element in all sampling. On the other hand, if the sampling x" had been 15, say, we see from Figure 5.31 that sampling variation almost never produces such a large value of x"- We would then conclude that the sex ratio was not 50:50. It is clear that such conclusions as these are valid only if a representative sample has been drawn. If just a few guinea pigs in one particular laboratory have been the basis for the sample, the conclusions drawn would not apply, without more sampling evidence, to guinea pigs in general. As a matter of fact, Figure 5.31 can be used as above for samples of various sizes just as long as there are only two classes of attributes involved. Under these circumstances, x" is said to have one degree of freedom. In this connection it should be noted that when the expected number has been calculated for one of the two classes, the other number follows automatically so as to keep the sum of the expected numbers ecjual to the sum of the observed numbers. Like- wise, if there are 3 too many males compared to expectation there must be 3 too few females; that is, there really is but one basic dif- ference between the observed numbers and the expected numbers. Basically, that is the reason there is only one degree of freedom for the x^. Table V makes it possible to determine more easily and accurately the probability that a sample x^ will exceed the observed value when the Ho is correct. Actually this mathematical distribution is not exactly right for the x" as defined in this chapter, but the loss of accuracy is negligible for most sample sizes which would cause peo- ple to have faith in the conclusions drawn therefrom. Problem 5.31. It was stated in Chapter 3 that the A-B blood groups were considered to be inherited in a simple Mendelian manner so that AO X AO should produce ofifspring three-fourths of whom test to be type A and one- fourth are type 0. Suppose that among a random sample of 400 children from such parents, 312 are A and 88 are O ; that is, 78 per cent are A and 22 per cent are O. Does this sampling evidence justify rejection of the hypothesis that 75 per cent should be A, or, in more symbolic terminology, should the hy- pothesis Hq(p = 3/4 for A) be rejected? The expected numbers are 300 A and 100 0; therefore, x^ = (12) V300 -\- (12)2/100 = 1.92, with 1 D/F. By Table V it is found by interpolation that P = .17. By any usual standards Hoip = 3/4 for A) is accepted, especially if the hypothesis seems to be well founded theoretically. In some circumstances we would wish to Sec. 5.3 PREDETERMINED HYPOTHESES REGARDING p 135 place a confidence interval on p. For example, the 95 per cent con- fidence interval could be obtained (from a larger table than 5.21 on page 127) . Such an interval would include p — 3/4, because that hy- pothesis was accepted far above the 5 per cent level, but would also include other possible values of p. If this interval included other defensible hypotheses about p, they also would be acceptable as far as this sample evidence is concerned. Larger samples then could be taken with the hope of so narrowing the confidence interval that only one theoretically defensible hypothesis would be acceptable upon the basis of the sampling evidence. PROBLEMS 1. According to Table 2.61, 50 female and 59 male guinea pigs were born during the period from January to April, inclusive. If these guinea pigs can be considered as a random sample of all guinea pigs as regards the sex ratio, is the observed difference in numbers of each sex sufficient to cause you to reject the hypothesis that the sex ratio actually is 1:1, if you wish to set the probability of committing an error of the first kind at .05? 2. Use the data for May to August, inclusive, to answer the question of problem 1. Atis. No, PCx^ ^ 1.12) ^ .30. 3. Solve as in problem 1 for the data for September to December. 4. Table 2.62 contains data from those guinea pigs which survived long enough to produce 4-day gains. Do the data for Januaiy to July, inclusive, indicate that the sex ratio is 1 : 1 for guinea pigs in that more select population which lives at least 4 days? Ans. Yes, P > .53. 5. Solve as in problem 4 for the data for August to December. 6. According to genetic theory, if a so-called heterozygous red-eyed fruit fly is mated with a white-eyed fruit fly, one-half the offspring are expected mathe- matically to be white-eyed. The reasoning is analogous to that given earlier for a mating of O and AB blood types. Suppose that among 500 offspring of such fruit flies, 240 are white-eyed. Does the x^-test indicate that such a sam- ple result would occur rarely {P < .05) while sampling from a binomial pop- ulation with p = 1/2, or not? Ans. No, P= .37. 7. If you assume (as is reasonable from Figure 5.31) that x~ must be at least 3.8 in problem 6 before the hypothesis that p = 1/2 should be rejected, how small can the number of white-eyed flies be among 500 offspring before that would occur? 8. Suppose that it were agreed that you should not seriously doubt the hypothesis that p = 1/2 unless x" exceeds a value xo^ which is such that PCx^^Xo^) — -01 • How small can the number of white-eyed flies among 500 become before you would reject the hypothesis that p — 1/2? Ans. 221 or 222. 9. Suppose that a sample of 100 college students showed that 40 opposed a certain proposal regarding student government. Does that result contradict the hypothesis that 48 per cent of the student body oppose the proposed change? 136 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 10. Use a confidence interval approach to answer the question in problem 9, and discuss the difference between the two methods. Ans. CI95: 30-50 per cent opposed. CI99: 27-54 per cent opposed. Both inchide 48 per cent. 11. Suppose that a poll of Topekans (Kansas) shows that one candidate re- ceived 135 votes to 115 for the other candidate for a certain public office. Use both the x^-test and confidence intervals to determine the probable winner, if the election is to be held very soon so that no appreciable change in opinion is expected. 5.4 TESTING THE HYPOTHESIS THAT TWO RANDOM SAMPLES CAME FROM THE SAME BINOMIAL POPULATION The type of problem to which the tests described in this section apply arises when two groups of observations have been taken under somewhat difierent circumstances. The question to be answered is: Did the difference in circumstances produce two distinct binomial populations as far as can be told from these samples? For example, consider a simulated test of two house-fly sprays, one made from lethane the other from pyrethrum. Suppose that 500 house flies have been placed in each of two wire cages, identical in all respects. The lethane spray is applied to one cage, the pyrethrum spray to the other, with the following results: Spray Dead Alive Sums Lethane 475 25 500 Pyrethrum 450 50 500 Totals 925 75 1000 Actually, the lethane spray killed 95 per cent of the flies in its cage, whereas the pyrethrum killed only 90 per cent. However, if both cages had been sprayed with the same spray, different per- centages would have been killed in the two cages in all probability. How rarely would they have been as different as they were found to be in this experiment? The x^-test introduced in section 5.3 can be employed successfully in the solution of this problem. However, there is no predetermined hypothesis regarding the magnitude of p like that available before. Hence some other method must be used to calculate the expected numbers needed in the x^-test. There is no theory regarding insecticides which will furnish an expected proportion "dead" in the population; but it was observed Sec. 5.4 TEST OF THE HYPOTHESIS Fq (pi = P2) 137 that among the 1000 flies sprayed 92.5 per cent were later classified as dead. If these two sprays are equally toxic to the house flies, they should tend to kill equally many flies per 500 sprayed. Therefore, the probability of death can be taken as .925 on the general hypoth- esis that the two sprays are equally toxic. This is equivalent to the hypothesis, Hoipi = p-2) ■ Then the expected number dead out of 500 in a cage is Eir) = .925(500) = 462.5. That leaves 37.5 as the expected number of survivors for each spray since 500 flies were sprayed with each spray. We then can extend formula 5.31 to obtain the following: ;^2 = (475 _ 462.5)7462.5 + (450 - 462.5)V462.5 + (25 - 37.5)2/37.5 + (50 - 37.5)V37.5 = 9.01. This X" has only one degree of freedom as before because there is only one chance difference between the observed and expected num- bers. Note that only one expected number need be calculated before all the rest follow automatically from the border totals of the table. Figure 5.31 and Table V clearly indicate that x" rarely would attain a size of 9.01, or more, purely from sampling variations; therefore it is concluded that the lethane spray is superior to the pyrethrum spray, that is, the hypothesis that pi = po is rejected, where pi = true pro- portion which would be killed by lethane and po is the same for pyrethrum over many trials. The technique just described also can be used to decide if two random samples supposedly drawn from the same binomial popula- tion actually are from a common population. For example, suppose that two separate random samples were taken on the toxicity of a lethane spray, with the following results: Dead Alive Sums Sample 1 480 20 500 Sample 2 380 20 400 Sums 860 40 900 If p remained constant during this sampling it is best estimated as p = 860/900 = .956 or 95.6 per cent. In the absence of any logical predetermined hypothesis, the hypothesis Hoipi = P2) is tested, where pi = true probability of death during the taking of the first sample, and similarly for p2 and the second sample. If the prob- ability of death for any randomly designated fly stays fixed, Pi = p-y. 138 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 As usual, the expected number killed during the first sampling is computed to be E{r) = .956(500) = 478, which deviates from the observed number by only 480 — 478 = +2. It follows that the other observed numbers also differ from their expected numbers by 2 in one direction or the other. Hence , (+2)2 (-2)2 (-2)2 (+2)2 x' = -^-^ + ^ + ^ + -^^-^ = 0.422, 478 382 22 18 with 1 D/F. It is learned from Table V that P = .52 ; hence Hq is accepted readily, and it is considered that the two samples were taken under conditions which kept the probability of death constant. It is not always true that the population can be kept the same under repeated sampling; hence it is well to check this matter before dif- ferent conditions (such as use of different insecticides) are purposely introduced so that their effects can be studied. Problem 5.41. Suppose that two sample polls of votes for two candidates for a public office are taken, one from among residents of cities with at least 25,000 population, the other from among residents not in any incorporated town or city. If the results were as given below would you accept the statement that place of residence was unrelated to voting preference in this election? If so, the two samples are from a common binomial population. Votes for A B Sums Rural 620 380 1000 Urban 550 450 1000 Sums 1170 830 2000 Over both the rural and urban samples 58.5 per cent voted for A. If both samples are from the same binomial population, p = .585 is the best available estimate of p, the true fraction who favor A. Hence the hypothesis Ho(pr = Vu) will be tested by means of the x^ distri- bution. The expected number of rural residents out of 1000 who favor A is .585(1000) = 585. It deviates from the observed number by 620 - 585 = +35; hence x' = (35)^ 1 1 1 1 + + + 585 585 415 415. = 10.09, 1 D/F. It is apparent from Table V that Ho should be rejected because P ^ .002. It is concluded that pr actually is > p,,; that is, the resi- Sec. 5.4 TEST OF THE HYPOTHESIS Hq (pi = V2) 139 dents of rural areas favor candidate A more strongly than do the urban residents because the observed results are very unlikely to be a sampling accident. PROBLEMS 1. Lerner and Taylor, University of California, published the following data on chick mortality in the Journal of Agricultural Gcience, Volume 60: Number Pr ogeny Which Sire Died Lived G14 22 65 G36 44 35 G52 17 45 H8 22 39 How would you rate these sires as regards low progeny mortality after taking account of samphng variability? 2. Compute x^ foi" the following practice data and obtain from Table V the probability that sampling variation alone would produce a x^ at least this large. Also explain how information of this sort is used to test a hypothesis about a binomial population. Answered Yes No Sum First Sample 187 113 300 Second Sample 213 187 400 Sum 400 300 700 Ans. Pix^ > 5.16) ^ .002. 3. Given the following x^'s, each with one degree of freedom, classify each as probably due to chance alone, or not, if an event which is as unlikely to occur as 1 time in 20 is considered to be purely a chance occurrence: 3.9, 7.1, 0.95, 2.1, 15.2, 8.7, and 1.2. 4. Within what approximate limits do the lower 75 per cent of all sampling values of x^ vvith one degree of freedom lie when the hypothesis being tested is correct? Ans. to 1.31. 5. For a population of x"'s each with one degree of freedom, the mean = fj, — 1, and the standard deviation = tr = V2. Approximately what proportion of the population of x" with 1 D/F (see Table V) lies in each of the following ranges: /u ± Iff, /u. ± 2a-, and /i =fc So-? How do these proportions compare with the corresponding snes for a normal distribution, as shown in Table III? What information does this set of comparisons give about the shape of the chi-square frequency distribution curve when x^ has one degi-ee of freedom? 140 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 6. Samples from records of male and female White Rock chicks up to 8 weeks of age raised at Kansas State College during 1945 gave the following data: Sex Died Lived Male 46 227 Female 30 290 Use the x^-test to decide if there probably is a fundamental difference in chick mortality due to sex. Ans. P{x~ — 7.35) =.007. 7. The following data are derived from a publication by Atkeson et al. {Jour- nal oj Economic Entomology, 37:428-35) on the effectiveness of 5 sprays in killing flies around dairy barns: Number of Flies Spray Killed Not Killed A 22 84 B 49 90 C 89 28 D 39 63 E 44 24 Which sprays do you consider as essentially equal in killing power, considering sprays which do not differ beyond reasonable sampling variation as being tied? 5.5 THE X--TEST WHEN MORE THAN ONE DEGREE OF FREEDOM IS REQUIRED There are many problems in sampling which require the use of the sampling distribution of a x" with more than one degree of free- dom, but only a few will be considered in this book. For example, suppose that both parents have the specific blood types AO and Rhrh, in the symbols of Chapter 3. Each parent produces four types of gametes: ARh, Arh, ORh, and Orh, with equal frequencies it is believed. Hence it can be deduced that such parents will produce offspring of four blood types: ARh + , ARh — , ORh + , and ORh—, with associated probabilities 9/16, 3/16, 3/16, and 1/16, respectively. Therefore, if a large number of such parents is obtained for a random sample we can test the hypothesis suggested by the above argument, namely, /fo(9 ARh+ : 3 ARh- : 3 ORh+ : 1 ORh-). To illus- trate, suppose that out of 1600 such families in a random sample, the children were classified as follows with respect to the A-B and Rh blood groups: 885 ARh+, 310 ARh-, 292 ORh+, and 113 ORh-. Do these observed numbers deviate enough from the corresponding Sec. 5.5 X- WITH OVER ONE DEGREE OF FREEDOM 141 theoretical numbers 900, 300, 300, and 100, respectively, to justify the rejection of //o? In these circumstances 2 (885 - 900)2 (310 - 300)2 ^292 - 300)^ (113 - 100)^ ^ ~ 900 300 300 100 = 2.49, by analogy with previous problems when x" had only one degree of freedom. How many degrees of freedom does this x" have, that is, how many of the deviations from theoretical expectations can be considered due to chance? The observed numbers and the expected numbers each must add to 1600, hence there cannot be more than 3 degrees of freedom. As this is the only such restriction, there are just 3 degrees of freedom. Naturally a sampling chi-square which results from 3 chance deviations usually will be larger than one based on fewer degrees of freedom because more "room" must be left for sampling fluctuations. It is seen in Table V that a x" with 3 degrees of freedom will exceed the observed value, 2.49, about one-third of the time when Ho is correct; that is, P = .33. Therefore, the hy- pothesis Hoi9 ARh+ : 3 ARh- : 3 ORh+ : 1 ORh-) is quite ac- ceptable in view of this sample evidence. Another circumstance which produces a x" with more than one degree of freedom is encountered when the same hypothesis is tested more than once by means of successive but independent random samples believed to have been taken under the same conditions. It may not be reasonably possible to obtain a convincingly large sample during any one experiment or survey so that some means of ac- cumulating statistical evidence from two or more studies is needed. This problem can be solved with the aid of the following theorem. Theorem. If two or more sample chi-squares are obtained from independent random samples, the sum of these chi-squares fol- lows the chi-square distribution for a number of degrees of freedom equal to the sum of those for the chi-squares so added. Obviously, the process to which the above theorem refers would make no practical sense unless each x" were obtained while the same hypothesis was being tested. It also is important to be assured that all the samples have been drawn from the same binomial population, regardless of the truth of the hypothesis Hq because nothing is ac- complished by such a study if several different populations are in- volved. We are trying to test one predetermined hypothesis which supposedly applies to a fixed set of conditions. To illustrate these 142 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 ideas, suppose that some respected group of persons has conjectured that 60 per cent of the voters in a certain area will vote yes on a given economic question, and that this conjecture is to be tested by- means of three samples taken in the three districts in the area in- volved. It will be assumed that these districts contain equally many voters. Before the hypothesis that p = .60 for the whole area is tested, it is of interest to determine if the three districts are the same bi- nomial population with respect to yes and no votes on the economic question which is to be asked the voters. Hence it is supposed that a poll of 200 randomly chosen voters in each district gave these results : Number Voting District Yes No Sum 1 105 95 200 2 100 100 200 3 125 75 200 Sum 330 270 600 If the whole of each district has the same fraction, p, of yes votes, the best estimate of p is p = 330/600 = .55 or 55 per cent. If this is used as the probability that a randomly chosen voter in a given district will vote yes, the expected number of yes votes in each dis- trict is .55(200) = 110. That leaves 90 as the expected number of noes; hence ^ ri05 - 110^2 C75 _ 90)2 x' = + •••+ = 7.07. 110 90 It is seen that the observed number of yeses in the first district is 5 below expectation; hence the number of noes is 5 above expectation, and only one chance difference between observation and theoretical expectation exists. The same can be said for district 2; but since we know that the yes vote in district 2 was 10 below expectation and that in district 1 is 5 below expectation, it follows that the number of yeses from district 3 must have been 15 above expectation. Hence only two chance deviations are involved basically, and this x^ has 2 degrees of freedom. The specific hypothesis being tested is that the true proportions of yes votes in the three districts are equal. Table V indicates that it is rather uncommon during sampling experience for a x" with 2D/F to become as large as the 7.07 observed for these samples if the hy- Sec. 5.5 x^ WITH OVER ONE DEGREE OF FREEDOM 143 pothesis used is correct. In fact P(x- — 7.07) = .03. If we have decided in advance to reject a hypothesis when P < .05, Ho(pi = y., = pg) would be rejected, and we would say that the true fraction of yes votes is not the same in all three districts. It is clear that after such a decision it would not be valid to conduct a separate survey in each district and then combine the evidence from these samples on the assumption that we have three independent x"'s testing the same hypothesis, as is supposed in the theorem stated earlier in this section. It appears from the samples given above that ipx does equal p2, but that ps is greater than pi or p2- This hypothesis could be tested by the method just illustrated; but for the purposes of this discussion districts 1 and 2 will be used to test the original conjecture that p = .6 for the area covered by districts 1 and 2. For district 1, the expected number of yes votes is E{r) — .6(200) = 120 votes. Therefore, x^ = (15)7120+ (15)V80 = 4.69 with 1 D/F so that P = .030 by Table V. On the basis of this sample evidence Hoip = .6) is rejected at the 3 per cent level. For district 2, the same expected numbers are used because 200 votes were recorded in this sample also. Therefore, x^ = (20) V120 + (20)780 = 8.33 with 1 D/F so that P ^ .003. This time Hoip = -6) is rejected more decisively. By the theorem of this section, x^ = 4.69 + 8.33 = 13.02, with 2 D/F so that P ^ .002 by Table V. Therefore Hoip = -6) is re- jected at the 0.2 per cent level upon the basis of the evidence in the two 200-vote samples. The chi-square distribution with more than one degree of freedom may be useful when the data are classified in a two-way table of r rows and c columns. For example, a random sample of Republicans and Democrats in a certain city might each be grouped on the basis of three income brackets as follows: Annual Income Party Under $5000 $5000-$9999 $10,000 and Over Sums Republican 200 50 8 258 Democrat 120 20 3 143 Sums 320 70 11 401 This will be described as a 2 by 3 contingency table. Earlier in this cliapter 2 by 2 contingency tables were analyzed by means of the chi-square distribution. 144 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 It is likely that a person might wish to answer the following ques- tion with the aid of the data in the above table. Is the announced party affiliation of a voter in this city associated with that voter's economic status? Or, in statistical terminology, the question can be rephrased as follows: Let pi be the proportion of Republicans in this city with incomes under $5000, po be the same for Republicans with incomes in the middle income group, ps the same for those Repub- licans in the highest income bracket; and let p/ = the corresponding proportions in the population for the Democrats in this city. The subscript i takes the values 1,2, and 3. Then we wish to test the more complex hypothesis: Hoipi — p/, i = 1, 2, 3). As usual, 2pi = 2p/ = 1 for I = 1, 2, and 3. The Pi and p/ are unknow^n and will be estimated from the sample observations on the assumption that Hq is correct. These estimates of parameters will be obtained as before: pi — pi — 320/401; p2 — P2 = 70/401; and pg = ps' = 11/401. It follows that the expected number of Republicans in the lowest income stratum is (320/401) (258) = 205.9. The other expected numbers are computed in a similar manner and are shown in parentheses in the following table: Annual Income Party Under $5000 $5000-$9999 $10,000 and Over Sums Republican 200(205.9) 50(45.0) 8(7.1) 258(258.0) Democrat 120(114.1) 20(25.0) 3(3.9) 143(143.0) Sums 320(320.0) 70(70.0) 11(11.0) 401(401.0) „ 2 (200 - 205.9)^ , (120 - 114.1)^ , , (3-3.9)^ ^^^^^ ^ ^ 205:9 + n44^ +• • •+ -^:9— = 2.35. How many degrees of freedom does this sampling chi-square have? In the process of estimating the pi and the p/ the expected numbers in a column were required to add to the same sum as the observed numbers for the same columns. This causes the deviations from expectation in a column to be the negatives of each other. For exam- ple, 200 - 205.9 = -(120- 114.1). Therefore, both these devia- tions of observation from expectation cannot be chance occurrences. There are, then, at most three chance deviations among the six which go into the computation of the chi-square. Furthermore, the expected numbers of Republicans in the three income classes must add to 258, Sec. 5.5 X- WITH OVER ONE DEGREE OF FREEDOM 145 the total number of Republicans in the sample. A similar statement holds for the Democrats, but this is not an independent requirement because the six expected numbers have been forced to total 401 by- making the column totals add to the observed numbers 320, 70, and 11. Hence the number of chance differences between the observed numbers and those expected mathematically upon the basis of Hq is reduced to 2. This, then, is the number of degrees of freedom. In general, the number of degrees of freedom for a chi-square cal- culated for an r X c contingency table is (r — 1) (c — 1). In the example above, r = 2 and c = 3; hence, (r — 1) (c — 1) =2. Having decided that the x^ of 2.35 has 2 D/F, it remains to deter- mine from Table V if this is an unusual size for a sample chi-square. Table V shows that PCx^ ^ 2.35, 2 D/F) = .31, approximately; there- fore it is entirely reasonable that this x' occurred while sampling from a population for which the hypothesis, Hq, is true. With this sampling result at hand, we would accept the proposed hypothesis. PROBLEMS 1. Suppose that of 300 salmon which went up a fish ladder in a certain river 185 were chinooks, 65 were silver salmon, and 50 were humpbacks. At another ladder farther south suppose that the following numbers were recorded: chinooks, 150; silvers, 80; and humpbacks, 20. Do these samples (if satisfac- torily random, and such is assumed) support the belief that the proportions of these three species are the same at the two locations which were sampled? 2. Referring to problem 1, what matters would cause you to consider them as truly rantlom samples? What factors might cause you to think they were not? 3. Suppose that three independent samplings at one fish ladder led to the following records: Number Which Were Not Sampling Chinook Chinook Sum First 60 25 85 Second 70 30 100 Third 52 18 70 Sum 182 73 255 Is the hypothesis that the percentage of chinooks stayed the same during the time of the sampling an acceptable one according to these data? 4. Suppose that some entomologists investigated yellow, short-leaved, and spruce pines in a certain forest to see how many were being seriously attacked 146 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 by insects. Assume the following data were obtained from random samples of 250 of each species: Seriously Not Species Damaged Damaged Sum Yellow 58 192 250 Short-leaved 80 170 250 Spruce 78 172 250 Do the insects studied attack one of these species more than another? Or is the assumption that the percentage of seriously damaged trees is the same for all these acceptable? Ans. Chi-square = 11.47, 2D/F, P = .03. Reject the assumption. 5. For each species of pine studied in problem 4, test the hypothesis that one-third of the trees in each species population are seriously damaged. Then combine these tests by adding the chi-squares, and draw appropriate conclusions. 5.6 CONTROL CHARTS Sampling techniques appropriate to binomial populations have some important applications in industry in addition to those con- sidered previously in this chapter. During a manufacturing process designed to produce marketable goods it is important to check fre- quently upon the quality of these products. Quality control charts provide a simple but effective means for watching both the general level of quality and the consistency with which this level is being maintained. No attempt will be made herein to discuss all the various methods in use because books devoted solely to industrial statistics or to quality control are available on this subject. How- ever, it can be seen rather easily that some of the topics presented earlier in this book are fundamental to this subject. The subsequent remarks in this section are intended to point out some of these funda- mentals. Consider first a manufactured item which could be classified as either defective or non-defective wdth respect to predetermined stand- ards of production. Clearly, a binomial frequency distribution must be involved with some unknown proportion, p, of defective products being manufactured. The number of items inspected and classified as defective or non-defective is the n in the previous discussions of sampling from binomial populations. As indicated earlier the stand- ard deviation of a proportion derived from a sample of n observations is Vp(1 — P) /n. If the manufacturing process is running smoothly with p = .05, say, and then something occurs to increase the fraction defective to .15, this occurrence will reveal itself in two ways: (a) the Sec. 5.6 CONTROL CHARTS 147 observed fraction defective will tend to increase rather soon, although it might not do so for several samples; and (6) the variability among samples will increase if p changes toward 1/2, and this is the case cited above. Both these points are illustrated by means of Table 5.61 and Figure 5.61. TABLE 5.61 Samples with n = 50, p Starting at .05 and Increasing .002 per Sample FROM THE Twenty-sixth to the Seventy-fifth Samples, Inclusive, After Which It Remains at .15 Sample Fraction Sample Fraction Sample Fraction Number Defective Number Defective Number Defective 1 .10 35 .10 68 .28 2 .06 36 .10 69 .14 3 .08 37 .02 70 .08 4 .10 38 .06 71 .12 5 .04 39 .10 72 .12 6 .06 40 .04 73 .14 7 .06 41 .06 74 .16 8 .10 42 .14 75 .12 9 .06 43 .16 76 .22 10 .04 44 .12 77 .20 11 .06 45 .06 78 .12 12 .00 46 .08 79 .14 13 .06 47 .10 80 .14 14 .06 48 .06 81 .12 15 .04 49 .04 82 .14 16 .04 50 .04 83 .10 17 .04 51 .08 84 .20 18 .02 52 .14 85 .12 19 .02 53 .12 86 .14 20 .08 54 .04 87 .04 21 .06 55 .12 88 .16 22 .10 56 .08 89 .18 23 .04 57 .16 90 .18 24 .06 58 .10 91 .14 25 .04 59 .18 92 .16 26 .02 60 .14 93 .08 27 .06 61 .04 94 .18 28 .04 62 .08 95 .20 29 .02 63 .10 96 .24 30 .08 64 .10 97 .16 31 .04 65 .15 98 .10 32 .10 66 .12 99 .06 33 .00 67 .12 100 .14 34 .08 148 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 Table 5.61 and Figure 5.61 were obtained in the following manner: (a) 7> was fixed at .05 by combining 950 green and 50 red beads in a receptacle, and considering the red beads as defective manufactured items; (6) 25 samples with n = 50 were drawn and the fraction de- fective was plotted over the order number of the sample; (c) start- ing with the twenty-sixth sample and continuing through the seventy- fifth, two green beads in the receptable were replaced by two red beads after each successive sample was drawn and recorded; and (d) starting with the seventy-sixth sample, no additional changes were made. In brief, p = .05 for the first 25 samples; p increased .30 UCL .30 I \ I \ 1 I I I I 1 \ \ TTi I 1 \ 1 r 9> .25 h- .20 I — o o o UCL = .143 >< °° ° .15 .10 .05 X X xxxx OOOOO <l X XX xxxxxoOO •• • XXXXx XXX o O- ' "XX XxXXX o _••••••• ••X XXX Xx • • ••• •^xXx X XXXX O ±.J I I "I ,. I " I I I I 10 20 30 40 50 60 70 80 90 100 Sample number Figure 5.61. Control chart for Table 5.61. .002 per sample (to simulate slippage or excessive wear) for the next 50 samples so that it finally was .15 for the last 25 samples. The horizontal lines in Figure 5.61 marked UCL (upper control limit) were obtained from p ± S\/p{l — p)/50 with p = .05 until the seventy-sixth sample, and p = .15 thereafter. Unless something has occurred unknowingly to change the size of p the fraction defec- tive rarely will go above the UCL; hence, when the observed fraction defective frequently exceeds this limit, it is suspected that the manu- facturing process has broken down to some degree. It can be seen in Figure 5.61 that when the "machine" had "slipped" and p began to increase, the fraction defective soon started an upward trend. Shortly, it exceeded the UCL which had been set on the supposition that p = .05. Then when p ceased to increase and a new UCL was figured with p = .15, the fraction defective again stayed below the UCL. Generally, there also is a lower control limit (LCL) , but in this situation it would have been negative and was taken as zero, as is customary. In practice when the percentage of defectives is unknown the frac- tion defective observed on at least 25 samples is used in place of p Sec. 5.6 CONTROL CHARTS 149 in the procedure described above. As more samples accumulate a better estimate of y can be computed and used. It may be that the quality of a manufactured item is judged by means of a measurement such as length, diameter, weight, or a volume which is likely to be a member of a near-normal population rather than a binomial population. The principles involved are the same but now X and Sx (see Chapter 6 for definitions) must be used instead of esti- mates of p and its standard deviation. In such situations the upper and lower control limits are given, respectively, by: X rt 3V(S.r2)/n(n - 1) Aside from this change, the control charts are constructed and inter- preted as before. Of course, sample means, Xi, are plotted against order of draw. In view of the fact that the s^ is somewhat tedious to compute, it has been found to be both satisfactory and economical to use control limits which employ the range as the measure of variation instead of the standard deviation. This procedure and the necessary tables are given and discussed in publications on quality control or on industrial statis- tics, and will not be given here. PROBLEMS 1. Another gi'oup of samples taken under conditions described for Table 5.61 gave the following results. Make a control chart similar to Figure 5.61 and draw appropriate conclusions. Sample Fraction Sample Fraction Sample Fraction Number Defective Number Defective Number Defective 1 .02 17 .04 33 .06 2 .00 18 .06 34 .08 3 .06 19 .02 35 .10 4 .06 20 .08 36 .04 5 .08 21 .00 37 .04 6 .06 22 .00 38 .02 7 .08 23 .06 39 .08 8 .06 24 .02 40 .08 9 .02 25 .02 41 .00 10 .06 26 .04 42 .10 11 .06 27 .02 43 .10 12 .02 28 .06 44 .08 13 .02 29 .00 45 .16 14 .06 30 .00 46 .14 15 .00 31 .04 47 .10 16 .02 32 .06 48 .16 150 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 Sample Fraction Sample Fraction Sample Fraction Number Defective Number Defective Number Defective 49 .12 67 .06 85 .12 50 .16 68 .16 86 .14 51 .02 69 .12 87 .12 52 .12 70 .10 88 .22 63 .12 71 .08 89 .08 54 .14 72 .12 90 .18 55 .04 73 .18 91 .10 56 .08 74 .06 92 .12 57 .14 75 .10 93 .18 58 .12 76 .24 94 .06 59 .10 77 .10 95 .10 60 .08 78 .14 96 .06 61 .12 79 .18 97 .06 62 .12 80 .16 98 .10 63 .06 81 .18 99 .22 64 .08 82 .18 100 .06 65 .14 83 .14 66 .08 84 .08 2. Calculate an estimate of p from the first 25 samples of problem 1, for the next 25, and for the last 25 samples. Discuss the effect their differences would have on the control chart. 3. Use the estimate of p from the first 25 samples of problem 1 to recompute the UCL for Figure 5.51. 4. Draw 50 successive samples of 50 each from the laboratory population on fraction defective (furnished by the instructor) and construct a control chart from those observations. 5. Perform the operations required in problem 4 for a near-normal population furnished by the instructor. REVIEW PROBLEMS 1. Which of the following bridge hands are you the more likely to receive on one future random deal? (a) A, K, 10, 4, 3, and 2 of hearts; A, Q, and 10 of diamonds; K, 10, and 3 of spades; and the ace of clubs. (6) No card larger than a 6. 2. Suppose that a coin is so biased that it turns up heads 3 times for each 2 times that it shows tails, on the average. What is the probability that on 8 flips there will be fewer heads than tails? Ans. .17. 3. Suppose that you have taken the bid in a bridge game and that you and the dummy have all the trumps except Q, 8, 5, and 2. What is the probability that you would get out all the trumps on successive leads of the A and K of trumps? Ch. 5 REVIEW PROBLEMS 151 4. The prices of barley in the North Central States during 1945 are given below in cents. {Agricultural Statistics, 1946, USDA.) Compute two different averages of these prices and discuss their meanings and their limitations. State: Ohio Indiana Illinois Michigan Wisconsin Iowa Minnesota Price: 106 111 111 118 119 103 107 State: Missouri N. Dakota S. Dakota Nebraska Kansas Price: 116 102 103 96 97 5. Determine and interpret the coefficient of variation for the prices of problem 4. 6. Calculate the mean deviation for the prices of problem 4 and discuss its meaning. Ans. AD = 6.32. 7. If the egg weights for Rhode Island Red hens are considered to be nor- mally distributed with fi = 60.5 grams and cr — 4.0 grams, what range of egg weights would you expect to include the middle 90 per cent of all weights? 8. The following table (taken from A. S. Weiner, Blood Groups and Trans- jusions, Thomas, with the consent of the author and the publisher) records the results of a study of the inheritance of the P factor. Number Children's Parents' of Blood Types Type Families P + P- Total P+ X P + 249 677 79 756 P+ XP- 134 286 179 465 P- XP- 34 (4)* 94 98 * Definite doubt established regarding legitimacy. Recalling that P+ is genetically PP or Pp, and that P— is only pp, test statistically the agreement between the above data on children's blood types and the numbers expected if P^- is assumed to be Pp twice as frequently as it is PP. Consider the (4) * entry as zero. Ans. Pj(x-S2.0.33) > .53; P.Ax-^5.57) = .018 on P+ X P+ and P+ X P-, respectively. 9. Make up a set of numbers which has an arithmetic mean of 10 and a standard deviation of 2. 10. Is there any evidence in the table of problem 8 for or against the assump- tion that P+ = Pp twice as frequently as P+ = PP? Explain. 11. Suppose that a large, deep pool in a mountain stream contains a great many trout of just two kinds, rainbow and brook. You wish to learn what percentage are rainbows. Two methods of sampling have been suggested thus far. (a) Fish the pool until 50 trout are caught, and then use this sample evidence as the basis for estimating p. (b) Devise a trap into which the trout will go and be caught, and secure a sample of 50 this way. 152 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 Assuming that b can be done, which method of sampling, if either, do you recommend as statistically best? Why? Can you suggest a better method than either of these? 12. Referring to the situation of problem 11, suppose that the true proportion of rainbow trout is .40 and that 8 per cent of the rainbows in this area are known to be afflicted with a certain disease. What is the probability that a trout caught at random is a rainbow trout without the disease? Ans. P = .37. 13. Referring again to problem 11, assume that 60 per cent of the trout in another stream in this same region are brook trout. If on a random sample of 50, 22 are brook trout and the other 28 are rainbows, would you accept or reject the hypothesis that p = .60 for brook trout in this stream? Explain. 14. Suppose that there are two mountain streams which run quite close to- gether in a certain area but whose head waters are far apart. You wish to know if the trout populations of these two streams are the same, in a certain well-defined area, as regards proportions of the four species: rainbow, cutthroat, brook, and dolly varden trout. Given the following random sampling data, what would you conclude? Number of Trout Rain- Cut- Dolly bow Brook throat Varden Sum Stream 1 73 68 49 10 200 Stream 2 70 85 80 15 250 Sum 143 153 129 25 450 Ans. Chi-square = 5.65 3D/F, P ^ .13 15. Referring to problem 14, compute and interpret the CI95 on the true per- centage of brook trout in stream 2. REFERENCES Dixon, Wilfrid J., and Frank J. Massey, Jr., Introduction to Statistical Analysis, McGraw-Hill Book Company, New York. 1951. Grant, Eugene L., Statistical Quality Control, Second Edition. McGraw-Hill Book Company, New York, 1952. Neyman, Jerzy, First Course in Probability and Statistics, Henry Holt and Company, New York, 1950. Snedecor, George W., Statistical Methods Applied to Experiments in Agriculture and Biology, Fourth Edition, Iowa State College Press, Ames, Iowa, 1946. Tippett, L. H. C, Technological Applications of Statistics, John Wiley and Sons, New York, 1950. CHAPTER 6 Introductory Sampling Theory for a Normal Population Involving Only One Variable When the population being sampled has a normal frequency dis- tribution with unknown i)arameters fj. and a, the problems of estima- tion and of testing hypotheses by means of samples are fundamentally much the same as those considered in Chapter 5 for binomial pop- ulations. Two differences are immediately apparent, (a) There now are two unknown parameters instead of one, as for the binomial pop- ulation, and (5) the measurements, X, have a continuous scale of measurement and a continuous frequency distribution. These dif- ferences between the normal and the binomial types of populations will appear in the discussions below as the causes of some changes in the mechanics of estimation and of testing hypotheses; but the reader should not lose sight of the fact that the problems and their solutions are much the same as in Chapter 5. 6.1 OBTAINING THE SAMPLE The process for obtaining good samples from a normal population is similar to that discussed in Chapter 5 for randomization and the avoidance of biases. Here, as there, the population to be sampled must be clearly defined, and the measurement to be taken on the units of this population must be stipulated precisely. After the population is specified and the units (persons, prices, pigs, plots of land, pots of plants, families, etc.) have been designated unambiguously, it is necessary to devise a method for obtaining the particular units which are to constitute the sample. The sampling situations which come within the scope of this chapter should be handled by completely randomized samples. To illustrate, suppose that a person who is interested in the production of raw rubber wishes 153 154 SAMPLING NORMAL POPULATIONS Ch. 6 to estimate the percentage rubber in a certain variety of guayule grown under specified environmental conditions. Suppose also that he wishes to select and to analyze 25 plants as a basis for this esti- mate. The population parameter which is to be estimated is the true average percentage of rubber in plants of the given variety. Assum- ing that there is a large number of guayule plants from which to select a sample, how should the particular 25 of the sample be chosen? If the 25 tallest, sturdiest, or most thrifty-looking plants were to be chosen they surely would not be representative of the population. If a person were to stroll about among the available plants and choose 25 in what he considered a random manner, he might unconsciously bias the sample. A better way to choose the sample is to assign location numbers (such as row and plant-in-row numbers) to the plants and then effectively "just draw 25 numbers out of a hat." He can use tables of random numbers and similar devices if he chooses. The main point is to see that every plant in the population had at the start an equal and independent chance to be included in the sample. If two varieties of guayule were compared for percentage of rub- ber, it might be best to start with a suitable area of land staked off for tree spacings and then assign the varieties at random to the various planting positions. This would make it true that each variety initially had an equal chance for any good, or bad, land among the possible planting positions. The subject of this section is very broad and complex partly be- cause there are many different sampling situations and a consequent need to devise different sampling procedures to fit these different cir- cumstances. However, as in Chapter 5, only enough is said here to give the reader some general ideas and, perhaps, induce him to do more reading on this subject if he is interested. At the least, the reader can be critical in accepting sampling results presented as information, advertising, or propaganda. 6.2 THE STATISTICAL DISTRIBUTION OF SAMPLE MEANS, X,, DRAWN FROM A NORMAL POPULATION Each sample drawn from a normal population of numerical meas- urements will nearly always differ from any other sample from the same population in one or more details. Yet certain features of samples from a population, as, a group, will tend to conform to a predictable pattern. For example, if 10 observations are to be taken Sec. 6.2 DISTRIBUTION OF THE SAMPLE MEAN 155 on a normal population with fi — 60 and a = 10, no one can say what the arithmetic mean of the sample will be; but a good estimate can be made of its probable size because sample means from such a population will have a frequency distribution over the long run of experience. Therefore, it should be expected that probability state- ments like those previously discussed herein can be made. The sample mean, to distinguish it from the unchanging population mean, will be designated by Xi, where the subscript refers to the zth sample. The frequency distribution of an approximately normal population with ^i = 60 and o- = 10 is presented in Table 6.21. Six hundred and forty-eight random samples, each containing 10 measurements, were drawn from that population. The arithmetic means of these samples were then computed. The frequency distribution for the 648 sample means also is given in Table 6.21, with the calculated mean (x) and standard deviation (%') of the Xi being given at the bottom of the table. TABLE 6.21 A Feequency Distribution Table for a Near-Normal Population of Measurements Xi, with /x = 60 and o- = 10; and the Frequency Distri- bution of 648 Sample Means, xj, Taken from That Population with 10 Measurements per Sample Distribution of Population Distribution of Sample Means Class Interval / r.c.f. Class Interval / r.c.f. 86.1 -90.0 7 1.00 69.1-71.0 1.00 82.1 -86.0 13 1.00- 67.1-69.0 7 1.00 78.1 -82.0 31 .99 65.1-67.0 27 .99 74.1 -78.0 64 .97 63.1-65.0 63 .95 70.1 -74.0 113 .92 61.1-63.0 140 .85 66.1 -70.0 169 .85 59.1-61.0 161 .63 62.1 -66.0 218 .74 57.1-59.0 136 .39 58.1 -62.0 241 .59 55.1-57.0 72 .18 54.1 -58.0 222 .43 53.1-55.0 34 .06 50.1 -54.0 176 .28 51.1-53.0 8 .01 46.1 -50.0 120 .16 49.1-51.0 .00 42.1 -46.0 69 .08 38.1 -42.0 34 .04 Total 648 34.1 -38.0 15 .02 30.0*-34.0 8 .01 Total 1500 ju = 60 0- = 10 X = 59.98 sj' = 3.14 * This interval was extended by 0.1 to include the remaining measurement in the population. 156 SAMPLING NORMAL POPULATIONS Ch. 6 The frequency distribution in the right-hand part of Table 6.21 is an approximation to an infinite population of Xi which would result if this sampling with n = 10 observations were continued indefinitely. Any one random sample from the normal population described above neces- sarily would be a member of the population of Xi. As an illustration of the preceding discussion, suppose that an agri- cultural economist is interested in learning if the per-acre income on a certain type of farm employing good (recommended by an agricultural experiment station, for example) farming practices is greater, on the average, than that for farmers not following those practices. He takes a sample of n farms on which the recommended practices are employed and calculates the mean per-acre income. The same is done for a comparable random sample of farms on which these recommendations are not followed. Some measurement of the consistency of income on each of the two groups of farms also would be needed. If the newer practices are worth recommending to replace those currently in use, they must produce a new population of per-acre incomes with a larger mean, a smaller variance, or both. To obtain information on these points, the economist must have adequate information regarding the manner in which sample means are distributed; that is, where their region of concentration will be, and how they will tend to be dispersed about that region of concentration. Hence, the first objective of this chapter will be to provide that sort of information about x's drawn from the same normal population of numerical measurements — such as per-acre incomes. Figure 6.21 presents the graphs of the frequency distributions shown in Table 6.21. The larger curve is for the near-normal parent popula- tion of X's, while the smaller curve is taken as a good approximation to the distribution of the population of ;f's obtained from samples of ten observations taken from the population of X's. It appears from Figure 6.21 that the tw^o frequency distributions are much alike in general form, and seem to be approximately normal about the same mean. The major difference lies in the fact that the Xi exhibit much less variability than the X's of the population from which the samples were taken. This is to be expected because one important reason for combining a number of individual X's into one sample is to achieve a smoothing out of the individual differences among those X's. It should be noted from the bottom of Table 6.21 that the mean of the Xi is 59.98, which is quite near to 60, the size of this population mean, m- Also, the standard deviation of the 648 sample means is 3.14, which is a bit less than one-third of a. As a matter of fact, 3.14 DISTRIBUTION OF THE SAMPLE MEAN 157 30 34 38 42 50 54 78 58 62 Xorx Figure 6.21. Frequency distribution curve for a normal variate, X, and also for the sampling mean, x , for samples with n = 10. is very nearly equal to a/vn = 10/ v 10 = 3.1G, to two decimals, where the symbol n is used to denote the number of observations taken in the sample. The preceding discussion has suggested three features which are exhibited by a large number of sample means obtained from a normal population of numerical measurements. These features are: (a) Although it is impossible to predict the actual content of a particular future sample it may be possible to predict the type of fre- quency distribution which the sample means will follow, for example, a normal distribution. (6) The average sample mean will be of essentially the same magni- tude as the mean of the population sampled. (c) The sample means, Xi, will display less variability than the X's of the population. It is logical that the variabihty of the sample means — from sample to sample — should decrease as the size of the samples increases. It was suggested that a factor l/-\/n is involved here. The following theorem is given without proof because that proof is inappropriate to this book. The theorem is stated here for the purpose of replacing the indefiniteness of statements (a), (h), and (c) above with precise information which can be used in practice. Theorem. If a population of numerical measurements, X, con- forms to a normal frequency distribution with mean, m, and stand- 158 SAMPLING NORMAL POPULATIONS Ch. 6 ard deviation, a, and if a very large number of random samples of n observations each is drawn from that population: (a) The population of Xi thus formed will have a normal fre- quency distribution. (6) The mean of the xi will be m also. (c) The standard deviation of the Xi will be a/-\/n. Note from this theorem that, if the X's are normally distributed we automatically know the form of the distribution of the sample means and hence can write down specifically the formula for their distribu- tion curve, namely; 1 (x-m)2 (6.21) 2/1 = -j= ^ • e """/'' V27r(cr/Vn) Formula 6.21 can be transformed into the standard normal formula of Chapter 4 by means of the substitutions y = ?/i'0'/V^ and X = (x — ii)l{(jl\/n) whereupon Table III can be employed as shown earlier. If ten measurements are taken per sample and the parent population has a mean of 60 and a standard deviation of 10, as above, the mean of the :r's is also 60, and the standard deviation is 10/\/lO = \/T0 = 3.16. Table 6.21 and Figure 6.21 furnish approximate verification of these statements from actual experience. If ri = 15 and the samples are dra\Mi from a normal population mth /x = 60 and a = 10, the mean of the resulting population of Xi also will be 60, and the standard deviation will be 10/\/l5. On 200 such samples, their mean was 60.22 and their standard deviation was 2.53 instead of the expected 60 and 2.58, respectively. Two hundred is a relatively small number of samples from which to seek empirical verifi- cation of mathematical theory, but these results do agree quite well with the theorem given above. Problem 6.21. If chemical determinations of the percentage of protein in samples of a certain variety of wheat are known to have a normal frequency distribution with /x = 14 and a = 2, what is the probability that five random samples will have a mean per cent protein above 16? In the following discussion, o-^ will be used to denote the standard deviation of the population of sampling means. In this problem, n = 5, M = 14, and a^ = 2/^5 = 0.90. Therefore, X = (16 - 14)/0.90 = 2.22; and P(X > 2.22) = .013, approximately. In other words, only about 13 times in 1000 sets of 5 observations like these would you Sec. 6.3 ESTIMATION OF ^ AND a^ 159 have the mean per cent of protein at or above 16. The rarity of such an occurrence might cause you to doubt the accuracy of the protein analyses and cause you to ask that they be done over. PROBLEMS 1. Given that a certain population of measurements is normally distributed about a mean of 30 and with a standard deviation of 8. If a sample of 16 members is to be drawn at random, what is the probability that its mean will be below 28? 2. Under the conditions of problem 1, what is the probability that a sample of 9 numbers taken from that population will have an arithmetic mean below 28? Ans. .23. 3. Solve problem 2 with the standard deviation changed to 12. 4. If in some particular area the daily wages of coal miners are normally distributed with n = $15 and <j = $1.50 what is the probability that a representa- tive sample of 25 miners will have an average daily wage below $14.25? Ans. .006. 5. Suppose that a thoroughly tested variety of corn has been found to yield an average of 35 bushels per acre with a standard deviation of 6, and that these yields have a normal frequency distribution. If a random sample of 25 yields for a new variety gives x = 40, show that there is good reason to believe that the yields of the new variety are from a population with a mean higher than the 35 bushels per acre for the population of the older variety. 6.3 ESTIMATION OF THE UNKNOWN MEAN AND VARIANCE OF A POPULATION FROM THE INFORMATION CONTAINED IN A SAMPLE If the parameters ju, and o- are unknown for a particular normal population which is being sampled (and they usually are or there would be no occasion for sampling) it becomes necessary to estimate them from the X's taken in the sample. How should this be accom- plished? Although this really is a mathematical problem whose solution lies beyond the scope of this book, certain desirable re- quirements for sampling estimates of fx and a^ can be considered. First, it seems logical that an acceptable estimate should have a mean equal to the corresponding population parameter after many samples have been taken. Even though only one sample of n measure- ments is to be taken, we usually would like to know that the x and's^ we shall obtain as estimates of lu. and a^ are from populations whose means are n and a'^, respectively. Sampling estimates which satisfy this requirement are called unbiased estimates, as noted in Chapter 5. The second — and more important — requirement which we should impose on a sampling estimate is that it be as reliable as possible in 160 SAMPLING NORMAL POPULATIONS Ch. 6 the sense that it have a relatively small variance from sample to sample. For example, suppose two methods of estimating /* each will produce an unbiased estimate but, over many samples, one has a variance of 100 whereas the other has a variance of only 25. The latter estimate obviously is more consistently near /a in size, and hence less allowance need be made for sampling error in this estimate. This second estimate would be considered a more efficient estimate than the one whose variance was 100. The estimate x of ju, which already has been mentioned, and whose symbolic definition is 2(X) (6.31) X = — — . n gives an unbiased and highly efficient estimate of n. It has been pointed out earlier in a theorem that the variance of x under repeated sampling is only one nth of the population variance, when a normal population is being sampled. (As a matter of fact, the variance of x is a^/n for any population if a^ is finite.) Hence the x is widely used as an estimate of ^l. The variance a^ will be estimated by means of the formula , S(X - xf (6.32) s2 = — -' n — 1 This estimate is unbiased and is considered to be about as efficient as any estimate of a^ as long as the sample is not extremely small. The usefulness of this estimate in practice will be illustrated repeatedly in subsequent discussions. By comparison with the methods used in Chapter 2 to compute c or 0-^, it is seen in formula 6.32 that two changes have been made. The M is replaced by x and the denominator is now (ji — 1) instead of n. Logically, the x must be used because n is unknown; but it also must be recognized that the differences, {Xi — .f), are more dependent upon chance events which occur in the process of sampling than were the quantities, (Xj — ju). The x itself is subject to sampling error whereas the /x is a fixed number for a given population. This matter is taken into account in sampling theory. One step in this direction is to associate with each estimated variance a number of degrees of freedom. The estimate s^ of formula 6.32 is said to be based on w — 1 degrees of freedom because only (n — 1) of the 7i differences {Xi — x) are actually chance differences. This follows from the fact that 2(X — :r) = SX — Xx = nx — nx = 0. Hence, given any n — \ oi Sec. 6.3 ESTIMATION OF tx AND a^ 161 the deviations of the sample X's from their mean, x, the other devia- tion can be computed without any risk of error. If the true mean, (jl, were known, the n {Xi — hYb would all be quantities whose specific sizes depended on chance, and o-^ could be estimated with n degrees of freedom, which is one more than s^ has. Also, the estimate made with /x known would be more reliable than s^, a fact which is asso- ciated with its greater number of degrees of freedom. As soon as a satisfactory method is available for the estimation of 0-^, it follows that the standard deviation of x — which is (j/\/n and is symbolized as a ^ — also can be estimated from the following quantity : (6.33) s^ = s/Vn = VS(X - x)'^/n{7i - 1) , which is calculated from the observations taken in the sample. It still is true — as for all sampling estimates — that s^ is variable from sample-to-sample. Although X is the best specific estimate of the population mean, /x, it is preferable to calculate from the sample an interval in which we can expect the true mean to lie, with a measurable degree of confi- dence in this expectation. The so-called 'point estimate, x, is almost never exactly right, but an interval can be defined in such a way that we can attach a measure of confidence to the statement that n lies in this interval. This problem can be solved by means of a ratio which is analogous to the (X — /x)/o- which was studied in Chapter 4. That ratio involves only one variable, X, and follows the normal distribu- tion. So also does the ratio (x — ^1 a^. If the standard deviation, 0-, is not known — which is the usual sampling situation — the corre- sponding ratio (6.34) t= {x- n)/Si involves a variable denominator and is not normally distributed. Its degree of departure from normality depends on the size of the sample, n, because the denominator is much less variable for the larger samples. Mathematicians have derived a formula for the frequency distribu- tion of the ratio, t, for a sample of any size. Although that derivation is not appropriate to this book, sampling experience will provide an approximation to this distribution, and then mathematical tables will be provided which give the same information more accurately and more easily. Table 6.31 presents the frequency and the r.c.f. distributions of 580 sampling t's obtained from random samples drawn from the near- normal population of Table 6.21. All samples contained n — 10 SAMPLING NORMAL POPULATIONS Ch. 6 162 observations. The ^'s were computed from formula 6.34, using n = 60 and the x's obtained from the samples. For example, if the x = 58.2 and Si is calculated from formula 6.33 to be 2.61, t = (58.2 - 60)/2.61 = -0.69. TABLE 6.31 Observed Frequency Distribution op 580 U Obtained from Samples of 10 Members Each Drawn from a Normal Population with ^t = 60 and (T = 10 Class Interval / r.c.f. Class Interval / r.c.f. >3.60 1 1.00 -2.80 to -2.01 17 .04 2.80 to 3.59 4 1.00- -3.60 to -2.81 3 .01 2.00 to 2.79 16 .99 <-3.60 1 .00 1.20 to 1.99 54 .96 0.40 to 1.19 127 .87 Total 580 -0.40 to 0.39 187 .65 -1.20 to -0.41 115 .33 Arithmetic mean = +0.015 -2.00 to -1.21 55 .13 Standard deviation = 1.10 Figure 6.31 presents the frequency and the relative cumulative fre- quency distribution curves corresponding to Table 6.31. The r.c.f. == 1.00 -3.60-2.80-2.00-1.20-0.40 0.40 1.20 2.00 2.80 3.60 t Figure 6.31. Frequency distribution of 580 sample values of t drawn normal population with /j. = 60, <x = 10, and n = 10. from a curve of Figure 6.31 furnishes information concerning the population of ti for n = 10 which is entirely analogous to that to be had from Table III for normal frequency distributions. Figure 6.31 shows that: (a) The point where ^ = on the horizontal axis divides the population of t's into two equal portions, each containing 0.50, or 50 per cent, of the whole population (as with the normal distribution Sec. 6.3 ESTIMATION OF fi AND a^ 163 at A = 0). (6) Approximately 95 per cent (as nearly as can be told from the graph) of the ti are less than or equal to +2 in magnitude, (c) The middle 80 per cent of the ^'s with n = 10 fall within the limits —1.5 to +1.5, approximately. Such information will be seen to be needed in arriving at the interval estimate for /a described above. It should be noted that conclusions (6) and (c) of the preceding paragraph referred only to samples with n = 10. The general effect 1.00 .90 ^ .80 c i-.70 (U "^ .60 _> I .50 I .40 ~ .30 ^ .20 .10 1 1 1 1 1 1 |-_— T- ^'^^"^""^ ' — // — // l/ jf - >7 // - // / f / / — -L— ..rrrrrnr 1 1 1 1 " -4.0 •3.0 -2.0 1.0 t 1.0 2.0 3.0 4.0 Figure 6.32. Relative cumulative frequency distributions for t when n = 5 (solid line) and for n = 25 (broken line). of the magnitude of n on the frequency distribution of the i's is illustrated in Figure 6.32 for n = 5 and n — 25. The larger the sample size, the less dispersed are the fs. In fact, after n becomes as large as 25 it is difficult to detect much difference between the r.c.]. curve for t and that for a normally distributed measure- ment. Also, if Ux is smaller than no, the ogive for samples with n^ observations will be above that for samples with 712 observations for negative ^'s and below it for positive f s. This is just a graphic verifi- cation of the fact that the ^s are more dispersed for the smaller- sized samples. In line with the earlier discussion of degrees of freedom for the estimate of the standard deviation, the t is said to have the same number of degrees of freedom as the standard deviation in the denominator of this ratio. The fs considered so far have one less degree of freedom than the size of the sample, that is, n — 1. 164 SAMPLING NORMAL POPULATIONS Ch. 6 Table IV provides an r.c.j. distribution of the sampling ratio, t, for most of the commonly used sizes of samples. This table is in the form of those r.c.j. distributions discussed in Chapter 2. This form is different from that found in most statistical tables, but the form of Table IV fits the purposes of this book better than the traditional table. However, the values in the more usual table can be derived from Table IV quite easily. For example, by Table IV the prob- ability that a random sampling t will have a size below —2 from a sample of 15(14Z)/.F) is seen to be .033. Because the f-distribution is symmetrical about ^ = the probability that a t computed with 14 D F will exceed 2 numerically is twice .033, or .066. This is the probability given in the usual table for ^ = 2 and 14 degrees of free- dom. To obtain such a number as .066 for P in those tables we must interpolate because they give the sampling fs which correspond to specified values of P. Table IV will be employed in subsequent discussions instead of the r.c.j. curve because it is both more accurate and more convenient to do so. However, the reader should remember that the two methods are basically the same. The use of tables for the ^-distribution is especially advantageous because there would have to be a different r.c.j. graph for each number of degrees of freedom. Suppose, now, that the true population mean, /a, is not known. In spite of our ignorance of the size of p. it remains true that sampling values of t will conform to the ^-distribution. For example, for n = 10(9 degrees of freedom), it wull be true that 92 per cent of the fs will lie between —2 and +2 (see Table IV). Or, put in terms of a mathematical inequality, it remains true that the following state- ment is correct for 92 per cent of a very large number of samples with n = 10: (6.35) -2 < ~ < +2. Si Approximate empirical verification of the truth of this inequality is found in Table 6.31 above. In view of the information just given, the following can be said: If we are about to take a random sample of 10 numerical measurements from a normal population, the probability is .92 that the t for this sample will satisfy inequality (6.35) because 92 per cent of all samples with 9 degrees of freedom do lie within the limits —2 to +2. When the /i is not known this statement still is true but we can compute the t only in terms of the /x. To illustrate, suppose that a random sample Sec. 6.3 ESTIMATION OF fi AND a^ 165 of 10 observations taken from a normal population has given x = S and Si = 2. Then t = {8 — m)/2, a function of m- Before the sample was taken it could be reasoned that there were 92 chances in 100 that the t to be obtained would have some size between —2 and +2. Like- wise, after the sample is taken, the assumption that the t does lie within these limits runs a risk of 8 in 100 of being wrong as a result of sampling variation. What does the assumption that the t obtained from the sample satisfies the inequality (6.35) require of fx now that t = {S — /a)/2? The quantity (S — /a)/2 must be at least as large as —2 but no larger than +2; therefore, (8 — /x) must be at least as large as —4 but not larger than +4. It follows that /a must be some number from 4 to 12 unless an 8-in-lOO event has occurred. We never actu- ally know in practice if such a t has been got; but we do know that the odds against it are 92:8. The probability, .92, associated with the expression (6.35) is called the confidence coefficient for the confidence inte^-val 4 to 12 because it measures the confidence we can put in the inference that jx lies within these limits. This usage is identical with that of Chapter 5. That is, a method for basing decisions on sampling evidence has been presented; and, although we know it is not infallible, we know what risk of error we run when we use the method. Obviously, other confidence coefficients besides .92 could be used. For example, 95 and 99 per cent confidence limits are quite common. They require the use of the following inequalities for 9 degrees of freedom : X — jJ. — 2.26 < < +2.26 for 95 per cent confidence limits, CI95. X — n —3.25 < < +3.25 for 99 per cent confidence limits, CI99. These two inequalities and that of (6.35) can be put into a more con- venient form simply by multiplying through each (all three members) by Sx and then transferring the x to the outer members of the inequali- ties. The final results for 92, 95, and 99 per cent confidence intervals are as follows for 9 degrees of freedom: (6.36) (x - 2Si) < ^l< ix + 2s^) for a CI92; (6.37) {x - 2.266-) < fji < ix-\- 2.26s^) for a CI95; (6.38) (x - 3.25Si) < fx < (x + 3.25Si) for a CI99. 166 SAMPLING NORMAL POPULATIONS Ch. 6 The latter two are used quite commonly for estimates with 9 degrees of freedom; the first inequality is used here chiefly for convenience of illustration. As an application of the above-described methods, suppose that you wish to learn what the average life of a certain type of light bulb is. Suppose that ten sample bulbs of this type are left burning until all have burned out, and the time it took each to burn out is recorded. The TABLE 6.32 Outline of Some Samples from a Near-Normal Population, with ^ = 60 (Samples were taken by statistics classes from the population of Table 6.21.) Sample Confidence Limits on Number x s t 80% 90% 95% 1 58.4 6.04 -0.86 55.7-61.0 54.8-61.8 54.0-62.7 2 60.9 56.4 12.64 5.81 0.24 -1.96 55.4-66.5 53.6-68.2 51.5-70.4 4 53.8-59.0* 53.0-59.8* 52.2-60.6 5 59.0 11.09 -0.27 54.2-63.9 52.6-65.4 51.1-67.0 6 53.4 10.33 -2.02 48.9-57.9* 47.4-59.4* 46.0-60.8 7 52.9 6.13 -3.66 50.2-55.6* 49.3-56.5* 48.5-57.3* 8 67.7 10.21 2.38 63.2-72.2* 61.8-73.6* 60.4-75.0* 9 54.0 10.52 -1.80 49.4-58.6* 47.9-60.1 46.5-61.5 10 54.1 8.75 -2.13 50.3-57.9* 49.0-59.2* 47.8-60.4 301 58.4 7.22 -0.70 302 61.8 12.67 0.45 303 63.2 7.69 1.32 304 61.6 8.04 0.63 305 58.4 10.69 -0.47 306 64.6 13.39 1.09 307 64.5 6.33 2.25 * * 308 55.9 7.65 -1.69 * 309 60.0 15.19 310 55.5 12.95 -1.10 578 60.2 7.94 0.08 Summary of 578 Confidence Intervals Limits Did Limits Did Not Confidence Include /i Include m Coefficient Number % Number % .80 460 79.6 118 20.4 .90 517 89.4 61 10.6 .95 559 96.7 19 3.3 Sec. 6.3 ESTIMATION OF ^ AND a^ 167 following results will be assumed to have been obtained: x — 1400 hours and s^ = 70 hours. The inequalities above become the follow- ing after simpHfication : 1260 < m < 1540, 1242 < m < 1558, and 1172 < ;n < 1628 hours, respectively, if the computations are rounded to the nearest whole hour. If you act on the assumption that the true average life of this type of bulb is between 1260 hours and 1540 hours, you run a risk of 8 in 100 that the sample has misled you. However, if the widest limits, 1172 to 1628, are used, the risk of an erroneous assumption is only 1 in 100. Table 6.32 has been included to illustrate further and to clarify the idea of confidence intervals. It contains some sampling results obtained from a normal population with /x = 60, and the n = 10. A summary of 578 samples is shown at the bottom of the table. Not all the sampling results are given; just enough to satisfy the pur- poses of this discussion. The asterisks indicate those intervals which fail to include /x. Some of the points which are illustrated by Table 6.32 are the following: (a) The confidence coefficients are long-run relative frequencies which are verified only after a large number of samples. If atten- tion were confined to samples 4 to 10, the confidence coefficients would seem to be wrong; but over the set of 578 confidence intervals, they are verified quite satisfactorily. (5) The determination of a confidence interval is doubly depend- ent on chance: once as regards the mean, and again regarding the magnitude of the standard deviation. For example, samples 306 and 307 had essentially the same mean but the standard deviations were, by chance, so dift'erent that even the 80 per cent limits from sample 306 included the true mean, 60. Only the 95 per cent confidence interval from sample 307 includes /x. On the other hand, samples 303 and 308 have practically the same standard deviation, but the sample means are so different that the 80 per cent limits from sample 308 failed to include the true mean. (c) The confidence interval is wider for the larger confidence co- efficients, that is, the more certain we choose to be in our conclu- sions, the more room we must leave for sampling variations. Problem 6.31. Suppose that a highway commission is interested in the strength of concrete which it wishes to make for highway projects, and that it concludes that the 7-day tensile strengths of standard samples will be the best 168 SAMPLING NORMAL POPULATIONS Ch. 6 criterion of quality. Suppose also that ten of the standard testing models gave these results: X = 439.0 pounds per square inch, s = 47.0 pounds per square inch. What valid and useful conclusions could they draw concerning the true average tensile strength of this concrete? Although the true average strength, /x, is a hypothetical strength rarely possessed by an actual sample, it does provide a useful de- scription of the tensile strength of a type of concrete. Before a con- fidence interval can be put on /a a confidence coefficient must be chosen. Such matters as the seriousness of committing an error, and the added cost of demanding narrower limits, are involved in this decision. However, for purposes of illustration it will be assumed that a risk of 1 in 20 of obtaining a confidence interval not including fjL is appropriate to these circumstances. Then, using inequality (6.37) because n = 10 and 95 per cent limits are sought, we obtain the following: 439 - 2.26(14.9) < m < 439 + 2.26(14.9) because s^ = 47.0/\/10 = 14.9 pounds per square inch. When this inequality is simplified it is found that the 95 per cent confidence inter- val is 405 pounds per square inch < ju < 475 pounds per square inch, to the nearest 5 pounds. Therefore, the true average tensile strength of this concrete will be considered to be somewhere between 405 and 475 pounds per square inch; but, at the same time, it will be kept in mind that there is 1 chance in 20 that this sample has been "wild" and hence has led to an incorrect conclusion. If the reader thinks a bit about the material in this section as compared to the corresponding section in the preceding chapter on binomial populations, it should become apparent that these two sec- tions have a great deal in common. In both, a sampling distribution was studied, and we were concerned with the relative frequencies with which certain sampling phenomena would occur. In particu- lar, we were interested in the relative frequencies with which inter- vals determined from samples would include the unknown population parameter. This probability is the confidence coefficient. There also are some differences which could be pointed out. A major one is that owing to the discontinuity of the binomial frequency distribution, the confidence coefficient is the lower limit on the rela- Sec. 6.3 ESTIMATION OF /. AND a^ 169 tive frequency with which the confidence interval will include the parameter. Basically, however, the methods of these two chapters involve the same kind of statistical inference. It may have occurred to the reader to wonder why the confidence interval is taken in the center of the sampling distribution. Al- though it is true that 92 per cent of all sampling fs with 9 degrees of freedom will have sizes between —2 and +2, it is also true that 92 per cent of all sampling f s with 9 degrees of freedom will lie be- tween — 5 and -|-1.54 (see Table IV). Therefore, the inequality X — u -5.0 < < +1.54 Si also will be true for 92 per cent of all samples with 9 degrees of free- dom. Why not use this inequality as the basis for computing the 92 per cent confidence interval instead of the one suggested earlier? Suppose the inequality above is used on the example used previously in which :c = 8 and s^ = 2. The 92 per cent confidence interval now is from 5 to 18 instead of the shorter interval, 4 to 12, obtained previ- ously. It always will be longer when a non-centrally located interval on t is used. It should be clear that the shorter the confidence interval for a given confidence coefficient, the better the interval estimate. Why be more indefinite than is necessary? PROBLEMS 1. Verify the 80, 90, and 95 per cent confidence intervals given in Table 6.32 for samples 1 and 2. 2. Compute 99 per cent confidence limits on yu for samples 8 and 9 of Table 6.32 and interpret them. yln.s;. 57.2 ^ ^ ^ 78.2. 43.2 ^ /x ^ 64.8. 3. Given that .f = 35 and s = 10 for a sample of ten observations compute and interpret the 95 per cent confidence interval on ii. Do the same for n = 15 and n = 20 and compare them. What is the implication regarding the relation be- tween the size of the sample and the width of the confidence interval, every- thing else being equal? 4. Given that the t was computed to be —2.08 for sample number 525, Table 6.32, determine whether or not the 90 per cent confidence interval includes yit. Do likewise for 95 per cent limits. Ans. CI9Q does not include /x = 60; CIg- does. 5. Use Figure 6.31 to determine the 86 per cent confidence interval on 11 from sample 6 of Table 6.32. 6. Suppose that an improved method of cultivating wheat has produced an average of 5 bushels per acre more yield than an older method on a sample of 21 plots. Also assume that the standard deviation on this sample is s = 5 bushels per acre. What are the 95 per cent confidence limits on the true aver- 170 SAMPLING NORMAL POPULATIONS Ch. 6 age additional yield produced by the new method? Suppose that the new method costs $5 per acre more to use than the older method. What can you say about the probable economic advantage obtained from the new method if wheat is currently bringing $2.25 per bushel? Ans. Clg-j : 2.65 — ;U — 7.35 bushels. Gain ^96 cents per acre. 7. Suppose that in problem 6, s had been 10 bushels per acre. Show how this increase in samphng variability among the 21 plots changes the answers to the questions asked in problem 6. 8. Suppose that in problem 6, the sample had involved but 10 plots. Show how this decrease in the size of the sample changes the answers to the questions asked in problem 6. 9. Suppose that chemical analysis shows that the mean per cent protein for 16 wheat samples is 14.28 and that the estimated standard deviation for the popula- tion of x's being sampled is sj = 2.00. What conclusions can you draw from the 99 per cent confidence interval on the true mean m? 10. If basal metabolisms determined for a random sample of 25 sixteen-year- old Kansas girls produced x = 45.80 calories per square meter per hour, with^^s = 0.50, what are the 80 per cent confidence limits, and what information do they pro- vide in setting up a standard for sixteen-year-old Kansas girls? Ans. CIgo: 45.67 < jj. < 45.93 calories per square meter per hour. 11. Suppose that during a recent period of strong prices twenty-five 450-pound choice steer calves were purchased, October 15th, wintered on silage and one pound of cottonseed meal per day, and then sold on April 15th as choice stocker steers. If the average net income per animal was x = $25 with s = $10, place a 90 per cent confidence interval on the true average net income per animal for the popu- lation so sampled. A similar sampling of choice 600-pound yearling steers pro- duced a 90 per cent confidence interval of $105 to $130 net income per steer. What conclusions can you draw regarding the most profitable choice for a cattleman to make between these two systems? 6.4 A STATISTICAL TEST OF A HYPOTHESIS THAT A GIVEN SAMPLE CAME FROM A NORMAL POPULATION WITH A SPECIFIED MEAN The general problem of deciding whether or not a particular sample came from a normal population whose mean, fi, is specified but whose standard deviation can be estimated only by means of s has received consideration earlier in this chapter. In practice, the speci- fication of fi is based upon a hypothesis about the population under study. For example, if a new method of cultivating wheat does not produce higher average yields the population of differences in yield between the new and old methods grown in a series of paired plots of land will have a true mean }i — because, on the average, there is no advantage to the new method. If the hypothesis that /x = is found from statistical analysis to be unreasonable in view of the Sec. 6.4 TEST OF HYPOTHESIS Hq{,i = w) 171 sampling results, that hypothesis should be rejected. However, if the evidence in the sample is in reasonable accord with that hypothe- sis, it should be accepted. This is the idea behind the methods to be presented in this section, and also of all so-called tests of signifi- cance. A generally satisfactory solution to the problem of this section can be obtained from the /-distribution when normal or near-normal popu- lations are being sampled. As the reader already knows, t = {x — ij)/Si and has n — 1 degrees of freedom if the sample contains n observa- tions. When the m is specified by the hypothesis to be tested, the t can be calculated. Thereafter, 'we can determine from Table IV how Figure 6.41. Illustration of the effects on the (-distribution of a false hypothesis regarding fi. uncommon such a t is when the samples are drawn from the supposed population. For example, if n = 14 and t turns out to be 0.90 on the assumption that /x = 0, we learn from Table IV that about 38 per cent of all sampling Ts with 13 degrees of freedom are numerically larger than 0.90. Therefore, this value of t is not at all unusual and hence we would have no reason to doubt the hypothesis that m = 0. But, suppose that t had been 3.0. It is seen in Table IV that only about one t m 100 from samples of this size ever gets as large as 3, numeri- cally. Hence, we might reasonably doubt that m really is zero because t rarely attains such a size when the hypothesis being tested is true. To illustrate the above discussion graphically, suppose that the true mean, /i, of a normal population of measurements actually is 2 but owing to some error in reasoning jj. is considered to be 0. What effect does this have on the frequency distribution of /? For this situation, t really is (x — 2)/Sj. but because of the error regarding /x the values of t are calculated from the formula t — x/s^. In view of the fact that ti = {x — 0)/S;g is just 2/ Si units larger than ^2 = (^ — 2)/Si, we are actually sampling population B of Figure 6.41, but we think that we are sampling from population A. The discrepancy should, and would, 172 SAMPLING NORMAL POPULATIONS Ch. 6 show up through an excess of large t's beyond the proportions pre- dictable from Table IV. To be more specific, suppose that attention is centered upon a particular t such that p per cent of the ^-distribution lies to the right of this point. Such a point is indicated in Figure 6.41 as tp. It is noted from this figure that a much larger fraction of the true t's (figure B) lie to the right of tp than is true for the population resulting from the calculations with the false value for ^u. This dis- crepancy between the hypothetical and the actual situation will show up in the sampling. Obviously, the greater the discrepancy the more easily it is detected by sampling. In practice it is not feasible, efficient, or economical to continue to draw samples from a population until the evidence for or against a certain hypothesis is so overwhelming that there is virtually no doubt of its truth or of its falsity. Instead, it is common to take what is considered an adequate number of observations on the population, choose the risk we shall take of rejecting a true hypothesis, and then reject the hypothesis being tested if t goes beyond that predetermined limit. To illustrate, suppose that a sample of 15 observations is to be taken under conditions which specify the population being sampled, and that it is decided that it is appropriate to take 1 chance in 20 of rejecting a true hypothesis. For 14 degrees of free- dom, a t which is at, or above, 2.15 numerically (see Table IV) will occur about 1 time in 20 when the choice of /a is entirely correct. If we decide to regard all ^'s which are outside the interval —2.15 ^t ^ -|-2.15 as being the result of a false hypothesis regarding /a, we run a risk of 1 in 20 of rejecting a true hypothesis as a result of sampling variations. Problem 6.41. Suppose that some educators test two proposed teaching pro- cedures in the following way : (1) All available records and the opinions of teachers are applied to the selection of 20 students who, as a group, do a good job of representing students who will be studying the materials upon which the test is to be based. (2) Two equally difficult sections of subject matter are carefully chosen. (3) The group of 20 students is taught one section by method A, the other by method B. (4) Two equally difficult examinations, one on each section of the subject matter, are formulated by competent teachers and given to the 20 students. (5) The average difference, student-by-student, between the two test scores is to be used as the measure of the difference in efficiency between the two methods of instruction. Sec. 6.4 TEST OF HYPOTHESIS Hq{ix = mo) 173 It will be assumed herein that the following test scores were made under the two methods: Student Number Grade on A Method B (A -B) 1 90 85 5 2 72 73 -1 3 86 80 6 4 78 75 3 5 97 95 2 6 85 81 4 7 64 50 14 8 69 65 4 9 76 70 6 10 79 70 9 11 81 78 3 12 83 83 13 75 71 4 14 85 80 5 15 72 69 3 16 100 90 10 17 88 82 6 18 77 65 12 19 80 70 10 20 73 65 8 , ^fel What conclusions can we draw validly from these results? In the usual manner it is found that x = 5.65 in favor of A, and that s^ = 0.87. Therefore, t = (5.65 - m)/0.87, with 19 degrees of freedom. What is a reasonable hypothesis regarding the magnitude of n in the population of Xi = Ai — Bi assumed to follow a normal distribution? The purpose of this study was to determine if one method of instruction is better than the other, and, perhaps, assess the magnitude of the difference if one exists. If one method is superior to the other, ^ is not equal to zero; however, there appears to be no logical way to decide ahead of the test just what the size of /x might be. The problem therefore is attack by assuming that /x = and then de- termining statistically just how satisfactory such a hypothesis is. If /, = 0, i = (5.65 - 0)/0.87 = 6.49, with 19 degrees of freedom. It is clear from Table IV that a i of this size is an extremely rare occurrence, a fact which leads us to reject decisively the hypothesis that /A = 0. In other words, method A most certainly is some better than method B. If there is any benefit to be derived from an estimate 174 SAMPLING NORMAL POPULATIONS Ch. 6 of the magnitude of /x, now that the hypothesis Hoifi — 0) is rejected, confidence intervals can be obtained at any appropriate level of confidence. PROBLEMS L Suppose that 20 pairs of college students have been so selected that the members of each pair can be considered equal in intelligence, scholastic records, and in other factors associated with the making of good scholastic records in a college. Suppose one member of each pair is enrolled in a class in social science which is to be taught by a discussion method emphasizing analysis of problems and reflective thinking; whereas the other member of each pair is put in a class taught by a more formal lecture-recitation procedure. The subject to be studied is the same in each class, and the teachers are considered to be equal in teaching abilities. At the end of the teaching period all 40 students are given the same examination with the following results: Grade Grade Pair Discussion Lecture Pair Discussion Lecture 1 120 110 11 115 108 2 79 75 12 103 91 3 65 70 13 75 70 4 67 75 14 92 95 5 80 75 15 105 102 6 85 80 16 82 78 7 98 90 17 78 76 8 110 95 18 87 90 9 108 92 19 131 120 10 86 80 20 50 51 The discussion method appears to be the better for producing good test scores, but there is considerable variation. Test the hypothesis that the two methods actually are equal on the average (that is, fi = 0) and draw appropriate conclusions. 2. A study was made to determine if tomatoes high up on a plant have more ascorbic acid (vitamin C) than those lower down on the same staked plant. To study this matter, 10 pairs of red-ripe tomatoes were taken from 10 plants, with one member of each pair being from the fifth cluster and the other from the seventh cluster from the bottom of each plant. Each of the tomatoes from the seventh cluster had more vitamin C than the corresponding tomato from the fifth cluster by the following respective amounts: X: 6.6, 11.6, 10.9, 7.4, 8.8, 10.3, 7.4, 7.8, 5.8, and 4.0 milligrams/100 grams. Given that 'ZX = 80.6 and SX2 = 700.46, compute a 90 per cent confidence inter- val on the true average amount by which the ascorbic acid in tomatoes at the seventh cluster exceeds that at the fifth cluster on the same plant, and draw appropriate conclusions regarding /i. Ans. Clor,: 6.7^^^9.4 milligrams/100 grams. Sec. 6.4 TEST OF HYPOTHESIS Ho{(i = «,) 175 3. Solve as in problem 2, using the following data from the sixth and eighth clusters of 10 plants: X: 7.0, 13.3, 8.6, 6.4, 8.3, 9.9, 2.6, 9.1, 6.6, and 1.6. You are given that SX = 73.4 and 2Z2 = 643.40. 4. During the winter tomatoes often are shipped green and allowed to ripen in the package. Aside from matters of flavor and appearance, it is of interest to know what effect this practice has on the vitamin C concentration in the fruit. Two tomatoes were picked from each of 18 plants and at the same cluster on the plant. One was red-ripe, the other was gi-een (no red or yellow coloring). The red-ripe member of each pair was analyzed immediately for vitamin C; the other was ripened at room temperatume out of the sun until red-ripe before its vitamin content was determined. Then the differences in vitamin C between members of pairs was determined with the following results: 2X = 49.37 milligrams/ 100 grams, favoring vine-ripened tomatoes, and SZ2 = 387.5911. Determine statistically if there probably is a loss in ascorbic acid which is due to picking tomatoes green and letting them ripen on the way to market or on the shelf. 5. Suppose that a sociologist has conjectured that the average rent for two-room furnished apartments in a certain section of a city is $90 per month. A sample of 20 apartments had x = $82.50, with s = $8. Use the i-test to determine if the hypothesis Hq (h = 90) is acceptable when sampling variance is taken into account. 6. Suppose that a timber cruiser has judged that the average breast-high diam- eter of a certain stand of timber is 2 feet. Is the timber cruiser's estimate reason- able if 31 trees are selected at random with these results: x = 2.3 feet and s = 0.8 feet? 7. Suppose that a store conducts a study of the comparative net profits from roasting ears sold in cellophane packs as compared to the loose ears in the husks. The experiment is conducted for the 26 business days of a month. At the end of each day, the net profit per ear is figured for each way of selling the corn. The average advantage of using the cellophane pack on these 26 daily comparisons was 2 cents, with a standard deviation s = 0.5 cent. When sampling variation is con- sidered, was the average advantage of the cellophane pack enough to justify the conclusion that it is really more profitable? 8. If, in a certain investigation, x = 10.5 and s = 3, how large must n be to cause the rejection of the hypothesis: Ho {fx = 0) at the one per cent level? 9. If in a sampling study to which the t-test is appropriate the n is 28 and s = 5, how large must x be before the hypothesis Ho (/x = 0) will be rejected at the 5 per cent level? 10. Suppose that 27 pairs of plants of a certain species have been selected for close similarity and are planted in pairs as close together as is appropriate for this species. One member of each pair has had some boron added to the fer- tilizer; otherwise the plants are treated identically. If the 30 plants having boron outgrew their partners by an average of 3.6 centimeters, with the standard deviation of the difference being s = 1.2 centimeters, is this sufficient evidence for the statement that the addition of the boron produces some additional growth ? 176 SAMPLING NORMAL POPULATIONS Ch. 6 6.5 A STATISTICAL TEST OF THE HYPOTHESIS THAT TWO SAMPLES OF OBSERVATIONS HAVE BEEN DRAWN FROM THE SAME NORMAL POPULATION OF NUMERICAL MEASUREMENTS If two samples have been taken under the same conditions but with some one important feature changed, we usually wish to learn if this change has produced a new population of measurements. For example, if two groups of Duroc-Jersey pigs have been fed two dif- ferent rations, the experimenter wants to know if the difference in ration has produced an important difference in average daily gains. That is, has the difference in ration created different populations of average daily gains? Fundamentally, the method to be employed in the solution of this problem is the same as that described in the preceding section, but the mechanics of the procedure need to be altered to fit the new sampling situation. The following symbols will be employed: di = Xii — X2i = the difference between the ith. sample mean from samples from group 1 and the corresponding sam- ple from group 2, and Sd = the standard deviation of the di. Before the general method for attacking the problem just posed is described, some actual sampling experiences will be presented in tabu- lar form, and discussed. Table 6.51 shows a summary of 403 di ob- tained from pairs of samples, each with n = 10 drawn from the near- normal population of Table 6.21. It is recalled that the standard deviation of that population is o- = 10. TABLE 6.51 Frequency and r.c.f. Distributions for 403 Sample Values of di with n = 10 Drawn from a Near-Normal Population with /x = 60 and <t = 10 Class Interval / r.c.J. Class Interval / r.c.J. 16.5-19.4 1 1.000 - 4.5 to - 1.6 88 .342 13.5-16.4 2 .998 - 7.5 to - 4.6 33 .124 10.5-13.4 3 .993 -10.5 to - 7.6 15 .042 7.5-10.4 14 .985 -13.5 to -10.6 1 .005 4.5- 7.4 36 .950 -16.5 to -13.6 1 .002 1.5- 4.4 80 .861 -1.5- 1.4 129 .663 Total 403 Arithmetic mean of di — +0.06; standard deviation = 4.43. Sec. 6.5 TEST OF HYPOTHESIS //o(mi = M2) 177 The frequency distribution in Table 6.51 displays one notable con- trast to that of the x's of Table 6.21, namely, the di are more variable. As a matter of fact, the standard deviation of the di is greater than that of the Xi by a factor of about 1.4 in this instance in which n = 10. It can be shown mathematically that the factor theoretically is \/2, which = 1.414, approximately; hence the empirical results of Table 6.51 agree quite well with the theory. The following theorem summarizes some of the above information and makes it more precise: Theorem. If a very large number of pairs of independently drawn samples of n observations is taken from a normal population with standard deviation = a, then: (a) The population of differences di = xn — X2i will conform to the normal distribution. (6) The arithmetic mean of the population of di is 0. (c) The standard deviation of the population of di is (7d = <T\' ^/n . For the situation summarized in Table 6.51, o^ = 10 v 2/10 = 4.47, an amount which agrees quite well with the 4.43 shown in that table as the observed standard deviation for 403 rf's. In practice, the standard deviation, a, nearly always is unknown so that an estimate must be made from the sample. When a pair of samples has been taken it has been determined by mathematical analysis that the best procedure to follow is this: Lump together, or pool, the sums of the squares of the Xi in each sample taken sepa- rately and divide that sum by 2{n — 1) before taking the square root. In symbols, the following is the recommended estimate of o-: (6.51) s = ]{x^') + i:{x2^) 2{n - 1) where ^{xi^) = the sum of squares of the deviations of the X's of the first sample from their mean; and likewise for 2(x2^). When the theorem above is applied, we obtain the following formula for the sampling estimate of o-^: /S(xi2) + SCa-a^) (6.52) S-, = sV2/n = J ' ' ^^ // ^ n{n — 1) 178 SAMPLING NORMAL POPULATIONS Ch. 6 It turns out from mathematical analysis that the sampling ratios U = {di — iJ.)/sdi follow the same sampling distribution as the t previ- ously discussed, if ^ = the true average di] hence Table IV can be used here provided we employ 2{7i — 1) degrees of freedom for t in- stead of the (n — 1) used before. The way is now open to solve the type of problem proposed at the beginning of this section. To illustrate, suppose that 20 steers of the same breed, weight, and previous history are divided into two equal lots by some impartial means such as drawing numbers from a hat. Thereafter, one group is fed a ration 50 per cent of which is peanut meal and 50 per cent is a standard ration. The other group of steers is fed only 20 per cent peanut meal, the remainder being the same standard ration. After an adequate period of time, the average daily gains of the steers were obtained as follows, with A standing for the group of steers whose diet contained 50 per cent of peanut meal: Group A B 1.55 lb 1.66 lb 1.68 1.82 1.42 1.71 1.45 1.78 1.52 1.69 1.58 1.73 1.56 1.75 1.61 1.61 1.54 1.90 1.48 1.72 r - / -^ "^ 15.39 17.37 X(xi^) = 0.0531, SCro^) = 0.0608. Z-"? ^ / 0.0531 + 0.0608 ^'^ V 10(10 - 1) = VO.001266 = 0.036, approximately. t = (0.20 - m)/0.036 = 5.56 if m = 0. t has 18 degrees of freedom. We learn from Table IV that less than one-half of one per cent of the sample t'& with 18 degrees of freedom are numerically as large as 5.56; therefore, the hypothesis that /a = is rejected and the two samples are regarded as having been drawn from different normal Sec. 6.5 TEST OF HYPOTHESIS Hoifn = M2) 179 populations of average daily gains. It is concluded that the steers on a diet containing 50 per cent peanut meal will, on the average, produce lower gains than those on only 20 per cent peanut meal. Ordinarily the experimenter would wish to carry the statistical analysis farther than this by means of confidence intervals. If the steers on 20 per cent peanut meal do not gain enough more to pay for the added expense of using more of the standard ration which costs more than the peanut meal, it still may not pay to use the diet B. If 95 per cent confidence limits are chosen here, they are deter- mined by the usual methods from -2.10 < (0.20 - m)/0.036 < +2.10; or 0.12 < M < 0.28. Therefore, it can be concluded with considerable confidence (asso- ciated with odds of 19 to 1) that the average advantage due to feed- ing 20 per cent peanut meal instead of 50 per cent is at least 0.12 pound of gain per day but not over 0.28 pound. Given the current price of steers of the sort under study, we can decide which ration is economically preferable. Obviously, other factors would be con- sidered in practice, but they are separate considerations. Although it seems preferable in studies such as those illustrated in this section to have equally many observations in each group, this is not always an attainable goal. If the sample sizes are unequal, say 7?i and ^2 instead of n each, the above methods are applicable but the formulas are changed to fit these new circumstances. For- mula 6.51 is replaced by (6.53) .= /5(fl!l±^M),,„d \ Hi -\- n2 — 2 formula 6.52 is replaced by / /2(.ri2) + S(a-22) (6.54) S2 = sVl/n, + l/n^ = J~-^ -^ (1/ni + IM). \ Wi + n2 — 2 Formulas 6.51 and 6.53 are fundamentally the same in all important respects; each is an estimate based on the deviations {Xn — Xi) and (X2i — £2) in both samples. Likewise, formulas 6.52 and 6.54 are fundamentally alike; each comes from the theorem of mathematical statistics that the variance of the difference between the means of pairs of random samples is the sum of the variances of the two means con- 180 SAMPLING NORMAL POPULATIONS Ch. 6 sidered separately. The reader should verify the fact that if in for- mulas 6.53 and 6.54 Ui = n2 = n, these formulas become 6.51 and 6.52, respectively. Many other applications of the ^-distribution, and accompany- ing statistical techniques, could be cited; but the fundamental prin- ciples are essentially the same as those already explained. PROBLEMS L Suppose that 5 experimental concrete cylinders of each of two types of concrete have been tested for breaking strength, with the following results in hundreds of pounds per square inch: Type 1 : 40, 50, 48, 46, and 41 ; and Type 2: 65, 57, 60, 70, and 55. Use the ^-distribution to determine if the difference in average breaking strength between the two types of concrete can be assigned reasonably to mere sampling variation. 2. Suppose that two gi'oups of 10 steers have been fed two different rations (one to each group) and that the steers are of the same age, breed, and initial weight. Given the following computations determine the 99 per cent confidence interval on the true difference between the means of the average daily gains under the two rations: Ration A Ration B m = 10 712 = 10 xi = 1.90 lb/day X2 = 1.55 lb/day s = 0.20, 18 D/F; t = 3.92 Ans. CI99: 0.1 < I Ml - P2 I < 0.6 lb/day. 3. Suppose that an experiment has been set up at an engineering laboratory to determine the difference in average breaking load between oak and fir beams of the dimensions: 2 inches x 2 inches x 28 feet. The data from tests on 10 beams of each wood are as follows, in pounds: Oak: 725, 1015, 1750, 1210, 1435, 1175, 1320, 1385, Fir: 1205, 810, 1110, 530, 765, 1075, 1475, 950, Oak: 1505, and 1340. Sum = 12,860: SX^ = 17,243,550. Fir: 1020, and 1070. Sum = 10,010: SX^ ^ io,625,400. If you can afford a risk of an error of only 1 in 100 what confidence limits do you set on the true difference in average breaking load for these two materials? 4. Draw 5 pairs of samples, each with n = 10, from the laboratory population furnished you by the instructor, and compute t = d/sa for each pair of samples. Then obtain from Table IV the probability that a numerical value of t that size or larger would be obtained while pairs of samples are drawn from the same normal population. Sec. 6.5 TEST OF HYPOTHESIS //o(mi = M2) 181 5. An experiment designed to find out if supplemental lighting with incan- descent lights will increase the vitamin C content of greenhouse tomatoes pro- duced the following results in milligrams per 100 grams for tomatoes on the bottom two clusters of the plants: No extra light: 25.92, 28.08, 21.27, 22.53, 26.27, 22.57, 22.57, 30.19, and 20.35. SX = 219.75, 2X2 ^ 5454.8279. Incandescent: 20.30, 29.21, 20.50, 21.50, 23.71, 29.34, 26.32, 15.55, and 29.56. 2A' = 215.98, 2A'2 = 5378.5612. Use the t-test to decide whether or not the incandescent lights changed the average ascorbic acid concentration in the greenhouse tomatoes. 6. Given the following two sets of simulated data, assume first that the obser- vations are paired (vertically) and compute and interpret the t. Then assume that the observations are not paired and again do a ^-test. Compare these results and the hypotheses tested. A: 85 72 28 59 75 46 39 68 53; 2X = 525. SX^ = 33,369. B: 80 65 24 58 65 40 38 60 42; 2X = 472. SX^ = 27,198. Ans. (a) t = 4.97, 8 D/F, reject Hoin = 0). (b) t = 0.78, 18 D/F, accept Ho(^a = mb). 7. The antibiotic, aureomycin, has been found to be a growth stimulant for certain animals. The discovery is illustrated by the following two sets of data obtained at Kansas State College under the direction of Dr. E. E. Bartley of the Department of Dairy Husbandry. The measurement of growth used is the total gain during a 12-week period, expressed as a percentage of birth weight. No Aureomycin Had. AureomyciE 77.6 125.6 81.3 135.5 109.2 122.9 124.1 144.8 101.4 103.3 106.0 142.9 81.7 70.6 SX2 = 775.0 54.8 X2 = 129.17 43.3 2(X22) = 101,298.36 119.2 100.0 2X1 = 1069.2 Xl = 89.10 2(xi2) = 6954.96 Obtain the 95 per cent confidence interval on the true difference between the two means fi^ and /.t2 ^^^ tell what information this interval makes available. 8. Suppose that 31 rainbow trout and 31 brook trout are taken at random 182 SAMPLING NORMAL POPULATIONS Ch. 6 from a mountain stream and are measured for length. The rainbows averaged 9.2 inches, with s = 2 inches; the brook trout averaged 8.7 inches, with standard deviation = 2.1 inches. Test the hypothesis Hfyi/j..^ = n^) and draw appropriate conclusions. 9. If from a certain study, xi = 32.7 and X2 = 35.9, and the pooled estimate of o- is s = 7.5. Both samples contained 12 observations. Test Ho | mi — /i2 | = 1. 10. Suppose that 15 samples of each of two varieties of tomatoes have been analyzed for vitamin C, with these results: Variety 1 Variety 2 xi = 28.5 X2 = 30.4 SCxi^) = 50 2(x22) = 60 Test the hypothesis that the true average ascorbic acid concentration in these two varieties is the same. 6.6 USE OF THE SAMPLE RANGE INSTEAD OF THE STANDARD DEVIATION IN CERTAIN TESTS OF STATISTICAL HYPOTHESES The most difficult computational part of the t-test is the determina- tion of either s^ or s^, as the case may be. Another method of testing hypotheses can be used in some situations without the need to com- pute these standard deviations at all. It uses the sample range as its measure of variation. The loss of precision is not serious for smal] samples, becomes greater as the size of the sample is increased, and renders the method useless for large samples. The trouble is that the sampling variability of the range is almost as low as that of the stand- ard deviation for small samples but increases quite rapidly with n. The ratio (6.61) G= {x- y)/R, where R = sample range can be used in a manner analogous to the tA.est procedure. When G has been calculated, Table IX gives the probability that such a sampling | G \ will occur by chance for samples of size n if the hypothesis regarding (jl is exactly right. Thereafter the reasoning is just as it was in section 6.4. When two random samples, each of size n, have been drawn from what is assumed to be the same normal population, the ratio a;i - a;2 (6.62) G = mean range where mean range = arithmetic mean of the ranges of the two samples can be used on problems like those in section 6.5. Table X now is used Sec. 6.6 THE G-TEST WITH SMALL SAMPLES 183 instead of Table IX. Again the tables give /^(| (j | > Gq), where Go is the observed numerical size of G. To illustrate the application of formulas 6.61 and 6.62 reference is made to the problems solved in sections 6.4 and 6.5. First consider problem 6.41. The sample mean is x = 5.65 and the range is 15; therefore, for /fo(M = 0): G = (5.65 - 0)/15 = 0.377, with the sam- ple size = n = 20. By Table IX, the probability that G would be so large if n actually were zero is much less than .001 ; hence the hypothesis that At = is rejected decisively, as it was from the ^-test. The next example is from section 6.5 and involves two diets fed to steers. In fact, Xi = 1.54 and Xa = 1.74, Ri = 0.26, R2 = 0.29, and hence the average sample range = 0.275. Then G = 0.20/0.275 = 0.727, with each n = 10. By Table X a G larger than 0.727 would occur by chance less than 0.1 per cent of the time if both samples were from the same normal population. The hypothesis is rejected; that is, the second diet, which produced the higher average gain in the sampling is considered to produce higher average gains than the first diet. Given the tables and formulas above, we can derive confidence intervals on /x as before when n is small. This interval would not be expected to be identical with one obtained from the i-distribution for the same confidence coefficient; but it has been shown that, on the average, the two intervals are very close to the same length as long as n is small. (Specifically, K. S. C. Pillai has shown in the September, 1951, number of the Annals of Mathematical Statistics that the ratio of the average lengths of the Clgs's by the two meth- ods still is 0.97 when w = 20.) To illustrate, consider again the problem of section 6.4 just used above to illustrate the G-test when there is one set of n observations. In this problem the two confi- dence intervals are obtained as follows: 5.65 — u -2.1 < < +2.1 ~ 0.87 ~ and the 95 per cent confidence interval is 3.82 < ix< 7.48. Using the ratio G, we have 5.65 — fjL -0.126 < < +0.126 ~ 15 ~ 184 SAMPLING NORMAL POPULATIONS Ch. 6 and hence the 95 per cent confidence interval is 3.76 < M ^ 7.54, which is very much like that derived from the ^-distribution. PROBLEMS L Solve problem 1, section 6.4, with the G-test instead of the i-test. 2. Solve problem 3, section 6.5, by means of the G-test. .4«.s. G = 0.289; P^ .064; accept H^, tentatively. 3. Draw 25 samples, each with n = 10 from a near-normal population, and compute the G for each sample. How many of these G's fall beyond 0.186 in numerical size? How do your results check with Table IX? 4. Suppose that a college is attempting to learn if instruction of a certain type improved in one year a student's ability to think analytically. Also as- sume that tests exist which reliabh^ measure such ability, and that these tests are given at the beginning and at the end of the school j^ear. If the following differences between the last and the first score of each student were obtained, would the G-test cause you to accept or to reject the hypothesis that the teach- ing procedures emploA'ed jailed to improve analytical thinking? X: 5, 0, 10, -4, -6, 8, 1, 7, -10, 0, 3, 5, -1, 8, 4, 0, -3, 7, 7, and 9. Ans. G = 0.125; P = .05; reject Hoijj. = 0). 5. Make up, and solve, a problem like problem 4, which has the same x but for which G is twice as large. Half as large. 6. Suppose that information is sought analogous to that in problem 4, but there are two separate cla.'sses of 15 students being taught by each method. The two classes are supposed to be equal at the start of the teaching period. Given the following gains (4-) or losses ( — ) in score during the year, draw appro- priate conclusions by means of the G-test and Table X: Method I: 10, 3, -2, 5, 0, -8, 14, 1, -12, 5, 5, 9, 7, -1, and 9. Method II: -2, 5, 5, 4, 0, 7, 6, -1, 4, 10, 8, 11, 10, 0, and 13. Ans. G = 0.114; F> .10; accept H^ifi-^ = a^o)- 7. An experiment intended to discover if blue fluorescent lights will increase the vitamin C concentration in tomatoes on the seventh and eighth clusters from the bottom of the plant gave these results, in milligi'ams per 100 grams: No extra Hght: 38.57, 39.39, 33.44, 34.32, and 38.01. Blue fluorescent: 33.72, 37.85, 39.07, 31.16, and 35.69. Test the hypothesis that the blue light does not change the vitamin C con- centration, and draw valid conclusions. 8. Suppose that two methods of computing basal metabolism for the same 11 subjects produced the following pairs of records, in calories per square meter per hour. 31.42, 30.90, 34.92, 30.59, 30.53, 33.08, 32.61, 30.46, 30.73, 31.44, 32.82, 31.80, 29.16, 32.96, 32.32, 30.76, 30.55, 33.19, and 29.22. 2Xi = 347.47. 27.65, 32.54, and 29.30. 2^2 = 341.48. Sec. 6.7 THE CENTRAL LIMIT THEOREM 185 Use the G-test to decide if one method tends to produce higher metaboHsm records than the other, and explain your decision in terms of sampHng phe- nomena. Ans. G = 0.100; P> .10; accept H^i/j.-^ = fi^). 9. Some varieties of wheat produce flour which typically takes longer to mix into proper doughs than others. Decide by the G-test if Kharkof actually has (as appears from the samples) a longer mixing time than Blackhull : Kharkof: 3.00, 1.88, 1.62, 1.50, 1.75, 1.38, 1.12, 1.88, 2.50, 1.62, 2.88, 2.50, 3.88, and 2.75. Mean = 2.16. Blackhull: 1.25, 2.38, 1.62, 1.50, 1.25, 1.38, 2.25, 2.12, 1.84, 2.38, 2.25, 1.50, 2.00, and 1.62. Mean = 1.84. 10. Compute the 90 per cent confidence intervals for the two varieties of problem 9 and compare them. Draw appropriate conclusions. Ans. CIgo: - 0.04 ^ | /x^ -/j.^ \ ^ + 0.69. 6.7 THE CENTRAL LIMIT THEOREM AND NON- NORMAL POPULATIONS The statistical methods which have been discussed in this chap- ter are based on the assumption that the populations involved are normal. In practice this requirement rarely is met rigorously; hence we may wonder if the subject matter of this chapter is chiefly of academic interest because it does not fit actual conditions. This is not the situation because of the truth of the central limit theorem. This theorem states essentially that if any population of numerical measurements has a finite mean and variance, fi and a^, respecti\ely, the frequency distribution of the sampling mean, x, will be essentially a normal distribution with mean = ix and variance = a^/n if the n is very large. As a matter of fact, the necessary size of n depends on the degree of non-normality of the original population. Tables 6.71^, B, C, and D summarize a decidedly non-normal population of counts of flies on dairy cattle, and show some observed distributions of x's for samples with n = 9, 16, and 25. Figure 6.71 displays these same dis- tributions visually. It is rather obvious that none of these sample sizes is very large, and therefore the distributions of x are still notice- ably non-normal. However, the meaning of the central limit theorem is illustrated. It can be seen from Tables 6.71 and from Figures 6.71: (a) That the parent population is extremely non -normal. (b) That even with only nine observations per sample, the distri- bution of X has gone a long step towards fulfilling the ideal expressed by the Central Limit Theorem. 186 SAMPLING NORMAL POPULATIONS Ch. 6 (c) As n was increased, the distribution of x, and its mean and vari- ance, approached more and more closely to those features which the Central Limit Theorem assures us will be attained if n is sufficiently large. The foregoing discussions are not intended to make us ignore the non-normality of distributions met in practice, but they do indicate that a great many moderately non-normal distributions can be studied by means of the techniques explained in this chapter. In this chapter the ratio {x — m)/^! was said to follow the ^dis- tribution with the same number of degrees of freedom that s^ has as a sampling estimate of o-^. Actually any ratio {w — fi)/s will follow TABLE Q.71A Summary of Counts of Flies on Dairy Cattle Tethered in a Field at Kansas State College After They Were Sprayed with an Effective Fly Repellent Normal r.c.f., Class Interval / r.c.f. Same fx and a Difference 168-175 1 1.000 1.000 16{>-167 1.000 1.000 152-159 1 1.000 1.000 144-151 .999 1.000 -.001 136-143 .999 1.000 -.001 128-135 .999 1.000 -.001 120-127 1 .999 1.000 -.001 112-119 3 .999 1.000 -.001 104-111 6 .998 1.000 -.002 9&-103 2 .995 1.000 -.005 88- 95 7 .994 1.000 -.006 80- 87 8 .992 1.000 -.008 72- 79 10 .988 1.000 -.012 64- 71 11 .984 .999 -.015 56- 63 31 .980 .999 -.019 48- 55 26 .968 .994 -.026 40- 47 59 .957 .976 -.019 32- 39 102 .934 .932 + .002 24- 31 206 .893 .837 + .056 16- 23 392 .812 .687 + .125 8- 15 771 .656 .500 + .156 0- 7 869 .348 .309 + .039 2(/) = 2506 M = 15.37, (t2 = 257.28 TABLE 6.715 Distribution of Means of Random Samples with n = 9 Drawn from THE Population of Table 6.71A Normal r.c.f., X Interval / r.c.f. Same ix and <r Difference 36-38.99.. 3 1.000 1.000 33-35.99.. 8 .997 1.000 -.003 30-32.99.. 4 .990 1.000 -.010 27-29.99.. 11 .986 .998 -.012 24-26.99.. 30 .975 .988 -.013 21-23.99.. 73 .947 .954 -.007 18-20.99.. 139 .877 .866 + .011 15-17.99.. 221 .744 .701 + .043 12-14.99.. 268 .533 .480 + .053 9-11.99.. 211 .278 .266 + .012 6- 8.99.. 74 .076 .113 -.037 3- 5.99.. 6 .006 .037 -.031 N :(/) = 1048 MJ = 15.27, 0-^2 = 26.72 By Central Limit Theorem (if n is large enough) fix = 15.37 (7x- = 28.59 TABLE 6.71C Distribution of Means of Random Samples with n = 16 Drawn from THE Population of Table 6.71A Normal r.c.f., X Interval f r.c.f. Same fj. and a Difference 32-33.99.. 2 1.000 1.000 30-31.99.. 1 .998 1.000 -.002 28-29.99.. 6 .997 1.000 -.003 26-27.99.. 13 .991 .999 -.008 24-25.99.. 13 .979 .994 -.015 22-23.99.. 40 .966 .978 -.012 20-21.99.. 74 .928 .938 -.010 18-19.99.. 106 .856 .855 + .001 16-17.99.. 162 .754 .719 + .035 14-15.99.. 209 .597 .540 + .057 12-13.99.. 206 .395 .352 + .043 10-11.99.. 145 .196 .195 + .001 8- 9.99.. 51 .056 .091 -.035 6- 7.99.. 7 .007 .035 -.028 ■^ ^ :(/) = 1035 Mx = 15.56, o-j2 = 17.32 By Central Limit Theorem (if n is large enough) fjLs = 15.37 ai^ = 16.08 187 188 SAMPLING NORMAL POPULATIONS Ch. 6 TABLE 6.7 ID BUTiON OF Means ; OF Random S.\mples with n = 25 Drawn from THE Population of T ABLE Q.71A Normal r.c.f., X Interval / r.c.f. Same /x and (t Difference 26.00-27.49 3 1.000 1.000 24.50-25.99 3 .997 1.000 -.003 23.00-24.49 16 .994 .997 -.003 2L50-22.99 22 .978 .990 -.012 20.00-2 L49 43 .956 .968 -.012 18.50-19.99 85 .914 .917 -.003 17.00-18.49 121 .830 .821 + .009 15.50-16.99 180 .710 .673 + .037 14.00-15.49 188 .532 .492 + .040 12.50-13.99 168 .346 .313 + .033 11.00-12.49 133 .179 .169 + .010 9.50-10.99 41 .048 .077 -.029 8.00- 9.49 5 .007 .029 -.022 6.50- 7.99 2 .002 .009 -.007 2(/) = 1010 M2 = 15.51, (ri2 = = 10.07 By Central Limit Theorem (if n is large enough) /xj = 15.37 (Tj^ = 10.29 the ^-distribution as long as w is normally distributed with mean fi, and s is calculated as described earlier. Hence if w is a sample mean drawTi from a non-normal population which satisfies the few require- ments of the Central Limit Theorem, and if n is large enough, the ratio (w — n)/s can be considered quite accurately to follow a ^-dis- tribution. Thereafter the methods introduced in this chapter for estimating parameters and for testing hypotheses regarding parameters become applicable. One word of warning is in order, however, before this subject is left. In any particular sampling situation, the standard deviation, o-^, needs to be estimated from the sample. This is done by means of s^. What happens to the quality of this estimate when the parent population is radically non-normal? Under such circumstances the beginner is advised to seek the advice of a statistician. Sec. 6.7 THE CENTRAL LIMIT THEOREM 189 CM ^^ U3 F II (M C II II II c c C C <1> IH < \H IH IH o j; t<j »o fl s 5 H ^;3 CD c 2 o -3 -2 ^ >. a c o A3uanb9Jj ^H T! I>. d CD 03 QJ d ts b(l (M kn -v 190 SAMPLING NORMAL POPULATIONS Ch. 6 REVIEW PROBLEMS 1. Who was Student, and how was his work connected with the development of present-day methods of statistical analysis? 2. Calculate the arithmetic mean and the standard deviation for a set of num- bers, Yi, given that SF = 900 grams, and 2F- = 55,465 grams^, where the Y's are the weights of female rats at 28 days of age. There are 20 rats in the sample. A71S. ?/ = 45 grams, s = 28.1 grams. 3. Compute the 80 per cent confidence interval for problem 2 on the true mean 28-day weight of such rats, and draw appropriate conclusions. 4. What would be the general change in the confidence interval of problem 3 if 95 per cent limits instead of 80 per cent limits had been computed? What would be the effect if the SF^ had been smaller, the remainder of the numbers staying the same? 5. Graph the binomial frequency distribution of the numbers of sums of 6 thrown with two unbiased dice on sets of 8 throws. 6. Compute for problem 5 the probability that on any particular future set of 8 throws at least 3 sums of 6 will be thrown. Ans. .087. 7. Take any newspaper which lists prices of bonds and determine the median price and also the range. 8. Calculate the coefficient of variation for problem 2, using y in place of n and s in place of <t, and tell what sort of information it provides about the weights of the rats in the sample. Ans. CV = 62.4 per cent. 9. Draw 10 samples of 12 members each from the laboratory population and compute t and G for each sample, using the correct hypotheses regarding fi. 10. Determine the upper limits of the 20th and 85th percentiles for the fre- quency distribution of Table 6.31 and state what information they give. Ans. Upper limit of 20th percentile = 0.90 by interpolation, = 0.86 by Figure 6.31. Upper limit of 85th percentile = 1.12 by interpolation, = 1.10 by Figure 6.31. 11. If 100 t- were to be drawn at random from among those summarized in Table 6.31, what is the expected number of them falling between t = and t = 1.50? 12. Following are some experimental results from tests of the breaking strengths of the wet warp of rayon and wool fabrics in pounds: Rayon: 29.5, 31.0, 28.7, 29.1, 28.4, 28.9, 30.9, and 29.0. Wool: 25.3, 28.9, 19.2, 25.1, 21.1, 31.4, 25.6, and 19.0. Does the difference in average breaking strength lie beyond the bounds of reasonable sampling variation according to the i-test? Solve problem also by the G-test, and compare the conclusions. SX^ = 235.5, 2X2^ = 6939.33. ^X^y = 195.6, 2X2^^ = 4921.48. Ans. t = 3.10, 14 D/F, P = .008; reject i/o^-^i = ^"2)- ^ = 0.665, n = 8, P = .002; reject HQifi^ = fi^). 13. Suppose that twelve 2 inch x 12 inch x 8 inch wood blocks were tested for strength with the following results in thousands of pounds: 6.5, 17.0, 10.0, 15.1, 13.5, 16.4, 19.8, 7.7, 11.5, 14.5, 12.7, and 12.9. Place 95 per cent confi- dence limits on the true average strength of such blocks, and interpret these limits. Ch. 6 REFERENCES 191 14. You are given the following hypothetical data from an experimental study of the average daily gains (in pounds) of two groups of 10 steers each: For group A: xa = 2.35, s"a = 12. For group B: xb = 1.75, Sx^b = 180. Is the difference in mean average daily gain, d = 0.60 pound, beyond the bounds of reasonable sampling variation; that is, is it statistically significant? Ans. t = 0.34; 18 D/F, P ^ .63; accept Ho (^i = ^2)- 15. Suppose that you have taken the bid in a bridge game and that you and your partner have all the trumps but J, 10, 7, 4, and 3. Before you have led at all, what do you compute as the probability that you would get all the trumps out within 3 leads? 16. Suppose that you have been told that when 6 unbiased coins were tossed at least 3 of them showed heads. What is the probability that exactly 4 of the coins turned up heads. Ans. P(r — 4 heads) = 15/42. 17. Suppose that a large jug contains the following numbers of each de- nomination of paper currency, and that you are to withdraw a bill without look- ing and keep it: 50 one-dollar bills, 25 five-dollar bills, 10 ten-dollar bills, 5 twenty-dollar bills, 2 fifty-dollar bills, and 1 one-hundred-dollar bill. What is your mathematical expectation on such a game? 18. If 2 cards are drawn simultaneously from a bridge deck, what is the probability that one will be a spade, the other a heart? Ans. 13/102. REFERENCES Dixon, Wilfrid J., and Frank J. Massey, Jr., Introduction to Statistical Analysis, McGraw-Hill Book Company, New York, 1951. Hald, A., Statistical Theory with Engineering Applications, John Wiley and Sons, New York, 1952. Neyman, Jerzy, First Course in Probability and Statistics, Henry Holt and Com- pany, New York, 1950. Snedecor, George W., Statistical Methods Applied to Experiments in Agriculture and Biology, Fourth Edition, Iowa State College Press, Ames, Iowa, 1946. CHAPTER 7 Linear Regression and Correlation It often is advantageous to consider two types of numerical meas- urements simultaneously because they are related to each other. For example, the following table records the mean monthly temperatures from January to July at Topeka, Kansas, along with the month of the year: Month of the Year: Jan. Feb. Mar. Apr. May June July Mean Temperatures (degrees Fahrenheit) 38.0 41.7 54.0 66.0 74.4 83.8 88.7 If the month to which each temperature applies were to be ignored, these temperatures simply would be seven numbers which might fall in the following random order (obtained by drawing them at ran- dom) : 88.7, 54.0, 66.0, 38.0, 83.8, 74.4, and 41.7. In this form the numbers seem to be quite variable about their arithmetic mean, 63.8°F. However, when considered in conjunction with the month as the second variable, these temperatures follow an orderly pattern. This point is illustrated graphically in Figures 7.01 A and B, in which temperatures are first plotted against the random order in which they were drawn, and then against the month to which they apply. Figure 7.01A merely re-emphasizes the remarks made above about the excessive variability about the mean, 63.8, and suggests that such an average would be of doubtful utility because the temperatures are too inconsistent. But it appears from Figure 7.015 that the mean temperatures for the first six months of the year increased in quite an orderly manner from month to month, with little deviation from a linear upward trend. Hence, a better analysis of these data can be obtained by taking proper account of the second variable, time. A straight line is drawn into Figure 7.01 A, 63.8 units above the horizontal axis, to represent the arithmetic mean of the tempera- tures whose individual magnitudes are indicated by the ordinates of 192 Ch. 7 LINEAR REGRESSION AND CORRELATION 193 the points on the graph. The amounts by which the monthly mean temperatures are above or below the mean of all the temperatures are shown as vertical distances above or below the horizontal line. 90 80 a. 70 - 1.60 E ^50 40 30 i 1 1 1 1 1 - • - - • - — • _ - • ~ • ~ 1 I 1 I 1 1 63.8° 2 3 4 5 Order of draw A May July Apr. Month B Figure 7.01. Mean monthly temperatures at Topeka, Kansas, plotted first at random and then according to the corresponding month. As far as Figure 7.01A is concerned these deviations are simply the consequences of unexplained variations in temperature. However, when each temperature is associated with the month to which it belongs (as in Figure I.OIB) it is apparent that all but a small amount of the variability among these temperatures is associated with a definite tendency to increase rather uniformly with succeed- ing months of the season. 194 LINEAR REGRESSION AND CORRELATION Ch. 7 The trend line drawn into Figure 7.015 was determined just "by- eye"; but it usually is preferable to have a standard method of de- termining where the line should be drawn. This matter will be dis- cussed in the following four sections. 7.1 SCATTER DIAGRAMS AND TYPES OF TREND LINES A number of the statistical methods with which the reader is al- ready familiar can be employed in the analysis of data involving two variables. One additional matter must be studied, however, namely, the relationship between the two variables. A little graphic analysis usually is worth while before the numerical analyses are undertaken. There are many ways in which one variable, Y, can change with respect to another variable, X, as successive pairs of observations are taken with the X, say, increasing in magnitude. The size of Y may tend to increase as X increases; Y may tend to decrease as A^ increases; or some of both may occur over the range of values studied. In addition there are numerous ways in which Y can increase as X increases; and similarly for the other possibilities just mentioned. To illustrate, consider the following tables of pairs of values for X and Y: (A) (B) (C) (D) (E) (F) (G) X Y X Y X Y X Y X Y X Y X Y 1 7 1 50 1 25 0.0 0.75 -3 8.5 1 2.5 1.0 40 3 12 2 39 2 40 0.5 0.80 -2 5.2 2 8.0 1.5 25 5 14 3 24 3 28 1.0 1.20 -1 0.5 3 32.2 2.0 44 7 20 4 21 4 42 1.5 2.60 0.5 4 60.0 2.5 20 9 22 5 9 5 22 2.0 3.80 1 0.7 5 127.2 3.0 35 11 30 2.5 5.75 3.0 10.40 2 3 4.8 8.9 3.5 28 4.0 47 4.5 26 5.0 34 It is helpful to a mathematical study of the relationship between two variables if the pairs of corresponding numerical measurements, X and Y, are represented by points on a graph, as they were in ele- mentary algebra. This has been done in Figures 7.11A, B, . . . , G for the data immediately above. Such graphs are called scatter diagrams. It is noted from these figures that the points may not exactly fit any simple curve, but they sometimes do exhibit a general pattern which may make it possible to study the relationship between Y and X. It is necessary here to Sec. 7.1 SCATTER DIAGRAMS AND TREND LINES 195 — • — — • — g - • - — • — __J \ \ L_._ - :i4:^ — • — — • — — • — — • — ro !>< !^ S — • — — • — s — • — 2 ^ ^ s bO a (1) s> -D o ca -C bD '-3 •- o !^ I-H 03 196 LINEAR REGRESSION AND CORRELATION Ch. 7 think in terms of general rather than precise curves as was done in algebra, where all points which belonged with a certain graph fell exactly on that graph. Data to which statistical analysis is applied are not so well behaved as that. It will be necessary later to learn how to decide which curve to choose as best describing the relation- ship between X and Y suggested by a scatter diagram ; and it will not be expected that all the points will fall perfectly on the line finally chosen. The following information can be derived from a careful inspection of Figures 7.11: From (A): Y definitely tends to increase uniformly (linearly) as A' increases. On the average, Y increases about 13/6 units for each unit increase in X. From (B) : Y decreases in proportion to the increase in X. Again the relationship can be briefly described as linear. More specifically, Y tends to decrease about 10 units for each unit increase in A". As a result the slope of the straight line which indicates the linear trend is said to be —10. From (C) .• Y has no apparent relation to X; hence the X measure- ment may as well be ignored in the statistical analysis of the meas- urements, Y. From {D) : Y increases with A^, but the increase is not uniform. In fact, Y increases more rapidly for large A's than for the smaller A's. This relationship between Y and A' is called curvilinear. In this in- stance, it follows the non-linear mathematical law: Y = 0.5e^, where e is the base for natural logarithms. From (E) : As the measurement represented by X increases from — 3 toward 0, the corresponding measurement, Y, tends to decrease in a non-linear manner. Thereafter, Y increases non-uniformly. As a matter of fact, the points on this scatter diagram tend to follow the curve, Y — X^. From {.F) : Y tends to increase non-uniformly with X, as in (D) , but the curve rises more sharply here. Froin (G) : There is no apparent relationship between X and Y, as in (C). Another point should be noted regarding the scatter diagrams of Figures 7.11. If the concomitant measurement, A^, were to be ignored during an analysis of the data of Y corresponding to any of the sit- uations except (C) and (G) , a considerable portion of the variability of the F's about their means would represent unnecessary variation in this sense. We know from (B) , say, that if A = 1, the corre- Sec. 7.1 SCATTER DIAGRAMS AND TREND LINES 197 spending Y is necessarily about 40 units larger than if X = 5 because there is definitely a downward trend of Y as A' increases. If the X were ignored, all that observed difference of 40 units must be as- signed to errors of measurement and/or to sampling accidents when, in fact, only about one unit should be so assigned. Methods will be presented later on in this chapter by which the apparent variation among the Y's can be reduced by taking account of the statistical relation between X and Y. However, nothing extensive will be done with non-linear trends. PROBLEMS 1. Construct a scatter diagram for the following pairs of measurements, and draw in by eye a straight line which seems to you to best depict the trend of Y with X. Is the assumption that Y and A' are linearly related a good one in your opinion? X: 2 4 6 8 10 12 14 16 18 20 Y: 100 140 200 235 280 325 370 415 450 500 2. From the following sampling data estimate how much }' changes, on the average, for each unit increase in X. X: 36 43 50 40 42 45 40 45 39 48 Y: 1.35 1.70 1.90 1.55 1.65 1.80 1.63 1.75 1.60 1.93 A71S. About 0.04. 3. Make a scatter diagram of the following pairs of observations and draw in a freehand line to summarize the way 1' changes with X: X, Y: 1,1; 2,2.5; 3,4.5; 4,6.0; 5,10.0; 6,14.5; 7,23.0; 8,35.0. 4. Would you approve of the assumption that the carotene and the nitrogen- free extract contents of pasture grasses are linearly related if assured that the following pairs of such values form a representative sample for pasture grasses of a given sort? Justify your decision. X(NFE): 50 48 53 51 49 53 51 48 F (Carotene): .44 .26 .20 .24 .44 .23 .26 .34 5. The following are means and corresponding standard deviations obtained from samples of 10 observations, each drawn from an approximately normal popu- lation. Construct a scatter diagram and decide what, if any, relationship exists between the sampling mean and standard deviation from a normal population if these samples are representative of such populations. Plot x on the horizontal axis. X 58.4 6.04 53.4 10.33 63.0 9.33 60.4 11.40 55.7 9.61 60.9 12.64 52.9 6.13 61.2 6.45 62.0 6.70 56.7 8.03 58.1 8.15 67.7 10.21 61.7 11.40 55.0 11.51 53.6 9.83 56.4 5.81 54.0 10.52 59.3 9.05 57.8 10.92 54.4 10.78 59.0 11.09 54.1 8.75 60.0 8.17 62.8 8.43 198 LINEAR REGRESSION AND CORRELATION Ch. 7 6. Grades in elementary statistics and in mathematics of finance for the same students are given below. What do you conclude is the relation between a student's grades in these two subjects? Give evidence upon which your con- clusion is based. X(statistics): 94 83 91 98 80 82 61 81 58 90 85 75 75 70 92 62 F(finance): 89 90 91 97 85 87 41 88 60 85 86 83 87 72 97 64 7. The following are weights of the larvae of honey bees at different ages: Set A. Z(days): 12 3 4 5 6 y (milligrams): 2.0 4.3 23.1 93.1 148.7 295.5 Sets. X(days): 12 3 4 5 6 F(log milligrams): 0.30 0.63 1.36 1.96 2.17 2.47 Construct scatter diagrams for each set separately and decide for which, if either, the assumption of a linear relation between X and Y appears to be justi- fied. If either set produces a satisfactorily linear trend, estimate the slope of the best-fitting freehand line and state what information it provides. Should you make some allowance for the fact that you used a freehand line in a posi- tion with which others might disagree? 8. Make a scatter diagram for the following pairs of observations. Draw in a freehand line which appears to you to be the best-fitting straight trend line, and derive from this line an estimate of the Y which should correspond, on the average, to X = 0, 4.5, and 7.5 : X: 1 2 3 4 5 6 7 8 9 10 Y: 21 20 17 15 14 14 12 9 6 5 Ans. About 23.3, 15.1, and 9.7, respectively. 7.2 A METHOD FOR DETERMINING LINEAR REGRESSION (OR TREND) LINES It is quite customary to use the term regression line to describe the line chosen to represent the relationship between two variables when this decision is based on sample points, as in a scatter diagram. The origin of the term regression probably lies in genetic studies of the tendency for offspring of parents who are well above, or below, the group average to go back, or "regress," toward that group aver- age. The term trend line will be used interchangeably with regres- sion line, even though the former is frequently associated with dis- cussions of time series. Previously, in this chapter, freehand lines have been used to depict the average change of one variable with respect to another. Such Sec. 7.2 DETERMINING LINEAR TREND LINES 199 a procedure, however, obviously is somewhat subjective because it depends quite a bit upon personal opinion. One of the chief purposes of numerical measurement and statistical analysis of such measure- ments is to free decisions based on relatively precise numbers from distortions which might result from the exercise of personal tastes and opinions. It is for this reason that it is desirable to be able to describe such a line by a method which will produce the same result no matter who uses it. The types of relationships between two kinds of numerical meas- urements which were discussed in the preceding section are illustra- tive of sampling experiences involving errors of observation and measurement. The dots of the scatter diagram usually fail to fall exactly on any simple curve for one of two reasons: (a) Sampling errors or chance variations cause the values of Y, say, to be partially inaccurate, (b) There are real variations from the general trend of Y and X which, however, are of minor importance compared to the general trend and should be smoothed out in order that the general trend may be studied more effectively. The data of Table 7.21 and the corresponding scatter diagram of Figure 7.21 help to illustrate these points. The data in the table are considered to be a population of pairs of observations, that is, a bivariate population. For con- venience these data have been grouped by 16-week weights (A") to the nearest pound in the scatter diagram of Figure 7.21. The bivariate population of Table 7.21 possesses several character- istics which are of statistical interest and importance. These fea- tures are exhibited by Figure 7.21, from which it is learned: (a) There is a general upward trend of the 28-week weight, with increas- ing 16-week weight of the same bird. (5) Within each 16-week- weight class there is a frequency distribution of 28-week weights, and this distribution is relatively symmetrical about the mean 28- week weight for the class, (c) The means of the six 16-week-weight classes lie perfectly on a straight line with a slope of 1/2. Thus the true linear regression line passes through the points representing the true average Y's for the given X's. The slope of this true trend line is denoted by the Greek letter ft (beta). In a study based on samples the /? is unknown, as is the exact loca- tion of the true linear trend line, and only the n pairs of sample measurements are available as a basis for making decisions about the linear trend line. For example, a random sample of 30 pairs {X, Y) was drawn from the bivariate population of Table 7.21, with the 200 LINEAR REGRESSION AND CORRELATION Ch. 7 TABLE 7.21 Pairs of Observations of the 16-Week Weights and Corresponding 28-Week Weights of Turkeys Raised on the Kansas State College Poultry Farm (X is the 16-week weight in pounds; Y is the 28- week weight.) X Y X Y X Y X Y X Y 4.9 13.3 5.4 14.6 5.2 13.4 5.4 14.0 4.6 13.0 4.7 12.7 4.8 12.9 4.9 12.4 5.0 12.6 5.1 12.5 5.2 13.1 5.3 13.5 5.4 13.6 4.7 14.3 4.9 13.8 5.1 13.7 5.2 13.2 4.6 14.0 4.7 14.3 4.8 14.1 4.9 13.6 5.0 14.0 5.1 13.6 5.2 13.5 5.3 13.5 6.2 13.5 6.5 14.8 6.4 14.3 6.5 13.0 6.4 14.4 6.5 13.0 5.5 13.5 6.1 15.3 5.5 13.0 6.0 14.3 6.4 14.6 6.5 13.8 6.5 13.3 5.5 14.3 6.2 14.4 6.1 13.4 6.3 13.9 6.5 13.7 6.4 14.0 6.4 14.4 6.5 15.4 5.9 14.4 6.5 14.9 5.9 14.8 5.5 14.6 5.5 13.1 5.6 13.2 5.7 12.8 5.8 13.5 5.9 13.5 6.0 13.6 6.1 13.7 6.2 14.0 6.3 14.0 6.4 14.0 6.5 14.1 5.5 12.5 5.6 13.3 5.7 13.2 5.8 13.9 5.9 13.8 6.0 13.8 6.1 15.0 6.2 14.7 6.3 15.2 6.4 15.1 6.6 13.6 6.7 13.8 6.8 13.7 6.9 13.7 7.0 14.5 7.1 14.2 7.2 14.2 7.3 14.3 7.4 14.3 6.6 14.3 6.7 14.4 6.8 14.4 6.9 14.4 7.0 14.5 7.1 14.5 7.2 14.5 7.3 14.5 7.1 14.7 7.2 13.3 6.7 15.5 7.0 15.9 7.4 14.7 7.3 14.7 6.6 14.1 6.6 14.4 7.1 15.3 7.1 14.9 7.0 15.6 7.4 14.0 7.0 15.0 7.2 15.5 7.2 15.2 6.8 15.3 7.3 15.4 7.1 13.9 7.0 14.7 6.9 14.8 6.6 15.5 7.0 14.7 6.8 15.1 7.1 13.2 7.3 14.0 7.2 14.9 6.7 14.6 6.7 14.8 6.8 15.0 7.3 12.8 7.3 14.3 7.3 13.6 7.0 13.5 7.0 15.0 6.8 13.9 7.4 14.1 6.9 15.0 7.2 13.3 7.3 15.3 6.6 14.0 6.6 14.8 7.4 14.1 7.3 14.0 6.9 14.0 7.0 14.1 6.8 14.2 7.4 15.6 7.3 16.3 7.0 15.8 6.6 14.7 6.8 12.9 7.0 13.4 7.2 14.8 7.4 14.9 7.5 14.7 7.6 14.7 7.7 13.7 7.8 14.7 7.9 14.8 8.0 15.7 8.1 15.1 8.2 15.1 8.3 15.7 8.4 15.7 8.5 15.8 7.5 14.5 8.2 15.1 8.3 15.5 7.8 15.3 7.9 16.6 8.3 17.1 8.3 15.3 7.5 15.4 7.6 14.7 8.5 14.5 8.5 15.0 7.5 14.8 8.4 15.6 7.5 15.6 7.8 16.1 7.8 15.5 7.7 15.9 7.8 16.0 8.3 13.4 8.2 14.9 8.1 14.4 8.2 14.0 8.3 14.5 8.2 13.5 7.5 16.0 8.0 15.0 7.5 15.5 7.7 15.8 7.5 14.4 7.6 15.4 8.0 16.7 7.5 14.3 8.0 16.4 8.1 16.7 8.5 15.1 7.6 14.6 8.5 15.2 7.8 14.3 7.8 15.2 7.6 17.0 7.9 15.2 8.1 14.2 Sec. 7.2 DETERMINING LINEAR TREND LINES 201 TABLE 7.21 {Continued) Pairs of Observations of the 16-Week Weights and Corresponding 28-Week Weights of Turkeys Raised on the Kansas State College Poultry Farm X (A" is the 16- week weight in pounds; Y is the 28-week weight.) Y X Y X Y X Y X 7.5 13.8 8.5 14.5 7.9 15.9 8.1 14.4 8.3 15.0 8.2 15.6 8.4 14.8 7.8 13.7 8.2 15.0 7.6 13.3 7.6 14.2 8.1 14.1 8.0 13.5 8.1 15.2 8.2 15.3 7.5 15.5 7.9 15.8 7.7 16.3 7.8 17.0 7.8 13.5 8.0 15.6 8.3 16.2 8.4 14.8 8.4 14.4 7.8 13.2 8.5 16.9 8.0 13.7 8.2 15.5 8.3 15.0 8.2 15.7 8.0 13.6 8.0 14.3 8.4 15.0 8.5 15.1 8.2 15.4 8.0 14.9 8.1 15.0 8.5 15.2 7.8 14.0 7.6 15.2 7.6 14.1 8.5 15.5 7.7 14.3 7.5 14.6 8.1 14.3 8.1 14.4 8.0 13.8 8.2 13.8 7.6 14.9 8.2 15.3 8.1 15.3 7.5 14.1 8.5 15.3 8.1 15.4 7.5 15.0 8.0 15.0 7.8 14.0 8.5 16.0 7.9 15.5 8.1 16.5 8.5 15.6 7.5 13.9 7.5 14.3 7.5 14.0 7.6 14.0 7.8 14.0 8.8 14.2 9.0 15.5 8.9 15.0 8.8 14.4 9.0 15.5 9.5 17.2 9.4 16.0 9.0 15.8 9.1 16.2 8.7 16.3 9.4 16.9 9.4 16.0 9.0 15.1 8.7 16.3 9.2 15.7 9.1 15.2 9.2 16.4 9.4 16.5 8.6 15.2 8.7 15.7 9.0 15.9 8.9 14.6 9.0 16.3 9.0 16.1 8.6 16.4 9.2 14.3 8.6 14.1 9.2 15.1 8.6 13.9 8.6 15.0 8.7 16.2 9.0 16.1 9.1 15.2 9.2 15.1 9.2 14.0 9.0 14.4 9.0 15.5 8.6 14.4 8.6 15.0 8.7 15.2 9.1 16.5 8.8 15.5 8.8 14.5 9.1 15.3 8.6 15.9 8.6 15.1 9.2 15.3 9.2 15.5 8.6 14.6 8.6 15.3 9.2 15.3 9.0 15.6 9.1 15.4 8.7 17.0 8.8 14.8 9.0 15.5 9.2 15.8 9.0 15.5 9.0 15.5 8.6 14.1 8.7 15.6 9.1 16.2 8.8 15.1 8.6 14.9 8.7 14.9 8.7 14.7 8.8 14.8 8.9 14.8 9.0 14.9 9.1 14.6 9.2 15.8 9.3 15.8 9.4 15.9 8.6 16.0 8.7 16.0 8.8 16.1 8.9 16.6 9.0 16.6 9.1 16.7 9.2 16.8 9.3 16.9 9.4 17.0 8.8 17.2 9.3 17.6 8.6 14.5 8.7 14.5 8.8 14.2 8.9 15.0 9.0 15.0 9.1 14.8 9.2 14.4 8.9 15.1 8.6 17.4 10.1 16.7 9.8 15.4 9.5 14.8 9.6 15.0 9.8 15.2 9.8 16.0 9.6 15.5 9.8 15.2 9.5 17.0 9.5 17.8 10.4 16.7 9.5 16.7 10.5 17.1 9.8 16.4 9.5 15.5 9.6 16.0 9.7 16.2 9.8 16.2 9.9 15.8 10.0 16.4 10.1 16.5 9.5 15.7 10.2 16.8 10.3 16.9 10.4 15.9 9.5 14.8 9.7 15.6 9.9 14.5 10.1 15.8 10.3 14.9 10.3 15.3 9.6 14.4 10.5 17.5 202 LINEAR REGRESSION AND CORRELATION Ch. 7 - 1 1 1 1 1 ) - • • ^ '"*yprv' "x - • >• •••——f»»»»i<«f«|— ■«»■■■■■ •— — > — - - ••'^i^^^^Pri* • - - • • •• — f •j»— ^ — • — 1 1 1 1 J _ o (3 ® t^ 0) ft s? =s 5 -Tf X r/3 X! I 00 >^ -o -^ m M Si >, « ^D in «* I— I t— I t— I Sec. 7.2 DETERMINING LINEAR TREND LINES 203 results shown specifically in Table 7.22 and graphically as X's in Figure 7.21. The decision regarding linearity of trend and the esti- mation of any desired features of the true trend line (such as slope) must be accomplished from the information contained in the sample. If the trend of one variable with another is linear, the relationship between the two kinds of measurements, X and Y, is of the form Y = A -\- BX, in which, for illustration, Y stands for the 28- week weight of a certain breed of turkey and A" = the 16-week weight of TABLE 7.22 A Random Sample op w = 30 Pairs (X, Y) from Table 7.21 X Y X Y X Y X Y 4.8 12.9 7.2 13.3 7.9 15.9 8.8 14.4 6.5 14.8 7.0 15.0 8.0 14.9 9.2 15.7 6.4 14.4 6.9 14.7 8.1 15.0 9.0 15.9 5.5 13.5 6.8 15.1 8.0 13.8 8.6 14.4 6.1 13.4 7.2 13.3 7.6 14.9 9.1 15.3 6.0 13.6 7.5 14.3 7.5 14.1 9.0 16.6 7.4 14.3 8.5 15.1 7.5 13.9 9.2 16.8 7.0 14.5 8.5 15.2 the same turkey. If all the observed pairs of measurements {X, Y) in Table 7.21 satisfy a linear equation perfectly, all the points of Figure 7.21 will lie exactly on the same straight line; and the rela- tionship between the two variables will be perfectly linear. More- over, the equation of the line can be determined from the coordinates of any two distinct points. Such obviously is not the case in Figure 7.21 because errors of measurement and uncontrollable fluctuations in the 28-week weights of turkeys which weighed the same at 16 weeks of age must be averaged out before the trend appears to be linear. By contrast with situations met in elementary algebra, where the equation is given and all appropriate points lie on the line, the present situation starts with the points given from sample observa- tions, and the problem is to determine which straight line best fits these observations, and, it is hoped, best estimates the true linear regression line. Assuming that a set of observations really does follow a linear trend quite well, how can a specific equation of the form Y = A -\- BX be determined and also defended as the best straight line to be em- ployed under the circumstances? The answer to this question de- pends upon the interpretation of the word "best." One interpreta- tion, and the one most frequently accepted, can be illustrated by 204 LINEAR REGRESSION AND CORRELATION Ch. 7 means of the line drawn into Figure 7.22. Some points lie above this line, some lie below, at distances whose magnitudes can be measured by the lengths of the vertical lines which could be drawn connecting the points of the scatter diagram to the regression line. In a useful sense, the goodness of fit achieved by any line drawn among the points to depict their trend should be measured somehow in terms of the amounts by which the proposed line misses the points of the scatter diagram. y 18 17 - i,i6 - a> •g 15 ffiUl- 12 1 1 1 1 1 1 1 1 1 1 • ^^^^^^••^'^ — • • ^^ t^^"^ * — • • % tftf*' • • i^ • • • ^^ "^^^^^ • • ^^-^""^^^"^^ •• <»*^ ^^^^ 1 1 1 1 1 1 13 - ^^ 10 5 6 7 8 9 16-Week weight Figure 7.22. A random sample of 30 pairs of observations from the population of Figure 7.21 and Table 7.2L Free-hand line to indicate the trend as it might appear to the eye ( ). Line determined by the method of least squares ( ). It will facilitate the discussion to introduce some symbolism before presenting the specific methods to be used in the determination of the equation of the regression line. For a given value Xj of the measure- ment X, let the corresponding value of Y be called Yi if it was observed with Xi when the sample was taken. It will be designated as Y i if it is calculated from the equation of the regression line. Also, let the general linear equation relating Yi and Xi be written in the form (7.21) Yi = a + h{Xi - x),* where a and h are the two constants which must be determined in order to have a specific trend line for a particular scatter diagram. * This form and the notation do not agree entirely with some other textbooks, but they are used here for convenience. The {Xi — x) is Xi so that the subsequent formulas and discussions come quite simply from this form of the equation for t. Some authors use other letters than a and 6; and several others use h as herein, but their a = (above a) — bx. Sec. 7.2 DETERMINING LINEAR TREND LINES 205 As stated above, the a and b will be calculated in terms of the col- lective amount by which a line misses the points of the scatter dia- gram. The a and the b also are considered estimates of the popula- tion parameters a and (3 in the true linear regression equation, (7.22) F = a + ^{X - m), where n = the true mean of the X's. For reasons given earlier for using 2(X — x)^ to measure variation about the mean instead of either 2(X — x) or w| X — ^ |, it is found advisable to use 2(F — Y)^ to measure the scatter of the Y's about the trend line. Therefore, the best-fitting straight line has been chosen as that one for which the 2(F — 1')" has the least possible size. This action makes the standard deviation about the trend line as small as possible. The mathematical process of achieving this goal produces formulas from which the a and the b can be computed. When these values are substituted into formula 1 .21 a. specific equation of a regression line is obtained. This line will have the property that the standard deviation about it is as small as it is possible to make it for any straight line. In other words, the variability of the F's has been reduced as much as it can be in consideration of their linear trend with X. The formulas for a and b are as follows : (7.23) a = y, and h = = — . 2(X,— J)2 2(1-2) where y = mean of the F's in the sample and y = the deviation of a F from the mean, y. The '^{xy) — which the student has not met before in this book — is (Xi — x)(Yi — y) + (X2 — ^) ( F2 — y) + • • • + {Xn - x)(Yn -y) = xxyi -f .T22/2 H h Xnyn* For the data of Table 7.22, a = ^{Y)/n = 439.0/30 = 14.63, b = 2(a;2/)/2(x2) = 23.0200/37.912 = 0.6072, and x = 2(X)/w = 226.8/30 = 7.56. Therefore, since y -\- b{X - x) =bX + {y -bx), (7.24) f = 0.6072X -f 10.04. Students in a statistics course are in an unusually fortunate position because when they take samples from laboratory populations they can see readily how well, or poorly, certain features of their samples agree with the corresponding features of the populations being sampled. * Experience shows that beginners in this field tend to think that S(x?/) = S(x) -2(2/). If the reader will recall that S(j) = 2(A' — x) = for any set of data — and likewise for 2(i/) — it becomes apparent that S(x2/) is not the same as S(x) • 2(?/) or it always would be zero. This obviously is untrue. 206 LINEAR REGRESSION AND CORRELATION Ch. 7 For example, the slope of the above sample estimate of the linear re- gression line is calculated to be 6 = 0.6072, whereas the true slope is known to be jS = 0.5000. In actual practice, only the h is known, and it is necessary to measure its reliability as an estimate of /3. This will be done later when the necessary techniques have been discussed; but it can be stated here that if the sample has been taken with the X's fixed — ^as suggested in Figure 7.21 — so that there is no sampling error in X or in x, the h as defined is an unbiased estimate of the parameter ^. The value Y which is obtained from formula 7.24 by substituting a particular value for X is described as the estimated average Y for that X. For example, if X is taken as 5, Y = 0.6072(5) + 10.04 = 13.1, ap- proximately. By reference to Figure 7.21 we learn that this estimate is somewhat low because the true average Y for turkeys weighing 5 pounds at 16 weeks of age is 13.5 pounds. If X is taken = 8, F = 14.9 pounds, which is nearer to the true average Y of 15 pounds than was obtained when X = 5 and the true Y was 13.5 pounds. It will be seen in a later discussion that greater accuracy in estimating the true average Y is to be expected for X's near the mean X. There often are more sample data near the mean ; but also errors in estimating the j8 will cause the ends of the trend line to be swung farther from the true position than is the middle of the line. In the above example the slope was h = 0.6072 instead of /3 = 0.5000; hence the line deter- mined from the sample is too steep and therefore too low at the left- hand end. This appears to be the major reason why the estimate of the true average F for X = 5 was too small. Of course, the general height of the sample line must be in error to some extent, and this also contributes to the inaccuracy of any estimate made from the sample trend line. The method described for obtaining the straight line which fits a linear trend best is called the method of least squares because it makes the sum of squares of the vertical deviations of the points of the scatter diagram from the regressioTi line the least it can be made for any straight line. Table 7.23 has been prepared to illustrate specifically the meaning of this minimization. Columns 6, 7, and 8 were obtained from the equation given over the right-hand side of the table. This equation represents a straight line which appears to the eye to fit the trend of the scatter diagram about as well as the line obtained by the method of least squares, as can be verified from Figure 7.22, which shows both lines. It should be noted that the total of the fifth column of Table 7.23 is less than that of the eighth column. This will always be true no mat- ter which straight line is used to obtain ?y as long as the equation is Sec. 7.2 DETERMINING LINEAR TREND LINES 207 TABLE 7.23 Illustration of Some Features of the Method of Least Squares Using Data of Table 7.22 Met hod. of Least Squares Freehand Straight Line f = 0.6072Z + 10.04 F. = 0.88Z + 7.92 X Y Y Y- Y (F - fy- T ^'i Y - Yj (F - Yjy (1) (2) (3) (4) (5) (6) (7) (8) 4.8 12.9 12 .95 -0.05 0.0025 12 .14 +0.76 0.5776 6.5 14.8 13 .99 +0.81 0.6561 13 .64 + 1.16 1.3456 6.4 14.4 13 .93 +0.47 0.2209 13 .55 +0.85 0.7225 5.5 13.5 13 .38 +0.12 0.0144 12 .76 +0.74 0.5476 6.1 13.4 13 .74 -0.34 0.1156 13 .29 +0.11 0.0121 6.0 13.6 13 .68 -0.08 0.0064 13 .20 +0.40 0.1600 7.4 14.3 14 .53 -0.23 0.0529 14 .43 -0.13 0.0169 7.0 14.5 14 .29 +0.21 0.0441 14.08 +0.42 0.1764 7.2 13.3 14 .41 -1.11 1.2321 14 .26 -0.96 0.9216 7.0 15.0 14 .29 +0.71 0.5041 14.08 +0.92 0.8464 6.9 14.7 14 .23 +0.47 0.2209 13 .99 +0.71 0.5041 6.8 15.1 14 .17 +0.93 0.8649 13 .90 + 1.20 1.4400 7.2 13.3 14 .41 -1.11 1.2321 14, .26 -0.96 0.9216 7.5 14.3 14 .59 -0.29 0.0841 14, ,52 -0.22 0.0484 8.5 15.1 15.20 -0.19 0.0361 15.40 -0.30 0.0900 8.5 15.2 15, ,20 15.40 -0.20 0.0400 7.9 15.9 14, ,84 + 1.06 1.1236 14, ,87 + 1.03 1.0609 8.0 14.9 14, ,90 14, ,96 -0.06 0.0036 8.1 15.0 14, ,92 +0.08 0.0064 15.05 -0.05 0.0025 8.0 13.8 14, ,90 -1.10 1.2100 14. ,96 -1.16 1.3456 7.6 14.9 14, ,65 +0.25 0.0625 14. ,61 +0.29 0.0841 7.5 14.1 14, ,59 -0.49 0.2401 14. 52 -0.42 0.1764 7.5 13.9 14, ,59 -0.69 0.4761 14. ,52 -0.62 0.3844 8.8 14.4 15, ,38 -0.98 0.9604 15. ,66 -1.26 1.5876 9.2 15.7 15. ,63 +0.07 0.0049 16.02 -0.32 0.1024 9.0 15.9 15, ,50 +0.40 0.1600 15. ,84 +0.06 0.0036 8.6 14.4 15.26 -0.86 0.7396 15.49 -1.09 1.1881 9.1 15.3 15.57 -0.27 0.0729 15. 93 -0.63 0.3969 9.0 16.6 15.50 + 1.10 1.2100 15. 84 +0.76 0.5776 9.2 16.8 15.63 + 1.17 1.3689 16.02 +0.78 0.6084 Sums 12.9226 - 2(F = Y)' 15.8926 not obtained by the method of least squares, and as long as sufficient accuracy is kept in the calculations to pick up small differences. This is the basis for the statement that the method of least squares makes the standard deviation of the F's from the trend line as small as possible for any straight line, which is a strong argument for the use of this line in practice. 208 LINEAR REGRESSION AND CORRELATION Ch. 7 PROBLEMS 1. Obtain the linear equation whose graph fits the points of Figure 7.1 IB best in the sense of the method of least squares. Graph the line on the scatter diagram and indicate graphically those deviations whose sum of squares is the least possible for any straight line. 2. Do as in problem 1, for Figure 7.1 lA. Also compute S(y — 1')^- Ans. Y = 2.16X + 4.54; S(F - Yf = 9.77. 3. By what average amount would you expect Y to increase for a unit increase in A' if the data corresponding to Figure 7.11 A constitute a representative sample of some two-variable population? 4. Compare the 2(F — Y)~ and S(F — y)"^ for the data of Figures 7.11B and G. What conclusions can you draw? 5. Use the method of least squares to estimate for the data of Figure 7.1 IB the average value of Y for A' = 1.5, 2.5, 3.5, and 4.5, respectively. 6. Make up two sets of 10 pairs of observations each and such that h is about 2 in one set and about —3 in the other. 7. Write down the equation of a trend line with slope = 5 and for which Y = 10 when X = 4. Graph this line, and then construct a scatter diagram which fits the trend and has 2(F - F)^ = 50. 8. Do as in problem 7, with slope = — 3 and everything else the same. 9. Assign row and column numbers to the data in Table 7.21. Then draw two random samples of 30 pairs each — as in Table 7.22 — and obtain the least- squares regression line from each sample. Plot these lines on their correspond- ing scatter diagrams and discuss their differences. (Round off each X to the nearest pound before doing your computations.) 10. "Cull" the flock of Table 7.21 at 16 weeks of age by eliminating all turkeys which weighed under 6 pounds at that time, then do as in problem 9. Would § still be 0.5 for this population? 7.3 MEASUREMENT OF THE VARIATION ABOUT A LINEAR TREND LINE DETERMINED BY THE METHOD OF LEAST SQUARES If measurements are taken on but one normally distributed variable, Y, the variability, or dispersion, of the Yi should be measured by means of the standard deviation about the mean, and estimated from sy = V 2(Fi — y)'^/{yi — 1) because sy^ is an unbiased and highly efficient sampling estimate of ay^. The variation measured by sy is then considered to be sampling variation. However, if for each Yi there is an associated measurement, Xi, and if the X's and F's tend to be linearly related, not all the apparent variability among the Yi should be assigned to mere sampling errors. Part of it can be ac- counted for in terms of the varying Xi associated with the Yi. For example, if Y tends to increase about 5 units for each unit increase in the magnitude of X the Y associated with X = 10 is expected to be Sec. 7.3 STANDARD DEVIATION ABOUT TREND LINE 209 about 15 units greater than the 1' for X = 7; hence some of any ob- served difference between the Y for X = 7 and the Y for X = 10 can be accounted for and need not be considered as sampling error. Graphically the preceding remarks mean that if Y and X can be considered to be Hnearly related the standard deviation of the F,; should be calculated from the trend line rather than from the horizon- tal line: Y = y. This means that the quantity 2(F — F)^ should be employed in this calculation instead of 2(F — y)^. However, the divi- sor in this calculation will not be (n — 1) as it is for sy, above. The divisor needed in the computation of the standard deviation about the trend line is {n — 2). The reason for this cannot be given conclusively without mathematical analysis which is not appropriate to this book; but it can be rationalized in the following manner. Suppose that a sample of 5 observations on X and F simultaneously were as follows: X: 12345 x = 3 F: 54687 y = 6 It is readily determined that F = 0.80.Y + 3.6; hence the following table can be set up for purposes of illustration : X 1 2 3 4 5 F 5 4 6 8 7 f 4.4 5.2 6.0 Y, Y, n 0.6 -1.2 ? ? (Yi What are the deviations from the trend line for X = 4 and A^ = 5, respectively? The fact that ^{Yi — Y i) = will be found to account for one of these deviations. The fact that h = 0.80 will allow the de- termination of the second unknown deviation. Let the unknown deviations F4 — Y^ and F5 — F5 corresponding to X = 4 and X = 5 be denoted by v and w, respectively. It follows that S(Fi - Yd = 0.6 + (-1.2) -f + y + w; = 0, which reduces easily to (7.31) v + w = 0.6. The slope of a straight line can be computed by determining the amount by which F changes for any chosen change in X, and by taking the ratio of the former to the latter. For example, if it is determined by measurement on the graph or by substitution into a mathematical formula that for the interval from X = 1 to X = 5 the height of the / 210 LINEAR REGRESSION AND CORRELATION Ch. 7 straight line above the horizontal axis increases from 10 to 30, the slope of this line is measured by 20/4 = 5. Hence, in the situation of the preceding paragraph the slope is given by (}^5 — 3'4)/(X5 — X^). But (X5 — X4) = 1, and the slope is known to be 0.80; hence (F5 — Y4) -^ 1 = 0.80. In order to transform this equation into one involving v and w consider the following two equations: F5 — F4 = F5 — F4 F5 - F4 = 0.80. When the left and right members of the second equation are sub- tracted from the corresponding members of the first equation, it is found that (Fs - F5) - (F4 - F4) = F5 - F4 - 0.80; but F5 — 1^5 = w, F4 — F4 = V, and F5 — F4 = — 1 ; therefore, (7.32) V -w = 1.8. When equations 7.31 and 7.32 are solved simultaneously it is found that y = 1.2 and w = —0.6. Hence, two of the deviations from the trend line can be calculated from the size of b and from the fact that 2(F — F) =0. Although there are five actual deviations from the linear trend line, only three (any three) of them actually should be considered chance deviations from the regression line determined from the sample. Hence, in the present problem, n — 2 = 5 — 2 = 3 will be used as the divisor of S(Fj — F^)^ in the computation of the stand- ard deviation about the linear trend line. This divisor, n — 2, is generally called the number of degrees oj free- dom for the estimated standard deviation about the linear trend line, just as the number, n — 1, is the number of degrees of freedom for the estimated standard deviation, sy, about the mean. With the above discussion as a background, the formula for the estimated standard deviation of the Yi about the trend line becomes (7.33) Sy., = VS(F,- - F,)V(w - 2) , wherein the symbol, Syx, is read "s sub y dot .t." For the data used as illustration in this section, 2(Fj — F^)^ = 3.60, n — 2 = 3; hence Syx = V 3.60/3 = 1.10. This is a measure of the variation among the F-measurements which remains unexplained even after the linear trend with X has been taken into account. When the trend with X is ignored, sy = v 10/4 = 1.58; so Sy.x is 0.48 of a unit smaller than sy. In other words, the variability of the Yi (as measured Sec. 7.3 STANDARD DEVIATION ABOUT TREND LINE 211 by the standard deviation) has been reduced 100(0.48)/!. 58 = 30.4 per cent by taking the linear relation between the two measurements into account statistically. Such success in accounting for part of the variation among the measurements, Yi, clearly is important in statisti- cal analyses because the only occasion for such analyses arises as a result of variability among numerical measurements. The standard deviation about the trend line, Sy^, also is specifically useful in certain applications of linear trend analysis, two of which will be considered. The regression coefficient, 6, estimates the average change in the F-measurement for each unit increase in the X-measure- ment. Its accuracy as such a measure is of interest, and its accuracy is measured by its standard deviation. The standard deviation of h is shown in more advanced statistics courses to be (7.34) Sb For the data of Table 7.22: S(F - Yf = 12.9226, Xx^ = 37.9120, n = 30, and hen ce Sy:, = \/l2.9226/28 = 0.679. Therefore, st = 0.679/V37.9120 = 0.110, approximately. It can be shown that the ratio, (7.35) t= {b - I3)/Sb, where {3 = the true regression coefficient which is estimated by b, follows the same ^-distribution as that summarized in Table IV with n — 2 degrees of freedom. Therefore, a confidence interval can be computed for f3, and it can be interpreted in the manner previously shown. For Table 7.22, the 95 per cent confidence interval is ob- tained as follows: -2.05 < (0.6072 - ^)/0.110 < +2.05 will be a true inequality for 95 per cent of all samples with 28 de- grees of freedom. Hence the 95 per cent confidence interval is found to be as follows after some simplification of the preceding inequality: (7.36) 0.38 < i3 < 0.83. It would be concluded in practice that the slope of the true linear regression line is some value between 0.38 and 0.83, but it is recog- nized that there are 5 chances in 100 that the sample has led us to a false statement. A more useful statement might be that it is esti- mated from (7.36) that a turkey which is one pound heavier than another at 16 weeks of age will, on the average, be 0.38 to 0.83 pound 212 LINEAR REGRESSION AND CORRELATION Ch. 7 heavier at 28 weeks of age. That is, the lighter turkeys at 16 weeks tend to catch up some, but they usually remain 0.38 to 0.83 pound lighter at 28 weeks for each pound that they were lighter at 16 weeks of age. Another application of linear trend analysis which makes use of Syx is one in which Y is to be estimated for some unobserved value of X; for instance, for X = 9.5 pounds at 16 weeks. If X is set equal to 9.5 in formula 7.24, f = 0.6072(9.5) + 10.04 = 15.8 pounds at 28 weeks of age. How reliable is this estimate? A look at the scatter diagram leaves only the impression that this estimate should be fairly reliable; hence a more specific measure of its accuracy is needed. The standard deviation of 1' is given by the following formula: (7.37) sf = Sy.^Vl/n + {X - x)yi:x^ , where X is the value used to calculate the Y. This estimate of the standard deviation of Y is based on n — 2 degrees of freedom, as ex- plained earlier. It will be convenient in subsequent discussions to add "with n — 2 D/F" after an estimate of this sort to indicate the num- ber of chance deviations upon which the estimate is based. In the example considered in this paragraph, sy = 0.679Vl/30 + (9.5 - 7.56)737.912 = 0.679(.364) = 0.247, with 28 D/F. This standard deviation applies when the X's have been chosen in advance and are not subject to sampling error. As noted earlier, the b is then an unbiased sample estimate of the population parameter, fi. Under these circumstances the formula 7.37 can be partially ration- alized in lieu of a more rigorous demonstration of its validity. The Y for a particular X, say Xj, is obtained from Y = y -]r [Xi — x)b = y + Xib. Hence the variance of Yi is obtained from the variance of a sum, y + Xib, in which the Xi is a fixed number. The variance of y for this particular X will be one-nth of the variance about the trend line, or Syx^/n. The variance of b is s^/.^^/Sx^, as noted earlier, and Xi is a constant; hence the variance of Xib = Xi'^ •Sy.J^/'^{x^). If the variance of the sum, y -\- Xib, is just the sum of the variances of those two terms, it follows that n ^(^ ) so that Sy = Sy.x'\/\/n + Xi~/'Ei{x^) , as in formula 7.37 for a particu- Sec. 7.3 STANDARD DEVIATION ABOUT TREND LINE 213 lar X = Xi. It is true that the variance of ^ + Xih is the sum of the variances of the two terms, but this will not be proved here. It can be shown that the ratio •& (7.38) t = —^ Sy where m^x is the true average Y for the given X, follows the ^-distribu- tion with n — 2 degrees of freedom. This fact makes it possible to place a confidence interval on Hyx with any appropriate confidence coefficient. The meaning of the M)/x can be made clearer by reference to Figure 7.21. For each particular X there is a frequency distribution of the corresponding F's. This distribution of F's has a true arithmetic mean, which is the Hyx for that X. If we wish to make an interval estimate which applies to an indi- vidual rather than to a group mean, we must take account of the greater variation exhibited by such individuals as compared to the group. For example, suppose that a study has been made of the rela- tionship between the ages of Kansas females and their basal metabolism rates as expressed in calories per square meter of surface area per hour. It is supposed that the age interval chosen is such that a linear relation- ship exists between these two variables, and that the least-squares equation for Y has been obtained from a sample. Suppose, further- more, that the equipment needed to determine the basal rate is not available in a certain area, and a Kansas woman 25 years of age wishes an estimate of her basal metabolism rate as a matter of interest. The best point estimate is the Y calculated for X = 25 ; but when an inter- val estimate is needed — and it is more useful in the present problem — account must be taken of the fact that this woman is not supposed to be an average person representing all those who are Kansas females 25 years of age. She is regarded as one particular person who wishes an estimate of her own basal rate. In this circumstance the variance of Y used earlier in this section is not correct because it includes only two sources of variation : one from the mean and one from the sampling regression coefficient. In the present problem a third source must be included, namely, individual variation about the mean. When the particular X has been taken into account, this additional variance is just Syx^; hence — again it turns out that this can be added to the other two components — we obtain the following formula for the vari- ance of the Y for an individual : ..2 _ 2 1 (X - x)^ 1 H \-- ^ n 2(a- ) . 214 LINEAR REGRESSION AND CORRELATION Ch. 7 When this change in sf is made in formula 7.37 we obtain the formula for Sy which is employed in the following ^-ratio: (7.38a) t = ^^^, where Hy.xi = the true F-value for the ^th individual for whom X = Xi. Formula 7.38a and the usual methods make it possible to obtain a confidence interval on fXyx^ with any specified confidence coefficient. It should be both clear and reasonable that such a confidence interval will be longer than a corresponding one from formula 7.38 because the standard deviation is larger. Problem 7.31. You are about to buy one turkey which weighs 6.5 pounds at 16 weeks of age, and you are going to keep it until it is at least 28 weeks of age. What is the 94 per cent confidence interval on its 28-week weight, assuming it comes from the population sampled in Table 7.22? It was seen in the discussion of Table 7.22 that Y = 0.6072X + 10.04, which = 13.99 pounds for X = 6.5 pounds. Also, Sy.^, = 0.679 pound, and 2(a:^) = 37.9120; hence by formula 7.39, after taking the square root, sp = O.679V1 + 1/30 + (6.5-7.56)737.9120 = 0.700 is the standard deviation of F f or X = 6.5 pounds at 16 weeks of age. Therefore, t = (13.99 — iJiy.x^)/0.700, and the 94 per cent confidence interval is derived from -2.19(0.700) < (13.99 - fiy.^) < 2.19(0.700). It is found that CI94 is 12.5 < Mj/xi < 15.5, to the nearest one-half pound. Notice that the situation just considered clearly is one in which the confidence interval for a particular turkey is required. You do not have a group of turkeys so that the high 28-week weights of some of them can be expected to offset the low 28-week weights of others. Therefore, you must face the fact that this particular turkey's weight at 28 weeks of age may be quite low, as well as quite high, for turkeys weighing 6.5 pounds at 16 weeks of age. Problem 7.32. Suppose that problem 7.31 is changed to state that you have bought a rather large flock of 6.5-pound turkeys each 16 weeks of age. Compute the CI94 appropriate to this new situation. Sec. 7.3 STANDARD DEVIATION ABOUT TREND LINE 215 The only change in the computations is that the standard deviation of Y now is Sy = Sy.^Vl/n-]- x^/'2{x^) = 0.679(0.251) = 0.170 pound instead of the 0.700 pound obtained for the individual. It follows that the required confidence interval is: CI94: 13.5 pounds < Hyx ^ 14.5 pounds, to the nearest one-half pound. This is a narrower interval than is obtained for problem 7.31, as should be expected. PROBLEMS 1. Compute Sy and Sy.x for the data for Figures 7.11^ and G and relate their comparative sizes to the scatter diagrams. 2. Work problem 1 for Figures 7.11fi and G. Does the downward trend of the points on a scatter diagram, as contrasted with an identical upward trend, have anything to do with the comparison between sy and s^.x? Ans. B: sy = 16.0, Vx = 3.1; G: sy = 9.2, Vx = 9.8. 3. Referring to Figure 7. HE and the associated data, compute and compare SY and Sy.x as in problem 1. Could you have predicted from the scatter diagram that they would be of essentially the same magnitude? Why? 4. By visual inspection of Figures 7.11C, D, and F what do you conclude about the comparative sizes of sy and Sy.x for each figure? 5. For the two sets of data in problem 7, section 7.1, compute the percentage reduction in the standard deviation of the Fj achieved if variability is measured about the linear trend line rather than about the line Y = y for each set. Discuss the two results obtained for the two sets in terms of the curvilinear trend in one set. 6. Use the data of problem 1, section 7.1, to estimate the average Y for A' = 13. Also compute the standard deviation of this estimate, first considering a group with X = 13 and then for an individual with A' = 13. Ans. Y = 345.7 when X = 13; sy = 0.53; 1.01. 7. Use the data of set B, problem 7, section 7.1, to place 92 per cent confi- dence Hmits on the log(weight) of the 7-day-oId bee larvae of the kind repre- sented by this sample. Interpret these limits. 8. Compute the 99 per cent confidence interval on j3 for problem 1, section 7.1, and draw appropriate conclusions. Ans. Cl^g: 21.2^/3^^23.0. 9. The following data express the farm population (as defined for the 1950 census) as a percentage of the total U. S. population: Year: 1940 1941 1942 1943 1944 1945 1946 1947 1948 Per Cent: 21.8 21.5 20.6 18.8 1949 16.7 17.7 1950 16.0 17.3 1951 15.0 17.9 17.9 16.9 These are not sampling data, but the fitting of a trend line to these data may be useful anyway. For example, if the war years, 1943 to 1945, inclusive, are 216 LINEAR REGRESSION AND CORRELATION Ch. 7 ignored, the downward trend in percentage farm population is quite closely represented by a straight line. Make a scatter diagram of the above data, omit 1943, 1944, and 1945 from further consideration, fit a linear trend line by the method of least squares, and then read from the line the approximate per- centages for the years omitted. Would tlie discussions of estimates of the standard deviation and the formulas given in this book for them be appropriate here? Give reasons for your answer. 10. Referring to problem 9, could you use the equation obtained there to predict satisfactorily the percentage farm population for 1953? For 1960? Justify your answers. 7.4 COEFFICIENTS OF LINEAR CORRELATION It is not always desirable — or even appropriate — to obtain an equa- tion for the linear relation between the two types of measurements being studied, as was done earlier in this chapter. It may be better to describe the relationship as linear, and to give a standard, unitless, measure of its strength, or closeness. This is the purpose of a co- efficient of linear correlation. Although correlation coefficients are widely used, and often with- out attention to the satisfaction of necessary assumptions, it should be kept in mind that, strictly speaking, both X and Y must be random variables which follow normal frequency distributions. This will be assumed to be true in the following discussion of this section. It has been seen that the variance of the observed F's about the least-squares regression line depends on the size of w(F — Y)^, in which Y = y -\- bx. Hence the magnitude of this variance depends on ^^Y - y - hx)^ = ^{y - hx)^ = 2(r - 2bxy + 6V). But the last summation can be computed in three parts as follows: 2(2/2 - 2hxy + 6V) = 2(^/2) - 2b^ixy) + b^i:x^ = 2(7/2) _ 2 Kry) V/'^2 (X^)J [^(xy)] + -,•■> 2(.r2)J .^r^-2 2(.r^), hence, ^ ^ _ 2[^(xy)f [2(:ry)]2 r2(.ry)]2 (7.41) S(F - Yf = 2(2/2) Lj\X ) Sec. 7.4 COEFFICIENTS OF LINEAR CORRELATION 217 Therefore, it is clear that the observed variabihty of the F's about the regression Hne will be large or small according to the size of [^(xy)^ -7- 2(x^) compared with the size of I^(i/). If [2(.r?/)]^/2(a:^) is multi- plied by 2(y^)/w(?/^) — which equals 1 and only changes the form of the quantity by which it is multiplied — it follows from (7.41) that the 2(F — Y)^ can be expressed as follows: s(F - yy = 2(2/^) J _ [^(^y)f V/.2 (•r^)-2(2/^)J [2(x2/)]2 Clearly, the quantity — has the following statistical features : 2M-2(?/^) (a) Its value cannot be less than zero nor more than one because it is essentially zero or positive, and if it exceeded one the sum of squares of deviations from the trend line would be negative, which is absurd. (6) If this quantity is near zero there is about as much scatter of the sample points about the trend line as about the horizontal line Y= y; hence there is little or no linear trend. (c) If this quantity is near one there is very little scatter about the regression line; hence the sample points lie quite close to that line. (d) As the size of this quantity varies— for different samples — from zero to one the scatter of the sample points about the least- squares regression line varies from a completely trendless, shot-gun, pattern to a perfect fit to a linear trend line. (e) This quantity is unitless so that the features noted above are true regardless of the units in which Y and A' are measured. (/) In its present form this quantity cannot distinguish between positive and negative slopes of trend lines, but its square root would have the same sign as b and would make this distinction if the square root of the denominator were always taken as positive. It follows then that the square root noted in /, ^(xy) ^^•''^ ' ^ VWyW)' is a unitless number within the range —l^r^-\-l which indicates the direction and strength of the observed linear trend. This number, r, is called the product-moment coefficient of linear correlation be- tween two measurements A' and Y. It obviously is subject to sam- pling variations and therefore has a sampling distribution. It is a sampling estimate of a corresponding population parameter indicated 218 LINEAR REGRESSION AND CORRELATION Ch. 7 by the Greek letter p (rho), which succinctly describes the degree of scatter of the population points about the true linear trend line as, for example, in Figure 7.21. If p = 1, all the points will lie on the regression line. Since they do not — in Figure 7.21 — there is sampling error in the estimation of p, and hence the r varies from sample to sample. This is similar to the situation when the true regression coefficient, p, was being estimated from samples. If the p is zero, all the sampling estimates ri will not be zero but will have a sampling distribution which is nearly normal in form. In such circumstances it can be shown that the following ratio follows the ^-distribution with n — 2 D/F. Thus r rwn — 2 (7.43) i = -,,=, = ^7^= can be used in the usual manner to test the hypothesis B.^{p = 0). As was seen in earlier discussions, //o will be rejected whenever the size of t becomes so great that it is unreasonable — according to some predetermined standard — to believe that this t is the product of sam- pling variation. It is more difficult to place a confidence interval on p than on /? because r is not nearly normally distributed when p 7^ 0. This process of computing a confidence interval on p will be discussed and developed somewhat heuristically by means of the empirical data found in Table 7.41. These data were obtained by drawing random TABLE 7.41 Observed Sampling Distributions of r and z = 1/2 Log^ [(1 + r)/(l — r)] FOR n = 12 AND p = +.749 r-interval / 0-interval / .890-1.000 7 1.70-1.89 1 .790- .889 32 1.50-1.69 3 .690- .789 48 1.30-1.49 9 .590- .689 52 1.10-1.29 22 .490- .589 23 0.90-1.09 43 .390- .489 12 0.70-0.89 56 .290- .389 8 0.50-0.69 29 .190- .289 5 0.30-0.49 19 .090- .189 2 0.10-0.29 7 -.010- .089 1 -0.10-0.09 1 Total 190 Total 190 Sec. 7.4 COEFFICIENTS OF LINEAR CORRELATION 219 samples of pairs of values of A' and Y from Table 7.21 for which p = .749. This population was considered to be approximately a normal bivariate population. The 190 sampling r's thus obtained are summarized in Table 7.41 along with the corresponding 2's (see discussion of z below Figure 7.41). The distribution of r for a p so large as this is definitely skewed, as can be seen to some extent in Table 7.41 and in Figure 7.41. ou 1 ' T 1 1 1 1 ' ' 1 1 1 1 1 1 1 1 1 1 50 — / \ / \ _ / \ \ ~ / / / / \ \ ''~ Q,40 — / / \ \ — / / \ \ / / \ \ — ;*5 / / \ \ >^ 30 "'^ — \j \ \ — \ \ Z3 — /) \ — 0) i 20 - / / / / / / \ - 10 x* y / X \(r) "^^.«- 1 f-^ H 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ~i — -.20 .20 .40 .60 .80 rorz 1.00 1.20 1.40 1.60 1.80 Figure 7.41. Sampling frequency distribution of the correlation coefficient, r, and of the corresponding z = (1/2) log^ [(1 + r)/{\ — r)]. n = 12. It was found by R. A. Fisher that under these circumstances it is helpful to use the following function of r: 1 + r Ll - rJ (7.44) z = (1/2) loge (|-3^) = (2.30259/2) logio r 1 + r1 = 1.1513 logioh^ • Ll — r. because its sampling distribution is essentially normal in all important features even when p is definitely 9^ 0. Moreover, its variance is given by a^ = \/{n — 2>). This is not a sampling estimate but the true variance of z. It follows that, as a good approximation, the quantity y = {z — Zp)/az, where Zp is the z corresponding to p in (7.44), is normally distributed. Hence, Table III gives the probabili- ties needed in tests of hypotheses regarding p or in the calculation of 220 LINEAR REGRESSION AND CORRELATION Ch. 7 confidence intervals on p. For example, consider the first sample drawn for Table 7.41. The n = 12 and r = .668; therefore, by formula 7.44, /1.668\ z = L1513 logio = 1.1513 logio 5.024 = 0.807, and \0.332/ a, = 1/V9 = 0.333. Then, since y = (0.807 — 5:p)/0.333 is a member of a standard normal population, the probability distribution of Table III can be used. If a confidence coefficient .95 is chosen, the inequality 0.807 - Zp -1.96 < < +1.96 0.333 ~ requires that 0.154 <Zp< 1.460 unless a 1 in 20 chance has occurred in this sample. The corresponding 95 per cent confidence interval on p is obtained by using formula 7.44 and solving for the p, which now replaces r. Thus, 1 /I + M 2pj = lower limit = 0.154 = - loge I 1 ; or 2 \1 — p\/ i±Zi = g0.308 jB^^^^ iQg^^ g0.308 ^ 308 logj^ e 1 - Pi = 0.308(0.4343) = 0.134; and 1+Pi Anti-log 0.134 = 1.36 = 1 -pi Hence 2.36pi = 0.36 so that pi = .155. Similarly p2 = upper limit of the 95 per cent confidence interval = .898; therefore, the 95 per cent confidence interval on p is .155 < p < .898, which is a very wide interval but does include the true p, known in this case to be .749. If a relatively narrow confidence interval is needed, it is apparent that a rather large sample must be taken. Figure 7.41 shows that the sample correlation coefficient varies over a considerable range even when p is as large as .749. As a matter of fact one sample r out of 190 was negative in spite of the relatively high positive correlation. This figure shows also, to a useful degree, the normalizing effect of the transformation z = -- loge ( :, ) • Given ^•''^'(r^i)- Sec. 7.4 COEFFICIENTS OF LINEAR CORRELATION 221 a large number of sample correlations, the s-curve would become approximately normal in shape, as can be imagined from Figure 7.41. Problem 7.41. Finney and Barmore (Cereal Chemislry, Vol. 25 [1948], page 299) have reported that the linear correlation between the per cent of protein in Nebred wheat flour and the loaf volume of bread baked therefrom was r = .94 on a sample of 30 pairs of measurements. What useful information does this provide? The mere fact that r^ = (.94)2 ^ gggg ^^jj^ ^^ ^^^^ gg 3^ p^^. ^.^^^ of the original sum of squares of the loaf volumes {Y) about their mean, y, can be associated with the linear increase of that measure- ment with increasing protein concentration in the flour {X). Loaf volume is an important factor when the quality of bread is judged, and it is important to know what affects it. It is inconceivable that such a large correlation coefficient would be obtained accidentally on thirty random observations; but, to illustrate the method, the hypothesis //o(p = 0) will be tested. It is seen that .94 .94 t = I = = 14.5, 28 D/F. '1 - .883G 0.065 28 Such a large t would occur by chance almost never; hence the hy- pothesis Hoip — 0) is decisively rejected. We know without even seeing the scatter diagram that the sample points lie closely about a linear regression line which has an upward trend. It also is ap- parent that the loaf volume from Nebred flour meeting the condi- tions of this experiment could be predicted quite accurately from a knowledge of its protein concentration. There are some circumstances under which it is desirable to deter- mine if two random samples probably were drawn from the same bivariate population as regards one, or both, of yS and p. For example, it might be of interest to learn if one method of raising turkeys pro- duces a more consistent relationship between the 16-week and the 28-week weights so that we could cull at 16 weeks of age with more confidence. Such an improvement in the relationship between these variables would indicate that the true coefficient of linear correla- tion, p, had been increased by the new methods. It also might be that superior poultry husbandry could increase the amount by which a weight advantage at 16 weeks of age would be followed by a weight advantage at 28 weeks of age. In the population considered earlier in this chapter, a one-pound advantage in weight at 16 weeks of aj wi/ 222 LINEAR REGRESSION AND CORRELATION Ch. 7 was reduced, on the average, to only a one-half-pound advantage at 28 weeks of age. Thus ^ was 1/2. It might be that the size of /? could be increased by superior breeding and handling. If two samples — from two methods of breeding and raising turkeys, for example — have resulted in the computation of 6i, 62, ri, and r2, the testing of the two hypotheses i/o(/8i = ^2) and Ho (pi = po) can be carried out as follows: For//o(i8i =^2): (a) Pool the S(Fi - F,)^ and the 2(Fj - Yj)^ from the two sam- ples, pool the degrees of freedom, and calculate 2(F, - n' + S(Fy - Yjf pooled Sy.^ = ^ ■ ni + n2 — 4 . 1 1 (6) Compute .. / — + S(x,2) S(x/) (c) Multiply the standard deviation from a by the result obtained in h. This is the estimated standard deviation of (61 — 62), w^hich will be called Sbi_b2- (d) Compute t = {bi — h2)/sb^_h^, assign it ni -f 712 — 4 degrees of freedom, and interpret as before with respect to the acceptance or the rejection of i/o(/3i = 182)- For Ho(pi = P2): (a) Transform the ri and the r2 to Z\ and 22, respectively, in the manner described earlier in this chapter. 1 1 (6) Compute Oz^-zi = \l h ^1 — 3 W2 — 3 (c) Calculate y = \ Zi — Z2 \laz^-z^ and consider this ratio as a normally distributed quantity in deciding whether or not it is so large that the hypothesis i/o(pi = P2) should be rejected. If it seems appropriate after a hypothesis has been rejected, con- fidence intervals can be determined for the difference /?i — ^2, but not for pi — p2. It is useful at times to have a convenient tabular procedure for computing h and r when the data are sufficiently numerous to justify the use of frequency distribution tables. Such data rarely would Sec. 7.4 COEFFICIENTS OF LINEAR CORRELATION 223 Oh IS O [i^ o n o O O OJ -^^ CJ 00 o CO GO -^ ci 00 .2P t-: -^ 00 <u i>. -*-' i>. I CD ^ CO -^ O 00 o ^ lO -^ CO 2S & CDOCOOOOOOOOOCOO 1— iCJOiCC 'OOClCiO I V I I I ^ CO <M ^ O --I <M CO -* lO r-( CO Tj< -* r^ (M o ^ CJi C5 CO CO ca (M l^ t^ ^ 00 (M lO ^ (MTfi'-iCO^iOC^'— I T-H CO 1— ( ^H CO ^H C<l ■-H -^ TfH C^ O >— I 0"*oqc<icooTj^oq(Ncq o6t-^CDCD^OlO-<:l^COCO(^^ 1— 1 J. 1— 1 s T-H 1 1 jj J I— 1 1 CO 1 1 I— 1 CO CO lo T— 1 T-H CO 1— 1 CO I— 1 I— 1 (M Oi 00 o 00 -H lO CO ^ (M 00 ,-H Tt< 00 CO to CO 1— 1 CO (>J CO (M 00 t^ 1— 1 t^ .-H lO lO CO o o o C5 ^ ^ 1 1 CO 1 o o 1 ^ (M CO ^ 1 CO 00 CO o 00 -fi 1 (M 00 CO (M 1 ^ 1 lO lO 1 ^ I I .^ L 224 LINEAR REGRESSION AND CORRELATION Ch. 7 come from sampling studies, but perhaps they occur in practice often enough to justify the inclusion here of a method for obtaining the r and the b. As in Chapter 2, the computations will be carried out in units of the class intervals. A two-way frequency distribution table is needed because two variables are involved. These matters, and others, are illustrated and discussed by means of 280 pairs of observations of 16- and 28-week weights of female turkeys similar to those studied earlier in this chapter. The symbol, A', is used to denote the 16-week weights and Y will stand for the 28-week weights, as before. Now that two variables are being considered simultaneously, frequencies in the A-classes will be symbolized by fx, those for the F-classes by Jy. When it is desirable to indicate both the X and the Y for a class of data, frx will denote the frequency in that "cell" in the two-way table. Also, there may be two different lengths of class interval, 7^ and Iy for X and Y, respectively. With these symbols in mind, the following formulas are seen to be analogous to those used previously for b and r: • , WxY-dx-dy) - [{^Sx-dx){'^fY-dY)]/^f ^ h = r :; ' and ^ifx-dx') - (2/x-rfx)V2/x same numerator as that above for h f ^ — — — • ■\/(same as denominator above) (same with Y replacing X) The data of Table 7.42 are arranged in a two-way frequency dis- tribution table to provide a relatively easy basis for calculating b and r from their formulas as given above. The following computations are derived from the summaries in Table 7.42: ^(fY-dy^) - {^jY-dyf/^fY = 800.5714, and its square FOot = 28.65. ^{fx-dx^) - {^jx-dxf/^Jx = 924.5679, and its square root = 30.41. ^{fxY-dx-dY) - [{^fx-dx){^fydY)]m = 448.2857. PROBLEMS 1. Calculate the r for the data of problem 1, section 7.L 2. Calculate as in problem 1 for the data of problem 2, section 7.L Given SA'2 = 18,484, SXF = 727.99, Sl^ = 28.6918. Ans. r = +.9Q. 3. Compute 2(F — Y)^ for the data of problem 3, section 7.1 by using the for- mula: 2(F - f)2 = (1 - r2) -2(2/2). Sec. 7.4 COEFFICIENTS OF LINEAR CORRELATION 225 4. In the formula of the previous problem, take 2(!/2) = 100 and plot the left member of this equation on the vertical scale against r on the horizontal scale. Take r from —1 to +1 by increments of 0.2. 5. Reynolds, Bond, and Kirkland {USD A Tech. Bull. 861) give the following information on the relation between the cost of hauling logs and the length of the haul in miles over high-grade dirt or gravel roads: Miles hauled: Cost/ 1000 cu ft ($): 1 0.35 2 0.44 3 0.53 4 0.62 5 0.71 6 0.81 7 0.90 8 0.99 9 1.08 Miles hauled: Cost/ 1000 cu ft ($): 10 1.17 11 1.26 12 1.36 13 1.45 14 1.54 15 1.63 16 1.72 17 1.82 18 1.90 Miles hauled: Cost/1000 cu ft ($): 19 2.00 20 2.09 Compute a coefficient of linear correlation between length of haul and cost per 1000 cubic feet of volume, and draw conclusions. Is this really a proper use of correlation analysis? Would a regression analysis be better? 6. The persons mentioned in problem 5 gave the following data on the cost of producing 1000 cubic feet of hardwood logs in relation to the breast-high diameter of the logs: Diam. (in.): 10 11 12 13 14 15 16 17 Cost ($): 12.70 12.63 12.38 12.03 11.62 11.32 11.10 10.84 Diam. (in.): 18 19 20 21 22 23 24 25 Cost($): 10.63 10.49 10.40 10.28 10.13 10.04 9.96 9.88 Make a scatter diagram of these data, compute r, and discuss it in terms of the scatter diagram. Given: Sx^ = 340; SXF = 3019.59; S(y2) = 14.3129. Ans. r = —.97. 7. Compute Sy.x and sy for the data of problem 5, with Y = cost per 1000 cubic feet. 8. Calculate as in problem 7 for the data of problem 6. Ans. Sy.x = 0.23; sy = 0.98. 9. The Yearbook of Labour Statistics for 1943-1944 gives the following average daily wages of Chilean copper workers, in pesos: Year: 1929 1930 1931 1932 1933 1934 1935 1936 1937 1938 1939 Wage: 11.89 11.26 11.75 11.33 12.80 13.31 14.77 16.37 21.31 23.20 25.34 S(x2/) = 155.74; SF = 173.33; 2(7)- = 2996.1887. Construct a scatter diagram of these data, calculate sy and Sy.x, and discuss their sizes relative to the graph. 10. Compute r for the data of problem 9 — ignoring the fact that the year is not a random variable — and relate the size of the r to the appearance of the scatter diagram. Let A" = 1 for 1929, 2 for 1930, etc. Ans. r = +.91. 11. Estimate the average wage for the year 1940 from the data of problem 9, using r in the computation of the standard deviation of this estimate. 226 LINEAR REGRESSION AND CORRELATION Ch. 7 12. Solve problem 1, and then test Hq{p = 0) and draw appropriate conclu- sions. Ans. r = .999, ,v ^ 10, reject Hq decisively. 13. For the data of problem 6 compute the CI,,-, on i3, and then interpret this interval in a practical way. 14. For the data of problem 6 compute the CIqq on p and interpret this in- terval. Ans. - .96 ^ p ^ - .80. 15. Suppose that two random samples of 15 observations each have resulted in the computation of i\ = .75 and r^ = .65. Test H^ip-^ = p^ and draw appro- priate conclusions. Also compute the Clc,r, for each parameter, p-^ and p^, and interpret these intervals. Can these interpretations be related to the test of 16. Draw a random sample of 30 observations from Table 7.21, compute the Cl^,, on p, and discuss the meaning of this interval. 17. Draw a random sample of 30 from Table 7.21 and test the hypothesis: H^ip = 0). How frequently would this procedure result in the rejection of H^^ when p^ (as in this case) at the 5 per cent level of rejection? 18. Draw two random samples of size 30 from Table 7.21 and test H^^ip^ ~ p^), assuming that the first sample is from a bivariate normal population with p = p^, and similarly for the second sample and p = p^- 7.5 RANK CORRELATION Sometimes it is either necessary or convenient to correlate the ranks of X's with those of their corresponding Y's. It may be that the A^'s and the F's are only ranks in the first place, or it may be merely convenient to use ranks instead of four- or five-digit decimals, for example. The practice of correlating ranks is both older and broader in its applications than is sometimes realized. Karl Pearson apparently was of the opinion that the idea of correlating ranks originated with Francis Galton during his studies of inheritance. Sometimes C. Spearman is credited with doing much to develop rank-correlation methods, especially as applied in psychological studies. It is his coefficient, Vg, which will be discussed specifically below. The works of M. G. Kendall, and others, recently have increased the use of ranks in statistics to a considerable degree, but no attempt will be made herein to give an exhaustive treatment of this subject. The interested reader is referred to Kendall's book, Rank Correlation Methods, published by Charles Griffin and Company, London. The calculation of the Spearman, or rank-difference, coefficient of linear correlation (r^.) will be illustrated by means of the following pairs of ranks of students in two mathematics courses. Each pair gives the respective ranks of that student in statistics (X) and in mathematics of finance (F). For example, the first student listed Sec. 7.5 RANK CORRELATION 227 ranked second in his class in statistics on the final examination, but ranked fifth in the final examination in mathematics of finance. Student 12 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Z (statistics) : 2 7 4 1 10 8 15 9 16 5 6 12 11 13 3 14 F(finance): 5 4 3 2 9 7 16 6 15 12 8 11 10 13 1 14 Change in rank (d):33111113 172 11020 It is seen from these data that there is a general but imperfect tendency for a student's grades to rank about the same in both sub- jects, that is, a student's grade in statistics has some relation to his grade in mathematics of finance. If the relationship is basically lin- ear, it can be measured rather simply and satisfactorily by means of the following formula for what is called the Spearman, or rank- difference, coefficient of correlation : 6S(rf2) (7.51) r« = 1 - 7i{n^ — 1) where d is the difference between successive pairs of ranks (in the above illustration) or, in general, between the ranks of Xi and Yi, i varying from 1 to n. For the data on ranks in statistics and in mathe- matics of finance, di = 2 - 5 = -3, ^2 = 7 - 4 = -1-3, • • •, di6 = 14 - 14 = 0; hence, ^(d^) = 92, and r, = I - 553/15(224) = 0.865. If there are ties for ranks, each .Y (or Y) so tied is given the mean of the ranks involved in the tie. For example, if two X's are tied for ranks 1 and 2, each X is given a rank of 1.5; if three F's are tied among themselves for ranks 1, 2, and 3, each is considered to have rank 2. It can be shown that r^ never has a size outside the range —1 to -f-1, regardless of the types of measurements involved or their sizes. It is seen from formula 7.51 that, if each Y has exactly the same rank as its corresponding A", all of the d's are and hence 2(<i") = and Ts = 1. If the ranks are perfectly reversed (1 with 16, 2 with 15, etc.), Ts = -1. Kendall discusses such matters as confidence intervals for rank- correlation coefficients in his book (reference above) as well as intro- ducing the coefficient tau (t), which he prefers to the Spearman co- efficient, Tg. These matters will not be discussed further here, but 228 LINEAR REGRESSION AND CORRELATION Ch. 7 the reader again is invited to consult Kendall's book on this subject if interested, PROBLEMS 1. Solve problem 5 of section 7.4, using the Spearman coefficient, r^. 2. Solve problem 6 of section 7.4, using the Spearman coefficient, r^. Ans. Vg = —1. 3. Compute ?•, for the data of problem 9, section 7.4, letting X = I for 1929, X = 2 for 1930, etc., and setting Y = wage. 4. Compute the rank-difference coefficient of linear correlation for the pairs of observations in Table 7.22. Ana. r^ = .68. 5. Make up a problem for which r^ = +1, also for r^ = —1, and r^ = +-5. Then make up another set for each case different from each of the others. 6. A sampling study in cereal chemistry gave the following product-moment linear correlations: Sample 1: ?ii = 44, ri = —0.93 Sample 2: 712 = 44, r2 = -0.81. Test Hq(pi = P2) and draw appropriate conclusions. Ans. V = 2.40; P^ .017; reject i/^. 7. Referring to problem 6, how small could the sample size become and still result in the rejection of Hq at the 5 per cent point if the r's stayed the same size? 8. If r^ = —.93, as in problem 7, could r^ = —.90 ever result in the rejection of the Hq of problem 6 at the 5 per cent level for any sized sample? If so, what size would n-^ and «o have to be if they were equal? Ans. n^ = n.y — Tib. r 9. It has been stated that each of the ratios (6 — /3)/s6 and / - V (1 — r)/(n — 2) follows the i-distribution with (n — 2) degrees of freedom under random sampling with a given n. Show that these two quantities are algebraically identical if ^ = 0; and hence that testing Hq (/3 = 0) is identical to testing H^ (p = 0). 10. Suppose that the following results were obtained from two samples (in- volving different methods of some sort), each containing 20 observations: Method 1 : ri = .40, Hq (pi = 0) accepted. Method 2: n = .60, i/o (p2 = 0) rejected, P < .01. Yet Hq (pi = P2) is accepted readily because P > .40. Explain how such results are not contradictory. Also, determine what sizes n must have in order that each of these three hypotheses will be rejected at the 5 per cent point if the correlations stay as they are. Ans. First H^, n-^ = 25; second //q, rig = 11; third Hq, n.^ = n.^ = 109. You are given the following two-variable frequency distribution table as the basis for solving problems 11 to 15 below. These data are derived from records of heights and weights of 9-year-old Kansas girls in certain schools. These data Sec. 7.5 RANK CORRELATION 229 were obtained from the Department of Home Economics, Kansas Agricultuial Experiment Station, through the courtesy of Dr. Abby Marlatt. Height in Centimeters (X) Weight in kilo- 123 127 131 135 139 143 147 151 grams to to to to to to to to (F) 126 130 134 138 142 146 150 154 650-689 1 610-649 1 570-609 3 530-569 4 490-529 1 4 2 3 450-489 1 4 9 4 410-449 1 3 7 12 6 370-409 1 5 6 14 13 5 330-369 1 2 9 23 17 14 4 290-329 1 1 17 25 32 13 4 250-289 1 15 32 19 5 210-249 5 8 5 2 Mean weight 246.6 259.9 285.4 306.8 336.5 378.7 409.9 458.1 Standard deviation 29.1 27.2 35.6 38.6 57.9 109.6 82.8 11. Plot the mean weights above against the midpoints of the height classes, and decide therefrom if the assumption of a linear relationship between these two variables seems acceptable. 12. Ignore any indication of non-Unearity of trend and compute r and b by the methods of this chapter. What conclusions can you draw from these estimates? Ans. b = 0.77, r = .72. 13. Compute the standard deviation not given in the above table by the method of Chapter 2 adjusted so as to take account of the fact that this is supposed to be a sample. 14. Each height class has some kind of one-variable frequency distribution of the weights within the height class. Hence the above data constitute several samples of weights within height classes. Theoretically, these weight distributions within height classes must have equal population variances. Plot the standard deviations against the midpoints of the corresponding height classes and decide therefrom — if you can — whether or not that is a good as- sumption in this case. 15. For the weight class 290 to 329 kilograms compute the coefficient of varia- tion for the heights, taking the point of view maintained in Chapter 2. 230 LINEAR REGRESSION AND CORRELATION Ch. 7 REVIEW PROBLEMS L Define the term percentile and explain liow it can be associated with the relative cumulative frequency distribution of a group of measurements. 2. Calculate the arithmetic mean, standard deviation, and coefficient of varia- tion for the following data on the carotene content of pasture grasses, in milli- grams per gram : X: 0.22 0.13 0.23 0.36 0.44 0.26 0.11 0.23 0.26 0.26 0.20 0.16 and 0.20 3. What is the median carotene content for the data of problem 2? The mean deviation? 4. Compute the geometric mean of 32 and 90. Of 2, 7, and 30. Of the X's of problem 2. Ans. 53.8; 7.49; 0.22. 5. You are given the following information regarding a group of 2000 weights (in grams) : m = 800, md = 700, Q^ = 500, Q^ = 900, extreme weights are 350 and 1300; and the upper limits of the 15th, 35th, 80th, and 90th percentiles are 400, 540, 950, and 1050, respectively. Sketch a graph of the r.c.j. curve. 6. Given the following scores made on an Ohio Psj'chological Test, construct a frequency distribution with equal class intervals and compute fi and cr. These scores are necessarily integers. 83 69 30 26 53 60 44 36 68 71 55 52 45 62 42 47 70 62 28 46 42 45 38 45 75 79 73 105 80 81 68 65 48 52 38 77 26 71 31 24 51 55 67 41 36 67 106 37 60 48 74 98 62 33 83 108 74 35 38 35 38 112 66 85 48 44 100 55 77 78 21 94 35 75 71 69 61 50 70 47 65 103 100 70 60 30 97 86 54 71 87 68 64 54 45 30 52 49 78 51 91 63 45 46 90 42 68 34 79 76 39 38 64 46 34 43 57 76 31 60 34 105 17 31 67 73 53 99 68 54 37 99 43 24 50 58 104 64 54 38 96 53 57 35 52 73 66 39 59 70 91 88 60 44 82 72 56 76 71 30 59 50 100 77 129 46 86 88 36 78 61 58 40 37 65 72 103 63 46 70 48 48 57 83 51 29 51 32 37 100 43 47 53 41 107 115 64 59 26 48 40 61 37 70 49 62 88 42 69 49 71 57 87 63 101 69 50 75 69 48 59 49 96 67 63 71 75 56 78 40 81 59 74 110 57 28 50 68 63 55 61 30 95 116 75 71 31 34 77 60 84 68 70 36 65 27 63 49 41 79 66 73 53 99 98 79 89 27 87 37 48 75 80 109 43 46 91 77 61 44 58 53 45 87 96 64 84 87 116 35 105 43 75 22 37 49 56 60 74 38 38 28 57 29 57 34 61 27 62 71 53 44 88 76 61 45 45 41 33 57 58 83 82 67 75 29 71 77 50 47 102 83 47 64 57 75 94 38 38 107 65 25 51 28 53 80 79 55 47 57 76 49 92 32 39 89 70 52 34 41 31 77 57 44 56 41 39 42 81 70 68 69 80 48 46 38 83 65 33 57 14 42 32 78 51 55 50 52 75 57 65 74 40 63 44 59 38 60 64 35 50 65 37 76 82 100 48 69 47 54 33 35 61 74 37 37 35 42 128 35 47 57 59 46 91 80 81 78 74 53 39 66 Ch. 7 REVIEW PROBLEMS 231 58 63 40 55 46 46 40 38 58 63 32 42 56 30 85 50 41 74 43 55 93 33 60 72 54 81 66 56 36 60 92 39 31 81 41 38 28 62 51 86 38 61 48 85 53 82 26 32 48 46 40 51 54 28 66 72 48 75 69 69 82 56 30 57 96 87 63 43 45 38 82 43 62 31 66 80 97 78 36 60 91 97 59 40 45 78 89 28 67 79 53 82 37 98 56 68 66 33 36 43 80 72 51 54 30 34 36 77 54 63 66 45 29 29 59 70 83 45 108 78 37 48 36 33 97 43 58 89 60 67 55 64 72 99 91 75 46 52 59 39 18 54 91 76 29 63 95 41 28 45 44 94 57 34 86 36 36 69 55 58 67 86 82 42 48 62 109 48 81 Am. ^ ^59.7; cr^21.9. 7. Construct an r.c.f. curve for the data of problem 6 and obtain from it evi- dence regarding the normality of the distribution of these test scores. 8. In what percentile would a person who made a score of 101 rank in the test of problem 6? Ans. 97th. 9. What are the modal and the median test scores, respectively, for prob- lem 6? 10. Calculate the median for the Ohio test scores of problem 6 by grouping them into about 12 classes of ecjual length. Ans. md ^ 51.Q. 11. The Year Book of Labour Statistics for 1943-1944 gives the following per- centages of unemployed in the United States and Sweden during 1941, by monthly averages: U.S.: 15.3 14.0 12.1 8.8 5.3 2.9 2.2 1.9 Sweden: 17.1 16.4 15.1 13.1 10.6 9.3 7.8 7.5 U.S.: 0.5 1.3 3.7 5.3 Sweden : 7.3 8.2 10.0 13.0 In which country was the level of unemployment relatively more stable during that year? Justify answer statistically. 12. The USDA publication, Agricult^iral Statistics, 1946, lists the following tax levies for the 48 states, in dollars per acre: 1.01 0.89 0.59 2.73 1.77 2.19 1.06 2.26 1.05 0.73 0.76 1.06 0.54 0.96 0.83 1.18 0.33 0.24 0.24 0.37 0.40 0.33 0.81 0.27 0.16 0.40 0.24 0.18 0.62 0.37 0.43 0.22 0.45 0.32 0.33 0.25 0.15 0.11 0.52 0.06 0.20 0.04 0.08 0.35 0.17 0.43 0.27 1.00 Compute the median tax per acre from an array of these data. Also compute the range and the midrange. Compare the latter with the median and draw any possible conclusions. Ans. md = 40; range = 2.69; MR = 1.38. 13. Referring to the data of problem 12, in what decile would a state rank if its tax rate were 0.40 dollars per acre? What percentage of the states would have a higher rate? 14. Suppose that a sample of 15 differences in yield between two varieties of corn grown side by side on 15 pairs of plots has been found to have an arithmetic 5 10 15 20 25 30 35 40 45 2.2 3.6 4.8 6.3 7.7 9.1 10.8 12.2 13.8 50 55 60 15.5 17.5 19.8 232 LINEAR REGRESSION AND CORRELATION Ch. 7 mean = 10 bushels per acre, with the standard deviation = s = 13 bushels per acre. Is one variety superior to the other or not? Give reasons for answer. Ans. t ^ 3.00; UD/F, P ^ .01. 15. Suppose that a large number of tractor gears has been produced and that 90 per cent of them are classifiable as acceptable. If a sample of 10 gears is taken at random from this group, which of the following is more likely to occur? (a) The sample will contain less than 90 per cent "acceptable" and will, therefore, give a pessimistic picture of the quality of the whole batch. Or ib) the sample will contain at least 90 per cent "acceptable" and hence will, if anything, overestimate the quality of the batch. 16. The depth of deterioration (in inches) is used as an index of the mer- chantable volume of timber remaining in fire-killed Douglas fir. Kimmey and Furniss (USD A Tech. Bull. 851) report a study made in western Oregon and Washington on such timber. The following data on old-growth trees were read from one of their graphs: Years after fire: Depth in inches: Years after fire: Depth in inches: What do you conclude from a regression analysis is the average increase in depth of deterioration per decade? Given 2XF = 5123.0, SF^ = 1617.09. Ans. 3.0 inches = point estimate; CI95: 0.25-/3^0.35. 17. Economists sometimes speak of commodities with elastic or inelastic prices, meaning generally that a commodity which is slow to change price in the face of changes in demand has an elastic price. If you adjust prices for inflation and for depression, and if demand is measured by per capita consump- tion of a given commodity, the definition of an elastic, or inelastic, price can be made more specific. For example, if the slope of the linear trend line re- lating adjusted price (Y) to consumption per capital (X) is less than unity, the price can be called elastic. If /3 is greater than 1, the price then is called inelastic. Given the following data regarding whole milk and cream, would the price be classified as elastic according to the above definition after due allowance for sampling error? Adjusted price ($/cwt) 1.88 2.06 2.07 2.26 2.36 2.29 2.48 2.19 Consumption per capita (cwt) 3.43 3.50 3.72 3.93 4.12 4.32 4.20 4.00 18. In the preceding problem a definition of an elastic price was based upon the size of the regression coefficient, /3. What information would it add to this discussion to include the size of the correlation coefficient? 19. Following are adjusted farm beef prices (per himdredweight) and con- sumption per capita (hundredweight) for the 10-year period indicated: Year: 1937 1938 1939 1940 1941 1942 1943 1944 1945 1946 Price: 6.67 6.67 7.52 7.79 8.32 8.63 8.62 7.94 8.71 9.06 Con- sump- tion: 0.55 0.54 0.54 0.55 0.60 0.61 0.53 0.55 0.59 0.61 Ch. 7 REVIEW PROBLEMS 233 Compute b and r, describe the price as elastic or inelastic, and bring the size of r into your discussion as suggested in problem 18. 20. Given the following data on turkeys, solve as in problem 19: Year: 1930 Price (cents/lb) Y: 15.9 Consumption (lb) X: 1.8 Year: 1938 Price (cents/lb) Y: 17.9 Consumption (lb) X: 2.7 Year: 1946 Price (cents/lb) Y: 22.6 Consumption (lb) X: 4.5 Given: SZ^ = 175.38, SXF = 1041.46, ^(y^) = 203.7361. Ans. Gigs: 1.0 < < 3.4, r = .58. 21. Referring to problems 19 and 20, were the beef or the turkey prices rela- tively more stable during the period 1937 to 1946? Give statistical evidence for your answer. 22. The earliness with which chickens obtain their feathers is economically important to persons who raise broilers because it affects the rapidity and cleanness of dressing. Early feathering, a sex-linked characteristic, is chiefly dependent upon one gene locus on the sex chromosome. Its inheritance can be described diagrammatically as follows: = late-feathering female, 1931 1932 1933 1934 1935 1936 1937 18.4 14.8 13.8 16.1 20.1 15.4 17.2 1.7 2.1 2.4 2.2 2.1 2.7 2.7 1939 1940 1941 1942 1943 1944 1945 16.5 15.9 18.8 22.2 23.6 25.0 24.2 3.0 3.5 3.5 3.7 3.3 3.3 4.3 1947 18.7 4.5 1, none t 2. none L 3. L or L 4. I 5. I = early-feathering female, = late-feathering male, = early-feathering male. If late-feathering females are mated to early-feathering males, what is the expected number of: (a) late-feathering females among the offspring, ib) late- feathering males, (c) early-feathering chicks of either sex among 1000 offspring? Ans. (a) none; (b) 500; (c) 500, all females. 23. Suppose that the late-feathering males in a flock can be assumed to have two-thirds of type LI and one-third of type LL. If these males are mated to 234 LINEAR REGRESSION AND CORRELATION Ch. 7 early-feathering females, what is the probability that a fertile egg selected at random will hatch into an early-feathering chick, sex disregarded? 24. Referring to problem 22, suppose parents 2 and 4 are mated. If 4 fertile eggs are to be incubated and all can be assumed to produce a live chick, what is the probability that at least one early-feathering chick of each sex will be hatched so that you could hope to develop a line of early-feathering chickens? Ans. 55/128. 25. Some educators believe that tests can be developed which measure a persons general ability to think critically. No particular field of subject matter is involved. The following data are test scores derived from such a testing program. The students were asked — but not required — to indicate the level of their father's annual income for the preceding year. All samples have n — 20 and are assumed to be from normal populations. Freshmen Juniors and Seniors Men Women Men Women (1) (2) (3) (1) (2) (3) (1) (2) (3) (3) 41 44 34 43 41 24 39 36 38 31 31 42 39 44 36 38 35 42 47 45 36 31 39 42 20 32 34 32 26 33 36 ' 31 41 26 26 35 25 40 34 29 26 34 36 33 25 35 27 23 45 28 25 28 35 41 34 20 31 41 24 39 34 29 36 41 38 26 44 31 39 25 24 35 33 28 42 32 45 32 43 41 28 29 21 35 33 20 48 35 35 44 29 34 31 28 34 34 39 40 33 18 28 38 33 28 28 31 37 36 31 35 37 27 26 24 34 33 29 37 27 46 35 20 28 30 20 22 36 32 28 38 32 42 35 36 45 24 35 45 29 49 33 28 30 38 37 35 41 32 32 30 35 36 32 38 32 29 35 43 33 42 30 34 30 31 35 28 26 38 37 40 49 38 21 45 40 21 39 44 32 37 38 42 25 40 24 37 37 35 40 36 34 26 39 37 34 41 43 46 38 13 x: 33.0 33.4 32.2 35.4 32.9 29.9 36. 2 37.0 34.6 35.0 (1) = not over $5000. (2) = $5001 to $10,000. (3) = none stated. Use any or all of these data for problems 25 to 32 below. Do the two samples for freshmen men and women whose fathers earned not more than $5000 indicate that freshmen women think more critically than freshmen men, if this test is assumed to be reliable? 26. Referring to problem 25, how do freshmen men compare with junior and senior men whose fathers are in the lower income group? Show how the G-test helps answer this question. Ans. G = 0.13; P > .10; accept H^ifip = Mjan^g). Ch. 7 REVIEW PROBLEMS 235 27. Solve as in problem 26 for the higher income group of fathers. 28. You might wonder if the group which refused (or neglected) to reveal the fathers income are from a different population as regards scores in critical thinking. Make a study of this matter for freshmen women by means of the G-test. Alls. Low income vs. undeclared: G = 0.26, P < .010. High income vs. undeclared: G = 0.09, P» .10. 29. Do freshmen women from the higher income group belong to a population with a lower mean score than that of the freshmen women from the lower income group? Use the G-test. 30. Solve problem 29, using the t-test instead of the G-test. (See other tables for these D/F.) Given: i:x~ for (1) = 804.80; for (2) = 953.80. Ans. t = 1.16, 38 D/F, P > .05. 31. Use the G-distribution to set a 90 per cent confidence interval on the true mean score of freshmen men whose fathers make $5000 or less per year. Does this interval make it possible to test the hypothesis Hoifi = 0)? 32. Compute CIgo's on the true means of the populations sampled by columns 4 and 6 (from left) in the table above, and draw all appropriate conclusions. Ans. Col. 4: 33.2^^^^37.6. Col. 6: 27.7 ^ ^2 — 32.1 by G-distribution. The following data record the thiamin-content, in micrograms per gram of meat (dry fat-free basis) in raw pork loin after various periods of storage (tem- perature not over 10°F). (These data made available through the courtesy of Dr. Beulah Westerman, Department of Foods and Nutrition, Kansas State College.) Period of Storage 12 (Weeks) 24 40 56 72 126.88 81.47 91.51 104.78 76.99 93.57 98.83 69.14 69.58 69.04 79.22 94.25 106.55 119.44 98.17 84.84 74.28 114.03 91.73 75.65 81.49 105.20 121.34 99.65 68.35 65.41 77.05 70.06 83.58 88.49 95.41 111.89 102.43 111.17 97.14 77.19 111.67 80.93 91.87 86.30 97.21 116.62 78.30 76.94 88.62 71.01 72.91 87.38 118.50 111.26 102.31 100.85 65.03 91.94 Mean 99.58 88.01 89.23 89.25 85.30 95.90 range 58.53 54.03 25.38 42.13 56.31 39.43 33. Compute through the G-distribution the CI^q on the true thiamin con- centration (micrograms per gram) in raw, unstored, pork loin produced under the conditions maintained during the sampling which produced the above data. Draw all appropriate conclusions. 34. Solve as in problem 33 for raw pork loin stored 12 weeks. Ans. CIjjq: 76.9 ^ /i ^ 99.1 micrograms per gram. 35. Does cold storage (at, or below, 10°F) of raw pork loin for 12 weeks re- duce the thiamin concentration, according to the evidence from the above data and the G-test? 236 LINEAR REGRESSION AND CORRELATION Ch. 7 36. Does the concentration of thiamin in raw pork loin increase between the fifty-sixth and the seventy-second week of cold storage (at, or below, 10°F), or is the observed average increase of 10.60 micrograms per gram probably just a sampling accident? Ans. G = 0.221, n = 9, P > .10; sampling accident. 37. It appears from the sampling data above that the thiamin concentration in raw pork loin decreases during the first 12 weeks of storage, stays about the same through the fifty-sixth week of such storage, and returns to about the original concentration by the end of the seventy-second week of storage. Is this actually the case, according to G-tests, or could the observed results rea- sonably be assigned to sampling error? The following data are from the same source as those immediately above, and were taken during the same general experimentation. They record the riboflavin concentration in raw pork loin instead of the thiamin content just studied. These data are to be employed in the solution of problems 38 through 43. Period of Storage 12 (Weeks) 24 40 56 72 3.42 4.31 5.98 5.17 4.08 5.39 2.86 3.52 4.84 4.19 3.22 5.02 2.99 3.47 5.14 4.87 4.03 5.51 2.24 3.47 4.72 4.55 4.19 5.03 2.02 3.43 4.52 4.58 3.35 4.25 2.17 4.07 2.91 4.28 3.35 4.18 1.69 3.52 3.73 4.33 5.23 4.80 2.09 3.48 3.61 4.29 4.91 3.51 1.57 3.84 3.60 5.14 5.80 4.63 Mean 2.34 range 1 . 85 3.68 0.88 4.34 3.07 4.60 0.98 4.24 2.58 4.70 2.00 38. Make a scatter diagram with Y = mean riboflavin concentration and X = weeks of storage. Is the trend in the bivariate population of X's and F's probably linear for these times of storage? 39. Would the above data cause you to accept, or to reject, the hypothesis that the riboflavin concentration in raw pork loin is increased by 12 weeks of storage at, or below, 10°F? 40. Can the apparent drop in riboflavin concentration between the fortieth and the fifty-sixth weeks of storage reasonably be assigned to sampling acci- dents? Ans. G = 0.202, n = 9, P > .10; accept Hq{/14o = -"se)- 41. Use the ^-distribution to set a 95 per cent confidence interval on the true riboflavin concentration in raw, unstored, pork loin of the kind sampled here. You are given that 2X = 21.05, and HX- = 52.3121. 42. Solve problem 41 by means of the G-distribution, and compare the result with that obtained from the ^-distribution. Ans. From G: 1.9^^0 — 2.8; from t: 1.8 ^mq — 2.9. Ch. 7 REVIEW PROBLEMS 237 43. According to the evidence above, 40 weeks might be an optimum storage period for increasing riboflavin. The means for and 40 weeks differ by 2.36. Use the G-distribution to place a CIgg on the true gain due to 40 weeks of storage, and draw conclusions. 44. A recessive lethal will destroy an organism only if carried by both chromo- somes of a pair. Suppose that l^ is such a lethal, and that the following mating has been made: Lj/^ X -^i^i- What is the probability that among the first 10 offspring none will be killed by this lethal? A)is. .057. 45. Suppose that a flock of chickens carries the lethal mentioned in problem 44, and that' the owner wishes to so select his future breeding stock that this lethal will disappear as rapidly as po.ssible from his flock. He knows that some of his chickens are carriers, that is, are i^j/j. New stock which he raises cannot be designated as L^Lj or as L^Z^ until they have produced some (perhaps many) offspring. Hence new members of the flock will be mated to known L^l^'s and then will be eliminated from the flock if any of their offspring are victims of the lethal because this will show that they are carrying that gene. How many offspring should the owner see from a chicken without the appearance of the lethal before accepting that chicken as being L^L^ and hence not a carrier of the lethal? Since he never can be absolutely positive, assume that he is willing to run a risk of 1 in 50 of reaching such a conclusion erroneously. 46. Suppose that a trait which is of economic interest to a sheep breeder is determined by two genes, R and S, believed to be carried on two different chromosomes. It also is believed that R is completely dominant to r and similarly for S with respect to s. It is supposed that only those animals showing both dominant characteristics are of special interest. If the breeder's hypotheses are correct, the mating RrSs X ^'"'Ss should produce 9/16 of its offspring with both the R and the S genes, 3/16 with R but not S, 3/16 with S but not R, and 1/16 with neither R nor S. Suppose that all four possibilities are distinguishable and that the following offspring have been recorded: 82 are R and S (called RS) ; 36 are R but not .S (called Rs) ; 28 are S but not R (called rS), and 14 are neither R nor S (called rs). Given these results, would you accept the hypothesis stated above, namely, HQi9RS:3Rs:3rS:lrs)? Give reasons. Ans. x^ = 3.644, 3 D/F, P» .11; accept H^. 47. What is the probabihty that both of two CI^j's on /i obtained from two random samples from the same normal population will include /i? Since the /J, would lie in the overlap of these two intervals (if both did include fi), and since this overlap would be shorter than either interval in many cases, and never longer, would you do a better job of estimating /x by using two random samples and considering this overlap? Would the probability of an error of the first kind be reduced if this process were used to test an Hq? Give reasons for answers. The following numbers are measurements of basal metabolism (in calories/ square meter of surface area/hour), and are to be used in answering problems 48 to 53 below. These data were derived from measurements provided through 238 LINEAR REGRESSION AND CORRELATION Ch. 7 the courtesy of Mrs. Ada Seymour and the Department of Home Economics, Kansas Agricultural Experiment Station. All ages are to the nearest birthday. CI95 on n Age Class n hyt mean, x Sx 10-11 45 42.96-44.72 43.84 0.441 12-14 46 37.27-39.33 0.516 15-16 52 34.59 0.350 17 65 33.69-35.11 .0.354 18 90 32.59-33.77 0.295 19 91 32.05-33.70 20 73 32.67-34.00 0.333 21-25 175 32.82 0.185 26-29 55 31.66-33.01 0.338 30-34 73 31.90-32.92 32.41 35-39 57 32.36-33.82 33.09 0.362 40-44 53 32.00-33.38 0.346 45-49 56 30.75-31.97 0.304 50-59 62 30.67-32.03 0.341 60 and over 33 30.06-32.12 48. Fill in the two CIg-'s omitted above and state what information they yield. 49. Graph the CIgg's versus age (on the horizontal axis) so as to produce a figure from which you could read, approximately, the confidence interval on true mean basal metabolism for any age, with a confidence coefficient .95. This is to be applied only to Kansans, of course. 50. Compute the two missing standard deviations in the above table. 51. Test the hypothesis that Kansas women between the ages of 35 and 39 have a higher average basal metabolism than those in the age interval from 30 to 34 years. 52. According to the "Mayo Foundation Normal Standards," published in July of 1936 in the American Journal 0} Physiology, the mean basal for 17-year- old females is 37.82 calories per square meter per hour. According to the table above do the Kansas girls fit that norm, or do they probably have a lower average metabolism rate? How confident can you be of your answer when allowance is made for sampling error in the above table, but none is allowed for the Mayo Standard? 53. Assuming that the records for those persons in the age group 21 to 25 years are normally distributed, estimate the range for this sample of 175. 54. Suppose that 147 freshmen, 18 years of age, have taken a test designed to measure their ability to think critically, and have taken this test at the beginning and also at the end of their freshmen year. Their progress during the year is measured by the difference between these two scores. Given that SF = 712 and 2t/2 = 5567.40, test the hypothesis that freshmen of the sort so sampled make some improvement in critical thinking during the year in so far as this is measured by the test administered. Consider that t has 30 D/F. Ans. t = 9.67, P nearly zero; fi ^ 0. Ch. 7 REVIEW PROBLEMS 239 55. Suppose that two varieties of corn have been grown at the same experi- mental farm during the same year, and that the following plot yields, in pounds, have been obtained: No. 1: 12.1 12.8 15.2 14.0 13.5 13.6 14.3 12.9 13.9 and 14.7 No. 2: 14.6 12.9 15.6 14.3 14.8 13.4 13.8 15.3 16.0 and 14.5 These field weights have been corrected for moisture content so that the variety yields per acre can be compared directly with these data. Use the G-test to test the hypothesis H^iii-y = /x^), where the yu's are the true means of the varieties. 56. The following data simulate those which might be obtained from an experimental comparison of the effectiveness of two fertilizers on the yield of orange trees in pounds per tree: Nitrogen (N): 74 89 90 72 78 76 84 79 81 76 and 80 N + Potash: 103 102 97 80 87 92 91 78 83 89 and 92 The two groups of trees (one for N and the other for N + P) were assumed with good reason to be on equivalent areas of land before the two fertilizers were applied. Test the hypothesis that the addition of potash does not affect yield. Ans. G = 0.488, n = 11, P ^.002; reject hypothesis. 57. Referring to problem 56, use the G-test to place a 92 per cent confidence interval on the true difference in average yield produced by adding potash under these circumstances, and draw appropriate conclusions. 58. The following numbers are the pounds of tobacco per acre yielded, on the average, in the United States during the years indicated. Make a scatter diagram and decide if the trend toward increasing yield can be reasonably considered as linear if this is taken to be a sample. Year: 1932 1933 1934 1935 1936 1937 1938 1939 Yield: 725 789 852 905 807 895 866 940 Year: 1940 1941 1942 1943 1944 1945 1946 1947 Yield: 1036 966 1023 964 1116 1094 1182 1142 59. Referring to problem 58, again assume that this is a sample from a bivariate population and compute, and interpret, the CIqq on /3, the true slope of the regression line. 60. Solve as in problem 59, after substituting p, the true coefficient of linear correlation, for /3. Arts. ClgQ: .86 ^ p ^ .98. 61. The following data are the numbers of sugar-maple trees tapped each year and the resulting pounds of sugar and sirup. If these data can be re- garded as a sample, did the production per tree change during this period in any orderly manner; and, if so, how? Year: 1929 1930 1931 1932 1933 1934 1935 Trees (lOOO's): 12,951 13,158 12,092 12,064 12,009 12,099 12,341 Pounds (lOOO's): 3724 5856 3589 3748 3269 3488 4673 240 LINEAR REGRESSION AND CORRELATION Ch. 7 Year: 1936 1937 1938 1939 1940 1941 Trees: 11,500 11,339 11,380 10,313 9,957 9,785 Pounds: 3122 3276 3475 2881 3031 2384 Year: 1942 1943 1944 1945 1946 1947 Trees: 9,847 9,281 8,681 7,336 8,000 8,568 Pounds: 3569 3133 3133 1228 1700 2344 62. The cumulative and r.c.f. distributions given below are those of the sizes of peach orchards in the Sandhills of North Carolina during 1946. (Data from Technical Bulletin 91, North Carolina Agricultural Experiment Station.) Do the sizes of these orchards follow a normal frequency distribution quite well, or is their distribution far from normal? Number of Trees in Cumula- Orchard tive/ r.c.f. 200-299 12 .03 200-399 23 .06 200-599 47 .13 200-799 64 .17 200-999 79 .22 Number of Trees in Cumula- Orchard tive/ r.c.f. 200-1999 121 .37 200-2999 153 .47 200-4999 188 .60 200-9999 225 .79 all orchards 257 1.00 63. Referring to problem 62, what is the median size of orchard? The lower limit of the second quartile? The following scores on certain academic aptitude tests and the student's grade point average (GPA) at the end of the indicated year are to be the basis for answering the questions in problems 64 to 70, inclusive. These data constitute samples from classes taking a natural science comprehensive course at Kansas State College. Freshmen ACE-T ACE-L ACE-Q GPA ACE-T ACE-L ACE-Q GPA 66 33 33 0.11 85 48 37 0.56 101 57 44 0.96 89 53 36 0.68 85 50 35 1.33 100 53 47 0.56 96 56 40 1.11 122 67 55 1.03 115 66 48 1.30 117 74 43 2.33 110 74 36 2.06 96 64 32 2.31 111 70 41 0.06 90 58 32 0.93 62 39 23 1.50 103 63 40 1.58 74 48 26 0.77 41 26 15 0.04 116 85 31 2.64 68 42 26 1.04 102 62 40 1.14 125 74 51 2.24 113 69 44 0.81 111 71 40 0.77 105 62 43 0.48 87 64 23 1.27 81 49 32 1.22 100 65 35 0.59 113 61 52 2.59 114 71 43 2.19 147 101 46 2.54 99 59 40 1.68 Ch. 7 REVIEW PROBLEMS 241 Freshmen {Continued) ACE-T ACE-L ACE-Q GPA ACE-T ACE-L ACE-Q GPA 93 58 35 0.97 115 60 55 1.39 59 31 28 0.50 77 46 31 0.50 75 52 23 0.83 89 53 36 1.27 106 63 43 1.24 137 81 56 1.38 37 27 10 0.41 42 30 12 0.96 139 72 67 1.97 125 67 38 1.41 126 80 46 1.83 Juniors 115 65 50 1.42 132 71 61 1.00 100 55 45 0.85 109 58 51 1.95 107 55 52 1.12 129 83 46 1.93 108 72 36 1.64 87 50 37 1.79 115 64 51 0.81 80 45 35 1.35 83 46 37 0.86 110 70 40 2.08 121 71 50 1.11 96 47 49 1.60 82 54 28 1.18 122 70 52 1.93 64. Make scatter diagrams of the total ACE scores (ACE-T's) on the hori- zontal axis and the GPA's on the vertical for freshmen and also for juniors, using the same coordinate system but different symbols for the two classes. 65. After solving the preceding problem, e.xplain why you agree or disagree with each of the following statements: (a) For freshmen, you would expect to find a positive and useful linear cor- relation between these two variables; but there also are other important factors affecting the grade point average of a college student. (b) For the juniors represented by this sample, there is little, or no, relation- ship between the ACE-T score and the grade point average. (c) The freshmen and the juniors fit the same general relationship between ACE-T and GPA; the persons with especially low ACE-T scores simply have been eliminated by the time of the junior year. (rf) Given that for freshmen the linear correlation between GPA and ACE-L score is .6, whereas that between GPA and ACE-Q is only .4 for these samples, it is concluded that whatever is measured by the L-score definitely is more important than whatever is measured by the Q-score. 66. Make a scatter diagram for the ACE-L scores of freshmen against their GPA's. And then do likewise for the juniors, using the same coordinate axes. Draw appropriate conclusions. 67. Solve as in problem 65, parts ia) to (c), but use the results of problem 66 and change ACE-T to 4^E-L wherever used. 68. Compute the Spearman rank-difference correlation, r^, for each scatter diagram of problem 64. Then consider problem 65 in the light of these cor- relations. Ans. For freshmen, r = .59; for juniors, r, = .14. 242 LINEAR REGRESSION AND CORRELATION Ch. 7 69. Compute the Spearman coefficient of linear correlation for each scatter diagram of problem 66 and then solve problem 67 in the light of these results. 70. Make a scatter diagram for the freshmen and for the juniors, as in prob- lem 64, but use the ACE-Q scores. Draw all appropriate conclusions. REFERENCES Arley, Niels, and K. Rander Buch, Introduction to the Theory of Probability and Statistics, John Wiley and Sons, New York, 1950. Dixon, Wilfrid J., and Frank J. Massey, Jr., Introduction to Statistical Analysis, McGraw-Hill Book Company, New York, 1952. Freund, John E., Modern Elementary Statistics, Prentice-Hall, New York, 1952. Hald, A., Statistical Theory with Engineering Applications, John Wiley and Sons, New York, 1952. Snedecor, George W., Statistical Methods Applied to Experiments in Agriculture and Biology, Fourth Edition, Iowa State College Press, Ames, Iowa, 1946. Tables I. Squares, Square Roots, and Reciprocals II. Mantissas for Common Logarithms III. Frequency and Relative Cumulative Frequency Distributions for the Standard Normal Population Given for the Abscissas from X = -3.00 to X = +3.00 IV. Relative Cumulative Frequency Distribution of t Showing the Proportions of All Sampling U with the Same Degrees of Free- dom Which Are Less Than the t Shown in Column 1 on the Left. V. Relative Cumulative Frequency Distribution of x^ Showing Proportion of All Sampling x^ with Same Degrees of Freedom Which Are Greater Than the x^ Shown on the Left VI. Values of the Function, y = (l/\/27r)-e""' VII. Binomial Coefficients: C„, r = n\/r\{n — r)\ VIII. Factorials and Their Logarithms IX. Probability Distribution of G = | 5 - m |/ (range) for a Sample of Size n from a Normal Population X. Probability Distribution of G = \ xi — X2 |/(mean range) for Two Samples Each of Size n from the Same Normal Population 243 244 TABLES TABLE I Squares, Square Roots, and Reciprocals n Vn VlOn n^ 1/n n Vn VlOn n- \/n 0.1 0.32 1.00 0.01 10.000 4.6 2.14 6.78 21.16 .217 0.2 0.45 1.41 0.04 5.000 4.7 2.17 6.86 22.09 .213 0.3 0.55 1.73 0.09 3.333 4.8 2.19 6.93 23.04 .208 0.4 0.63 2.00 0.16 2.500 4.9 2.21 7.00 24.01 .204 0.5 0.71 2.24 0.25 2.000 5.0 2.24 7.07 25.00 .200 0.6 0.77 2.45 0.36 1.667 5.1 2.26 7.14 26.01 .196 0.7 0.84 2.65 0.49 1.429 5.2 2.28 7.21 27.04 .192 0.8 0.89 2.83 0.64 1.250 5.3 2.30 7.28 28.09 .189 0.9 0.95 3.00 0.81 1.111 5.4 2.32 7.35 29.16 .185 1.0 1.00 3.16 1.00 1.000 5.5 2.35 7.42 30.25 .182 1.1 1.05 3.32 1.21 0.909 5.6 2.37 7.48 31.36 .179 1.2 1.10 3.46 1.44 0.833 5.7 2.39 7.55 32.49 .175 1.3 1.14 3.61 1.69 0.769 5.8 2.41 7.62 33.64 .172 1.4 1.18 3.74 1.96 0.714 5.9 2.43 7.68 34.81 .169 1.5 1.22 3.87 2.25 0.667 6.0 2.45 7.75 36.00 .167 1.6 1.26 4.00 2.56 0.625 6.1 2.47 7.81 37.21 .164 1.7 1.30 4.12 2.89 0.588 6.2 2.49 7.87 38.44 .161 1.8 1.34 4.24 3.24 0.556 6.3 2.51 7.94 39.69 .159 1.9 1.38 4.36 3.61 0.526 6.4 2.53 8.00 40.96 .156 2.0 1.41 4.47 4.00 0.500 6.5 2.55 8.06 42.25 .154 2.1 1.45 4.58 4.41 0.476 6.6 2.57 8.12 43.56 .152 2.2 1.48 4.69 4.84 0.455 6.7 2.59 8.19 44.89 .149 2.3 1.52 4.80 5.29 0.435 6.8 2.61 8.25 46.24 .147 2.4 1.55 4.90 5.76 0.417 6.9 2.63 8.31 47.61 .145 2.5 1.58 5.00 6.25 0.400 7.0 2.65 8.57 49.00 .143 2.6 1.61 5.10 6.76 0.385 7.1 2.66 8.43 50.41 .141 2.7 1.64 5.20 7.29 0.370 7.2 2.68 8.49 51.84 .139 2.8 1.67 5.29 7.84 0.357 7.3 2.70 8.54 53.29 .137 2.9 1.70 5.39 8.41 0.345 7.4 2.72 8.60 54.76 .135 3.0 1.73 5.48 9.00 0.333 7.5 2.74 8.66 56.25 .133 3.1 1.76 5.57 9.61 0.323 7.6 2.76 8.72 57.76 .132 3.2 1.79 5.66 10.24 0.312 7.7 2.77 8.77 59.29 .130 3.3 1.82 5.74 10.89 0.303 7.8 2.79 8.83 60.84 .128 3.4 1.84 5.83 11.56 0.294 7.9 2.81 8.89 62.41 .127 3.5 1.87 5.92 12.25 0.286 8.0 2.83 8.94 64.00 .125 3.6 1.90 6.00 12.96 0.278 8.1 2.85 9.00 65.61 .123 3.7 1.92 6.08 13.69 0.270 8.2 2.86 9.06 67.24 .122 3.8 1.95 6.16 14.44 0.232 8.3 2.88 9.11 68.89 .120 3.9 1.97 6.24 15.21 0.256 8.4 2.90 9.17 70.56 .119 4.0 2.00 6.32 16.00 0.250 8.5 2.92 9.22 72.25 .118 4.1 2.02 6.40 16.81 0.244 8.6 2.93 9.27 73.96 .116 4.2 2.05 6.48 17.64 0.238 8.7 2.95 9.33 75.69 .115 4.3 2.07 6.56 18.49 0.233 8.8 2.97 9.38 77.44 .114 4.4 2.10 6.63 19.36 0.227 8.9 2.98 9.43 79.21 .112 4.5 2.12 6.71 20.25 0.222 9.0 3.00 9.49 81.00 .111 TABLES 245 v; TABLE I (Continued) Squares, Square Roots, and Reciprocals 10?i n- 1/n n Vra VlOn 1/w 9.1 3.02 9.54 82.81 .110 13.6 3.69 11.66 184.96 .074 9.2 3.03 9.59 84.64 .109 13.7 3.70 11.70 187.69 .073 9.3 3.05 9.64 86.49 .108 13.8 3.72 11.75 190.44 .072 9.4 3.07 9.70 88.36 .106 13.9 3.73 11.79 193.21 .072 9.5 3.08 9.75 90.25 .105 14.0 3.74 11.83 196.00 .071 9.6 3.10 9.80 92.16 .104 14.1 3.76 11.87 198.81 .071 9.7 3.11 9.85 94.09 .103 14.2 3.77 11.92 201.64 .070 9.8 3.13 9.90 96.04 .102 14.3 3.78 11.96 204.49 .070 9.9 3.15 9.95 98.01 .101 14.4 3.79 12.00 207.36 .069 10.0 3.16 10.00 100.00 .100 14.5 3.81 12.04 210.25 .069 10.1 3.18 10.05 102.01 .099 14.6 3.82 12.08 213.16 .068 10.2 3.19 10.10 104.04 .098 14.7 3.83 12.12 216.09 .068 10.3 3.21 10.15 106.09 .097 14.8 3.85 12.17 219.04 .068 10.4 3.22 10.20 108.16 .096 14.9 3.86 12.21 222.01 .067 10.5 3.24 10.25 110.25 .095 15.0 3.87 12.25 225.00 .067 10.6 3.26 10.30 112.36 .094 15.1 3.89 12.29 228.01 .066 10.7 3.27 10.34 114.49 .093 15.2 3.90 12.33 231.04 .066 10.8 3.29 10.39 116.64 .093 15.3 3.91 12.37 234.09 .065 10.9 3.30 10.44 118.81 .092 15.4 3.92 12.41 237.16 .065 11.0 3.32 10.49 121.00 .091 15.5 3.94 12.45 240.25 .065 11.1 3.33 10.54 123.21 .090 15.6 3.95 12.49 243.36 .064 11.2 3.35 10.58 125.44 .089 15.7 3.96 12.53 246.49 .064 11.3 3.36 10.63 127.69 .088 15.8 3.97 12.57 249.64 .063 11.4 3.38 10.68 129.96 .088 15.9 3.99 12.61 252.81 .063 11.5 3.39 10.72 132.25 .087 16.0 4.00 12.65 256.00 .062 11.6 3.41 10.77 134.56 .086 16.1 4.01 12.69 259.21 .062 11.7 3.42 10.82 136.89 .085 16.2 4.02 12.73 262.44 .061 11.8 3.44 10.86 139.24 .085 16.3 4.04 12.77 265.69 .061 11.9 3.45 10.91 141.61 .084 16.4 4.05 12.81 268.96 .061 12.0 3.46 10.95 144.00 .083 16.5 4.06 12.85 272.25 .061 12.1 3.48 11.00 146.41 .083 16.6 4.07 12.88 275.56 .060 12.2 3.49 11.05 148.84 .082 16.7 4.09 12.92 278.89 .060 12.3 3.51 11.09 151.29 .081 16.8 4.10 12.96 282.24 .060 12.4 3.52 11.14 153.76 .081 16.9 4.11 13.00 285.61 .059 12.5 3.54 11.18 156.25 .080 17.0 4.12 13.04 289.00 .059 12.5 3.55 11.22 158.76 .079 17.1 4.14 13.08 292.41 .058 12.7 3.56 11.27 161.29 .079 17.2 4.15 13.11 295.84 .058 12.8 3.58 11.31 163.84 .078 17.3 4.16 13.15 299.29 .058 12.9 3.59 11.36 166.41 .078 17.4 4.17 13.19 302.76 .057 13.0 3.61 11.40 169.00 .077 17.5 4.18 13.23 306.25 .057 13.1 3.62 11.45 171.61 .076 17.6 4.20 13.27 309.76 .057 13.2 3.63 11.49 174.24 .076 17.7 4.21 13.30 313.29 .056 13.3 3.65 11.53 176.89 .075 17.8 4.22 13.34 316.84 .056 13.4 3.66 11.58 179.56 .075 17.9 4.23 13.38 320.41 .056 13.5 3.67 11.62 182.25 .074 18.0 4.24 13.42 324.00 .056 246 TABLES Vn TABLE I (Continued) Squakes, Squake Roots, and Reciprocals VlO/i n^ 1/n n \n VlOra 1/w 18.1 4.25 13.45 327.61 .055 21.6 4.65 14.70 466.56 .046 18.2 4.27 13.49 331.24 .055 21.7 4.66 14.73 470.89 .046 18.3 4.28 13.53 334.89 .055 21.8 4.67 14.76 475.24 .046 18.4 4.29 13.56 338.56 .054 21.9 4.68 14.80 479.61 .046 18.5 4.30 13.60 342.25 .054 22.0 4.69 14.83 484.00 .045 18.6 4.31 13.64 345.96 .054 22.1 4.70 14.87 488.41 .045 18.7 4.32 13.67 349.69 .053 22.2 4.71 14.90 492.84 .045 18.8 4.34 13.71 353.44 .053 22.3 4.72 14.93 497.29 .045 18.9 4.35 13.75 357.21 .053 22.4 4.73 14.97 501.76 .045 19.0 4.36 13.78 361.00 .053 22.5 4.74 15.00 506.25 .044 19.1 4.37 13.82 364.81 .052 22.6 4.75 15.03 510.76 .044 19.2 4.38 13.86 368.64 .052 22.7 4.76 15.07 515.29 .044 19.3 4.39 13.89 372.49 .052 22.8 4.77 15.10 519.84 .044 19.4 4.40 13.93 376.36 .052 22.9 4.79 15.13 524.41 .044 19.5 4.42 13.96 380.25 .051 23.0 4.80 15.17 529.00 .043 19.6 4.43 14.00 384.16 .051 23.1 4.81 15.20 533.61 .043 19.7 4.44 14.04 388.09 .051 23.2 4.82 15.23 538.24 .043 19.8 4.45 14.07 392.04 .050 23.3 4.83 15.26 542.89 .043 19.9 4.46 14.11 396.01 .050 23.4 4.84 15.30 547.56 .043 20.0 4.47 14.14 400.00 .050 23.5 4.85 15.33 552.25 .043 20.1 4.48 14.18 404.01 .050 23.6 4.86 15.36 556.96 .042 20.2 4.49 14.21 408.04 .050 23.7 4.87 15.39 561.69 .042 20.3 4.51 14.25 412.09 .049 23.8 4.88 15.43 566.44 .042 20.4 4.52 14.28 416.16 .049 23.9 4.89 15.46 571.21 .042 20.5 4.53 14.32 420.25 .049 24.0 4.90 15.49 576.00 .042 20.6 4.54 14.35 424.36 .049 24.1 4.91 15.52 580.81 .041 20.7 4.55 14.39 428.49 .048 24.2 4.92 15.56 585.64 .041 20.8 4.56 14.42 432.64 .048 24.3 4.93 15.59 590.49 .041 20.9 4.57 14.46 436.81 .048 24.4 4.94 15.62 595.36 .041 21.0 4.58 14.49 441.00 .048 24.5 4.95 15.65 600.25 .041 21.1 4.59 14.53 445.21 .047 24.6 4.96 15.68 605.16 .041 21.2 4.60 14.56 449.44 .047 24.7 4.97 15.72 610.09 .040 21.3 4.62 14.59 453 . 69 .047 24.8 4.98 15.75 615.04 .040 21.4 4.63 14.63 457.96 .047 24.9 4.99 15.78 620.01 .040 21.5 4.64 14.66 462.25 .047 25.0 5.00 15.81 625.00 .040 TABLES 247 TABLE II Mantissas for Common Logarithms iV0123456789 1.0 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 1.1 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 1.2 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 1.3 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 1.4 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 1.5 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 1.6 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 1.7 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 1.8 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 1.9 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 2.0 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 2.1 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 2.2 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 2.3 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 2.4 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 2.5 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 2.6 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 2.7 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 2.8 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 2.9 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 3.0 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 3.1 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 3.2 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 3.3 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 3.4 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 3.5 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 3.6 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 3.7 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 3.8 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 3.9 5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 4.0 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 4.1 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 4.2 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 4.3 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 4.4 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 4.5 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 4.6 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 4.7 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 4.8 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 4.9 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 5.0 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 5.1 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 5.2 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 5.3 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 5.4 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 248 TABLES TABLE II (Continued) Mantissas for Common Logarithms A^0123456789 5.5 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 5.6 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 5.7 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 5.8 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 5.9 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 6.0 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 6.1 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 6.2 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 6.3 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 6.4 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 6.5 8129 8136 8142 8149 8156 8162 8169 8176 8182 8489 6.6 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 6.7 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 6.8 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 6.9 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 7.0 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 7.1 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 7.2 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 7.3 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 7.4 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 7.5 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 7.6 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 7.7 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 7.8 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 7.9 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 8.0 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 8.1 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 8.2 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 8.3 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 8.4 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 8.5 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 8.6 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 8.7 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 8.8 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 8.9 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 9.0 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 9.1 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 9.2 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 9.3 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 9.4 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 9.5 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 9.6 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 9.7 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 9.8 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 9.9 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 TABLES 249 TABLE III Frequency and Relative Cumulative Frequency Distributions for the Standard Normal Population Given for the Abscissas from X = —3.00 TO X = +3.00 Abscissas Ordinates X y r.c.f. -3 00 .004 .001 -2 90 .006 .002 -2 80 .008 .003 -2 70 .010 .003 -2 60 .014 .005 -2 50 .018 .006 _2 40 .022 .008 -2 30 .028 .011 -2 25 .032 .012 -2 20 .035 .014 — 2 15 .040 .016 -2 10 .044 .018 -2 05 .049 .020 -2 00 .054 .023 — 1 95 .060 .026 — 1 90 .066 .029 — 1 85 .072 .032 — 1 80 .079 .036 -1 75 .086 .040 — 1 70 .094 .045 — 1 66 .101 .048 — 1 62 .107 .053 — 1 58 .114 .057 — 1 54 .122 .062 — 1 50 .130 .067 — 1 48 . 133 .069 — 1 46 .137 .072 — 1 44 .141 .075 — 1 42 .146 .078 — 1 40 .150 .081 — 1 38 .154 .084 — 1 36 .158 .087 — 1 34 .163 .090 — 1 32 .167 .093 — 1 30 .171 .097 — 1 28 .176 .100 — 1 26 .180 .104 — 1 24 .185 .107 — 1 22 .190 .111 — 1 20 .194 .115 — 1 .18 .199 .119 -1 .16 .204 .123 Abscissas Ordinates Abscissas Ordinates X y r.c.f. X y r.c.f. -1.14 .208 .127 -0.30 .381 .382 -1.12 .213 .131 -0.28 .384 .390 -1.10 .218 .136 -0.26 .386 .397 -1.08 .223 .140 -0.24 .388 .405 -1.06 .227 .145 -0.22 .389 .413 -1.04 .232 .149 -0.20 .391 .421 -1.02 .237 .154 -0.18 .393 .429 -1.00 .241 .159 -0.16 .394 .436 -0.98 .247 .164 -0.14 .395 .444 -0.96 .252 .169 -0.12 .396 .452 -0.94 .256 .174 -0.10 .397 .460 -0.92 .261 .179 -0.08 .398 .468 -0.90 .266 .184 -0.06 .398 .476 -0.88 .271 .189 -0.04 .399 .484 -0.86 .276 .195 -0.02 .399 .492 -0.84 .280 .200 0.00 .399 .500 -0.82 .285 .206 +0.02 .399 .508 -0.80 .290 .212 0.04 .399 .516 -0.78 .294 .218 0.06 .398 .524 -0.76 .299 .224 0.08 .398 .532 -0.74 .303 .230 0.10 .397 .540 -0.72 .308 .236 0.12 .396 .548 -0.70 .312 .242 0.14 .395 .556 -0.68 .317 .248 0.16 .394 .564 -0.66 .321 .255 0.18 .393 .571 -0.64 .325 .261 0.20 .391 .579 -0.62 .329 .268 0.22 .389 .587 -0.60 .333 .274 0.24 .388 .595 -0.58 .337 .281 0.26 .386 .603 -0.56 .341 .288 0.28 .384 .610 -0.54 .345 .295 0.30 .381 .618 -0.52 .348 .302 0.32 .379 .626 -0.50 .352 .309 0.34 .377 .633 -0.48 .356 .316 0.36 .374 .641 -0.46 .359 .323 0.38 .371 .648 -0.44 .362 .330 0.40 .368 .655 -0.42 .365 .337 0.42 .365 .663 -0.40 .368 .345 0.44 .362 .670 -0.38 .371 .3.52 0.46 .359 .677 -0.36 .374 .359 0.48 .356 .684 -0.34 .377 .367 0.50 .352 .691 -0.32 .379 .374 0.52 .348 .698 250 TABLES TABLE III {Continued) Frequency and Relative Cumulative Frequency Distributions for the Standard Normal Population Given for the Abscissas from X = —3.00 TO X = +3.00 Abscissas Ordinates X y r.c.f. 0.54 0.56 0.58 0.60 0.62 0.64 0.66 0.68 0.70 0.72 0.74 0.76 0.78 0.80 0.82 0.84 0.86 0.88 0.90 0.92 0.94 0.96 0.98 1.00 1.02 .345 ,341 .337 .333 .329 .325 .321 .317 .312 .308 .303 .299 .294 .290 .285 .280 .276 .271 .266 .261 .256 .252 .247 .241 .237 .705 .712 .719 .726 .732 .739 .745 .752 .758 .764 .770 .776 .782 .788 .794 .800 .805 .811 .816 .821 .826 .831 .836 .841 .846 Abscissas Ordinates X y r.c.f. 1.04 .232 .851 1.06 .227 .855 1.08 .223 .860 1.10 .218 .864 1.12 .213 .869 1.14 .208 .873 1.16 .204 .877 1.18 .199 .881 1.20 .194 .885 1.22 .190 .888 1.24 .185 .893 1.26 .180 .896 1.28 .176 .900 1.30 .171 .903 1.32 .167 .907 1.34 .163 .910 1.36 .158 .913 1.38 .154 .916 1.40 .150 .919 1.42 .146 .922 1.44 .141 .925 1.46 .137 .928 1.48 .133 .931 1.50 .130 .933 Abscissas Ordinates X y r.c.f. 1.54 .122 .938 1.58 .114 .943 1.62 .107 .947 1.66 .101 .952 1.70 .094 .955 1.75 .086 .960 1.80 .079 .964 1.85 .072 .968 1.90 .066 .971 1.95 .060 .974 2.00 .054 .977 2.05 .049 .980 2.10 .044 .982 2.15 .040 .984 2.20 .035 .986 2.25 .032 .988 2.30 .028 .989 2.40 .022 .992 2.50 .018 .994 2.60 .014 .995 2.70 .010 .997 2.80 .008 .997 2.90 .006 .998 3.00 .004 .999 TABLES 251 TABLE IV Relative Cumulative Frequency Distribution of t Showing the Pro- portions OF All Sampling U with the Same Degrees of Freedom Which Are Less Than the t Shown in Column 1 on the Left Degrees of Freedom t 8 9 10 11 12 14 16 18 20 22 26 30 -5.0 .001 .000 -4.6 .001 .001 .000 .000 .000 -4.2 .002 .001 .001 .001 .001 .000 .000 .000 .000 -3.8 .003 .002 .002 .001 .001 .001 .001 .001 .001 .000 .000 .000 -3.4 .005 .004 .003 .003 .003 .002 .002 .002 .001 .001 .001 .001 -3.0 .009 .007 .007 .006 .006 .005 .004 .004 .004 .003 .003 .003 -2.8 .012 .010 .009 .009 .008 .007 .006 .006 .006 .005 .005 .004 -2.6 .016 .014 .013 .012 .012 .010 .010 .009 .009 .008 .008 .007 -2.4 .022 .020 .019 .018 .017 .015 .014 .014 .013 .013 .012 .011 -2.2 .030 .028 .026 .025 .024 .023 .021 .021 .020 .019 .019 .018 -2.0 .040 .038 .037 .035 .034 .033 .031 .030 .030 .029 .028 .027 -L8 .055 .053 .051 .050 .049 .047 .045 .044 .043 .043 .042 .041 -L6 .074 .072 .070 .969 .068 .066 .065 .064 .063 .062 .061 .060 -1.4 .100 .098 .096 .095 .093 .092 .090 .089 .088 .088 .087 .086 -L2 .132 .130 .129 .128 .127 .125 .124 .123 .122 .121 .121 .120 -LO .173 .172 .170 .169 .169 .167 .166 .165 .165 .164 .163 .163 -0.8 .223 .222 .221 .220 .220 .219 .218 .217 .217 .216 .216 .215 -0.6 .283 .282 .281 .280 .280 .279 .278 .278 .278 .277 .277 .277 -0.4 .350 .349 .349 .348 .348 .348 .347 .347 .347- .347 .346 .346 -0.2 .423 .423 .423 .423 .422 .422 .422 .422 .422 .422 .421 .421 .500 .500 .500 .500 .500 .500 .500 .500 .500 .500 .500 .500 0.2 .577 .577 .577 .577 .578 .578 .578 .578 .578 .578 .579 .579 0.4 .650 .651 .651 .652 .652 .652 .653 .653 .653 .653 .654 .654 0.6 .717 .718 .719 .720 .720 .721 .722 .722 .722 .723 .723 .723 0.8 .777 .778 .779 .780 .780 .781 .782 .783 .783 .784 .784 .785 LO .827 .828 .830 .831 .831 .833 .834 .835 .835 .836 .837 .837 L2 .868 .870 .871 .872 .873 .875 .876 .877 .878 .879 .879 .880 L4 .900 .902 .904 .905 .907 .908 .910 .911 .912 .912 .913 .914 L6 .926 .928 .930 .931 .932 .934 .935 .936 .937 .938 .939 .940 L8 .945 .947 .949 .950 .951 .953 .955 .956 .957 .957 .958 .959 2.0 .960 .962 .963 .965 .966 .967 .969 .970 .970 .971 .972 .973 2.2 .970 .972 .974 .975 .976 .977 .979 .979 .980 .981 .981 .982 2.4 .978 .980 .981 .982 .983 .985 .986 .986 .987 .987 .988 .989 2.6 .984 .986 .987 .988 .988 .990 .990 .991 .991 .992 .992 .993 2.8 .988 .990 .991 .991 .992 .993 .994 .994 .994 .995 .995 .996 3.0 .991 .993 .993 .994 .994 .995 .996 .996 .996 .997 .997 .997 3.4 .995 .996 .997 .997 .997 .998 .998 .998 .999 .999 .999 .999 3.8 .997 .998 .998 .999 .999 .999 .999 .999 .999 : 1. 000 : 1. 000 : 1. 000 4.2 .998 .999 .999 .999 .999 ] 1. 000 : 1. 000 : 1. 000 : 1.000 4.6 .999 .999 ] 1. 000 J 1. 000 : 1. 000 5.0 .999 ] 1. 000 Some Frequently Used t's Corresponding to Pre-assigned Probabilities OF Occurrence during Random Sampling Degrees of Freedom P(I < I > to) 8 9 10 11 12 14 16 18 20 22 26 30 .100 1.86 1.83 1.81 1.80 1.78 1.76 1.75 1.73 1.72 1.72 1.71 170 .050 2.31 2.26 2.23 2.20 2.18 2.14 2.12 2.10 2.09 2.07 2.06 2 04 .010 3.36 3.25 3.17 3.11 3.06 2.98 2.92 2.88 2.84 2.82 2.78 2 75 .001 5.04 4.78 4.59 4.44 4.32 4.14 4.02 3.92 3.85 3.79 3.71 3 65 252 TABLES TABLE V Relative Cumulative Fbequency Distribution of x^ Showing Propor- tion OF All Sampling x^ with Same Degrees of Freedom Which Are Greater Than the x^ Shown on the Left Degrees of Freedom Degrees ) of Freedom X^ 1 2 3 x' 1 2 3 0.40 .53 6.00 .014 .050 .11 0.50 .48 6.25 .012 .044 .10 0.60 .44 6.50 .011 .039 .090 0.70 .40 6.75 .009 .034 .082 0.80 .37 7.00 .008 .030 .072 0.90 .34 7.50 .006 .023 .057 1.00 .32 8.00 .005 .018 .046 1.25 .26 .53 8.50 .003 .014 .036 1.50 .22 .47 9.00 .003 .011 .029 1.75 .19 .42 9.50 .002 .009 .024 2.00 .16 .37 10.00 .002 .007 .019 2.25 .13 .33 .50 10.50 .001 .005 .015 2.50 .11 .29 .47 11.00 .000 .004 .012 2.75 .10 .25 .43 11.50 .003 .010 3.00 .084 .22 .39 12.00 .002 .007 3.25 .071 .20 .35 12.50 .002 .006 3.50 .061 .18 .32 13.00 .002 .005 3.75 .053 .15 .29 13.50 .001 .004 4.00 .046 .13 .26 14.00 .001 .003 4.25 .039 .12 .24 14.50 .001 .002 4.50 .034 .10 .21 15.00 .001 .002 4.75 .029 .092 .19 15.50 .000 .002 5.00 .025 .082 .17 16.00 .001 5.25 .022 .073 .15 17.00 .001 5.50 .019 .064 .14 18.00 .000 5.75 .016 .057 .12 TABLES 253 TABLE VI Values of the Function, y = (l/v27r)-e~^ w y w y w y .399 1.8 .066 4.2 .006 0.1 .361 1.9 .060 4.4 .005 0.2 .327 2.0 .054 4.6 .004 0.3 .295 2.1 .049 4.8 .003 0.4 .267 2.2 .044 5.0 .003 0.5 .242 2.3 .040 5.2 .002 0.6 .219 2.4 .036 5.4 .002 0.7 .198 2.5 .033 5.6 .001 0.8 .179 2.6 .030 5.8 .001 0.9 .162 2.7 .027 6.0 .001 1.0 .147 2.8 .024 6.2 .001 1.1 .133 2.9 .022 6.4 .001 1.2 .120 3.0 .020 6.6 .001 1.3 .109 3.2 .016 6.8 .000 1.4 .099 3.4 .013 1.5 .089 3.6 .011 1.6 .081 3.8 .009 1.7 .073 4.0 .007 254 TABLES OOOiO-^OOOOcDOOOO-^iOOOO r-lcot^C^COOOcOCat^CO'-H OiT-IOiCOOOC<JOO(MOOGO(MOO(MOOCDOi'-ia) 1— it^CDt^C^IOOOOOOIr^t^OOOOCOCat^CDt^i— I i-HOJOOCO^HCOiOCOCOiOrO^HCDOOO'— I r-ICCllOI>-Oi05t^lO(Mi— I > o w w ^ H m ^; -< 1^ H 1^ fe H O O 1-5 <^ M 1$ o 15 m OOCOCOOQO'^i'^GOOOO'+i^OOOCDCOOO 1— iGOOLOiooor^cot^ooioioooOT— I COOOOO'-HCOGOCO^HGOOOfO .— I CO '^ •* rfH CO <—* t^COOOGOCOOOOOOOOOOOOOt^i-H rHCOGOOOOOt^^^H,— iTtit^OOOOOOCO'— I rHCDCO-— iCO'TtHCOCO'+iCO'-HCOCD'— I --<.-( (M (N r-i i-H C0OOO00Q0OOO0000OOO<:0 1— liOOOCOO'^GO'^OCOOOiO'— I 1— (lOiOiO>OCOtOiOiOiOCO>OiOiOiO r-HO'OOOOCOCOOOOiOO'— I '— l-^COOO'^-^OOCO^r-H 1— ICOiOOCOiOCO'-H 1— lOSCDOOOCOOOO'^OS'— I COOOO^OOOOO -— I (M CO CO CO (M I— I COOOCOtOt^COOt^lOCOOOCOr-l cq ^ t^ o t^ ^ cs r-HlOCOCOOCOCOOlO T-l CO "* ^ CO r-l O lO O O iM O O r-H TtH (M ^ lO -— I Ol '-I (M (M <M rH QOOOCOOOOOOO'-I CSJ lO t^ lO <M 1— I lO O O lO I— I TABLES 255 T-tHOlOr-HO-COOilO^t^-^Ot^-^Ot^ oooooooooooooooo XXXXXXXXXXXXXXXX C00005'*C<JCO<MCOOiOCOC^CO(M^QO c^coc5cooa3C3>OLO,-HQooc^cD'-HC<j C:iOt^CO(MO^CO'*iO'Xi'— iiOiO(MO (M^HCOr-ltOlMCiOCO'— ICDC<)r-llOC<l'— 1^ ■'^lOCOt^OOOO'-ltNCOTtliOCOt^OOOS w s o I— I M > s < HI H a < o H o OOi— (r-Hiy^OOiOt^CDCOO'— iiOI^O<Mt^ OiOCOOOO(M(MO'Ot^'*OCOiOOGO QoooooiO'-HOiaiOcoGOTticq^c^co ooocoi--;0^t^'-HcqoTjHOi'*iCi^Oi oooooooooooooooo XXXXXXXXXXXXXXXX -*lOO^^Ol^^lC'— lOiciOiooiooococo c^oc^O':t^iqT)H--Hc^oqoq'-;c^c^^_co o— icooc^odoiocoGO^'^idc^icood ^(M-^T— ii— iiOfMiOOOOOOOOlOO CD>— ilMiO'-ilMCO'-H'*,— ICOOO<NOO(MOO Q0050'-l(MCO':JHiOCOb-00050'-<IMCO --iT-lC^C^IC^fMC^OaiMC^C^lC^COCOCOOO O(M(M(MC0'-tHiO00C0<MC0C0'*iOC0'— I r:jooooit^c-5io05C55'-HO^ocoO'— I Ot^GOt'-iOOO'OiOOOOCiTt*!— i(MiO ^t^coooqi^O'OLOcDcqt^aji— looto ooT-^c<i<^^c6Ti^>ocDI>^o6aic5(^jl^6■^ ooooooo (MO'+HOOOOOO'-H'-H'-l'-l'-HrH'-l ^ ^ § g S §8 X X X X X X X T^COiMt^OOOOt^COOi roco^ot^i^t^(MO CO Oi 05 <>)■—( O Ci lO C5 t^ (M r^ CO o lo CO Tfl CO 00 T-( (M CO MCO-^iOOt^OOOO^lMCO-^iOOt^ 256 TABLES TABLE IX Probability Distribution of G = | .r — ju | /(range) for a Sample of Size n FROM A Normal Population Probability tliat G will be greater than table value Sample Size, n .100 .050 .020 .010 .002 .001 2 3.157 6.353 15.910 31.828 159.16 318.31 3 0.885 1.304 2.111 3.008 6.77 9.58 4 0.529 0.717 1.023 1.316 2.29 2.85+ 5 0.388 0.507 0.685+ 0.843 1.32 1.58 6 0.312 0.399 0.523 0.628 0.92 1.07 7 0.263 0.333 0.429 0.507 0.71 0.82 8 0.230 0.288 0.366 0.429 0.59 0.67 9 0.205 0.255 0.322 0.374 0.50 0.57 10 0.186 0.230 0.288 0.333 0.44 0.50 11 0.170 0.210 0.262 0.302 0.40 0.44 12 0.158 0.194 0.241 0.277 0.36 0.40 13 0.147 0.181 0.224 0.256 0.33 0.37 14 0.138 0.170 0.209 0.239 0.31 0.34 15 0.131 0.160 0.197 0.224 0.29 0.32 16 0.124 0.151 0.186 0.212 0.27 0.30 17 0.118 0.144 0.177 0.201 0.26 0.28 18 0.113 0.137 0.168 0.191 0.24 0.26 19 0.108 0.131 0.161 0.182 0.23 0.25+ 20 0.104 0.126 0.154 0.175- 0.22 0.24 The above table was derived from Table 9, page 66, vol. 34, Biometrika, in an article by E. Lord, with the permission of the publishers of Biometrika. TABLES 257 TABLE X Probability Distribution of G = \ xi — i^ |/(mean range) for Two Samples Each of Size n from the Same Normal Population Probability that G will be % ;reater than tabular value Sample Size, n .100 .050 .020 .010 .002 .001 2 2.322 3.427 5.553 7.916 17.81 25.23 3 0.974 1.272 1.715- 2.093 3.27 4.18 4 0.644 0.813 1.047 1.237 1.74 1.99 5 0.493 0.613 0.772 0.896 1.21 1.35+ 6 0.405 + 0.499 0.621 0.714 0.94 1.03 7 0.347 0.426 0.525+ 0.600 0.77 0.85- 8 0.306 0.373 0.459 0.521 0.67 0.73 9 0.275- 0.334 0.409 0.464 0.59 0.64 10 0.250 0.304 0.371 0.419 0.53 0.58 11 0.233 0.280 0.340 0.384 0.48 0.52 12 0.214 0.260 0.315+ 0.3.55 + 0.44 0.48 13 0.201 0.243 0.294 0.331 0.41 0.45- 14 0.189 0.228 0.276 0.311 0.39 0.42 15 0.179 0.216 0.261 0.293 0.36 0.39 16 0.170 0.205- 0.247 0.278 0.34 0.37 17 0.162 0.195+ 0.236 0.264 0.33 0.35+ 18 0.155+ 0.187 0.225+ 0.252 0.31 0.34 19 0.149 0.179 0.216 0.242 0.30 0.32 20 0.143 0.172 0.207 0.232 0.29 0.31 The above table was derived from the same article as Table IX, also with the permission of the publishers of Biometrika. Ind ex A-B blood groups, 56 ACE test scores, 10, 11, 16, 23, 25, 41 Addition theorem, for the chi-square distribution, 141 for probabilities, 53 Approximation of binomial distribution by normal, 87, 101 Arley, Niels, 75, 242 Atkeson, F. W., 140 Averages, arithmetic mean, 13 geometric mean, 20 harmonic mean, 21 median, 19 midrange, 12 mode, 20 properties of, 12 Barmore, Mark, 221 Bartley, E. E., 181 Batting averages, 47, 93 Binomial coefficients, 65, 254 Binomial frequency distribution, 77, 79 Binomial probability function, 68 Bivariate population, 198 Blood factors, A-B, 56 M-N, 60 P, 60 Rh, 60 Buch, K. Rander, 75, 242 Central 95 per cent, 124, 125, 126 Chi-square, empirical distribution for 1 D/F, 133 probability distribution table for 1, 2, and 3 D/F, 252 use in testing Hq{p = po), 132 use in testing i/o(pi = P2), 136 use in testing Ho(pi:p2'-P3 = Pi'-Pi'-Ps), 144 use with contingency tables, 143 CI95, 124 Classes of events, 51 Class interval, length of, 23, 28 midpoint of, 28 Clopper, C. J., 126 Coefficient of correlation, product-mo- ment, 217, 224 rank, 227 Coefficient of linear regression, 199, 224 Compound probability law, 55 Confidence coefficient, 122, 165, 214 Confidence interval, based on G-distri- bution, 183 length of, 169 observed examples of, 166 on the correlation coefficient, p, 220 on the mean, n, 165 on Ml — H2, 179 on fiy.x, 213 on Mj/.x-, 214 on the proportion, p, in a binomial population, 124, 127 on the regression coefficient, /3, 211 Contingency tables, 136, 143 Control limits, 148 Correlation coefficient, computation of, 217, 224 hypotheses about, 218 product-moment, 217 rank, 227 Cumulative distribution curve, 23, 25 Cumulative frequency table, 23, 25 Curvilinear trends, 196, 197 Decisions based on samples, 114 Degrees of freedom {D/F), for chi- square, 134, 142 for estimate of correlation, 209, 218 for estimate of regression, 209, 211 for estimate of standard deviation of regression, 6, 211 259 260 INDEX Degrees of freedom {D/F), for estimate of standard deviation of X, 162 for estimate of variance about trend line, 210 for <-test, 163, 178, 211, 218 Deviations from the mean, 14 Dice, classes of events, 52 Difference between two means, 176, 179 Distribution, binomial frequency, graph, 78 cumulative, 23, 25 normal frequency, curve, 91, 93 normal frequency, formula, 90, 92 normal probability, curve, 96 normal probability, table, 249 of chi-square, 133, 252 of correlation, r, 218 of difference between means, xi — X2, 176 of mean, x, 155, 157, 158 of t, 162, 251 ofz = (l/2)log,[(l +r)/(l -r)],219 standard normal, 92, 249 Dixon, Wilfrid J., 48, 112, 152, 191, 242 Dot (Scatter) diagram, 194 Efficient estimates, 123, 160 Elastic prices, 232 Empirical distribution, of chi-square, 133 of correlation coefficient, 218 of difference between sample means, 176 of sample means, 155, 157 of t, 9 D/F, 162 of z, 219 Error of the first kind, 116 Error of the second kind, 116 Estimated average Y for a given X, 204, 206 Estimation, of the linear correlation co- efficient, 217 of the linear regression coefficient, 205, 211 of the mean of a normal population, 160 of the mean Y for a given X, 204, 212 of the percentage, p, for a binomial population, 122 of the standard deviation, about the linear trend line, 210 Estimation, of the standard deviation, of the regression coefficient, b, 211 of the sample mean, 161 of X, 160 point, 122 unbiased, 122 Events, dependent, 55 exhaustive set of, 51 independent, 54 mutually exclusive, 53 Expected gain or loss, 71 Expected number, 70 Factorial n, 64 Finney, K. F., 221 Fisher, R. A., 5, 219 Foster, Jackson W., 45 Frequency curve for, chi-square with 1 D/F, 133 cumulative standard normal, 96 cumulative t with 4 and 24 D/F, 163 sample correlation coefficient, 10 D/F, 219 standard normal, 93 t with 9 D/F, 162 Frequency distribution tables, 23 Freund, John E., 48, 242 (?-test for, one random sample, 182 two random samples, 183 Galileo, 3 Galton, Sir Francis, 3, 226 Geometric mean, 20 Gosset, William Seely ("Student"), 4 Grading on the curve, 97 Grant, Eugene L., 152 Graunt, John, 2 Guinea pig gains, table, 39 Guinea pig weights, table, 38 Hald, A., 112, 191, 242 Harmonic mean, 21 Ibsen, H. L., 38 Independent events, 54 Inference, statistical, 114 Interval estimate, of average Y for a given X, 213 of correlation coefficient, 220 of population mean, 164, 166 INDEX 261 Interval estimate, of proportion, p, 123 of regression coefficient, 21 1 of true Y for the iih. individual with a given A', 214 Kendall, M. G., 226, 228 Kenney, John F., 28, 48, 75, 112 Law of compound probability, 55 Law of total probability, 53 Lerner, I. M., 139 Levy, H., 75 Lord, E., 256 Marlatt, Abby, 229 Massey, Frank J., 48, 112, 152, 191, 242 Mean, arithmetic, 13, 155 distribution of, 155, 157 geometric, 20 harmonic, 21 of binomial distribution, 84 of population, 13 of sample, 155 standard deviation of, 161 variance of, 161 Mean deviation, 18 Median, 19 Median, for binomial distribution, 82 for normal distribution, 92 Method of least squares, 205 Midpoint of class interval, 28 Midrange, 12 M-N blood groups, 60 Mode, 20 of normal distribution, 92 Multiplication of probabilities, depend- ent events, 55 independent events, 54 Mutually exclusive events, 53 n factorial, 64 Neiswanger, W. A., 48 Neyman, Jerzy, 5, 152, 191 Normal approximation to a binomial distribution, 86 Normal distribution, cumulative {r.c.f.) curve for, 96 curve, 91, 93 equation for, 90, 92 estimates of parameters, 159 Normal distribution, for any mean and standard deviation, 90 mean of, 90 median of, 90 mode of, 90 standard deviation of, 90 standardized, 92, 93, 96 Normal-arithmetic paper, construction of, 106 use in study of normality, 106 Observed confidence intervals on popu- lation mean, 166 Ogive, 26 Opinion polls, 113, 118 Ordered array, 19 p, confidence interval for, 123 tests of hypotheses regarding, 131 P factor in bloods, 60 Parameter, 114, 122, 153, 159, 199, 217 Pearson, E. S., 5, 126 Pearson, Karl, 3, 5, 226 Pillai, K. S. C, 183 Pine, W. H., Ill Point estimate, p, 122, 123 X, 160 Political arithmetic, 2 Probability, addition formula, 53 conditional, 55 determination of, 51 multiplication formula, 55 of error of the first kind, 116 of error of the second kind, 116 Questionnaire, mailed, 119 r, computation of, 217, 224 observed distribution with 10 D/F, 218, 219 Random sampling, 43, 156 Range, 149, 182 Rank correlation coefficient, 227 Ranked data, 226 Rectification of a logarithmic curve, 103 Region of rejection, 131 Regression coefficient, computation of, 205 interpretation of, 211 test of significance for, 211 J<' 262 INDEX Relative cumulative frequency curve, for binomial distribution, 81, 82 for chi-square distribution, 133 for normal distribution, 96 for t distribution, 162, 163 Relative variability, 40 Rh factor in bloods, 60 Roth, L., 75 Sample, 7, 43 Sampling distribution of, chi-square, 133, 252 correlation coefficient, 219 mean, 155, 157 t, 161, 162, 251 Sampling public opinion, 113, 118 Scatter (Dot) diagram, 194 Semi-log paper, construction of, 103, 104 use, 104, 105 Seymour, Ada, 238 Single events, defined, 50 with equal relative frequencies, 50 with unequal relative frequencies, 56 Skewed distributions, 32 Slope of true regression line, 199, 205 Snedecor, George W., 152, 191, 242 Spearman, C, 3, 226 Staatenkunde, 2 Standard deviation of, a measurement, X, 15 binomial distribution, 84 difference between two means, 177 estimated average Y for a given X, 212 estimated particular Y for a given X, 213 fraction, r/n, 84 regression coefficient, 211 sample mean, 157, 158, 161 Standard deviation of, Y after linear adjustment for A', 210 z, 219 Standard normal units, 92 "Student," 4 t, 161, 178, 211, 213, 214, 218, 222 degrees of freedom for, 163 observed distribution, 9 D/F, 162 r.c.f. distribution, 162, 163 table of probability distribution, 251 Tallying frequency distributions, 24 Tau, T, 227 Taylor, L. W., 139 Testing hypotheses regarding, correla- tion coefficient, 218, 222 difference between correlation coeffi- cients, 222 difference between means, 176 difference between regression coeffi- cients, 222 mean, 170 regression coefficient, 211 theoretical proportions in a binomial population, 130 Tippett, L. H. C, 152 Transformation of the correlation coef- ficient, 219 Unbiased estimate, 122, 161 Variance, 15 Waugh, Albert E., 48, 112 Weiner, A. S., 151 Westergaard, Harald L., 1 Westerman, Beulah, 235 Wise, George, 17 Woodruff, H. Boyd, 45 2-transformation, 219