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
-
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-
•
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-
• >• •••——f»»»»i<«f«|— ■«»■■■■■ •— — > —
-
-
••'^i^^^^Pri* • -
-
• • •• — f •j»— ^ — • —
1 1 1 1 J
_ o
(3
® t^
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s? =s
5 -Tf
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r/3 X!
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00
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m
M Si
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^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
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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
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