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5 Inferences for Two Population Means, Using Paired Samples

5 Inferences for Two Population Means, Using Paired Samples

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508



CHAPTER 10 Inferences for Two Population Means



EXAMPLE 10.15



Comparing Two Means, Using a Paired Sample

Ages of Married People The U.S. Census Bureau publishes information on the

ages of married people in Current Population Reports. Suppose that we want to

decide whether, in the United States, the mean age of married men differs from the

mean age of married women.

a. Formulate the problem statistically by posing it as a hypothesis test.

b. Explain the basic idea for carrying out the hypothesis test.

c. Suppose that 10 married heterosexual couples in the United States are selected

at random and that the ages, in years, of the people chosen are as shown in the

second and third columns of Table 10.13. Discuss the use of these data to make

a decision concerning the hypothesis test.



TABLE 10.13

Ages, in years, of a random

sample of 10 married couples



Couple



Husband



Wife



Difference, d



1

2

3

4

5

6

7

8

9

10



59

21

33

78

70

33

68

32

54

52



53

22

36

74

64

35

67

28

41

44



6

−1

−3

4

6

−2

1

4

13

8

36



Solution

a. To formulate the problem statistically, we first note that we have one variable—

namely, age—and two populations:

Population 1: All married men

Population 2: All married women.

Let μ1 and μ2 denote the means of the variable “age” for Population 1 and

Population 2, respectively:

μ1 = mean age of all married men

μ2 = mean age of all married women.

We want to perform the hypothesis test

H0: μ1 = μ2 (mean ages of married men and women are the same)

Ha: μ1 = μ2 (mean ages of married men and women differ).

b. Independent samples could be used to carry out the hypothesis test: Take independent simple random samples of, say, 10 married men and 10 married women

and then apply a pooled or nonpooled t-test to the age data obtained. However,

in this case, a paired sample is more appropriate. Here, a pair consists of a married (heterosexual) couple. The variable we analyze is the difference between

the ages of the husband and wife in a couple. By using a paired sample, we can

remove an extraneous source of variation: the variation in the ages among married couples. The sampling error thus made in estimating the difference between

the population means will generally be smaller and, therefore, we are more

likely to detect differences between the population means when such differences

exist.

c. The last column of Table 10.13 contains the difference, d, between the ages

of each of the 10 couples sampled. We refer to each difference as a paired



10.5 Inferences for Two Population Means, Using Paired Samples



509



difference because it is the difference of a pair of observations. For example, in

the first couple, the husband is 59 years old and the wife is 53 years old, giving

a paired difference of 6 years, meaning that the husband is 6 years older than

his wife.

If the null hypothesis of equal mean ages is true, the paired differences of

the ages for the married couples sampled should average about 0; that is, the

sample mean, d, of the paired differences should be roughly 0. If d is too much

different from 0, we would take this as evidence that the null hypothesis is false.

From the last column of Table 10.13, we find that the sample mean of the paired

differences is

di

36

d=

=

= 3.6.

n

10

The question now is, can this difference of 3.6 years be reasonably attributed to

sampling error, or is the difference large enough to indicate that the two populations have different means? To answer that question, we need to know the

distribution of the variable d, which we discuss next.



The Paired t-Statistic

Suppose that x is a variable on each of two populations whose members can be paired.

For each pair, we let d denote the difference between the values of the variable x on

the members of the pair. We call d the paired-difference variable.

It can be shown that the mean of the paired differences equals the difference between the two population means. In symbols,

μd = μ1 − μ2 .

Furthermore, if d is normally distributed, we can apply this equation and our knowledge

of the studentized version of a sample mean (Key Fact 8.5 on page 375) to obtain Key

Fact 10.6.



KEY FACT 10.6



Distribution of the Paired t-Statistic

Suppose that x is a variable on each of two populations whose members can

be paired. Further suppose that the paired-difference variable d is normally

distributed. Then, for paired samples of size n, the variable

t=



d − (μ1 − μ2 )



sd/ n



has the t-distribution with df = n − 1.



Note: We use the phrase normal differences as an abbreviation of “the paireddifference variable is normally distributed.”



Hypothesis Tests for the Means of Two Populations,

Using a Paired Sample

We now present a hypothesis-testing procedure based on a paired sample for comparing the means of two populations when the paired-difference variable is normally

distributed. In light of Key Fact 10.6, for a hypothesis test that has null hypothesis

H0: μ1 = μ2 , we can use the variable

t=



d



sd / n



as the test statistic and obtain the critical value(s) or P-value from the t-table, Table IV.



510



CHAPTER 10 Inferences for Two Population Means



We call this hypothesis-testing procedure the paired t-test. Note that the paired

t-test is simply the one-mean t-test applied to the paired-difference variable with null

hypothesis H0: μd = 0. Procedure 10.6 provides a step-by-step method for performing

a paired t-test by using either the critical-value approach or the P-value approach.



PROCEDURE 10.6 Paired t-Test

Purpose To perform a hypothesis test to compare two population means, μ1 and μ2

Assumptions

1.

2.



Simple random paired sample

Normal differences or large sample



Step 1



The null hypothesis is H0: μ1 = μ2 , and the alternative hypothesis is

Ha: μ1 = μ2

(Two tailed)



or



Ha: μ1 < μ2

(Left tailed)



Step 2



Decide on the significance level, α.



Step 3



Compute the value of the test statistic

t=



Ha: μ1 > μ2

.

(Right tailed)



or



d



sd / n



and denote that value t0 .



CRITICAL-VALUE APPROACH

Step 4



OR



P-VALUE APPROACH

Step 4 The t-statistic has df = n − 1. Use Table IV

to estimate the P-value, or obtain it exactly by using

technology.



The critical value(s) are



−tα



±tα/2

or

or

(Two tailed)

(Left tailed)

(Right tailed)

with df = n − 1. Use Table IV to find the critical

value(s).



P -value



P-value



Reject

H0



Do not

reject H0



Reject

H0



Do not reject H0 Reject

H0



Reject Do not reject H0

H0



−|t0| 0 |t0|

Two tailed



␣ /2

−t␣/2



0



t␣/2



Two tailed



t



t0



0



Left tailed



t



0



t0



t



Right tailed











␣ /2



t



P- value



−t␣



0



t



0



t␣



Right tailed



Left tailed



t



Step 5 If P ≤ α, reject H0 ; otherwise, do not

reject H0 .



Step 5 If the value of the test statistic falls in the

rejection region, reject H0 ; otherwise, do not reject H0 .

Step 6



Interpret the results of the hypothesis test.



Note: The hypothesis test is exact for normal differences and is approximately correct

for large samples and nonnormal differences.



Properties and guidelines for use of the paired t-test are the same as those given for

the one-mean z-test in Key Fact 9.7 on page 411 when applied to paired differences. In

particular, the paired t-test is robust to moderate violations of the normality assumption

but, even for large samples, can sometimes be unduly affected by outliers because the



10.5 Inferences for Two Population Means, Using Paired Samples



511



sample mean and sample standard deviation are not resistant to outliers. Here are two

other important points:

r Do not apply the paired t-test to independent samples, and, likewise, do not apply a

pooled or nonpooled t-test to a paired sample.

r The normality assumption for a paired t-test refers to the distribution of the paireddifference variable, not to the two distributions of the variable under consideration.



EXAMPLE 10.16



The Paired t-Test

Ages of Married People We now return to the hypothesis test posed in Example 10.15. A random sample of 10 married couples gave the data on ages, in years,

shown in the second and third columns of Table 10.13 on page 508. At the 5% significance level, do the data provide sufficient evidence to conclude that the mean age

of married men differs from the mean age of married women?

Solution First, we check the two conditions required for using the paired t-test,

as listed in Procedure 10.6.



FIGURE 10.15



Normal score



Normal probability plot of the paired

differences in Table 10.13



r Assumption 1 is satisfied because we have a simple random paired sample. Each

pair consists of a married couple.

r Because the sample size, n = 10, is small, we need to examine issues of normality

and outliers. (See the first bulleted item in Key Fact 9.7 on page 411.) To do so,

we construct in Fig. 10.15 a normal probability plot for the sample of paired

differences in the last column of Table 10.13. This plot reveals no outliers and is

roughly linear. So we can consider Assumption 2 satisfied.



From the preceding items, we see that the paired t-test can be used to conduct

the required hypothesis test. We apply Procedure 10.6.



3

2

1

0

–1

–2

–3



Step 1 State the null and alternative hypotheses.

Let μ1 denote the mean age of all married men, and let μ2 denote the mean age of

all married women. Then the null and alternative hypotheses are, respectively,

–3



0



3



6



9



H0: μ1 = μ2 (mean ages are equal)

Ha: μ1 = μ2 (mean ages differ).



12 15



Paired difference (yr)



Note that the hypothesis test is two tailed.

Step 2 Decide on the significance level, α.

We are to perform the test at the 5% significance level, so α = 0.05.

Step 3 Compute the value of the test statistic

t=



d

√ .

sd / n



The paired differences (d-values) of the sample pairs are shown in the last column of

Table 10.13. We need to determine the sample mean and sample standard deviation

of those paired differences. We do so in the usual manner:

d=



36

di

=

= 3.6,

n

10



and

sd =



di2 − ( di )2 /n

=

n−1



352 − (36)2 /10

= 4.97.

10 − 1



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CHAPTER 10 Inferences for Two Population Means



Consequently, the value of the test statistic is

t=



CRITICAL-VALUE APPROACH



d

3.6

√ = 2.291.

√ =

sd / n

4.97/ 10



OR



P-VALUE APPROACH



Step 4 The critical values for a two-tailed test

are ±tα/2 with df = n − 1. Use Table IV to find the

critical values.



Step 4 The t-statistic has df = n − 1. Use Table IV

to estimate the P-value, or obtain it exactly by using

technology.



We have n = 10 and α = 0.05. Table IV reveals that,

for df = 10 − 1 = 9, ±t0.05/2 = ±t0.025 = ±2.262, as

shown in Fig. 10.16A.



From Step 3, the value of the test statistic is t = 2.291.

The test is two tailed, so the P-value is the probability of

observing a value of t of 2.291 or greater in magnitude

if the null hypothesis is true. That probability equals the

shaded area shown in Fig. 10.16B.



FIGURE 10.16A

Reject H 0



Do not reject H 0



Reject H 0



FIGURE 10.16B

P-value



0.025

−2.262



0.025

0



2.262



t

t



0



Step 5 If the value of the test statistic falls in the

rejection region, reject H0 ; otherwise, do not

reject H0 .

From Step 3, the value of the test statistic is t = 2.291,

which falls in the rejection region depicted in Fig.10.16A.

Thus we reject H0 . The test results are statistically significant at the 5% level.



t = 2.291



Because n = 10, we have df = 10 − 1 = 9. Referring to

Fig. 10.16B and Table IV, we find that 0.02 < P < 0.05.

(Using technology, we found that P = 0.0478.)

Step 5 If P ≤ α, reject H0 ; otherwise, do not

reject H0 .

From Step 4, 0.02 < P < 0.05. Because the P-value is

less than the specified significance level of 0.05, we reject H0 . The test results are statistically significant at the

5% level and (see Table 9.8 on page 408) provide strong

evidence against the null hypothesis.



Step 6 Interpret the results of the hypothesis test.

Report 10.6

Exercise 10.159

on page 518



Interpretation At the 5% significance level, the data provide sufficient evidence

to conclude that the mean age of married men differs from the mean age of married

women.



Confidence Intervals for the Difference between the Means of

Two Populations, Using a Paired Sample

We can also use Key Fact 10.6 on page 509 to derive a confidence-interval procedure

for the difference between two population means. We call that confidence-interval procedure the paired t-interval procedure.



10.5 Inferences for Two Population Means, Using Paired Samples



513



PROCEDURE 10.7 Paired t-Interval Procedure

Purpose To find a confidence interval for the difference between two population

means, μ1 and μ2



Assumptions

1.

2.



Simple random paired sample

Normal differences or large sample



Step 1



For a confidence level of 1 − α, use Table IV to find tα/2 with df = n − 1.



Step 2



The endpoints of the confidence interval for μ1 − μ2 are

sd

d ± tα/2 · √ .

n



Step 3



Interpret the confidence interval.



Note: The confidence interval is exact for normal differences and is approximately

correct for large samples and nonnormal differences.



EXAMPLE 10.17



The Paired t-Interval Procedure

Ages of Married People Use the age data in the second and third columns of

Table 10.13 on page 508 to obtain a 95% confidence interval for the difference,

μ1 − μ2 , between the mean ages of married men and married women.

Solution We apply Procedure 10.7.

Step 1 For a confidence level of 1 − α, use Table IV to find tα/2 with

df = n − 1.

For a 95% confidence interval, α = 0.05. From Table IV, we determine that, for

df = n − 1 = 10 − 1 = 9, we have tα/2 = t0.05/2 = t0.025 = 2.262.

Step 2 The endpoints of the confidence interval for μ1 − μ2 are

sd

d ± tα/2 · √ .

n

From Step 1, tα/2 = 2.262 and n = 10. And, from Example 10.16, d = 3.6 and

sd = 4.97. So, the endpoints of the confidence interval for μ1 − μ2 are

4.97

3.6 ± 2.262 · √ ,

10

or 0.04 to 7.16.

Step 3 Interpret the confidence interval.



Report 10.7

Exercise 10.165

on page 518



Interpretation We can be 95% confident that the difference between the mean

ages of married men and married women is somewhere between 0.04 years and

7.16 years. In other words (see page 465), we can be 95% confident that the mean

age of married men exceeds the mean age of married women by somewhere between 0.04 years and 7.16 years.



Paired versus Independent Samples

Suppose that we want to perform a hypothesis test or obtain a confidence interval in

order to compare two population means. As you know, two sampling designs that are

candidates for the comparison are paired samples and independent samples.



514



CHAPTER 10 Inferences for Two Population Means



When a paired sample is possible, it is often preferable to independent samples.

Indeed, as we mentioned in the solution to part (b) of Example 10.15, by using a paired

sample, we can frequently remove extraneous sources of variation. The sampling error

thus made in estimating the difference between the population means will generally be

smaller and, therefore, we are more likely to detect differences between the population

means when such differences exist.

The paired-sample design is preferable when, for the variable under consideration,

there is a positive correlation between the population pairs; that is, when population

pairs tend to both have above-average values of the variable or below-average values

of the variable. This situation is precisely the case for ages of husbands and wives

(Example 10.15)—older husbands tend to be married to older wives.



THE TECHNOLOGY CENTER

Most statistical technologies have programs that automatically perform paired

t-procedures. In this subsection, we present output and step-by-step instructions for

such programs.



EXAMPLE 10.18



Using Technology to Conduct Paired t-Procedures

Ages of Married People The second and third columns of Table 10.13 on page 508

give the ages of 10 randomly selected married couples. Use Minitab, Excel, or the

TI-83/84 Plus to perform the hypothesis test in Example 10.16 and obtain the confidence interval required in Example 10.17.

Solution Let μ1 denote the mean age of all married men, and let μ2 denote the

mean age of all married women. The task in Example 10.16 is to perform the hypothesis test

H0: μ1 = μ2 (mean ages are equal)

Ha: μ1 = μ2 (mean ages differ)

at the 5% significance level; the task in Example 10.17 is to obtain a 95% confidence

interval for μ1 − μ2 .

We applied the paired t-procedures programs to the data, resulting in Output 10.4. Steps for generating that output are presented in Instructions 10.4. Note

to Excel users: For brevity, we have presented only the essential portions of the

actual output.

As shown in Output 10.4, the P-value for the hypothesis test is about 0.048.

Because the P-value is less than the specified significance level of 0.05, we reject H0 . Output 10.4 also shows that a 95% confidence interval for the difference

between the means is from 0.04 to 7.16.



OUTPUT 10.4

MINITAB



Paired t-procedures on the age data



10.5 Inferences for Two Population Means, Using Paired Samples



515



OUTPUT 10.4 (cont.) Paired t-procedures on the age data

EXCEL



TI-83/84 PLUS



Using T-Test



Using TInterval



INSTRUCTIONS 10.4 Steps for generating Output 10.4

MINITAB

1 Store the age data from the second and third columns

of Table 10.13 in columns named HUSBAND and WIFE

2 Choose Stat ➤ Basic Statistics ➤ Paired t. . .

3 Press the F3 key to reset the dialog box

4 Click in the Sample 1 text box and specify HUSBAND

5 Click in the Sample 2 text box and specify WIFE

6 Click the Options. . . button

7 Click in the Confidence level text box and type 95

8 Click the arrow button at the right of the Alternative

hypothesis drop-down list box and select Difference =

hypothesized difference

9 Click OK twice

EXCEL

1 Store the age data from the second and third columns

of Table 10.13 in columns named HUSBAND and WIFE

2 Choose XLSTAT ➤ Parametric tests ➤ Two-sample

t-test and z-test

3 Click the reset button in the lower left corner of the

dialog box

4 Click in the Sample 1 selection box and then select

the column of the worksheet that contains the

HUSBAND data

5 Click in the Sample 2 selection box and then select the

column of the worksheet that contains the WIFE data

6 In the Data format list, select the Paired samples

option button

7 Click the Options tab

8 Click the arrow button at the right of the Alternative

hypothesis drop-down list box and select Mean 1 –

Mean 2 = D

9 Type 5 in the Significance level (%) text box



10 Click OK

11 Click the Continue button in the XLSTAT – Selections

dialog box

TI-83/84 PLUS

Store the age data from the second and third columns of

Table 10.13 in lists named HUSB and WIFE.

FOR THE PAIRED DIFFERENCES:

1 Press 2nd ➤ LIST, arrow down to HUSB, and press

ENTER

2 Press −

3 Press 2nd ➤ LIST, arrow down to WIFE, and press ENTER

4 Press STO ➧

5 Press 2nd ➤ A-LOCK, type DIFF, and press ENTER

FOR THE HYPOTHESIS TEST:

1 Press STAT, arrow over to TESTS, and press 2

2 Highlight Data and press ENTER

3 Press the down-arrow key, type 0 for μ0 , and press

ENTER

4 Press 2nd ➤ LIST, arrow down to DIFF, and press

ENTER twice

5 Type 1 for Freq and then press ENTER

6 Highlight = μ0 and press ENTER

7 Arrow down to Calculate and press ENTER

FOR THE CI:

1 Press STAT, arrow over to TESTS, and press 8

2 Highlight Data and press ENTER

3 Press the down-arrow key

4 Press 2nd ➤ LIST, arrow down to DIFF, and press

ENTER twice

5 Type 1 for Freq and then press ENTER

6 Type .95 for C-Level and press ENTER twice



516



CHAPTER 10 Inferences for Two Population Means



Note to Minitab and Excel users: As we have noted, Minitab and Excel compute a

two-sided confidence interval for a two-tailed test and a one-sided confidence interval

for a one-tailed test. To perform a one-tailed hypothesis test and obtain a two-sided

confidence interval, apply Minitab’s or Excel’s paired t-procedure twice: once for the

one-tailed hypothesis test and once for the confidence interval specifying a two-tailed

hypothesis test.

Note to TI-83/84 Plus users: The paired t-procedures are just one-mean t-procedures

applied to the paired differences. Since, at the time of this writing, the TI-83/84 Plus

does not have built-in paired t-procedures, we applied its one-mean t-procedures to the

paired differences, as seen in Instructions 10.4.



Exercises 10.5

Understanding the Concepts and Skills

10.139 State one possible advantage of using paired samples instead

of independent samples.

10.140 What constitutes each pair in a paired sample?

10.141 State the two conditions required for performing a paired

t-procedure. How important are those conditions?

10.142 Provide an example (different from the ones considered in

this section) of a procedure based on a paired sample being more appropriate than one based on independent samples.

In Exercises 10.143–10.148, hypothesis tests are proposed. For each

hypothesis test,

a. identify the variable.

b. identify the two populations.

c. identify the pairs.

d. identify the paired-difference variable.

e. determine the null and alternative hypotheses.

f. classify the hypothesis test as two tailed, left tailed, or right tailed.

10.143 TV Viewing. The A. C. Nielsen Company collects data on

the TV viewing habits of Americans and publishes the information in

Nielsen Report on Television. Suppose that you want to use a paired

sample to decide whether the mean viewing times of married men

and married women differ.

10.144 Self-Reported Weight. The article “Accuracy of SelfReported Height and Weight in a Community-Based Sample of

Older African Americans and Whites” (Journal of Gerontology Series A: Biological Sciences and Medical Sciences, Vol. 65A, No. 10,

pp. 1123–1129) by G. Fillenbaum et al. explores the relationship between measured and self-reported height and weight. The

authors sampled African American and White women and men older

than 70 years of age. A hypothesis test is to be performed to decide

whether, on average, self-reported weight is less than measured

weight for the aforementioned age group.

10.145 Hypnosis and Pain. In the paper “An Analysis of Factors

That Contribute to the Efficacy of Hypnotic Analgesia” (Journal

of Abnormal Psychology, Vol. 96, No. 1, pp. 46–51), D. Price and

J. Barber examined the effects of hypnosis on pain. They measured

response to pain using a visual analogue scale (VAS), in centimeters,

where higher VAS indicates greater pain. VAS sensory ratings were

made before and after hypnosis on each of 16 subjects. A hypothesis test is to be performed to decide whether, on average, hypnosis

reduces pain.



10.146 Sports Stadiums and Home Values. In the paper “Housing Values Near New Sporting Stadiums” (Land Economics, Vol. 81,

Issue 3, pp. 379–395), C. Tu examined the effects of construction of

new sports stadiums on home values. Suppose that you want to use a

paired sample to decide whether construction of new sports stadiums

affects the mean price of neighboring homes.

10.147 Breastmilk and Antioxidants. There is convincing evidence that breastmilk containing antioxidants is important in the prevention of diseases in infants. Researchers A. Xavier et al. studied the

effects of storing breastmilk on antioxidant levels in the article “Total

Antioxidant Concentrations of Breastmilk—An Eye-Opener to the

Negligent” (Journal of Health, Population and Nutrition, Vol. 29,

No. 6, pp. 605–611). Samples of breastmilk were taken from women

and divided into fresh samples that were immediately tested and the

remaining samples that were stored in the refrigerator and tested after

48 hours. A hypothesis test is to be performed to decide whether, on

average, stored breastmilk has a lower total antioxidant capacity.

10.148 Fiber Density. In the article “Comparison of Fiber Counting

by TV Screen and Eyepieces of Phase Contrast Microscopy” (American Industrial Hygiene Association Journal, Vol. 63, pp. 756–761),

I. Moa et al. reported on determining fiber density by two different

methods. The fiber density of 10 samples with varying fiber density

was obtained by using both an eyepiece method and a TV-screen

method. A hypothesis test is to be performed to decide whether, on

average, the eyepiece method gives a greater fiber density reading

than the TV-screen method.

In Exercises 10.149–10.154, the null hypothesis is H0 : μ1 = μ2 and

the alternative hypothesis is as specified. We have provided data from

a simple random paired sample from the two populations under consideration. In each case, use the paired t-test to perform the required

hypothesis test at the 10% significance level.

10.149 Ha : μ1 = μ2

Observation from

Pair



Population 1



Population 2



1

2

3

4

5

6

7



13

16

13

14

12

8

17



11

15

10

8

8

9

14



10.5 Inferences for Two Population Means, Using Paired Samples



10.150 Ha : μ1 < μ2



517



10.154 Ha : μ1 > μ2

Observation from



Observation from

Pair



Population 1



Population 2



Pair



Population 1



Population 2



1

2

3

4

5

6

7



7

4

10

0

20

−1

12



13

9

6

2

19

5

10



1

2

3

4

5

6

7

8

9

10



40

30

34

22

35

26

26

27

11

35



32

29

36

18

31

26

25

25

15

31



10.151 Ha : μ1 > μ2



Applying the Concepts and Skills

Observation from

Pair



Population 1



Population 2



1

2

3

4

5

6

7

8



7

9

12

15

27

16

11

8



6

8

11

14

6

9

5

2



Preliminary data analyses indicate that use of a paired t-test is reasonable in Exercises 10.155–10.160. Perform each hypothesis test by

using either the critical-value approach or the P-value approach.

10.155 Behavioral Genetics. In the article “Growth references for

height, weight and BMI of Twins aged 0–2.5 years” (ACTA Pediatrica, Vol. 97, pp. 1099–1104), the researchers P. Dommelen et al.

determined the size of the growth deficit in Dutch monozygotic and

dizygotic twins aged between 0–2.5 years as compared to the singletons and to construct reference growth charts for twins. The following table shows the difference of the height of the twins at various age

groups from 0 to 2.5 years.



10.152 Ha : μ1 = μ2



5.2

5.0

4.7



4.5

4.3

3.9



3.2

2.7

1.9



4.1

3.4

2.3



4.8

3.8

1.0



Observation from

Pair



Population 1



Population 2



1

2

3

4

5

6

7

8



10

8

13

13

17

12

12

11



12

7

11

16

15

9

12

7



10.153 Ha : μ1 < μ2



Identify the variable under consideration.

Identify the two populations.

Identify the paired-difference variable.

Are the numbers in the table paired differences? Why or why not?

At the 5% significance level, do the data provide sufficient evidence to conclude that the mean heights of twins separately differ?

(Note: d = 3.65 cm and sd = 1.23 cm.)

f. Repeat part (e) at the 1% significance level.



a.

b.

c.

d.

e.



10.156 Sleep. In 1908, W. S. Gosset published “The Probable Error

of a Mean” (Biometrika, Vol. 6, pp. 1–25). In this pioneering paper,

published under the pseudonym “Student,” he introduced what later

became known as Student’s t-distribution. Gosset used the following

data set, which gives the additional sleep in hours obtained by

10 patients who used laevohysocyamine hydrobromide.



Observation from

Pair



Population 1



Population 2



1

2

3

4

5

6

7

8

9



15

22

15

27

24

23

8

20

2



18

25

17

24

30

23

10

27

3



1.9

4.4



0.8

5.5



1.1

1.6



0.1

4.6



−0.1

3.4



Identify the variable under consideration.

Identify the two populations.

Identify the paired-difference variable.

Are the numbers in the table paired differences? Why or why not?

At the 5% significance level, do the data provide sufficient evidence to conclude that laevohysocyamine hydrobromide is effective in increasing sleep? (Note: d = 2.33 and sd = 2.002.)

f. Repeat part (e) at the 1% significance level.

a.

b.

c.

d.

e.



518



CHAPTER 10 Inferences for Two Population Means



10.157 Anorexia Treatment. Anorexia nervosa is a serious eating

disorder, particularly among young women. The following data provide the weights, in pounds, of 17 anorexic young women before

and after receiving a family therapy treatment for anorexia nervosa.

[SOURCE: D. Hand et al. (ed.) A Handbook of Small Data Sets, London: Chapman & Hall, 1994; raw data from B. Everitt (personal communication)]

Before



After



Before



After



Before



After



83.3

86.0

82.5

86.7

79.6

87.3



94.3

91.5

91.9

100.3

76.7

98.0



76.9

94.2

73.4

80.5

81.6

83.8



76.8

101.6

94.9

75.2

77.8

95.2



82.1

77.6

83.5

89.9

86.0



95.5

90.7

92.5

93.8

91.7



Does family therapy appear to be effective in helping anorexic young

women gain weight? Perform the appropriate hypothesis test at the

5% significance level.

10.158 Measuring Treadwear. R. Stichler et al. compared two

methods of measuring treadwear in their paper “Measurement of

Treadwear of Commercial Tires” (Rubber Age, Vol. 73:2). Eleven

tires were each measured for treadwear by two methods, one based

on weight and the other on groove wear. The data, in thousands of

miles, are as follows.

Weight

method



Groove

method



Weight

method



Groove

method



30.5

30.9

31.9

30.4

27.3

20.4



28.7

25.9

23.3

23.1

23.7

20.9



24.5

20.9

18.9

13.7

11.4



16.1

19.9

15.2

11.5

11.2



At the 5% significance level, do the data provide sufficient evidence

to conclude that, on average, the two measurement methods give different results?

10.159 Glaucoma and Corneal Thickness. Glaucoma is a leading cause of blindness in the United States. N. Ehlers measured the

corneal thickness of eight patients who had glaucoma in one eye but

not in the other. The results of the study were published as the paper

“On Corneal Thickness and Intraocular Pressure, II” (Acta Opthalmologica, Vol. 48, pp. 1107–1112). The data on corneal thickness, in

microns, are shown in the following table.

Patient



Normal



Glaucoma



1

2

3

4

5

6

7

8



484

478

492

444

436

398

464

476



488

478

480

426

440

410

458

460



At the 10% significance level, do the data provide sufficient evidence

to conclude that mean corneal thickness is greater in normal eyes than

in eyes with glaucoma?



10.160 Cooling Down. Cooling down with a cold drink before exercise in the heat is believed to help an athlete perform. Researcher

J. Dugas explored the difference between cooling down with an ice

slurry (slushy) and with cold water in the article “Ice Slurry Ingestion Increases Running Time in the Heat” (Clinical Journal of Sports

Medicine, Vol. 21, No. 6, pp. 541–542). Ten male participants drank

a flavored ice slurry and ran on a treadmill in a controlled hot and humid environment. Days later, the same participants drank cold water

and ran on a treadmill in the same hot and humid environment. The

following table shows the times, in minutes, it took to fatigue on the

treadmill for both the ice slurry and the cold water.

Subject



Cold Water



Ice Slurry



1

2

3

4

5

6

7

8

9

10



52

37

44

51

34

38

41

50

29

38



56

43

52

58

38

45

45

58

34

44



At the 1% significance level, do the data provide sufficient evidence

to conclude that, on average, cold water is less effective than ice slurry

for optimizing athletic performance in the heat? (Note: The mean and

standard deviation of the paired differences are −5.9 minutes and

1.60 minutes, respectively.)

In Exercises 10.161–10.166, apply Procedure 10.7 on page 513 to

obtain the required confidence interval. Interpret your result in each

case.

10.161 Behavioral Genetics. Refer to Exercise 10.155.

a. Determine a 95% confidence interval for the difference between

the mean heights of the twins.

b. Repeat part (a) for a 99% confidence level.

10.162 Sleep. Refer to Exercise 10.156.

a. Determine a 90% confidence interval for the additional sleep that

would be obtained, on average, by using laevohysocyamine hydrobromide.

b. Repeat part (a) for a 98% confidence level.

10.163 Anorexia Treatment. Refer to Exercise 10.157 and find a

90% confidence interval for the weight gain that would be obtained,

on average, by using the family therapy treatment.

10.164 Measuring Treadwear. Refer to Exercise 10.158 and find a

95% confidence interval for the mean difference in measurement by

the weight and groove methods.

10.165 Glaucoma and Corneal Thickness. Refer to Exercise 10.159 and obtain an 80% confidence interval for the difference

between the mean corneal thickness of normal eyes and that of eyes

with glaucoma.

10.166 Cooling Down. Refer to Exercise 10.160 and find a

98% confidence interval for the difference between the mean times

to fatigue on a treadmill in a hot and humid environment after cooling down with cold water and after cooling down with an ice slurry.



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