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b. Obtain the diagonal elements of the hat

matrix (leverages). Identify the point with

the highest leverage. What is unusual about

the point? Given the instructor’s assertion that

there were no students in the class less than

five feet tall, would you say that there was an

error? Give another reason that this student’s

measurements seem wrong.

c. For the other points with high leverages, what

distinguishes them from the points with ordi-

nary leverage values?

d. Examining the residuals, find another student

whose data might be wrong.

e. Discuss the elimination of questionable points

in order to obtain valid regression results.

104. Here is a method for obtaining the variance of the

residuals in simple (one predictor) linear regres-

sion, as given by Equation (12.13).

a. We have shown in Equations (12.26) and

(12.27) that Covð

YÞ¼s

H and

CovðY 

YÞ¼ s

ðI  HÞ. Show therefore

that VðY



Þ¼s

 Vð

Þ.

b. Use part (a) and Vð

Þ from Section 12.4 to

show that for simple linear regression,

VðY



Þ¼s

 1 



ðx

 xÞ

Supplementary Exercises (105–121)

105. The presence of hard alloy carbides in high

chromium white iron alloys results in excellent

abrasion resistance, making them suitable for

materials handling in the mining and materials

processing industries. The accompanying data

on x ¼ retained austenite content (%) and y ¼

abrasive wear loss (mm

) in pin wear tests with

garnet as the abrasive was read from a plot in

the article “Microstructure-Property Relation-

ships in High Chromium White Iron Alloys”

(Internat. Mater. Rev., 1996: 59–82).

x 4.6 17.0 17.4 18.0 18.5 22.4 26.5 30.0 34.0

y .66 .92 1.45 1.03 .70 .73 1.20 .80 .91

x 38.8 48.2 63.5 65.8 73.9 77.2 79.8 84.0

y 1.19 1.15 1.12 1.37 1.45 1.50 1.36 1.29

SAS output for Exercise 105

Analysis of Variance

Source DF Sum of Squares Mean Square F Value Prob > F

Model 1 0.63690 0.63690 15.444 0.0013

Error 15 0.61860 0.04124

C Total 16 1.25551

Root MSE 0.20308 R-square 0.5073

Dep Mean 1.10765 Adj R-sq 0.4744

C.V. 18.33410

Parameter Estimates

Variable DF

Parameter

Estimate

Standard

Error

T for H0:

Parameter ¼ 0

Prob

> |T|

INTERCEP 1 0.787218 0.09525879 8.264 0.0001

AUSTCONT 1 0.007570 0.00192626 3.930 0.0013

718 CHAPTER 12 Regression and Correlation

a. What proportion of observed variation in

wear loss can be attributed to the simple lin-

ear regression model relationship?

b. What is the value of the sample correlation

coefficient?

c. Test the utility of the simple linear regression

model using a ¼ .01.

d. Estimate the true average wear loss when

content is 50% and do so in a way that con-

veys information about reliability and preci-

sion.

e. What value of wear loss would you predict

when content is 30%, and what is the value of

the corresponding residual?

106. An investigation was carried out to study the

relationship between speed (ft/s) and stride rate

(number of steps taken/s) among female mara-

thon runners. Resulting summary quantities

included n ¼ 11, S(speed) ¼ 205.4, S(speed)

¼ 3880.08, S(rate) ¼ 35.16, S(rate)

¼ 112.681, and S(speed)(rate) ¼ 660.130.

a. Calculate the equation of the least squares

line that you would use to predict stride rate

from speed.

b. Calculate the equation of the least squares

line that you would use to predict speed from

stride rate.

c. Calculate the coefficient of determination

for the regression of stride rate on speed of

part (a) and for the regression of speed on

stride rate of part (b). How are these related?

d. How is the product of the two slope esti-

mates related to the value calculated in (c)?

107. In Section 12.4, we presented a formula for

the variance Vð



Þ and a CI for

þ b



. Taking x

¼ 0 gives s

and a CI

for b

. Use the data of Example 12.12 to cal-

culate the estimated standard deviation of

and a 95% CI for the y-intercept of the true

regression line.

108. Show that SSE ¼ S



, which gives an

alternative computational formula for SSE.

109. Suppose that x and y are positive variables and

that a sample of n pairs results in r  1. If the

sample correlation coefficient is computed for

the (x, y

) pairs, will the resulting value also be

approximately 1? Explain.

110. Let s

and s

denote the sample standard devia-

tions of the observed x’s and y’s, respectively [so

ðx

 xÞ

=ðn  1Þ and similarly for s

a. Show that an alternative expression for the

estimated regression line

x is

y ¼

y þ r 

ðx  x Þ

b. This expression for the regression line can

be interpreted as follows. Suppose r ¼ .5.

What then is the predicted y for an x that

lies 1 SD (s

units) above the mean of the

’s? If r were 1, the prediction would be for

y to lie 1 SD above its mean

y, but since

r ¼ .5, we predict a y that is only .5 SD (.5s

unit) above y. Using the data in Exercise 62

for a patient whose age is 1 SD below the

average age in the sample, by how many

standard deviations is the patient’s predicted

DCBG above or below the average DCBG

for the sample?

111. In biofiltration of wastewater, air discharged

from a treatment facility is passed through a

damp porous membrane that causes contami-

nants to dissolve in water and be transformed

into harmless products. The accompanying data

on x ¼ inlet temperature (



C) and y ¼ removal

efficiency (%) was the basis for a scatter plot

that appeared in the article “Treatment of

Mixed Hydrogen Sulfide and Organic Vapors

in a Rock Medium Biofilter”(Water Environ.

Res., 2001: 426–435).

Obs Temp Removal

1 7.68 98.09 17 8.55 98.27

2 6.51 98.25 18 7.57 98.00

3 6.43 97.82 19 6.94 98.09

4 5.48 97.82 20 8.32 98.25

5 6.57 97.82 21 10.50 98.41

6 10.22 97.93 22 16.02 98.51

7 15.69 98.38 23 17.83 98.71

8 16.77 98.89 24 17.03 98.79

9 17.13 98.96 25 16.18 98.87

10 17.63 98.90 26 16.26 98.76

11 16.72 98.68 27 14.44 98.58

12 15.45 98.69 28 12.78 98.73

13 12.06 98.51 29 12.25 98.45

14 11.44 98.09 30 11.69 98.37

15 10.17 98.25 31 11.34 98.36

16 9.64 98.36 32 10.97 98.45

Calculated summary quantities are

¼ 384:26, Sy

¼ 3149:04,

5099:2412, Sx

¼ 37; 850:7762, and

309;892:6548.

Supplementary Exercises 719

a. Does a scatter plot of the data suggest appro-

priateness of the simple linear regression

model?

b. Fit the simple linear regression model, obtain

a point prediction of removal efficiency when

temperature ¼ 10.50, and calculate the value

of the corresponding residual.

c. Roughly what is the size of a typical deviation

of points in the scatter plot from the least

squares line?

d. What proportion of observed variation in

removal efficiency can be attributed to the

model relationship?

e. Estimate the slope coefficient in a way that

conveys information about reliability and pre-

cision, and interpret your estimate.

f. Personal communication with the authors of

the article revealed that one additional observa-

tion was not included in their scatter plot: (6.53,

96.55). What impact does this additional obser-

vation have on the equation of the least squares

line and the values of s and r

112. Normal hatchery processes in aquaculture inev-

itably produce stress in fish, which may nega-

tively impact growth, reproduction, flesh

quality, and susceptibility to disease. Such

stress manifests itself in elevated and sustained

corticosteroid levels. The article “Evaluation of

Simple Instruments for the Measurement of

Blood Glucose and Lactate, and Plasma Protein

as Stress Indicators in Fish”(J. World Aquacult.

Soc., 1999: 276–284) described an experiment

in which fish were subjected to a stress protocol

and then removed and tested at various times

after the protocol had been applied. The accom-

panying data on x ¼ time (min) and y ¼ blood

glucose level (mmol/L) was read from a plot.

x 2 2 5 7 12 13 17 18 23 24 26 28

y 4.0 3.6 3.7 4.0 3.8 4.0 5.1 3.9 4.4 4.3 4.3 4.4

x 29 30 34 36 40 41 44 56 56 57 60 60

y 5.8 4.3 5.5 5.6 5.1 5.7 6.1 5.1 5.9 6.8 4.9 5.7

Use the methods developed in this chapter to

analyze the data, and write a brief report sum-

marizing your conclusions (assume that the

investigators are particularly interested in glu-

cose level 30 min after stress).

113. The article “Evaluating the BOD POD for

Assessing Body Fat in Collegiate Football

Players” (Med. Sci. Sports Exercise, 1999:

1350–1356) reports on a new air displacement

device for measuring body fat. The customary

procedure utilizes the hydrostatic weighing

device, which measures the percentage of

body fat by means of water displacement.

Here is representative data read from a graph

in the paper.

BOD 2.5 4.0 4.1 6.2 7.1 7.0 8.3 9.2 9.3 12.0 12.2

HW 8.0 6.2 9.2 6.4 8.6 12.2 7.2 12.0 14.9 12.1 15.3

BOD 12.6 14.2 14.4 15.1 15.2 16.3 17.1 17.9 17.9

HW 14.8 14.3 16.3 17.9 19.5 17.5 14.3 18.3 16.2

a. Use various methods to decide whether it is

plausible that the two techniques measure on

average the same amount of fat.

b. Use the data to develop a way of predicting

an HW measurement from a BOD POD

measurement, and investigate the effective-

ness of such predictions.

114. Reconsider the situation of Exercise 105, in

which x ¼ retained austenite content using a

garnet abrasive and y ¼ abrasive wear loss

were related via the simple linear regression

model Y ¼ b

+ b

x + e. Suppose that for a sec-

ond type of abrasive, these variables are also

related via the simple linear regression model

Y ¼ g

+ g

x + e and that V(e) ¼ s

for both

types of abrasive. If the data set consists of n

observations on the first abrasive and n

on the

second and if SSE

and SSE

denote the two

error sums of squares, then a pooled estimate of

¼ SSE

þ SSE

ðÞ= n

þ n

 4ðÞ. Let

and SS

denote

ðx

 xÞ

for the data

on the first and second abrasives, respectively.

A test of H

: b

 g

¼ 0 (equal slopes) is based

on the statistic

T ¼



ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

When H

is true, T has a t distribution with

+ n

 4 df. Suppose the 15 observations

using the alternative abrasive give SS

¼ 7152.5578,

¼ :006845, and SSE

¼ .51350. Using this along with the data of

Exercise 105, carry out a test at level .05 to see

whether expected change in wear loss associated

with a 1% increase in austenite content is identi-

cal for the two types of abrasive.

720 CHAPTER 12 Regression and Correlation

115. Show that the ANOVA version of the model

utility test discussed in Section 12.3 (with test

statistic F ¼ MSR/MSE) is in fact a likelihood

ratio test for H

: b

¼ 0 versus H

: b

6¼ 0. [Hint:

We have already pointed out that the least

squares estimates of b

and b

are the mle’s.

What is the mle of b

when H

is true? Now

determine the mle of s

both in O (when b

not necessarily 0) and in O

(when H

is true).]

116. Show that the t ratio version of the model utility

test is equivalent to the ANOVA F statistic ver-

sion of the test. Equivalent here means that

rejecting H

: b

¼ 0 when either t  t

a/2,n2

t t

a/2, n2

is the same as rejecting H

when f

 F

a,1,n2

117. When a scatter plot of bivariate data shows a

pattern resembling an exponentially increasing

or decreasing curve, the following multiplicative

exponential model is often used: Y ¼ ae

 e.

a. What does this multiplicative model imply

about the relationship between Y

¼ ln(Y) and

x?[Hint: take logs on both sides of the model

equation and let b

¼ ln(a), b

¼ b, e

¼ ln

(e), and suppose that e has a lognormal distri-

bution.]

b. The accompanying data resulted from an

investigation of how ethylene content of let-

tuce seeds (y, in nL/g dry wt) varied with

exposure time (x , in min) to an ethylene

absorbent (“Ethylene Synthesis in Lettuce

Seeds: Its Physiological Significance,” Plant

Physiol., 1972: 719–722).

x 2 20 20 30 405060708090100

y 408 274 196 137 90 78 51 40 30 22 15

Fit the simple linear regression model to this

data, and check model adequacy using the

residuals.

c. Is a scatter plot of the data consistent with the

exponential regression model? Fit this model

by first carrying out a simple linear regression

analysis using ln(y) as the dependent variable

and x as the independent variable. How good a

fit is the simple linear regression model to the

“transformed” data [the (x, ln(y)) pairs]? What

are point estimates of the parameters a and b?

d. Obtain a 95% prediction interval for ethylene

content when exposure time is 50 min. [Hint:

first obtain a PI for ln(y) based on the simple

linear regression carried out in (c).]

118. No tortilla chip afficionado likes soggy chips, so

it is important to identify characteristics of the

production process that produce chips with an

appealing texture. The following data on x ¼

frying time (sec) and y ¼ moisture content (%)

appeared in the article “Thermal and Physical

Properties of Tortilla Chips as a Function of

Frying Time” (J. Food Process. Preserv., 1995:

175–189).

x 5 10152025304560

y 16.3 9.7 8.1 4.2 3.4 2.9 1.9 1.3

a. Construct a scatter plot of the data and com-

ment.

b. Construct a scatter plot of the [ln(x), ln(y)]

pairs (i.e. transform both x and y by logs) and

comment.

c. Consider the multiplicative power model Y ¼

e. What does this model imply about the

relationship between y

¼ ln(y) and x

¼ ln

(x) (assuming that e has a lognormal distribu-

tion)?

d. Obtain a prediction interval for moisture con-

tent when frying time is 25 s. [Hint: first carry

out a simple linear regression of y

on x

and

calculate an appropriate prediction interval.]

119. The article “Determination of Biological Matu-

rity and Effect of Harvesting and Drying Condi-

tions on Milling Quality of Paddy” (J. Agric.

Engr. Res., 1975: 353–361) reported the follow-

ing data on date of harvesting (x, the number of

days after flowering) and yield of paddy, a grain

farmed in India (y, in kg/ha).

x 16 18 20 22 24 26 28 30

y 2508 2518 3304 3423 3057 3190 3500 3883

x 32 34 36 38 40 42 44 46

y 3823 3646 3708 3333 3517 3241 3103 2776

a. Construct a scatter plot of the data. What

model is suggested by the plot?

b. Use a statistical software package to fit the

model suggested in (a) and test its utility.

c. Use the software package to obtain a predic-

tion interval for yield when the crop is har-

vested 25 days after flowering, and also a

confidence interval for expected yield in

situations where the crop is harvested

Supplementary Exercises 721

25 days after flowering. How do these two

intervals compare to each other? Is this result

consistent with what you learned in simple

linear regression? Explain.

d. Use the software package to obtain a PI and

CI when x ¼ 40. How do these intervals com-

pare to the corresponding intervals obtained

in (c)? Is this result consistent with what you

learned in simple linear regression? Explain.

e. Carry out a test of hypotheses to decide

whether the quadratic predictor in the model

fit in (b) provides useful information about

yield (presuming that the linear predictor

remains in the model).

120. The article “Validation of the Rockport Fitness

Walking Test in College Males and Females”

(Res. Q. Exercise Sport, 1994: 152–158) recom-

mended the following estimated regression equa-

tion for relating y ¼ VO

max (L/min, a measure

of cardiorespiratory fitness) to the predictors x

¼ gender (female ¼ 0, male ¼ 1), x

¼ weight

(lb), x

¼ 1-mile walk time (min), and x

heart rate at the end of the walk (beats/min):

y ¼ 3:5959 þ :6566x

þ :0096x

 :0996x

 :0080x

a. How would you interpret the estimated coef-

ficient .0996?

b. How would you interpret the estimated coef-

ficient .6566?

c. Suppose that an observation made on a male

whose weight was 170 lb, walk time was

11 min, and heart rate was 140 beats/min

resulted in VO

max ¼ 3.15. What would

you have predicted for VO

max in this situa-

tion, and what is the value of the

corresponding residual?

d. Using SSE ¼ 30.1033 and SST ¼ 102.3922,

what proportion of observed variation in

max can be attributed to the model rela-

tionship?

e. Assuming a sample size of n ¼ 20, carry out a

test of hypotheses to decide whether the cho-

sen model specifies a useful relationship

between VO

max and at least one of the pre-

dictors.

121. A sample of n ¼ 20 companies was selected, and

the values of y ¼ stock price and k ¼ 15 predic-

tor variables (such as quarterly dividend, previ-

ous year’s earnings, and debt ratio) were

determined. When the multiple regression

model using these 15 predictors was fit to the

data, R

¼ .90 resulted.

a. Does the model appear to specify a useful

relationship between y and the predictor vari-

ables? Carry out a test using significance level

.05. [Hint: The F critical value for 15 numer-

ator and 4 denominator df is 5.86.]

b. Based on the result of part (a), does a high R

value by itself imply that a model is useful?

Under what circumstances might you be sus-

picious of a model with a high R

value?

c. With n and k as given previously, how large

would R

have to be for the model to be

judged useful at the .05 level of significance?

Bibliography

Chatterjee, Samprit, Ali Hadi, and Bertram Price,

Regression Analysis by Example (4th ed.), Wiley,

New York, 2006. A brief but informative discus-

sion of selected topics.

Daniel, Cuthbert, and Fred Wood, Fitting Equations to

Data (2nd ed.), Wiley, New York, 1980. Contains

many insights and methods that evolved from the

authors’ extensive consulting experience.

Draper, Norman, and Harry Smith, Applied Regression

Analysis (3rd ed.), Wiley, New York, 1998. A com-

prehensive and authoritative book on regression.

Hoaglin, David, and Roy Welsch, “The Hat Matrix

in Regression and ANOVA,” American Statisti-

cian, 1978: 17–23. Describes methods for

detecting influential observations in a regression

data set.

Kutner, Michael, Christopher Nachtsheim, John

Neter, and William Li, Applied Linear Statistical

Models (5th ed.), McGraw-Hill, New York, 2005.

The first 14 chapters constitute an extremely

readable and informative survey of regression

analysis.

722

CHAPTER 12 Regression and Correlation

CHAPTER THIRTEEN

Goodness-of-Fit

Tests and

Categorical Data

Analysis

Introduction

In the simplest type of situation considered in this chapter, each observation

in a sample is classiﬁed as belonging to one of a ﬁnite number of categories

(For example, blood type could be one of the four categories O, A, B, or AB).

With

denoting the probability that any particular observation belongs in

category

(or the proportion of the population belonging to category

), we wish

to test a null hypothesis that completely speciﬁes the values of all the

’s (such

¼ .45,

¼ .35,

¼ .15,

¼ .05, when there are four categories).

The test statistic will be a measure of the discrepancy between the observed

numbers in the categories and the expected numbers when

is true. Because a

decision will be reached by comparing the computed value of the test statistic to

a critical value of the chi-squared distribution, the procedure is called a chi-squared

goodness-of-ﬁt test.

Sometimes the null hypothesis speciﬁes that the

’s depend on some

smaller number of parameters without specifying the values of these parameters.

For example, with three categories the null hypothesis might state that

¼ y

¼ 2y(1 – y), and

¼ (1 – y)

. For a chi-squared test to be performed, the

values of any unspeciﬁed parameters must be estimated from the sample data.

These problems are discussed in Section 13.2. The methods are then applied to test

a null hypothesis that states that the sample comes from a particular family of

distributions, such as the Poisson family (with

estimated from the sample) or the

normal family (with m and s estimated).

Chi-squared tests for two different situations are presented in Section 13.3.

In the ﬁrst, the null hypothesis states that the

’s are the same for several different

populations. The second type of situation involves taking a sample from a single

population and classifying each individual with respect to two different categorical

J.L. Devore and K.N. Berk, Modern Mathematical Statistics with Applications, Springer Texts in Statistics,

DOI 10.1007/978-1-4614-0391-3_13,

Springer Science+Business Media, LLC 2012

723

factors (such as religious preference and political party registration). The null

hypothesis in this situation is that the two factors are independent within the

population.

13.1

Goodness-of-Fit Tests When Category

Probabilities Are Completely Specified

A binomial experiment consists of a sequence of independent trials in which each

trial can result in one of two possible outcomes, S (for success) and F (for failure).

The probability of success, denoted by p, is assumed to be constant from trial to

trial, and the number n of trials is fixed at the outset of the experiment. In Chapter 9,

we presented a large-sample z test for testing H

: p ¼ p

. Notice that this null

hypothesis specifies both P(S) and P(F ), since if P(S) ¼ p

, then P(F ) ¼ 1–p

Denoting P(F )byq and 1 – p

by q

, the null hypot hesis can alternatively be

written as H

: p ¼ p

, q ¼ q

. The z test is two-tailed when the alternative of

interest is p 6¼ p

A multinomial experiment generalizes a binomial experiment by allowing

each trial to result in one of k possible outcomes, where k  2. For example,

suppose a store accepts three different types of credit cards. A multinomial experi-

ment would result from observing the type of credit card used—type 1, type 2, or

type 3—by each of the next n customers who pay with a credit card. In general, we

will refer to the k possible outcomes on any given trial as categories, and p

will

denote the probability that a trial results in category i. If the experiment consists of

selecting n individuals or objects from a population and categorizing each one, then

is the proportion of the population falling in the ith category (such an experiment

will be approximately multinom ial provided that n is much smaller than the

population size).

The null hypothesis of interest will specify the value of each p

. For example,

in the case k ¼ 3, we might have H

: p

¼ .5, p

¼ .3, p

¼ .2. The alternative

hypothesis will state that H

is not true—that is, that at least one of the p

’s has a

value different from that asserted by H

(in which case at least two must be

different, since they sum to 1). The symbol p

will represent the value of p

claimed

by the null hypothesis. In the example just given, p

¼ .5, p

¼ .3, and p

¼ .2.

Before the multinomial experiment is performed, the number of trials that

will result in category i (i ¼ 1, 2, ... ,ork) is a random variable—j ust as the

number of successes and the number of failures in a binomial experiment are

random variables. This random variable will be denoted by N

and its observed

value by n

. Since each trial results in exactly one of the k categories, SN

¼ n, and

the same is true of the n

’s. As an example, an experiment with n ¼ 100 and k ¼ 3

might yield N

¼ 46, N

¼ 35, and N

¼ 19.

The expected number of successes and expected number of failures in a

binomial experiment are np and nq, respective ly. When H

: p ¼ p

, q ¼ q

true, the expected numbers of successes and failures are np

and nq

, respectively.

Similarly, in a multinomial experiment the expected number of trials resulting in

category i is E(N

) ¼ np

(i ¼ l, ..., k). When H

: p

¼ p

, ... , p

¼ p

is true,

these expected values become E(N

) ¼ np

, E(N

) ¼ np

, ... , E(N

) ¼ np

For the case k ¼ 3, H

: p

¼ .5, p

¼ .3, p

¼ .2, and n ¼ 100, we have

E(N

) ¼ 100(.5) ¼ 50, E(N

) ¼ 30, and E(N

) ¼ 20 when H

is true. The n

’s

724 CHAPTER 13 Goodness-of-Fit Tests and Categorical Data Analysis

are often displayed in a tabular format consisting of a row of k cells, one for each

category, as illustrated in Table 13.1. The expected values when H

is true are

displayed just below the observed values. The N

’s and n

’s are usually referred to

as observed cell counts (or observed cell frequencies), and np

, np

, ..., np

are

the corresponding expected cell counts under H

The n

’s should all be reasonably close to the corresponding np

’s when

is true. On the other hand, several of the observed counts should differ

substantially from these expected counts when the actual values of the p

’s differ

markedly from what the null hypothesis asserts. The test proce dure involves

assessing the discrepancy between the n

’s and the np

’s, with H

being rejected

when the discrepancy is sufficiently large. It is natural to base a measure of

discrepancy on the squared deviations (n

–np

)

,(n

–np

)

, ... ,(n

–np

)

An obvious way to combine these into an overall measure is to add them together

to obtain S(n

–np

)

. However, suppose np

¼ 100 and np

¼ 10. Then if

¼ 95 and n

¼ 5, the two categories contribute the same squared deviations

to the proposed measure. Yet n

is only 5% less than what would be expected when

is true, whereas n

is 50% less. To take relative magnitudes of the deviations

into account, we will divide each squared deviation by the corresponding expected

count and then combine.

Before giving a more detailed description, we mus t discuss the chi-squared

distribution. This distribution was introduced in Section 4.4, discussed in

Section 6.4, and used in Chapter 8 to obtain a confidence interval for the variance

of a normal population. The chi-squared distribution has a single parameter,

called the number of degrees of freedom (df) of the distribution, with possible

values 1, 2, 3, ... . Analogous to the critical value t

a,n

for the t distribution, w

a; n

the value such that a of the area under the w

curve with n df lies to the right of w

a;n

(see Figure 13.1). Selected values of w

a;n

are given in Appendix Table A.6.

Table 13.1 Observed and expected cell counts

Shaded area

curve

Figure 13.1 A critical value for a chi-squared distribution

13.1 Goodness-of-Fit Tests When Category Probabilities Are Completely Speciﬁed 725

THEOREM

Provided that np

 5 for every i (i ¼ 1, 2, ..., k), the variable

i¼1

ðN

 np

all cells

ðobserved  expectedÞ

expected

has approximately a chi-squared distribution with k – 1 df.

The fact that df ¼ k – 1 is a consequence of the restriction SN

¼ n. Although there are k

observed cell counts, once any k – 1 are known, the remaining one is uniquely deter-

mined. That is, there are only k – 1 “freely determined” cell counts, and thus k –1df.

If np

is substituted for np

in w

, the resulting test statistic has approximately

a chi-squared distribution when H

is true. Rejection of H

is appropriate when

 c (because large discrepancies between observed and expected count s lead to

a large value of w

), and the choice c ¼ w

a;k1

yields a test with significance level a.

Null hypothesis: H

: p

¼ p

, p

¼ p

, ... , p

¼ p

Alternative hypothesis: H

: at least one p

does not equal p

Test statistic value: w

all cells

(observedexpected)

expected

i¼1

ðn

 np

i 0

Rejection region: w

 w

a; k1

Example 13.1 If we focus on two different characteristics of an organism, each controlled by a

single gene, and cross a pure strain having genotype AABB with a pure strain

having genotype aabb (capital letters denoting dominant alleles and small letters

recessive alleles), the resulting genotype will be AaBb. If these first-generation

organisms are then crossed among themselves (a dihybrid cross), there will be four

phenotypes depending on whether a dominan t allele of either type is present.

Mendel’s laws of inheritance imply that these four phenotypes should have prob-

abilities 9/16, 3/16, 3/16, and 1/16 of arising in any given dihybrid cross.

The article “Linkage Studies of the Tomato” ( Trans. Royal Canad. Institut.,

1931: 1–19) reports the following data on phenotypes from a dihybrid cross of tall

cut-leaf tomatoes with dwarf pota to-leaf tomatoes. There are k ¼ 4 categories

corresponding to the four possible phenotypes, with the null hypothesis being

: p

; p

The expected cell counts are 9n/16, 3n/16, 3n/16, and n/16, and the test is based on

k –1¼ 3 df. The total sample size was n ¼ 1611. Observed and expected counts

are given in Table 13.2.

Table 13.2 Observed and expected cell counts for Example 13.1

726 CHAPTER 13 Goodness-of-Fit Tests and Categorical Data Analysis

The contribution to w

from the first cell is

ðn

 np

ð926  906:2Þ

906:2

¼ :433

Cells 2, 3, and 4 contribute .658, .274, and .108, respectively, so

¼ .433 + .658 + .274 + .108 ¼ 1.473. A test with significance level .10

requires w

:10;3

, the number in the 3 df row and .10 column of Appendix Table

A.6. This critical value is 6.251. Since 1.473 is not at least 6.251, H

cannot be

rejected even at this rather large level of significance. The data is quite consistent

with Mendel’s laws.

■

Consider the special case of just two categor ies, k ¼ 2. The null hypothesis

in this case can be stated as H

: p

¼ p

, because the relations p

¼ 1–p

and

¼ 1–p

make the inclusion of p

¼ p

in H

redundant. The alternative

hypothesis is H

: p

6¼ p

. These hypotheses can also be tested using a two-tailed

z test with test statistic

Z ¼

ðN

=nÞp

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ð1  p

 p

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

Surprisingly, the two test procedures are completely equivalent. This is

because it can be shown that Z

¼ w

and ðz

¼ w

a;1

, so that w

 w

a;1

if and

only if (iff) |Z|  z

a/2

If the alternative hypothesis is either H

: p

> p

or H

< p

, the chi-squared test cannot be used. One must then revert to an upper- or

lower-tailed z test.

As is the case with all test procedures, one must be careful not to confuse

statistical significance with practical significance. A computed w

that exceeds

a;k1

may be a result of a very large sample size rather than any practical

differences between the hypothesized p

’s and true p

’s. Thus if

¼ p

, but the true p

’s have values .330, .340, and .330, a large

value of w

is sure to arise with a sufficiently large n. Be fore rejecting H

, the

’s should be examined to see whether they suggest a model different from that

of H

from a practical point of view.

P-Values for Chi-Squared Tests

The chi-squared tests in this chapter are all upper-tailed, so we focus on this case.

Just as the P-value for an upper-tailed t test is the area under the t

curve to the right

of the calculated t, the P-value for an upper-tailed chi-squared test is the area under

the w

curve to the right of the calculated w

. Appendix Table A.6 provides limited

P-value information because only five upper-tail critical values are tabulated for

each different n. We have therefore included Appendix Table A.10, analogous to

Table A.7, that facilitates making more precise P-value statements.

The fact that (z

a/2

)

¼ w

a;1

is a consequence of the relationship between the standard normal distribution

and the chi-squared distribution with 1 df; if Z  N(0, 1), then Z

has a chi-squared distribution with

n ¼ 1. See the first proposition in Section 6.4.

13.1 Goodness-of-Fit Tests When Category Probabilities Are Completely Speciﬁed

727