Devore J.L., Berk K.N. Modern Mathematical Statistics with Applications

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43. In an experiment designed to study the effects of

illumination level on task performance (“Perfor-

mance of Complex Tasks Under Different Levels

of Illumination,” J. Illumin. Engrg., 1976:

235–242), subjects were required to insert a fine-

tipped probe into the eyeholes of 10 needles in

rapid succession both for a low light level with a

black background and a higher level with a white

background. Each data value is the time (sec)

required to complete the task.

Subject 12345

Black 25.85 28.84 32.05 25.74 20.89

White 18.23 20.84 22.96 19.68 19.50

Subject 6789

Black 41.05 25.01 24.96 27.47

White 24.98 16.61 16.07 24.59

Does the data indicate that the higher level of

illumination yields a decrease of more than 5 s in

true average task completion time? Test the appro-

priate hypotheses using the P-value approach.

44. It has been estimated that between 1945 and 1971,

as many as 2 million children were born to

mothers treated with diethylstilbestrol (DES),

a nonsteroidal estrogen recommended for preg-

nancy maintenance. The FDA banned this drug

in 1971 because research indicated a link with

the incidence of cervical cancer. The article

“Effects of Prenatal Exposure to Diethylstilbestrol

(DES) on Hemispheric Laterality and Spatial

Ability in Human Males” (Hormones Behav.,

1992: 62–75) discussed a study in which 10

males exposed to DES and their unexposed

brothers underwent various tests. This is the sum-

mary data on the results of a spatial ability test:

x ¼ 12:6 (exposed), y ¼ 13:7, and standard error

of mean difference ¼ .5. Test at level .05 to see

whether exposure is associated with reduced

spatial ability by obtaining the P-value.

45. Cushing’s disease is characterized by muscular

weakness due to adrenal or pituitary dysfunction.

To provide effective treatment, it is important to

detect childhood Cushing’s disease as early as

possible. Age at onset of symptoms and age at

diagnosis for 15 children suffering from the

disease were given in the article “Treatment of

Cushing’s Disease in Childhood and Adolescence

by Transphenoidal Microadenomectomy” (New

Engl. J. Med., 1984: 889). Here are the values of

the differences between age at onset of symptoms

and age at diagnosis:

24  12 55 15 30 60 14 21

48  12 25 53 61 69 80

a. Does the accompanying normal probability

plot cast strong doubt on the approximate nor-

mality of the population distribution of differ-

ences?

b. Calculate a lower 95% confidence bound for

the population mean difference, and interpret

the resulting bound.

c. Suppose the (age at diagnosis)  (age at onset)

differences had been calculated. What would

be a 95% upper confidence bound for the

corresponding population mean difference?

46. Example 1.2 describes a study of children’s pri-

vate speech (talking to themselves). The 33 chil-

dren were each observed in about 100 ten-second

intervals in the first grade, and again in the second

and third grades. Because private speech occurs

more in challenging circumstances, the children

were observed while doing their mathematics.

Infant

Method 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Isotopic 1509 1418 1561 1556 2169 1760 1098 1198 1479 1281 1414 1954 2174 2058

Test 1498 1254 1336 1565 2000 1318 1410 1129 1342 1124 1468 1604 1722 1518

Difference 11 164 225 9 169 442 312 69 137 157 54 350 452 540

−1.5 −.5 .5 1.5

−10

−20

−30

−40

−50

−60

−70

−80

z percentile

Difference

518 CHAPTER 10 Inferences Based on Two Samples

The speech was classified as on task (about the

math lesson), off task, or mumbling (the observer

could not tell what was said). Here are the 33 first-

grade mumble scores:

20.8 24.4 19.4 33.3 26.0 56.6 39.5 24.7

21.6 32.1 48.1 19.5 19.2 43.0 26.3 22.7

49.4 35.4 56.8 45.4 28.7 42.2 20.3 20.0

34.0 26.9 48.4 27.6 52.6 5.9 38.5 22.1

22.2

and here are the third-grade mum ble scores:

28.8 57.0 23.9 46.9 50.0 64.6 54.2 55.3

21.4 38.3 78.5 38.1 44.3 11.7 58.6 76.1

76.4 48.6 37.2 69.8 29.1 60.4 57.8 38.7

46.5 50.0 69.6 69.8 59.4 22.7 84.9 42.0

67.2

The numbers are in the same order for each grade;

for example, the third student mumbled in 19.4%

of the intervals in the first grade and 23.9% of the

intervals in the third grade.

a. Verify graphically that normality is plausible

for the population distribution of differences.

b. Find a 95% confidence interval for the differ-

ence of population means, and interpret the

result.

47. Construct a paired data set for which t ¼1,so

that the data is highly significant when the correct

analysis is used, yet t for the two-sample t test is

quite near zero, so the incorrect analysis yields an

insignificant result.

10.4

Inferences About Two Population

Proportions

Having presented methods for com paring the means of two different populations,

we now turn to the comparison of two population proportions. The nota tion for this

problem is an extension of the notation used in the corresponding one-population

problem. We let

and

denote the proportions of individuals in populations 1

and 2, respectively, who possess a particular characteristic. Alternatively, if we use

the label S for an individual who possesses the characteristic of interest (does favor

a particular proposition, has read at least one book within the last month, etc.), then

and

represent the probabilities of seeing the label S on a randomly chosen

individual from populations 1 and 2, respectively.

We will assume the availability of a sample of m individuals from the first

population and n from the second. The variables X and Y will represent the number

of individuals in each sample possessing the char acteristic that defines

and

Provided the population sizes are much larger than the sample sizes, the distribution

of X can be taken to be binomial with parameters m and

, and similarly, Y is taken

to be a binomial variable with parameters n and

. Furthermore, the samples are

assumed to be independent of each other, so that X and Y are independent rv’s.

The obvious estimator for



, the difference in population proportions, is

the corresponding difference in sample proportions X /m  Y/n. With

¼ X=m and

¼ Y=n, the estimator of



can be expressed as



PROPOSITION

Let X ~ Bin(m,

)andY ~ Bin(n,

)withX and Y independent variables. Then

Eð



Þ¼p

 p



is an unbiased estimator of



, and

Vð



Þ¼

ðwhere q

¼ 1  p

Þð10:3Þ

10.4 Inferences About Two Population Proportions 519

Proof Since E(X) ¼ mp

and E(Y) ¼ np





EðXÞ

EðYÞ¼



¼ p

 p

Since V(X) ¼ mp

, V(Y) ¼ np

, and X and Y are independent,





¼ V



þ V



VðXÞþ

VðYÞ¼

■

We will focus first on situations in which both m and n are large. Then

because

and

individually have approximately normal distributions, the esti-

mator



also has approximately a normal distribution. Standardizing



yields a variable Z whose distribution is approximately standard normal:

Z ¼



ðp

 p

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

A Large-Sample Test Procedure

Analogously to the hypotheses for m

 m

, the most general null hypothesis an

investigator might consider would be of the form



¼ D

, where D

again a specified number. Although for population means the case D

6¼ 0 pre-

sented no difficulties, for population proportions the cases D

¼ 0 and D

6¼ 0 must

be considered separately. Since the vast majority of actual problems of this sort

involve D

¼ 0 (i.e., the null hypothesis

), we will concentrate on this case.

When



¼ 0 is true, let

denote the common value of

and

(and

similarly for q). Then the standardized variable

Z ¼



 0

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ



ð10:4Þ

has approximat ely a standar d normal distribution when

is true. However, this Z

cannot serve as a test statistic because the value of

is unknown—

asserts only

that there is a common value of

, but does not say what that value is. To obtain a

test statistic having approximately a standard normal distribution when

is true

(so that use of an appropriate z critical value specifies a level a test),

must be

estimated from the sample data.

Assuming then that

, instead of separate samples of size m and n

from two different populations (two different binomial distributions), we really

have a single sample of size m + n from one population with proportion

. Since the

total number of individuals in this combined sample having the characteristic of

interest is X + Y, the estimator of

p ¼

X þ Y

m þ n

ð10:5Þ

520 CHAPTER 10 Inferences Based on Two Samples

The second expression for

p shows that it is actually a weighted average of

estimators

and

obtained from the two samples. If we take (10.5) (with

q ¼ 1 

p) and substitute back into ( 10.4), the resulting statistic has approximately

a standard normal distribution when

is true.

Null hypothesis:



¼ 0

Test statistic value (large samples): z ¼



ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ



Alternative Hypothesis Rejection Region for Approximate

Level a Test



> 0 z  z



< 0 z z



6¼ 0 either z  z

a/2

or z z

a/2

A P-value is calculated in the same way as for previous z tests.

Example 10.11 Some defendants in criminal proceedings plead guilty and are sentenced without a

trial, whereas others who plead innocent are subsequently found guilty and then are

sentenced. In recent years, legal scholars have speculated as to whether sentences

of those who plead guilty differ in severity from sentences for those who plead

innocent and are subsequently judged guilty. Consider the accompanying data on

defendants from San Francisco County accused of robbery, all of whom had

previous prison records (“Does It Pay to Plead Guilty? Differential Senten cing

and the Functioning of Criminal Courts,” Law Soc. Rev., 1981–198 2: 45–69). Does

this data suggest that the proportion of all defendants in these circumstances who

plead guilty and are sent to prison differs from the proportion who are sent to prison

after pleading innocent and being found guilty?

Plea

Guilty Not guilty

Number judged guilty m ¼ 191 n ¼ 64

Number sentenced to prison x ¼ 101 y ¼ 56

Sample proportion

¼ .529

¼ .875

Let

and

denote the two population proportions. The hypotheses of

interest are



¼ 0 versus



6¼ 0. At level .01,

should be

rejected if either z  z

.005

¼ 2.58 or if z 2.58. The com bined estimate of the

common success proportion is

p ¼ 101 þ 56ðÞ= 191 þ 64ðÞ¼:616. The value of

the test statistic is then

z ¼

:529  :875

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ð:616Þð:384Þ

191



:346

:070

¼4:94

10.4 Inferences About Two Population Proportions 521

Since 4.94 2.58,

must be rejected.

The P-value for a two-tailed z test is

P-value ¼ 2[1  F(|z|)] ¼ 2[1  F(4.94)] < 2[1  F(3.49)] ¼ .0004

A more extensive standard normal table yields P-value ¼ .0000006. This P-value

is so minuscule that at any reasonable level a,

should be rejected. The data very

strongly suggests that

6¼ p

and, in particular, that initially pleading guilty may

be a good strategy as far as avoiding prison is concerned.

The cited article also reported data on defendants in several other counties.

The auth ors broke down the data by type of crime (burglary or robbery) and by

nature of prior record (none, some but no prison, and prison). In every case, the

conclusion was the same: Among defendants judged guilty, those who pleaded that

way were less likely to receive prison sentences.

■

Type II Error Probabilities and Sample Sizes

Here the determination of b is a bit more cumbersome than it was for other

large-sample tests. The reason is that the denominator of Z is an estimate of the

standard deviation of



, assuming that

. When

is false,



must be restandardized using



ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ð10:6Þ

The form of s implies that b is not a function of just



, so we denote it by

)

Alternative Hypothesis b(

)



> 0

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

p q



ðp

 p



< 0

1  F

z

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

p q



ðp

 p



6¼ 0

a=2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

p q



ðp

 p

 F

z

a=2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

p q



ðp

 p

where p ¼ðmp

þ np

Þ=ðm þ nÞ, q ¼ðmq

þ nq

Þ=ðm þ nÞ, and s is given

by (10.6).

522 CHAPTER 10 Inferences Based on Two Samples

Proof For the upper-tailed test (



> 0),

bðp

; p

Þ¼P



< z

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ



¼ P



ðp

 p

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ



ðp

 p

When m and n are both large,

p ¼

þ n

m þ n



þ np

m þ n

and

q 

q, which yields the previous (approximate) expressio n for b(

). ■

Alternatively, for specified

with



¼ d, the sample sizes neces-

sary to achieve b(

) ¼ b can be determined. For example, for the upper-tailed

test, we equate z

to the argument of F(·) (i.e., what’s inside the parentheses) in

the foregoing box. If m ¼ n, there is a simple expression for the com mon value.

For the case m ¼ n, the level a test has type II error probability b at the

alternative values

with



¼ d when

n ¼

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ðp

þ p

Þðq

þ q

Þ=2

þ z

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

þ p



ð10:7Þ

for an upper- or lower-tailed test, with a/2 replacing a for a two-tailed test.

Example 10.12 One of the truly impressive applications of statistics occurred in connection with

the design of the 1954 Salk polio vaccine experiment and analysis of the resulting

data. Part of the experiment focused on the efficacy of the vaccine in combating

paralytic polio. Because it was thought that without a control group of children,

there would be no sound basis for assessment of the vaccine, it was decided to

administer the vaccine to one group and a placebo injection (visually indistinguish-

able from the vaccine but known to have no effect) to a control group. For ethical

reasons and also because it was thought that the knowledge of vaccine administra-

tion might have an effect on treatment and diagnosis, the experiment was con-

ducted in a double-blind manner. That is, neither the individuals receiving

injections nor those administering them actually knew who was receiving vaccine

and who was receiving the placebo (samples were numerically coded)—remember,

at that point it was not at all clear whether the vaccine was benefic ial.

Let

and

be the probabilities of a child getting paralytic polio for the

control and treatment conditions, respectively. The objective was to test the

hypotheses



¼ 0 versus



> 0 (the alternative hypothesis

10.4 Inferences About Two Population Proportions 523

states that a vaccinated child is less likely to contract polio than an unvaccinated

child). Supposing the true value of

is .0003 (an incidence rate of 30 per 100,000),

the vaccine would be a significant improvement if the incidence rate was halved—

that is,

¼ .00015. Using a level a ¼ .05 test, it would then be reasonabl e to ask

for sample sizes for which b ¼ .1 when

¼ .0003 and

¼ .00015. Assuming

equal sample sizes, the required n is obtained from (10.7)as

n ¼

1:645

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ð:5Þð:00045Þð:199955Þ

þ 1:28

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ð:00015Þð:99985Þþð:0003Þð:9997Þ



ð:0003  :00015Þ

¼½ð:0349 þ :0271Þ=:00015

 171; 000

The actual data for this experiment follows. Sample sizes of approximately

200,000 were used. The reader can easily verify that z ¼ 6.43, a highly significant

value. The vaccine was judged a resounding success!

Placebo: m ¼ 201,229 x ¼ number of cases of paralytic polio ¼ 110

Vaccine: n ¼ 200,745 y ¼ 33

■

A Large-Sample Conﬁdence Interval for



As with means, many two-sample problems involve the objective of comparison

through hypothesis testing, but sometimes an interval estimate for



appropriate. Both

¼ X=m and

¼ Y=n have approximate normal distributions

when m and n are both large. If we identify y with



, then

y ¼



satisfies the conditions necessary for obtaining a large-sample CI. In particular, the

estimated standard deviation of

y is

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

=mÞþð

=nÞ

. The 100(1  a)%

interval

y  z

a=2



then becomes



 z

a=2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

Notice that the estimated standard deviation of



(the square-root expression)

is different here from what it was for hypothesis tes ting when D

¼ 0.

Recent research has shown that the actual confidence level for the traditional

CI just given can sometimes deviate substantia lly from the nominal level (the level

you think you are getting when you use a particular z critical value—e.g., 95%

when z

a/2

¼ 1.96). The suggested improvement is to add one success and one

failure to each of the two samples and then replace the

p’s and

q’s in the foregoing

formula by

p’s and

q’s where

¼ðx þ 1Þ=ðm þ 2Þ, etc. This interval can also be

used when sample sizes are quite small.

Example 10.13 The authors of the article “Adjuvant Radiotherapy and Chemotherapy in Node-

Positive Premenopausal Women with Breast Cancer” (New Engl. J. Med., 1997:

956–962) reported on the results of an experiment designed to compare treating

cancer patients with only chemotherapy to treatment with a combination of chemo-

therapy and radiation. Of the 154 individuals who received the chemotherapy-only

treatment, 76 survived at least 15 years, whereas 98 of the 164 patients who

received the hybrid treatment survived at least that long. With

denoting the

proportion of all such women who, when treated with just chemotherapy, survive at

524 CHAPTER 10 Inferences Based on Two Samples

least 15 years and

denoting the analogous proportion for the hybrid treatment,

¼ 76=154 ¼ :494 and

¼ 98=164 ¼ :598. A confidence interval for the dif-

ference between proportions based on the traditional formula with a confidence

level of approximately 99% is

:494  :598  2:58

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ð:494Þð:506Þ

154

ð:598Þð:402Þ

164

¼:104  :143 ¼ð:247;:039Þ

At the 99% confidence level, it is plausible that .247 <



< .039. This

interval is reasonably wide, a reflec tion of the fact that the sample sizes are not

terribly large for this type of interval. Notice that 0 is one of the plausible values of



suggesting that neither treatment can be judged superior to the other. Using

¼ 77=156 ¼ :494,

¼ 79=156 ¼ :506,

¼ :596,

¼ :404 based on sample

sizes of 156 and 166, respectively, the “improved” interval here is essentially

identical to the earlier interval.

■

Small-Sample Inferences

On occasion an inference concerning



may have to be based on samples for

which at least one sample size is small. Appropriate methods for such situatio ns are

not as straightforward as those for large samples, and there is more controversy

among statisticians as to reco mmended procedures. One frequently used test, called

the Fisher–Irwin test, is based on the hypergeometric distribution.

Exercises Section 10.4 (48–59)

48. Is someone who switches brands because of a

financial inducement less likely to remain loyal

than someone who switches without induce-

ment? Let

and

denote the true proportions

of switchers to a certain brand with and without

inducement, respectively, who subsequently

make a repeat purchase. Test



¼ 0

versus



< 0 using a ¼ .01 and the

following data:

m ¼200 number of successes ¼ 30

n ¼600 number of successes ¼ 180

(Similar data is given in “Impact of Deals and

Deal Retraction on Brand Switching,” J. Market-

ing, 1980: 62–70.)

49. A sample of 300 urban adult residents of a par-

ticular state revealed 63 who favored increasing

the highway speed limit from 55 to 65 mph,

whereas a sample of 180 rural residents yielded

75 who favored the increase. Does this data indi-

cate that the sentiment for increasing the speed

limit is different for the two groups of residents?

a. Test

versus

6¼

using

a ¼ .05, where

refers to the urban population.

b. If the true proportions favoring the increase

are actually

¼ .20 (urban) and

¼ .40

(rural), what is the probability that

will

be rejected using a level .05 test with

m ¼ 300, n ¼ 180?

50. It is thought that the front cover and the nature of

the first question on mail surveys influence the

response rate. The article “The Impact of Cover

Design and First Questions on Response Rates

for a Mail Survey of Skydivers” (Leisure Sci.,

1991: 67–76) tested this theory by experimenting

with different cover designs. One cover was

plain; the other used a picture of a skydiver.

The researchers speculated that the return rate

would be lower for the plain cover.

Cover

Number

Sent

Number

Returned

Plain 207 104

Skydiver 213 109

Does this data support the researchers’ hypothesis?

Test the relevant hypotheses using a ¼ .10byfirst

calculating a P-value.

10.4 Inferences About Two Population Proportions 525

51. Do teachers find their work rewarding and satisfy-

ing? The article “Work-Related Attitudes” (Psych.

Rep., 1991: 443–450) reports the results of a survey

of 395 elementary school teachers and 266 high

school teachers. Of the elementary school teachers,

224 said they were very satisfied with their jobs,

whereas 126 of the high school teachers were very

satisfied with their work. Estimate the difference

between the proportion of all elementary school

teachers who are satisfied and all high school tea-

chers who are satisfied by calculating a CI.

52. A random sample of 5726 telephone numbers

from a certain region taken in March 2002

yielded 1105 that were unlisted, and 1 year later

a sample of 5384 yielded 980 unlisted numbers.

a. Test at level .10 to see whether there is a

difference in true proportions of unlisted

numbers between the 2 years.

b. If

¼ .20 and

¼ .18, what sample sizes

(m ¼ n) would be necessary to detect such a

difference with probability .90?

53. Ionizing radiation is being given increasing

attention as a method for preserving horticultural

products. The article “The Influence of Gamma-

Irradiation on the Storage Life of Red Variety

Garlic” (J. Food Process. Preserv., 1983:

179–183) reports that 153 of 180 irradiated garlic

bulbs were marketable (no external sprouting,

rotting, or softening) 240 days after treatment,

whereas only 119 of 180 untreated bulbs were

marketable after this length of time. Does this

data suggest that ionizing radiation is beneficial

as far as marketability is concerned?

54. In medical investigations, the ratio y ¼

often of more interest than the difference



(e.g., individuals given treatment 1 are how

many times as likely to recover as those given

treatment 2?). Let

y ¼

. When m and n are

both large, the statistic lnð

yÞ has approximately a

normal distribution with approximate mean

value ln(y) and approximate standard deviation

[(m  x)/(mx)+(n  y)/(ny)]

1/2

a. Use these facts to obtain a large-sample 95%

CI formula for estimating ln(y), and then a CI

for y itself.

b. Return to the heart attack data of Example 1.3,

and calculate an interval of plausible values for

y at the 95% confidence level. What does this

interval suggest about the efficacy of the aspi-

rin treatment?

55. Sometimes experiments involving success or

failure responses are run in a paired or before/

after manner. Suppose that before a major policy

speech by a political candidate, n individuals are

selected and asked whether (S) or not (F) they

favor the candidate. Then after the speech the

same n people are asked the same question. The

responses can be entered in a table as follows:

After

Before

where X

+ X

¼ n. Let

and

denote the four cell probabilities, so that

¼ P(S before and S after), and so on. We wish

to test the hypothesis that the true proportion of

supporters (S) after the speech has not increased

against the alternative that it has increased.

a. State the two hypotheses of interest in terms

, and

b. Construct an estimator for the after/before

difference in success probabilities.

c. When n is large, it can be shown that the rv

 X

)/n has approximately a normal distri-

bution with variance [

 (



)

]/n.

Use this to construct a test statistic with

approximately a standard normal distribution

when

is true (the result is called

McNemar’s test).

d. If x

¼ 350, x

¼ 150, x

¼ 200, and

¼ 300, what do you conclude?

56. The Chicago Cubs won 73 games and lost 71 in

1995. This was described as a much more suc-

cessful season for them than 1994, when they

won only 49 and lost 64.

a. Based on a binomial model with

for 1994

and

for 1995, carry out a two-tailed test for

the difference. Based on your result, could the

difference in sample proportions be attributed

to luck (bad in 1994, good in 1995)?

b. Criticize the binomial model. Do baseball

games satisfy the assumptions?

57. Using the traditional formula, a 95% CI for



is to be constructed based on equal sample sizes

from the two populations. For what value of n (¼m)

526

CHAPTER 10 Inferences Based on Two Samples

will the resulting interval have width at most .1

irrespective of the results of the sampling?

58. Statin drugs are used to decrease cholesterol levels,

and therefore hopefully to decrease the chances of

a heart attack. In a British study (“MRC/BHF

Heart Protection Study of Cholesterol Lowering

with Simvastin in 20,536 High-Risk Individuals:

A Randomized Placebo-Controlled Trial,” Lancet,

2002: 7–22) 20,536 at-risk adults were assigned

randomly to take either a 40-mg statin pill or

placebo. The subjects had coronary disease, artery

blockage, or diabetes. After 5 years there were

1328 deaths (587 from heart attack) among the

10,269 in the statin group and 1507 deaths (707

from heart attack) among the 10,267 in the placebo

group.

a. Give a 95% confidence interval for the

difference in population death proportions.

b. Give a 95% confidence interval for the differ-

ence in population heart attack death propor-

tions.

c. Is it reasonable to say that most of the difference

in death proportions is due to heart attacks, as

would be expected?

59. A study of male navy enlisted personnel was

reported in the Bloomington, Illinois, Daily Pan-

tagraph, Aug. 23, 1993. It was found that 90 of

231 left-handers had been hospitalized for inju-

ries, whereas 623 of 2148 right-handers had been

hospitalized for injuries. Test for equal population

proportions at the .01 level, find the P-value for

the test, and interpret your results. Can it be con-

cluded that there is a causal relationship between

handedness and proneness to injury? Explain.

10.5

Inferences About Two Population Variances

Methods for comparing two population variances (or standard deviations) are

occasionally needed, though such problems arise much less frequently than those

involving means or proportions. For the case in which the populations under

investigation are normal, the procedures are based on the F distribution, as dis-

cussed in Section 6.4.

Testing Hypotheses

A test procedure for hypotheses concerning the ratio s

, as well as a CI for this

ratio are based on the following result from Section 6.4 .

THEOREM

Let X

, ...,X

be a random sample from a normal distribution with variance

, let Y

, ...,Y

be another random sample (independent of the X

’s) from a

normal distribution with variance s

, and let S

and S

denote the two sample

variances. Then the rv

F ¼

ð10:8Þ

has an F distribution with n

¼ m  1 and v

¼ n  1.

Under the null hypothesis of equal population variances, (10.8) reduces to the

ratio of sample variances. For a test statistic we use this ratio of sample variances;

and the claim that s

¼ s

is rejected if the ratio differs by too much from 1.

10.5 Inferences About Two Population Variances 527