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

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Let’s apply the b ootstrap to this problem. Begin with a sample from the boys,

standard deviation 5.264, and a sample from the girls, standard deviation 1.505,

with ratio 5.264/1.505 ¼ 3.498. We do this 999 times using the boot package in R,

and the resulting distribution of ratios is shown in Figure 10.13.

The bootstrap distribution is strongly skewed to the right. For a 95% confi-

dence interval, the percentile method uses the middle 95% of the bootstrap distri-

bution. The 2.5 percentile is 1.013 and the 97.5 percentile is 7.888, so the 95%

confidence interval for the population ratio of standard deviations is (1.013, 7.888).

The bias corrected and accelerated (BCa) refinement gives the interval (0.885,

7.382). These two intervals differ in an important respect, that the percentile

interval excludes 1 but the BCa refinement includes 1. In other words, the BCa

interval allows the possibility that the two standard deviations are the same, but the

percentile interval does not. We expect the BCa method to be an improvement, and

this is verified in the next example, where we see that the BCa result is consistent

with the results of a permutation test.

■

Consider using a permutation test for

: s

¼ s

Example 10.20 From Example 10.19 we know that the ratio of sample standard deviations for off-

task private speech, males versus females, is 8.72/2.85 ¼ 3.064. The idea of the

permutation tes t is to find out how unusual this value is if we blur the distinction

between males and females. That is, we remove the labels from the 18 males and

15 females and then consider all possible choices of 18 from the 33 children.

For each of these possible choices we find the ratio of the standard deviation of

the first 18 to the standard deviation of the last 15. The one-tailed P-value is the

fraction that are at least as big as the original value, 3.064. Because there are more

than a billion possible choices of 18 from 33, we instead selected 4999 random

choices. This gives a total of 5000 when the original select ion of males and females

is included. Of these, 432 are at least as big as 3.064, so the one-tailed P-value is

432/5000 ¼ .0864. For a two-tailed P-value we double this and get .1728. The

permutation test does not reject at the 5% level (or the 10% level) the null

hypothesis that the two population standard deviations are the same.

How does the permutation test result compare with the other results? Recall

that the F interval and the percentile interval ruled out the possibility that the two

standard deviations are the same, but the BCa refinement disagreed, because 1 is

included in the BCa interval. Taking it for granted that the permutation test is a valid

approach and the permutation test does not reject the equality of standard deviations,

the BCa interv al is the only one of the three CIs consistent with this result.

■

The Analysis of Paired Data

The bootstrap can be used for paired data if we work with the paired differences, as

in the paired t methods of Section 10.3.

Example 10.21 The private speech study was introduced in Examples 1.2 and 10.16. The study

included the percentage of intervals with on-task private speech for 33 children in

the first, second, and third grades. Here we will consider just the 15 girls in the first

and second grades. Is there a change in on-task private speech when the girls go

from the first to the second grade? Here are the percentages of intervals in which on

task private speech occurred, and also the differences.

538 CHAPTER 10 Inferences Based on Two Samples

Grade 1 Grade 2 Difference

25.7 18.6 7.1

36.0 17.4 18.6

27.6 2.6 25.0

29.7 0.9 28.8

36.0 1.5 34.5

35.1 14.1 21.0

42.0 3.3 38.7

7.6 1.6 6.0

14.1 0.0 14.1

25.0 1.5 23.5

20.2 0.0 20.2

24.4 2.1 22.3

10.4 18.4 8.0

21.1 2.6 18.5

5.6 26.0 20.4

Our null hypothesis is that the population mean difference between first- and

second-grade percentages is zero. Figure 10.14 shows a histogram for the differ-

ences, and it shows a negatively skewed distribution.

The paired t method of Section 10.3 requires normality, so the skewness

might invalidate this, but we will show the results here anyway for comparison

purposes. The mean of the differences is

d ¼ 16:66 with standard deviation

¼ 15.43, so the 95% confidence interval for the population mean difference is

d  t

:025;151

ﬃﬃﬃﬃﬃ

¼ 16:66  2:145

15:43

ﬃﬃﬃﬃﬃ

¼ 16:66  8:54 ¼ð8:12; 25:20Þ

What about the bootstrap for paired data? The bootstrap focuses on the 15

differences and uses the method of Section 8.5. Using Stata, we draw 999 samples

of size 15 with replacement from the 15 differences, and these 999 samples

constitute the bootstrap distribution. Figure 10.15 shows the histogr am.

Frequency

−20 −10 10 20 300

girls ontask1 - ontask2

Figure 10.14 Histogram of differences for girls from Stata

10.6 Comparisons Using the Bootstrap and Permutation Methods 539

The histogram shows negative skewness, which is expected becau se of the

negative skewness shown in Figure 10.14 for the original sample. The skewness

implies that a symmetric confidence interval will not be entirely appropriate, but

we show it for comparison with the other intervals. The standard deviation of the

bootstrap distribution is s

boot

¼ 3.994, compared to the estimated standard error

ﬃﬃﬃﬃﬃ

¼ 15:43=

ﬃﬃﬃﬃﬃ

¼ 3:984. The 95% boots trap confidence interval is nar-

rower because of using z

.025

instead of t

.025,151

d  z

:025

boot

¼ 16:66  1:96ð3:994Þ¼16:66  7:83 ¼ð8:83; 24:49Þ

This is slightly different from what Stata produces, because it uses t

.025,B1

.025,9991

.where B is the siz e of the bootstrap sample.

The 95% percentile interval uses the 2.5 percentile ¼ 7.91 and the 97.5

percentile ¼ 23.97 of the bootstrap distribution, so the confidence interval is

(7.91, 23.97). This interval is to the left of the t intervals because of the negative

skewness of the bootstrap distribution. The bias corrected and accelerated refine-

ment from Stata yields the interval (6.43, 23.12), which is even farther to the left.

All of the intervals agree that there is a substantial population difference

between first grade and second grade. There is a strong reduction in the on-task

private speech of girls between first and second grades.

■

A permutation test for paired data involves permutations within the pairs.

Under the null hypothesis, the two observations in a pair have the same population

mean, so the population mean difference is zero, even if the order is reversed.

Therefore, we consider all possible orderings of the n pairs. Because there are two

possible orderings within each pair, there are 2

arrangements of n pairs. The one-

tailed P-value is the fraction of the 2

differences that are at least as extreme as the

observed value, and the two-tailed P-value is double this.

100

Frequency

5 10 15 20 25

bootstrap mean girls ontask1 - ontask2

Figure 10.15 Histogram of bootstrap differences for girls from Stata

540

CHAPTER 10 Inferences Based on Two Samples

Example 10.22 To see how the permutation test works for paired data, consider a scaled-down version

of the data from Example 10.21 with only the first three pairs. These are (25.7, 18.6),

(36.0, 17.4), (27.6, 2.6). They give a mean difference of (7.1 + 18.6 + 25.0)/3 ¼16.9.

Here are all 8 ¼ 2

permutations with the corresponding means.

Arrangements Mean difference

(25.7, 18.6) (36.0, 17.4) (27.6, 2.6) 16.90

(25.7, 18.6) (36.0, 17.4) (2.6, 27.6) .23

(25.7, 18.6) (17.4, 36.0) (27.6, 2.6) 4.50

(25.7, 18.6) (17.4, 36.0) (2.6, 27.6) 12.17

(18.6, 25.7) (36.0, 17.4) (27.6, 2.6) 12.17

(18.6, 25.7) (36.0, 17.4) (2.6, 27.6) 4.50

(18.6, 25.7) (17.4, 36.0) (27.6, 2.6) .23

(18.6, 25.7) (17.4, 36.0) (2.6, 27.6) 16.90

Because the mean difference for the original sample is the highest value of eight,

the one-taile d P-value is

¼ :125, and the two-tailed P-value is 2



¼ :25.

■

Example 10.23 Let’s now apply the permutation test to the paired data for the 15 girls of Example

10.21. In principle it is no harder to deal with the 2

¼ 2

¼ 32,768 arrangements

when all 15 pairs are included, but this exact approach is generally approximated

using a random sample. We used Stata to draw an additional 4999 samples. Of the

4999, none yielded a mean difference as large as the value of 16.66 obtained for the

original sample of 15 differences. Therefore, the one-tailed P-value is

5000

¼ :0002, and the two-tailed P -value is 2(.0002) ¼ .0004. Rejection of the

null hypothesis at the 5% level was to be expected, given that none of the

confidence intervals in Example 10.21 included 0.

It is interesting to compare the permutation test result with the t test of

Section 10.3. For testing the null hypothesis of 0 population mean difference, the

value of t is

d  0

ﬃﬃﬃﬃﬃ

16:66

15:425=

ﬃﬃﬃﬃﬃ

¼ 4:183

The two-tailed P-value for this is .0009, not very different from the result of the

permutation test.

■

Exercises Section 10.6 (69–84)

69. A student project by Heather Kral studied

students on “lifestyle floors” of a dormitory in

comparison to students on other floors. On a life-

style floor the students share a common major,

and there are a faculty coordinator and resident

assistant from that department. Here are the grade

point averages of 30 students on lifestyle floors

(L) and 30 students on other floors (N):

L: 2.00, 2.25, 2.60, 2.90, 3.00, 3.00, 3.00, 3.00,

3.00, 3.20, 3.20, 3.25, 3.30, 3.30, 3.32, 3.50,

3.50, 3.60, 3.60, 3.70, 3.75, 3.75, 3.79, 3.80,

3.80, 3.90, 4.00, 4.00, 4.00, 4.00.

N: 1.20, 2.00, 2.29, 2.45, 2.50, 2.50, 2.50, 2.50,

2.65, 2.70, 2.75, 2.75, 2.79, 2.80, 2.80, 2.80,

2.86, 2.90, 3.00, 3.07, 3.10, 3.25, 3.50, 3.54,

3.56, 3.60, 3.70, 3.75, 3.80, 4.00.

10.6 Comparisons Using the Bootstrap and Permutation Methods 541

Notice that the lifestyle grade point averages

have a large number of repeats and the distribu-

tion is skewed, so there is some question about

normality.

a. Obtain a 95% confidence interval for the dif-

ference of population means using the method

based on the theorem of Section 10.2.

b. Obtain a bootstrap sample of 999 differences

of means. Check the bootstrap distribution for

normality using a normal probability plot.

c. Use the standard deviation of the bootstrap

distribution along with the mean and t critical

value from (a) to get a 95% confidence inter-

val for the difference of means.

d. Use the bootstrap sample and the percentile

method to obtain a 95% confidence interval

for the difference of means.

e. Compare your three confidence intervals. If

they are very similar, why do you think this is

the case?

f. Interpret your results. Is there a substantial dif-

ference between lifestyle and other floors? Why

do you think the difference is as big as it is?

70. In this application from major league baseball,

the populations represent an abstraction of what

the players can do, so the populations will vary

from year to year. The Colorado Rockies and the

Arizona Diamondbacks played nine games in

Phoenix and ten games in Denver in 2001. The

thinner air in Denver causes curve balls to curve

less and it allows fly balls to travel farther. Does

this mean that more runs are scored in Denver?

The numbers of runs scored by the two teams in

the nine Phoenix games (P) and ten Denver

games (D) are

P: 5.09 15.88 3 8.47 11.65 6.48 11.65 7.41 9.53

D: 10 18 15.56 19 8.1 14 13.76 10 20.12

10.59

The fractions occur because the numbers have

been adjusted for nine innings (54 outs). For

example, in the third Denver game the Rockies

won 10 to 7 on a home run with two out in the

bottom of the tenth inning, so there were 59 outs

instead of 54, and the number of runs is adjusted

to (54/59)(17) ¼ 15.56. We want to compare the

average runs in Denver with the average runs in

Phoenix.

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

ence of population means using the method

given in the theorem of Section 10.2.

b. Obtain a bootstrap sample of 999 differences

of means. Check the bootstrap distribution for

normality using a normal probability plot.

c. Use the standard deviation of the bootstrap

distribution along with the mean and t critical

value from (a) to get a 95% confidence inter-

val for the difference of means.

d. Use the bootstrap sample and the percentile

method to obtain a 95% confidence interval

for the difference of means.

e. Compare your three confidence intervals. If

you used a standard normal critical value in

place of the t critical value in (c), why would

that make this interval more like the one in

(d)? Why should the three intervals be fairly

similar for this data set?

f. Interpret your results. Is there a substantial

difference between the two locations? Com-

pare the difference with what you thought it

would be. If you were a major league pitcher,

would you want to be traded to the Rockies?

71. For the data of Exercise 70 we want to compare

population medians for the runs in Denver versus

the runs in Phoenix.

a. Obtain a bootstrap sample of 999 differences

of medians. Check the bootstrap distribution

for normality using a normal probability

plot.

b. Use the standard deviation of the bootstrap

distribution along with the difference of the

medians in the original sample and the t criti-

cal value from Exercise 70(a) to get a 95%

confidence interval for the difference of

population medians.

c. Use the bootstrap sample and the percentile

method to obtain a 95% confidence interval

for the difference of population medians.

d. Compare the two confidence intervals.

e. How do the results for the median compare

with the results for the mean? In terms of

precision (measured by the width of the con-

fidence interval) which gives the best results?

72. For the data of Exercise 69 now consider testing

the hypothesis of equal population variances.

a. Carry out a 2-tailed test using the method of

Section 10.5. Recall that this method requires

the data to be normal, and the method is sensi-

tive to departures from normality. Check the

data for normality to see if the F test is justi-

fied.

b. Carry out a 2-tailed permutation test for the

hypothesis of equal population variances (or

standard deviations). Why does it not matter

whether you use variances or standard devia-

tions?

c. Compare the two results and summarize your

conclusions.

542

CHAPTER 10 Inferences Based on Two Samples

73. For the data of Exercise 69 we want a 95%

confidence interval for the ratio of population

standard deviations.

a. Use the method of Section 10.5. Recall that

this method requires the data to be normal,

and the method is sensitive to departures from

normality. Check the data for normality to see

if the F distribution can be used for the ratio

of sample variances.

b. With a bootstrap sample of size 999 use the

percentile method to obtain a 95% confidence

interval for the ratio of standard deviations.

c. Compare the two results and discuss the rela-

tionship of the results to those of Exercise 72.

74. Can the right diet help us cope with diseases asso-

ciated with aging such as Alzheimer’s disease?

A study (“Reversals of Age-Related Declines

in Neuronal Signal Transduction, Cognitive, and

Motor Behavioral Deficits with Blueberry,

Spinach, or Strawberry Dietary Supplement,” J.

Neurosci., 1999; 8114–8121) investigated the

effects of fruit and vegetable supplements in the

diet of rats. The rats were 19 months old, which is

aged by rat standards. The 40 rats were randomly

assigned to four diets, of which we will consider

just the blueberry diet and the control diet here.

After 8 weeks on their diets, the rats were given a

number of tests. We give the data for just one of the

tests, which measured how many seconds they

could walk on a rod. Here are the times for the

ten control rats (C) and ten blueberry rats (B):

C: 15.00 7.00 2.44 5.60 3.63 6.24 4.12 8.21

3.90 0.95

B: 5.12 9.38 18.77 15.03 6.67 7.91 7.38

15.09 11.57 8.98

The objective is to obtain a 95% confidence

interval for the difference of population means.

a. Determine a 95% confidence interval for the

difference of population means using the

method based on the Theorem of Section 10.2.

b. Obtain a bootstrap sample of 999 differences

of means. Check the bootstrap distribution for

normality using a normal probability plot.

c. Use the standard deviation of the bootstrap

distribution along with the mean and t critical

value from (a) to get a 95% confidence inter-

val for the difference of means.

d. Use the bootstrap sample and the percentile

method to obtain a 95% confidence interval

for the difference of means.

e. Compare your three confidence intervals.

If they are very similar, why do you think

this is the case? If you had used a critical

value from the normal table rather than the

t table, would the result of (c) agree better

with the result of (d)? Why?

f. Interpret your results. Do the blueberries make

a substantial difference?

75. For the data of Exercise 74, we now want to test

the hypothesis of equal population means.

a. Carry out a 2-tailed test using the method

based on the theorem of Section 10.2.

Although this test requires normal data, it will

still work pretty well for moderately nonnor-

mal data. Nevertheless, you should check the

data for normality to see if the test is justified.

b. Carry out a 2-tailed permutation test for the

hypothesis of equal population means.

c. Compare the results of (a) and (b). Would you

expect them to be similar for the data of this

problem? Discuss their relationship to the

results of Exercise 74. Summarize your con-

clusions about the effectiveness of blueberries.

76. Researchers at the University of Alaska have

been trying to find inexpensive feed sources for

Alaska reindeer growers (“Effects of Two Barley-

Based Diets on Body Mass and Intake Rates of

Captive Reindeer During Winter,” Poster Presen-

tation: School of Agriculture and Land Resources

Management, University of Alaska Fairbanks,

2002). They are focusing on Alaska-grown barley

because commercially available feed supplies are

too expensive for farmers. Typically, reindeer lose

weight in the fall and winter, and the researchers

are trying to find a feed to minimize this loss.

Thirteen pregnant reindeer were randomly divided

into two groups to be fed on two different varieties

of barley, thual and finaska. Here are the weight

gains between October 1 and December 15 for the

seven that were fed thual barley (T) and the six that

were fed finaska barley (F).

T: 5.83 11.5 5.5 1.33 3.83 3.33

7.17

F: 0.17 0.67 4 3 1.33 0.5.

The weight gains are all negative, indicating that

all of the animals lost weight. The thual barley is

less fibrous and more digestible, and the intake

rates for the two varieties of barley were very

nearly the same, so the experimenters expected

less weight loss for the thual variety.

a. Determine a 95% confidence interval for the

difference of population means using the

method given in the theorem of Section 10.2.

10.6 Comparisons Using the Bootstrap and Permutation Methods 543

b. Obtain a bootstrap sample of 999 differences of

means. Check the bootstrap distribution for

normality using a normal probability plot.

c. Use the standard deviation of the bootstrap

distribution along with the mean and t critical

value from (a) to get a 95% confidence interval

for the difference of means.

d. Use the bootstrap sample and the percentile

method to obtain a 95% confidence interval

for the difference of means.

e. Compare your three confidence intervals. If

they are very similar, why do you think this is

the case?

f. Interpret your results. Is there a substantial dif-

ference? Is it in the direction anticipated by the

experimenters?

77. For the data of Exercise 76 we want to test the

hypothesis of equal population variances.

a. Carry out a 2-tailed test using the method of

Section 10.5. Recall that this method requires

the data to be normal, and the method is sensi-

tive to departures from normality. Check the

data for normality to see if the F test is justified.

b. Carry out a 2-tailed permutation test for the

hypothesis of equal population variances (or

standard deviations).

c. Compare the two results and summarize your

conclusions.

78. Recall the data from Example 10.4 about the exper-

iment in the low-level college mathematics course.

Here again are the 85 final exam scores for those in

the experimental group (E) and the 79 final exam

scores for those in the control group (C):

E: 34 27 26 33 23 37 24 34 22 23 32 5

30 35 28 25 37 28 26 29 22 33 31 23

3729 0303426282732293133

28 21 34 29 33 6 8 29 36 7 21 30

28 34 28 35 30 34 9 38 9 27 25 33

923322537282326343234 0

24 30 36 28 38 35 16 37 25 34 38 34

C: 37 22 29 29 33 22 32 36 29 6 4 37

036 03227 7193526222828

32 35 28 33 35 24 21 0 32 28 27 8

30 37 9 33 30 36 28 3 8 31 29 9

0 0 35 25 29 3 33 33 28 32 39 20

32 22 24 20 32 7 8 33 29 9 0 30

26 25 32 38 22 29 29

a. Determine a 95% confidence interval for the

difference of population means using the z

method given in Section 10.1.

b. Obtain a bootstrap sample of 999 differences

of means. Check the bootstrap distribution for

normality using a normal probability plot.

c. Use the standard deviation of the bootstrap

distribution along with the mean and t critical

value from (a) to get a 95% confidence interval

for the difference of means.

d. Use the bootstrap sample and the percentile

method to obtain a 95% confidence interval

for the difference of means.

e. Compare your three confidence intervals. If

they are very similar, why do you think this is

the case? In the light of your results for (c) and

(d), does the z method of (a) seem to work,

regardless of normality? Explain.

f. Are your results consistent with the results of

Example 10.4? Explain.

79. For the data of Example 10.4 we want to try a

permutation test.

a. Carry out a 2-tailed permutation test for the

hypothesis of equal population means.

b. Compare the results for (a) and Example 10.4.

Why should you have expected (a) and Exam-

ple 10.4 to give similar results?

80. For the data of Example 10.4 it might be more

appropriate to compare medians.

a. Find the medians for the two groups. With the

help of a stem-and-leaf display for each group,

explain why the medians are much closer than

the means.

b. Do a two-tailed permutation test to compare

the medians. Given what you found in (a),

explain why the result of the permutation test

was to be expected.

81. Two students, Miguel Melo and Cody Watson, did

a study of textbook pricing. They compared prices

at the campus bookstore and Amazon.com. To be

fair, they included the sales tax for the local store

and added shipping for Amazon. Here are the

prices for a sample of 27 books.

Campus Amazon

100.41 106.94

99.34 113.94

51.53 61.44

20.45 31.59

28.69 29.89

70.66 83.94

98.81 107.74

111.56 115.99

97.22 108.29

(continued)

544

CHAPTER 10 Inferences Based on Two Samples

(continued)

61.89 78.44

70.39 82.94

58.17 65.74

108.38 122.09

61.63 63.49

59.50 69.24

87.66 73.84

26.56 33.98

44.63 40.39

96.69 117.99

18.06 27.94

103.06 115.74

14.61 24.69

77.03 88.04

99.34 113.94

81.81 90.74

48.88 58.94

76.50 91.94

a. Determine a 95% confidence interval for the

difference of population means using the

t method of Section 10.3. Check the data for

normality. Even if the normality assumption is

not valid here, explain why the t method (or the

z method of Section 10.1) might still be appro-

priate.

b. Based on the 27 differences, obtain a bootstrap

sample of 999 differences of means. Check the

bootstrap distribution for normality.

c. Use the standard deviation of the bootstrap

distribution along with the mean and t critical

value from (a) to get a 95% confidence interval

for the difference of means.

d. Use the bootstrap sample and the percentile

method to obtain a 95% confidence interval

for the difference of means.

e. Compare your three confidence intervals. In the

light of your results for (d), does nonnormality

invalidate the results of (a) and (c)? Explain.

f. Interpret your results. Is there a substantial

difference between the two ways to buy

books? Assuming that the populations remain

unchanged and you have just these two sources,

where would you buy?

82. Consider testing the hypothesis of equal popula-

tion means based on the data in Exercise 81.

a. Carry out a 2-tailed test using the method of

Section 10.3. Is the normality assumption satis-

fied here? If not, why might the test be valid

anyway?

b. Carry out a 2-tailed permutation test for the

hypothesis of equal population means.

c. Compare the results for (a) and (b). If the two

results are similar, does it tend to validate (a),

regardless of normality?

83. Compare bootstrapping with approximate permu-

tation tests in which random permutations are

used. Discuss the similarities and differences.

84. Assume that X is uniformly distributed on (1, 1)

and Y is split evenly between a uniform distribu-

tion on (101, 100) and a uniform distribution

on (100, 101). Thus the means are both 0, but the

variances differ strongly. We take random sam-

ples of size three from each distribution and

apply a permutation test for the null hypothesis

: m

¼ m

against the alternative

: m

< m

a. Show that the probability is

that all three of

the Y values come from (100, 101).

b. Show that, if all three Y values come from

(100, 101), then the P-value for the permuta-

tion test is .05.

c. Explain why (a) and (b) are in conflict. What is

the true probability that the permutation test

rejects the null hypothesis at the .05 level?

Supplementary Exercises (85–113)

85. A group of 115 University of Iowa students was

randomly divided into a build-up condition group

(m ¼ 56) and a scale-down condition group

(n ¼ 59). The task for each subject was to build

his or her own pizza from a menu of 12 ingredients.

The build-up group was told that a basic cheese

pizza costs $5 and that each extra ingredient would

cost 50 cents. The scale-down group was told that a

pizza with all 12 ingredients (ugh!!!) would cost

$11 and that deleting an ingredient would save 50

cents. The article “A Tale of Two Pizzas: Building

Up from a Basic Product Versus Scaling Down

from a Fully Loaded Product” (Market. Lett.,

2002: 335–344) reported that the mean number

of ingredients selected by the scale-down group

was significantly greater than the mean number for

the build-up group: 5.29 versus 2.71. The calcu-

lated value of the appropriate t statistic was 6.07.

Would you reject the null hypothesis of equality in

favor of inequality at a significance level of .05?

.01? .001? Can you think of other products aside

from pizza where one could build up or scale

Supplementary Exercises 545

down? [Note: A separate experiment involved stu-

dents from the University of Rome, but details

were a bit different because there are typically

not so many ingredient choices in Italy.]

86. Is the number of export markets in which a firm

sells its products related to the firm’s return on

sales? The article “Technology Industry Success:

Strategic Options for Small and Medium Firms”

(Gongming Qian, Lee Li, Bus. Horizons, Sept.–

Oct. 2003: 41–46) gave the accompanying informa-

tion on the number of export markets for one group

of firms whose return on sales was less than 10% and

another group whose return was at least 10%.

Return

Sample

Size

Sample

Mean

Sample

Less than 10% 36 5.12 .57

At least 10% 47 8.26 1.20

The investigators reported that an appropriate test

of hypotheses resulted in a P-value between .01

and .05. What hypotheses do you think were

tested, and do you agree with the stated P-value

information? What assumptions if any are needed

in order to carry out the test? Can the plausibility of

these assumptions be investigated based just on the

foregoing summary data? Explain.

87. Suppose when using a two-sample t CI or test that

m < n, and show that df > m 1. This is why some

authors suggest using min(m  1, n  1) as df in

place of the formula given in the text. What impact

does this have on the CI and test procedure?

88. The accompanying summary data on compression

strength (lb) for 12  10  8 in. boxes appeared

in the article “Compression of Single-Wall

Corrugated Shipping Containers Using Fixed and

Floating Test Platens” (J. Testing Eval., 1992:

318–320). The authors stated that “the difference

between the compression strength using fixed and

floating platen method was found to be small

compared to normal variation in compression

strength between identical boxes.” Do you agree?

Method

Sample

Size

Sample

Mean

Sample

Fixed 10 807 27

Floating 10 757 41

89. The authors of the article “Dynamics of Canopy

Structure and Light Interception in Pinus elliotti,

North Florida” (Ecol. Monogr., 1991: 33–51)

planned an experiment to determine the effect of

fertilizer on a measure of leaf area. A number of

plots were available for the study, and half were

selected at random to be fertilized. To ensure that

the plots to receive the fertilizer and the control

plots were similar, before beginning the experi-

ment tree density (the number of trees per hectare)

was recorded for eight plots to be fertilized and

eight control plots, resulting in the given data.

MINITAB output follows.

Fertilizer plots 1024 1216 1312 1280

1216 1312 992 1120

Control plots 1104 1072 1088 1328

1376 1280 1120 1200

Two sample T for fertilize vs. control

N Mean StDev SE Mean

fertilize 8 1184 126 44

control 8 1196 118 42

95% CI for mu fertilize-mu control:

(144, 120)

a. Construct a comparative boxplot and comment

on any interesting features.

b. Would you conclude that there is a significant

difference in the mean tree density for fertilizer

and control plots? Use a ¼ .05.

c. Interpret the given confidence interval.

90. Is the response rate for questionnaires affected

by including some sort of incentive to respond

along with the questionnaire? In one experiment,

110 questionnaires with no incentive resulted in

75 being returned, whereas 98 questionnaires that

included a chance to win a lottery yielded 66

responses (“Charities, No; Lotteries, No; Cash,

Yes,” Public Opinion Q., 1996: 542–562). Does

this data suggest that including an incentive

increases the likelihood of a response? State and

test the relevant hypotheses at significance level

.10 by using the P-value method.

91. The article “Quantitative MRI and Electrophysiol-

ogy of Preoperative Carpal Tunnel Syndrome in a

Female Population” (Ergonomics, 1997: 642–649)

reported that (473.3, 1691.9) was a large-sam-

ple 95% confidence interval for the difference

between true average thenar muscle volume

(mm

) for sufferers of carpal tunnel syndrome

and true average volume for nonsufferers. Calcu-

late and interpret a 90% confidence interval for

this difference.

546

CHAPTER 10 Inferences Based on Two Samples

92. The following summary data on bending strength

(lb-in/in) of joints is taken from the article

“Bending Strength of Corner Joints Constructed

with Injection Molded Splines” (Forest Products

J., April 1997: 89–92). Assume normal distribu-

tions.

Type

Sample

Size

Sample

Mean

Sample

Without side coating 10 80.95 9.59

With side coating 10 63.23 5.96

a. Calculate a 95% lower confidence bound for

true average strength of joints with a side

coating.

b. Calculate a 95% lower prediction bound for the

strength of a single joint with a side coating.

c. Calculate a 95% confidence interval for the

difference between true average strengths for

the two types of joints.

93. An experiment was carried out to compare

various properties of cotton/polyester spun yarn

finished with softener only and yarn finished

with softener plus 5% DP-resin (“Properties

of a Fabric Made with Tandem Spun Yarns,”

Textile Res. J., 1996: 607–611). One particularly

important characteristic of fabric is its durability,

that is, its ability to resist wear. For a sample of

40 softener-only specimens, the sample mean

stoll-flex abrasion resistance (cycles) in the fill-

ing direction of the yarn was 3975.0, with a

sample standard deviation of 245.1. Another

sample of 40 softener-plus specimens gave a

sample mean and sample standard deviation of

2795.0 and 293.7, respectively. Calculate a con-

fidence interval with confidence level 99% for

the difference between true average abrasion

resistances for the two types of fabrics. Does

your interval provide convincing evidence that

true average resistances differ for the two types

of fabrics? Why or why not?

94. The derailment of a freight train due to the cata-

strophic failure of a traction motor armature bear-

ing provided the impetus for a study reported in

the article “Locomotive Traction Motor Armature

Bearing Life Study” (Lubricat. Engrg., Aug.

1997: 12–19). A sample of 17 high-mileage trac-

tion motors was selected, and the amount of cone

penetration (mm/10) was determined both for the

pinion bearing and for the commutator armature

bearing, resulting in the following data:

Motor 123456

Commutator 211 273 305 258 270 209

Pinion 226 278 259 244 273 236

Motor 789101112

Commutator 223 288 296 233 262 291

Pinion 290 287 315 242 288 242

Motor 13 14 15 16 17

Commutator 278 275 210 272 264

Pinion 278 208 281 274 268

Calculate an estimate of the population mean dif-

ference between penetration for the commutator

armature bearing and penetration for the pinion

bearing, and do so in a way that conveys informa-

tion about the reliability and precision of the esti-

mate. [Note: A normal probability plot validates

the necessary normality assumption.] Would you

say that the population mean difference has been

precisely estimated? Does it look as though popu-

lation mean penetration differs for the two types of

bearings? Explain.

95. The article “Two Parameters Limiting the Sensi-

tivity of Laboratory Tests of Condoms as Viral

Barriers” (J. Test. Eval., 1996: 279–286) reported

that, in brand A condoms, among 16 tears pro-

duced by a puncturing needle, the sample mean

tear length was 74.0 mm, whereas for the 14 brand

B tears, the sample mean length was 61.0 mm

(determined using light microscopy and scanning

electron micrographs). Suppose the sample stan-

dard deviations are 14.8 and 12.5, respectively

(consistent with the sample ranges given in the

article). The authors commented that the thicker

brand B condom displayed a smaller mean tear

length than the thinner brand A condom. Is this

difference in fact statistically significant? State

the appropriate hypotheses and test at a

¼ .05.

96. Information about hand posture and forces gener-

ated by the fingers during manipulation of various

daily objects is needed for designing high-

tech hand prosthetic devices. The article “Grip

Posture and Forces During Holding Cylindrical

Objects with Circular Grips” (Ergonomics, 1996:

1163–1176) reported that for a sample of 11

females, the sample mean four-finger pinch

strength (N) was 98.1 and the sample standard

deviation was 14.2. For a sample of 15 males,

the sample mean and sample standard deviation

were 129.2 and 39.1, respectively.

a. A test carried out to see whether true average

strengths for the two genders were different

Supplementary Exercises 547