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resulted in t ¼ 2.51 and P-value ¼ .019. Does

the appropriate test procedure described in this

chapter yield this value of t and the stated

P-value?

b. Is there substantial evidence for concluding

that true average strength for males exceeds

that for females by more than 25 N? State

and test the relevant hypotheses.

97. The article “Pine Needles as Sensors of Atmo-

spheric Pollution” (Environ. Monitor., 1982:

273–286) reported on the use of neutron-activity

analysis to determine pollutant concentration in

pine needles. According to the article’s authors,

“These observations strongly indicated that for

those elements which are determined well by the

analytical procedures, the distribution of concentra-

tion is lognormal. Accordingly, in tests of signifi-

cance the logarithms of concentrations will be

used.” The given data refers to bromine concentra-

tion in needles taken from a site near an oil-fired

steam plant and from a relatively clean site. The

summary values are means and standard deviations

of the log-transformed observations.

Site

Sample

Size

Mean Log

Concentration

SD of Log

Concentration

Steam

plant

8 18.0 4.9

Clean 9 11.0 4.6

Let m



be the true average log concentration at

the first site, and define m



analogously for the

second site.

a. Use the pooled t test (based on assuming nor-

mality and equal standard deviations) to decide

at significance level .05 whether the two con-

centration distribution means are equal.

b. If s



and s



, the standard deviations of the two

log concentration distributions, are not equal,

would m

and m

, the means of the concentra-

tion distributions, be the same if m



¼ m



Explain your reasoning.

98. Torsion during hip external rotation (ER) and

extension may be responsible for certain kinds of

injuries in golfers and other athletes. The article

“Hip Rotational Velocities during the Full Golf

Swing” (J. Sport Sci. Med., 2009: 296–299)

reported on a study in which peak ER velocity

and peak IR (internal rotation) velocity (both in

deg/s) were determined for a sample of 15 female

collegiate golfers during their swings. The follow-

ing data was supplied by the article’s authors.

Golfer ER IR diff z perc

1 130.6 98.9 31.7 1.28

2 125.1 115.9 9.2 0.97

3 51.7 161.6 109.9 0.34

4 179.7 196.9 17.2 0.73

5 130.5 170.7 40.2 0.34

6 101.0 274.9 173.9 0.97

7 24.4 275.0 250.6 1.83

8 231.1 275.7 44.6 0.17

9 186.8 214.6 27.8 0.52

10 58.5 117.8 59.3 0.00

11 219.3 326.7 107.4 0.17

12 113.1 272.9 159.8 0.73

13 244.3 429.1 184.8 1.28

14 184.4 140.6 43.8

1.83

15 199.2 345.6 146.4 0.52

a. Is it plausible that the differences came from a

normally distributed population?

b. The article reported that MeanðSD) ¼

145:368:0ðÞfor ER velocity and ¼

227.8(96.6) for IR velocity. Based just on

this information, could a test of hypotheses

about the difference between true average IR

velocity and true average ER velocity be car-

ried out? Explain.

c. Do an appropriate hypothesis test about the

difference between true average IR velocity

and true average ER velocity and interpret the

result.

99. The accompanying summary data on the ratio of

strength to cross-sectional area for knee extensors

is taken from the article “Knee Extensor and Knee

Flexor Strength: Cross-Sectional Area Ratios in

Young and Elderly Men” ( J. Gerontol., 1992:

M204–M210).

Group

Sample

Size

Sample

Mean

Standard

Error

Young 13 7.47 .22

Elderly men 12 6.71 .28

Does this data suggest that the true average ratio

for young men exceeds that for elderly men?

Carry out a test of appropriate hypotheses using

a ¼ .05. Be sure to state any assumptions neces-

sary for your analysis.

548

CHAPTER 10 Inferences Based on Two Samples

100. The accompanying data on response time

appeared in the article “The Extinguishment of

Fires Using Low-Flow Water Hose Streams—

Part II” (Fire Techn., 1991: 291–320). The

samples are independent, not paired.

Good

visibility

.43 1.17 .37 .47 .68 .58 .50 2.75

Poor

visibility

1.47 .80 1.58 1.53 4.33 4.23 3.25 3.22

The authors analyzed the data with the pooled

t test. Does the use of this test appear justified?

[Hint: Check for normality. The normal scores

for n ¼ 8 are 1.53, .89, .49, .15, .15, .49,

.89, and 1.53.]

101. The accompanying data on the alcohol content of

wine is representative of that reported in a study

in which wines from the years 1999 and 2000

were randomly selected and the actual content

was determined by laboratory analysis (London

Times, Aug. 5, 2001).

Wine 123456

Actual 14.2 14.5 14.0 14.9 13.6 12.6

Label 14.0 14.0 13.5 15.0 13.0 12.5

The two-sample t test gives a test statistic value

of .62 and a two-tailed P-value of .55. Does this

convince you that there is no significant difference

between true average actual alcohol content and

true average content stated on the label? Explain.

102. The article “The Accuracy of Stated Energy

Contents of Reduced-Energy, Commercially

Prepared Foods” (J. Am. Diet. Assoc., 2010:

116–123) presented the accompanying data on

vendor-stated gross energy and measured value

(both in kcal) for 10 different supermarket

convenience meals):

Meal 123456789 10

Stated 180 220 190 230 200 370 250 240 80 180

Measured 212 319 231 306 211 431 288 265 145 228

Obtain a 95% confidence interval for the differ-

ence of population means. By roughly what per-

centage are the actual calories higher than the

stated value?

Note that the article calls this a convenience

sample and suggests that therefore it should have

limited value for inference. However, even if the

ten meals were a random sample from their local

store, there could still be a problem in drawing

conclusions about a purchase at your store.

103. How does energy intake compare to energy

expenditure? One aspect of this issue was con-

sidered in the article “Measurement of Total

Energy Expenditure by the Doubly Labelled

Water Method in Professional Soccer Players”

(J. Sports Sci., 2002: 391–397), which contained

the accompanying data (MJ/day).

Player 1234567

Expenditure 14.4 12.1 14.3 14.2 15.2 15.5 17.8

Intake 14.6 9.2 11.8 11.6 12.7 15.0 16.3

Test to see whether there is a significant differ-

ence between intake and expenditure. Does the

conclusion depend on whether a significance

level of .05, .01, or .001 is used?

104. An experimenter wishes to obtain a CI for the

difference between true average breaking

strength for cables manufactured by company I

and by company II. Suppose breaking strength is

normally distributed for both types of cable with

¼ 30 psi and s

¼ 20 psi.

a. If costs dictate that the sample size for the

type I cable should be three times the sample

size for the type II cable, how many observa-

tions are required if the 99% CI is to be no

wider than 20 psi?

b. Suppose a total of 400 observations is to be

made. How many of the observations should

be made on type I cable samples if the width

of the resulting interval is to be a minimum?

105. An experiment to determine the effects of tempera-

ture on the survival of insect eggs was described in

the article “Development Rates and a Temperature-

Dependent Model of Pales Weevil” (Environ.

Entomol., 1987: 956–962). At 11



C, 73 of 91 eggs

survived to the next stage of development. At 30



102 of 110 eggs survived. Do the results of this

experiment suggest that the survival rate (propor-

tion surviving) differs for the two temperatures?

Calculate the P-value and use it to test the appro-

priate hypotheses.

106. The insulin-binding capacity (pmol/mg protein)

was measured for four different groups of rats:

(1) nondiabetic, (2) untreated diabetic, (3) diabetic

treated with a low dose of insulin, (4) diabetic

treated with a high dose of insulin. The accompa-

nying table gives sample sizes and sample stan-

dard deviations. Denote the sample size for the ith

treatment by n

and the sample variance by

ði ¼ 1; 2; 3; 4Þ. Assuming that the true variance

for each treatment is s

, construct a pooled esti-

mator of s

that is unbiased, and verify using rules

Supplementary Exercises 549

of expected value that it is indeed unbiased. What

is your estimate for the following actual data?

[Hint: Modify the pooled estimator S

from Sec-

tion 10.2.]

Treatment

1234

Sample Size 16 18 8 12

Sample SD .64 .81 .51 .35

107. Suppose a level .05 test of

: m

 m

¼ 0

versus

: m

 m

> 0 is to be performed,

assuming s

¼ s

¼ 10 and normality of both

distributions, using equal sample sizes (m ¼ n).

Evaluate the probability of a type II error when

 m

¼ 1 and n ¼ 25, 100, 2500, and 10,000.

Can you think of real problems in which the

difference m

 m

¼ 1 has little practical signif-

icance? Would sample sizes of n ¼ 10,000 be

desirable in such problems?

108. The following data refers to airborne bacteria

count (number of colonies/ft

) both for m ¼ 8

carpeted hospital rooms and for n ¼ 8 uncar-

peted rooms (“Microbial Air Sampling in a

Carpeted Hospital,” J. Environ. Health, 1968:

405). Does there appear to be a difference in

true average bacteria count between carpeted

and uncarpeted rooms?

Carpeted 11.8 8.2 7.1 13.0 10.8 10.1 14.6 14.0

Uncarpeted 12.1 8.3 3.8 7.2 12.0 11.1 10.1 13.7

Suppose you later learned that all carpeted rooms

were in a veterans’ hospital, whereas all uncarpeted

rooms were in a children’s hospital. Would you be

able to assess the effect of carpeting? Comment.

109. Researchers sent 5000 resumes in response to job

ads that appeared in the Boston Globe and Chicago

Tribune. The resumes were identical except that

2500 of them had “white sounding” first names,

such as Brett and Emily, whereas the other 2500

had “black sounding” names such as Tamika and

Rasheed. The resumes of the first type elicited 250

responses and the resumes of the second type only

167 responses (these numbers are very consistent

with information that appeared in a January 15,

2003, report by the Associated Press). Does this

data strongly suggest that a resume with a “black”

name is less likely to result in a response than is a

resume with a “white” name?

110. McNemar’s test, developed in Exercise 55, can

also be used when individuals are paired

(matched) to yield n pairs and then one member

of each pair is given treatment 1 and the other is

given treatment 2. Then X

is the number of pairs

in which both treatments were successful, and

similarly for X

, X

, and X

. The test statistic

for testing equal efficacy of the two treatments

is given by ðX

 X

Þ=

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

þ X

, which has

approximately a standard normal distribution

when

is true. Use this to test whether the

drug ergotamine is effective in the treatment of

migraine headaches.

Ergotamine

Placebo S 44 34

F 46 30

The data is fictitious, but the conclusion agrees

with that in the article “Controlled Clinical Trial

of Ergotamine Tartrate” (British Med. J., 1970:

325–327).

111. Let X

, ... , X

be a random sample from a

Poisson distribution with parameter l

, and let

, ... , Y

be a random sample from another

Poisson distribution with parameter l

. We wish

to test

: l

 l

¼ 0 against one of the three

standard alternatives. Since m ¼ l for a Poisson

distribution, when m and n are large the large-

sample z test of Section 10.1 can be used. How-

ever, the fact that Vð

XÞ¼l=n suggests that a

different denominator should be used in standar-

dizing

X  Y. Develop a large-sample test proce-

dure appropriate to this problem, and then apply

it to the following data to test whether the plant

densities for a particular species are equal in two

different regions (where each observation is the

number of plants found in a randomly located

square sampling quadrat having area 1 m

, so for

region 1, there were 40 quadrats in which one

plant was observed, etc.):

Frequency

01234567

Region 1 28 40 28 17 8 2 1 1 m ¼ 125

Region 2 14 25 30 18 49 2 1 1 n ¼ 140

550

CHAPTER 10 Inferences Based on Two Samples

112. Referring to Exercise 111, develop a large-

sample confidence interval formula for l

 l

Calculate the interval for the data given there

using a confidence level of 95%.

113. Let R

be a rejection region with significance

level a for testing

: y ∈ O

versus

: y =2

, and let R

be a level a rejection region for

testing

: y ∈ O

versus

: y =2 O

, where O

and O

are two disjoint sets of possible values of

y. Now consider testing

: y ∈ O

[ O

versus

the alternative

: y =2 O

[ O

. The proposed

rejection region for this latter test is R

. That

is,

is rejected only if both

and

can

be rejected. This procedure is called a union–

intersection test (UIT).

a. Show that the UIT is a level a test.

b. As an example, let m

denote the mean value

of a particular variable for a generic (test)

drug, and m

denote the mean value of this

variable for a brand-name (reference) drug. In

bioequivalence testing, the relevant hypoth-

eses are

: m

 d

or m

 d

(not

bioequivalent) versus

: d

< m

< d

(bioequivalent). The limits d

and d

are

standards set by regulatory agencies; for cer-

tain purposes the FDA uses .80 and 1.25 ¼

1/.8, respectively. By taking logarithms and

letting  ¼ ln(m), t ¼ ln(d), the hypotheses

become

: either 

 

 t

or  t

versus

: t

<

 

< t

. With this

setup, a type I error involves saying the

drugs are bioequivalent when they are not.

The FDA mandates a ¼ .05.

Let D be an estimator of 

 

with stan-

dard error S

such that standardized variable

T ¼ [D  (

 

)]/S

has a t distribution

with v df. The standard test procedure is

referred to as TOST for “two one-sided

tests,” and is based on the two test statistics

¼ (D  t

)/S

and T

¼ (D  t

)/S

If v ¼ 20, state the appropriate conclusion

in each of the following cases: (1) t

¼ 2.0,

¼1.5; (2) t

¼ 1.5, t

¼2.0; (3)

¼ 2.0, t

¼2.0.

Bibliography

See the bibliography at the end of Chapter 8.

Bibliography 551

CHAPTER ELEVEN

The Analysis

of Variance

Introduction

In studying methods for the analy‘sis of quantitative data, we ﬁrst focused on

problems involving a single sample of numbers and then turned to a comparative

analysis of two different samples. Now we are ready for the analysis of several

samples.

The analysis of variance, or more brieﬂy ANOVA, refers broadly to a collection

of statistical procedures for the analysis of quantitative responses. The simplest

ANOVA problem is referred to variously as a single-factor, single-classiﬁcation,or

one-way ANOVA and involves the analysis of data sampled from two or more

numerical populations (distributions). The characteristic that labels the populations

is called the factor under study, and the populations are referred to as the levels of

the factor. Examples of such situations include the following:

1. An experiment to study the effects of five different brands of gasoline on

automobile engine operating efficiency (mpg)

2. An experiment to study the effects of four different sugar solutions (glucose,

sucrose, fructose, and a mixture of the three) on bacterial growth

3. An experiment to investigate whether hardwood concentration in pulp (%) has

an effect on tensile strength of bags made from the pulp

4. An experiment to decide whether the color density of fabric specimens depends

on the amount of dye used

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

DOI 10.1007/978-1-4614-0391-3_11,

Springer Science+Business Media, LLC 2012

552

In (1) the factor of interest is gasoline brand, and there are ﬁve different

levels of the factor. In (2) the factor is sugar, with four levels (or ﬁve, if a control

solution containing no sugar is used). In both (1) and (2), the factor is qualitative in

nature, and the levels correspond to possible categories of the factor. In (3) and (4),

the factors are concentration of hardwood and amount of dye, respectively; both

these factors are quantitative in nature, so the levels identify different settings of

the factor. When the factor of interest is quantitative, statistical techniques from

regression analysis (discussed in Chapter 12) can also be used to analyze the data.

In this chapter we ﬁrst introduce single-factor ANOVA. Section 11.1 presents

the

test for testing the null hypothesis that the popul ation means are identical.

Section 11.2 considers further analysis of the data when

has been rejected.

Section 11.3 covers some other aspec ts of single-factor ANOVA. Many experimen-

tal situations involve study ing the simultaneous impact of more than one factor.

Various aspects of two-factor ANOVA are considered in the last two sections of the

chapter.

11.1

Single-Factor ANOVA

Single-factor ANOVA focuses on a comparison of two or more populations. Let

I ¼ the number of treatments being compared

¼ the mean of population 1 (or the true average response when

treatment 1 is applied)

¼ the mean of population I (or the true average response when

treatment I is applied)

Then the hypotheses of interest are

: m

¼ m

¼¼m

versus

: at least two of the m

’s are different

If I ¼ 4, H

is true only if all four m

’s are iden tical. H

would be true, for example,

if m

¼ m

6¼ m

¼ m

,ifm

¼ m

6¼ m

, or if all four m

’s differ from each

other.

A test of these hypotheses requires that we have available a random sample

from each population or treatment.

Example 11.1 The article “Compression of Single-Wall Corrugated Shipping Container s Using

Fixed and Floating Test Platens” (J. Test. Eval., 1992: 318–320) describes an

experiment in which several different types of boxes were compared with respect

to compression strength (lb). Table 11.1 presents the results of a single-fa ctor

ANOVA experiment involving I ¼ 4 types of boxes (the sample means and

standard deviations are in good agreement with values given in the article).

11.1 Single-Factor ANOVA 553

With m

denoting the true average compression strength for boxes of type i (i ¼ 1, 2,

3, 4), the null hypothesis is H

: m

¼ m

. Figure 11.1(a) shows a

comparative boxplot for the four samples. There is a substantial amount of overlap

among observations on the first three types of boxes, but compression strengths for

the fourth type appear considerably smaller than for the other types. This suggests

that H

is not true. The comparative boxplot in Figure 11.1(b) is based on adding 120

Table 11.1 The data and summary quantities for Example 11.1

Type of box Compression strength (lb) Sample mean Sample SD

1 655.5 788.3 734.3 713.00 46.55

721.4 679.1 699.4

2 789.2 772.5 786.9 756.93 40.34

686.1 732.1 774.8

3 737.1 639.0 696.3 698.07 37.20

671.7 717.2 727.1

4 535.1 628.7 542.4 562.02 39.87

559.0 586.9 520.0

Grand mean ¼

682.50

630

660 690 750720 780

550

600 650

700 750

Figure 11.1 Boxplots for Example 11.1: (a) original data; (b) altered data

554

CHAPTER 11 The Analysis of Variance

to each observation in the fourth sample (giving mean 682.02 and the same standard

deviation) and leaving the other observations unaltered. It is no longer obvious

whether H

is true or false. In situations such as this, we need a formal test

procedure.

■

Notation and Assumptions

In two-sample problems, we used the letters X and Y to designate the observations

in the two samples. Because this is cumbersome for three or more samples, it is

customary to use a single letter with two subscripts. The first subscript identifies the

sample number, corresponding to the population or treatment being sampled, and

the second subscript denotes the position of the observation within that sample. Let

¼ the random variable (rv) denoting the jth measurement from the ith

population

¼ the observed value of X

when the experiment is performed

The observed data is usually displayed in a rectangular table, such as

Table 11.1 . There samples from the different populations appear in different rows

of the table, and x

i,j

is the jth number in the ith row. For example, x

2,3

¼ 786.9 (the

third observation from the secon d population), and x

4,1

¼ 535.1. When there is no

ambiguity, we will write x

rather than x

i,j

(e.g., if there were 15 observations on

each of 12 treatments, x

112

could mean x

1,12

or x

11,2

). It is assumed that the X

’

within any particular sample are independent—a random sample from the ith

population or treatment distribution— and that different samples are independent

of each other.

In some experiments, different samples contain different numbers of obser-

vations. However, the concepts and methods of single-factor ANOVA are most

easily developed for the case of equal sample sizes. Unequal sample sizes will be

considered in Section 11.3. Restricting ourselves for the moment to equal sample

sizes, let J denote the number of observations in each sample (J ¼ 6 in Example

11.1). The data set consists of IJ observations. The individual sample means will

be denoted by

1

; X

2

; ..., X

I

. That is,

i

j¼1

i ¼ 1; 2; ...; I

The dot in place of the second subscript signifies that we have added over all values

of that subscript while holding the other subscript value fixed, and the horizontal

bar indicates division by J to obtain an average. Similarly, the average of all IJ

observations, called the grand mean,is



i¼1

j¼1

For the strength data in Table 11. 1 ,



1

¼ 713:00,



2

¼ 756:93,



3

¼ 698:07,



¼ 562:02, and





¼ 682:50. Additionally, let S

; S

; ...; S

represent the sample

variances:

11.1 Single-Factor ANOVA 555

j¼1

ðX

 X

i

J  1

i ¼ 1; 2; ...; I

From Example 11.1, s

¼ 46.55, s

¼ 2166:90, and so on.

ASSUMPTIONS

The I population or treatment distributions are all normal with the same

variance s

. That is, each X

is normally distributed with

EðX

Þ¼m

VðX

Þ¼s

In previous chapters, a normal probability plot was suggested for checking

normality. The individual sample sizes in ANOVA are typically too small for I

separate plots to be informat ive. A single plot can be constructed by subtracting



1

from each observation in the first sample,



2

from each observation in the second,

and so on, and then plotting these IJ deviations against the z percentiles. The

deviations are called residuals so this plot is the normal plot of the residuals.

Figure 11.2 gives the plot for the residuals of Example 11.1. The straightness of the

pattern gives strong support to the normality assumption.

At the end of the section we discuss Levene’s test for the equal variance

assumption. For the moment, a rough rule of thumb is that if the largest s is not

much more than twice the smallest s, it is reasonable to assume equal variances.

This is especially true if the sample sizes are equal or close to equal. In Example

11.1, the largest s is only about 1.25 times the smallest.

Sums of Squares and Mean Squares

If H

is true the J observations in each sample come from a normal population

distribution with the same mean value m, in which case the sample means



1

;



2

; ...



: should be reasonably close. The test procedure is based on comparing

−1.4 −.7 0 .7 1.4

−50

z percentile

Deviation

Figure 11.2 A normal probability plot based on the data of Example 11.1

556

CHAPTER 11 The Analysis of Variance

a measure of differences among these sample means (“between-samples” variation)

to a measure of variation calculated from within each sample. These measures

involve quantities called sums of squares.

DEFINITION

The treatment sum of squares SSTr is given by

SSTr ¼ J

ðX

i

 X



¼ J½ðX

1

 X



þþðX

I

 X





and the error sum of squares SSE is

SSE ¼

ðX

 X

i

ðX

 X

1

þþ

ðX

 X

I

¼ðJ  1ÞS

þðJ  1ÞS

þþðJ  1ÞS

¼ðJ  1Þ½S

þ S

þþS



Now recall a result from Section 6.4 :ifX

, ..., X

is a random sample from a normal

distribution with mean m and variance s

, then the sample mean X and the sample

variance S

are independent. Also, X is normally distributed, and (n1)S

[i.e.,

 XðÞ

] has a chi-squared distribution with n  1 df. That is, dividing

the sum of squares

 XðÞ

by s

gives a chi-squared random variable. Similar

results hold in our ANOVA situation.

THEOREM

When the basic assumptions of this section are satisfied, SSE/s

has a chi-

squared distribution with I(J – 1) df (each sample contributes J–1 df and df’s

add because the samples are independent). Furthermore, when H

is true,

SSTr/s

has a chi-squared distribution with I – 1 df [there are I deviations

1

 X



; ...; X

I

 X



but 1 df is lost because

ðX

i

 X



Þ¼0]. Lastly,

SSE and SSTr are independent rando m variables.

If we let Y

¼ X

i

; i ¼ 1; ...; I,thenY

, Y

, ..., Y

are independent and normally

distributed with the same mean under H

and with variance s

/J. Thus, by the key

result from Section 6.4, I  1ðÞS

= s

=JðÞhas a chi-squared distribution with I –1df.

Furthermore, I  1ðÞS

= s

=JðÞ¼J

i

 X



ðÞ

¼ SSTr=s

, so SSTr=s



I1

. Independence of SSTr and SSE follows from the fact that SSTr is based on the

individual sample means whereas SSE is based on the sample variances, and

i

independent of S

for each i.

The expected value of a chi-squared variable with

df is just

.Thus

SSE



¼ IðJ  1Þ)E

SSE

IðJ  1Þ



¼ s

11.1 Single-Factor ANOVA 557