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3 Inferences for Two Population Means, Using Independent Samples: Standard Deviations Not Assumed Equal

3 Inferences for Two Population Means, Using Independent Samples: Standard Deviations Not Assumed Equal

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10.3 Inferences for Two Population Means: σ s Not Assumed Equal



481



In light of Key Fact 10.3, for a hypothesis test with null hypothesis H0: μ1 = μ2 ,

we can use the variable

x¯1 − x¯ 2

t=

(s12 /n 1 ) + (s22 /n 2 )

as the test statistic and obtain the critical value(s) or P-value from the t-table, Table IV. We call this hypothesis-testing procedure the nonpooled t-test.† Procedure 10.3

provides a step-by-step method for performing a nonpooled t-test by using either the

critical-value approach or the P-value approach.



PROCEDURE 10.3 Nonpooled t-Test

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

Assumptions

1.

2.

3.



Simple random samples

Independent samples

Normal populations or large samples



Step 1



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

Ha: μ1 = μ2

(Two tailed)



Ha: μ1 < μ2

(Left tailed)



or



Ha: μ1 > μ2

.

(Right tailed)



or



Step 2



Decide on the significance level, α.



Step 3



Compute the value of the test statistic

x¯ 1 − x¯ 2

.

t=

(s12 /n1 ) + (s22 /n2 )



Denote the value of the test statistic t0 .

CRITICAL-VALUE APPROACH

Step 4



OR



P-VALUE APPROACH

Step 4



The critical value(s) are



The t-statistic has df =



−tα



±tα/2

or

or

(Two tailed)

(Left tailed)

(Right tailed)

with df =



=



, where

=



s12 /n1 + s22 /n2

2



2



Do not

reject H 0



Reject

H0



−t␣/2



0



t␣/2



Two tailed



t



Two tailed











␣ /2



−t␣



0



t



0



t␣



Right tailed



Left tailed



2



P-value

−|t 0| 0 |t0 |



␣ /2



2



s2 /n2

s12 /n1

+ 2

n1 − 1

n2 − 1



2



P -value



Do not reject H0 Reject

H0



Reject Do not reject H0

H0



+



, where



s22 /n2



rounded down to the nearest integer. Use Table IV to

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

technology.



2



s12 /n1

s2 /n2

+ 2

n1 − 1

n2 − 1

rounded down to the nearest integer. Use Table IV to

find the critical value(s).

Reject

H0



s12 /n1



t



t



P - value

t0



0



Left tailed



t



0



t0



t



Right tailed



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.



† The nonpooled t-test is also known as the two-sample t-test (with equal variances not assumed), the (nonpooled)



two-variable t-test, and the (nonpooled) independent samples t-test.



482



CHAPTER 10 Inferences for Two Population Means



Regarding Assumptions 1 and 2, we note that the nonpooled t-test can also be used

as a method for comparing two means with a designed experiment. In addition, the nonpooled t-test is robust to moderate violations of Assumption 3 (normal populations),

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

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



EXAMPLE 10.6



The Nonpooled t-Test

Neurosurgery Operative Times Several neurosurgeons wanted to determine

whether a dynamic system (Z-plate) reduced the operative time relative to a static

system (ALPS plate). R. Jacobowitz, Ph.D., an Arizona State University professor,

along with G. Vishteh, M.D., and other neurosurgeons obtained the data displayed

in Table 10.7 on operative times, in minutes, for the two systems. At the 5% significance level, do the data provide sufficient evidence to conclude that the mean

operative time is less with the dynamic system than with the static system?



TABLE 10.7



TABLE 10.8

Summary statistics for the

samples in Table 10.7



Dynamic



Static



x¯1 = 394.6

s1 = 84.7

n 1 = 14



x¯2 = 468.3

s2 = 38.2

n2 = 6



Dynamic

370

345



FIGURE 10.6

Boxplots of the operative times

for the dynamic and static systems



510

505



445

335



295

280



315

325



490

500



430

455



445

490



455

535



Solution First, we find the required summary statistics for the two samples, as

shown in Table 10.8. Because the two sample standard deviations are considerably

different, as seen in Table 10.8 or Fig. 10.6, the pooled t-test is inappropriate here.

Next, we check the three conditions required for using the nonpooled t-test.

These data were obtained from a randomized comparative experiment, a type of

designed experiment. Therefore, we can consider Assumptions 1 and 2 satisfied.

To check Assumption 3, we refer to the normal probability plots and boxplots

in Figs. 10.5 and 10.6, respectively. These graphs reveal no outliers and, given that

the nonpooled t-test is robust to moderate violations of normality, show that we can

consider Assumption 3 satisfied.



Normal score



FIGURE 10.5

Normal probability plots of the sample

data for the (a) dynamic system and

(b) static system



360

450



Static



3

2

1

0

–1

–2

–3



Normal score



Operative times, in minutes,

for dynamic and static systems



3

2

1

0

–1

–2

–3



250 300 350 400 450 500 550



400 425 450 475 500 525 550



Operative time (min.)



Operative time (min.)



(a) Dynamic system



(b) Static system



Dynamic



Static



300



350



400



450



Operative time (min.)



500



550



10.3 Inferences for Two Population Means: σ s Not Assumed Equal



483



The preceding two paragraphs suggest that the nonpooled t-test can be used to

carry out the hypothesis test. We apply Procedure 10.3.

Step 1 State the null and alternative hypotheses.

Let μ1 and μ2 denote the mean operative times for the dynamic and static systems,

respectively. Then the null and alternative hypotheses are, respectively,

H0: μ1 = μ2 (mean dynamic time is not less than mean static time)

Ha: μ1 < μ2 (mean dynamic time is less than mean static time).

Note that the hypothesis test is left tailed.

Step 2 Decide on the significance level, α.

The test is to be performed at the 5% significance level, or α = 0.05.

Step 3 Compute the value of the test statistic

t=



x¯ 1 − x¯ 2

(s12 /n1 ) + (s22 /n2 )



.



Referring to Table 10.8, we get

t=



CRITICAL-VALUE APPROACH



From Step 2, α = 0.05. Also, from Table 10.8, we see

that

df =



=



84.72 /14

14 − 1



+

2



38.22 /6



38.22 /6

+

6−1



(84.72 /14) + (38.22 /6)



OR



Step 4 The critical value for a left-tailed test is −tα

with df = . Use Table IV to find the critical value.



84.72 /14



394.6 − 468.3



2

2



,



which equals 17 when rounded down. From Table IV

with df = 17, we determine that the critical value is

−tα = −t0.05 = −1.740, as shown in Fig. 10.7A.



= −2.681.



P-VALUE APPROACH



Step 4 The t-statistic has df = . Use Table IV to

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

technology.

From Step 3, the value of the test statistic is t = −2.681.

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

observing a value of t of −2.681 or less if the null hypothesis is true. That probability equals the shaded area

shown in Fig. 10.7B.

FIGURE 10.7B

t-curve

df = 17

P-value



FIGURE 10.7A

Reject H 0 Do not reject H 0

t -curve

df = 17



t = −2.681



From Table 10.8, we find that



0.05

−1.740



t



0



0



t



df =



=



84.72 /14 + 38.22 /6

84.72 /14

14 − 1



2



38.22 /6

+

6−1



2

2



,



which equals 17 when rounded down. Referring to

Fig. 10.7B and Table IV with df = 17, we determine

that 0.005 < P < 0.01. (Using technology, we find that

P = 0.00789.)



484



CHAPTER 10 Inferences for Two Population Means



CRITICAL-VALUE APPROACH



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 = −2.681,

which, as we see from Fig. 10.7A, falls in the rejection

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



OR



P-VALUE APPROACH



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

reject H0 .

From Step 4, 0.005 < P < 0.01. 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

very strong evidence against the null hypothesis.



Step 6 Interpret the results of the hypothesis test.



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

to conclude that the mean operative time is less with the dynamic system than with

the static system.



Report 10.3

Exercise 10.79

on page 489



Confidence Intervals for the Difference between the Means

of Two Populations, Using Independent Samples

Key Fact 10.3 on page 480 can also be used to derive a confidence-interval procedure

for the difference between two means. We call this procedure the nonpooled t-interval

procedure.†



PROCEDURE 10.4 Nonpooled t-Interval Procedure

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

means, μ1 and μ2

Assumptions

1.

2.

3.



Simple random samples

Independent samples

Normal populations or large samples



Step 1

where



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

=



s12 /n1 + s22 /n2

2



s12 /n1

s2 /n2

+ 2

n1 − 1

n2 − 1



,



2

2



rounded down to the nearest integer.

Step 2



The endpoints of the confidence interval for μ1 − μ2 are

( x¯ 1 − x¯ 2 ) ± tα/2 ·



Step 3



EXAMPLE 10.7



(s12 /n1 ) + (s22 /n2 ).



Interpret the confidence interval.



The Nonpooled t-Interval Procedure

Neurosurgery Operative Times Use the sample data in Table 10.7 on page 482

to obtain a 90% confidence interval for the difference, μ1 − μ2 , between the mean

operative times of the dynamic and static systems.

† The nonpooled t-interval procedure is also known as the two-sample t-interval procedure (with equal variances



not assumed), the (nonpooled) two-variable t-interval procedure, and the (nonpooled) independent samples

t-interval procedure.



10.3 Inferences for Two Population Means: σ s Not Assumed Equal



485



Solution We apply Procedure 10.4.

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



.



For a 90% confidence interval, α = 0.10. From Example 10.6, df = 17. In Table IV,

with df = 17, tα/2 = t0.10/2 = t0.05 = 1.740.

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

( x¯ 1 − x¯ 2 ) ± tα/2 ·



(s12 /n1 ) + (s22 /n2 ).



From Step 1, tα/2 = 1.740. Referring to Table 10.8 on page 482, we conclude that

the endpoints of the confidence interval for μ1 − μ2 are

(394.6 − 468.3) ± 1.740 ·



(84.72 /14) + (38.22 /6)



or −121.5 to −25.9.

Step 3 Interpret the confidence interval.



Report 10.4

Exercise 10.85

on page 490



Interpretation We can be 90% confident that the difference between the

mean operative times of the dynamic and static systems is somewhere between

−121.5 minutes and −25.9 minutes. In other words (see page 465), we can be

90% confident that the dynamic system, relative to the static system, reduces the

mean operative time by somewhere between 25.9 minutes and 121.5 minutes.



Pooled Versus Nonpooled t-Procedures

Suppose that we want to perform a hypothesis test based on independent simple random samples to compare the means of two populations. Further suppose that either the

variable under consideration is normally distributed on each of the two populations or

the sample sizes are large. Then two tests are candidates for the job: the pooled t-test

and the nonpooled t-test.

In theory, the pooled t-test requires that the population standard deviations be

equal, but what if they are not? The answer depends on several factors. If the population standard deviations are not too unequal and the sample sizes are nearly the same,

using the pooled t-test will not cause serious difficulties. If the population standard deviations are quite different, however, using the pooled t-test can result in a significantly

larger Type I error probability than the specified one.

In contrast, the nonpooled t-test applies whether or not the population standard

deviations are equal. Then why use the pooled t-test at all? The reason is that, if the

population standard deviations are equal or nearly so, then, on average, the pooled

t-test is slightly more powerful; that is, the probability of making a Type II error

is somewhat smaller. Similar remarks apply to the pooled t-interval and nonpooled

t-interval procedures.



KEY FACT 10.4



Choosing between a Pooled and a Nonpooled t-Procedure

Suppose you want to use independent simple random samples to compare

the means of two populations. To decide between a pooled t-procedure and

a nonpooled t-procedure, follow these guidelines: If you are reasonably sure

that the populations have nearly equal standard deviations, use a pooled

t-procedure; otherwise, use a nonpooled t-procedure.



THE TECHNOLOGY CENTER

Most statistical technologies have programs that automatically perform nonpooled

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

such programs.



486



CHAPTER 10 Inferences for Two Population Means



EXAMPLE 10.8



Using Technology to Conduct Nonpooled t-Procedures

Neurosurgery Operative Times Table 10.7 on page 482 displays samples of neurosurgery operative times, in minutes, for dynamic and static systems. Use Minitab,

Excel, or the TI-83/84 Plus to perform the hypothesis test in Example 10.6 and

obtain the confidence interval required in Example 10.7.

Solution Let μ1 and μ2 denote, respectively, the mean operative times of the dynamic and static systems. The task in Example 10.6 is to perform the hypothesis

test

H0: μ1 = μ2 (mean dynamic time is not less than mean static time)

Ha: μ1 < μ2 (mean dynamic time is less than mean static time)

at the 5% significance level; the task in Example 10.7 is to obtain a 90% confidence

interval for μ1 − μ2 .

We applied the nonpooled t-procedures programs to the data, resulting in Output 10.2. Steps for generating that output are presented in Instructions 10.2. Note to

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

output.

As shown in Output 10.2, the P-value for the hypothesis test is about 0.008. Because the P-value is less than the specified significance level of 0.05, we reject H0 .

Output 10.2 also shows that a 90% confidence interval for the difference between

the means is from −121 to −26.



OUTPUT 10.2

MINITAB



Nonpooled t-procedures on the operative-time data



10.3 Inferences for Two Population Means: σ s Not Assumed Equal



487



OUTPUT 10.2 (cont.) Nonpooled t-procedures on the operative-time data

EXCEL



TI-83/84 PLUS



Using 2-SampTInt



Using 2-SampTTest



Note: For nonpooled t-procedures, discrepancies may occur among results provided by statistical technologies because some round the number of degrees of freedom and others do not.

INSTRUCTIONS 10.2 Steps for generating Output 10.2

MINITAB



EXCEL



Store the two samples of operative-time data from

Table 10.7 in columns named DYNAMIC and STATIC.



Store the two samples of operative-time data from

Table 10.7 in columns named DYNAMIC and STATIC.



FOR THE HYPOTHESIS TEST:

1 Choose Stat ➤ Basic Statistics ➤ 2-Sample t. . .

2 Press the F3 key to reset the dialog box

3 Select Each sample is in its own column from the

drop-down list box

4 Click in the Sample 1 text box and specify DYNAMIC

5 Click in the Sample 2 text box and specify STATIC

6 Click the Options. . . button

7 Click the arrow button at the right of the Alternative

hypothesis drop-down list box and select

Difference < hypothesized difference

8 Click OK twice



FOR THE HYPOTHESIS TEST:

1 Choose XLSTAT ➤ Parametric tests ➤ Two-sample

t-test and z-test

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

dialog box

3 Click in the Sample 1 selection box and then select

the column of the worksheet that contains the

DYNAMIC data

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

column of the worksheet that contains the STATIC data

5 Click the Options tab

6 Click the arrow button at the right of the Alternative

hypothesis drop-down list box and select Mean 1 –

Mean 2 < D

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

8 In the Population variances for the t-test list,

uncheck the Assume equality check box

(continued )



FOR THE CI:

1 Repeat steps 1–6 from the hypothesis-test instructions

2 Click in the Confidence level text box and type 90

3 Click OK twice



488



CHAPTER 10 Inferences for Two Population Means



EXCEL

9 Click OK

10 Click the Continue button in the XLSTAT – Selections

dialog box

FOR THE CI:

1 Repeat steps 1–5 from the hypothesis-test instructions

2 Type 10 in the Significance level (%) text box

3 In the Population variances for the t-test list, uncheck

the Assume equality check box

4 Click OK

5 Click the Continue button in the XLSTAT – Selections

dialog box

TI-83/84 PLUS

Store the two samples of operative- time data from

Table 10.7 in lists named DYNA and STAT.

FOR THE HYPOTHESIS TEST:

1 Press STAT, arrow over to TESTS, and press 4

2 Highlight Data and press ENTER

3 Press the down-arrow key



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

ENTER twice

5 Press 2nd ➤ LIST, arrow down to STAT, and press

ENTER twice

6 Type 1 for Freq1, press ENTER, type 1 for Freq2, and

press ENTER

7 Highlight < μ2 and press ENTER

8 Press the down-arrow key, highlight No, and press

ENTER

9 Arrow down to Calculate and press ENTER

FOR THE CI:

1 Press STAT, arrow over to TESTS, and press 0

2 Highlight Data and press ENTER

3 Press the down-arrow key

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

ENTER twice

5 Press 2nd ➤ LIST, arrow down to STAT, and press

ENTER twice

6 Type 1 for Freq1, press ENTER, type 1 for Freq2, and

press ENTER

7 Type .90 for C-Level and press ENTER

8 Highlight No and press ENTER

9 Press the down-arrow key and press ENTER



Note to Minitab and Excel users: As we have previously 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 nonpooled t-procedure twice:

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

a two-tailed hypothesis test.



Exercises 10.3

Understanding the Concepts and Skills



10.68 You know that the population standard deviations are not

equal.



In each of Exercises 10.67–10.70, suppose that you know that a variable is normally distributed on each of two populations. Further suppose that you want to perform a hypothesis test based on independent

random samples to compare the two population means. For each exercise, decide whether you would use the pooled or nonpooled t-test,

and give a reason for your answer.

10.67 You know that the population standard deviations are equal.



10.69 The sample standard deviations are 23.6 and 59.2.

10.70 The sample standard deviations are 45.4 and 43.7, and each

sample size is 20.

10.71 Each pair of graphs in Fig. 10.8 shows the distributions of a

variable on two populations. Suppose that, in each case, you want

to perform a small-sample hypothesis test based on independent



FIGURE 10.8

Figure for Exercise 10.71



(a)



(b)



(c)



(d)



10.3 Inferences for Two Population Means: σ s Not Assumed Equal



simple random samples to compare the means of the two populations.

In each case, decide whether the pooled t-test, nonpooled t-test, or

neither should be used. Explain your answers.

10.72 Discuss the relative advantages and disadvantages of using

pooled and nonpooled t-procedures.

In each of Exercises 10.73–10.78, we have provided summary statistics for independent simple random samples from two populations. In

each case, use the nonpooled t-test and the nonpooled t-interval procedure to conduct the required hypothesis test and obtain the specified

confidence interval.



10.80 Pancreatic Carcinoma. Pancreatic ductal adenocrarcinoma

is the 4th leading cause of cancer death in the U.S. Researchers

D. Delbeke et al. published the article “Optimal Interpretation of

FDGPET in the diagnosis staging and management of Pancreatic

Carcinoma” (The journal of Nuclear Medicine, Vol. 40, No. 11,

pp. 1784−1791) on optimizing semi-quantitative interpretation of

FDGPET scans. The following table provides summary statistics for

the standardized uptake value (SUV) level value with malignant and

benign pancreatic lesions.

SUV with malignant



SUV with benign



x¯ 1 = 5.1

s1 = 2.6

n 1 = 52



x¯2 = 0.85

s2 = 1.7

n 2 = 13



10.73 x¯1 = 10, s1 = 2, n 1 = 15, x¯2 = 12, s2 = 5, n 2 = 15

a. Two-tailed test, α = 0.05

b. 95% confidence interval

10.74 x¯ 1 = 15, s1 = 2, n 1 = 15, x¯2 = 12, s2 = 5, n 2 = 15

a. Two-tailed test, α = 0.05

b. 95% confidence interval

10.75 x¯ 1 = 20, s1 = 4, n 1 = 10, x¯2 = 18, s2 = 5, n 2 = 15

a. Right-tailed test, α = 0.05

b. 90% confidence interval

10.76 x¯1 = 20, s1 = 4, n 1 = 10, x¯2 = 23, s2 = 5, n 2 = 15

a. Left-tailed test, α = 0.05

b. 90% confidence interval



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

to conclude that, on average, SUV with malignance has higher value

than SUV with benign?

10.81 Acute Postoperative Days. Refer to Example 10.6 (page 482).

The researchers also obtained the following data on the number of

acute postoperative days in the hospital using the dynamic and static

systems.



10.77 x¯1 = 20, s1 = 6, n 1 = 20, x¯2 = 24, s2 = 2, n 2 = 15

a. Left-tailed test, α = 0.05

b. 90% confidence interval

10.78 x¯1 = 23, s1 = 10, n 1 = 25, x¯2 = 24, s2 = 12, n 2 = 50

a. Right-tailed test, α = 0.10

b. 95% confidence interval



Applying the Concepts and Skills

Preliminary data analyses indicate that you can reasonably use nonpooled t-procedures in Exercises 10.79–10.84. For each exercise, apply a nonpooled t-test to perform the required hypothesis test, using

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

10.79 Political Prisoners. According to the American Psychiatric

Association, posttraumatic stress disorder (PTSD) is a common psychological consequence of traumatic events that involve a threat to

life or physical integrity. During the Cold War, some 200,000 people in East Germany were imprisoned for political reasons. Many

were subjected to physical and psychological torture during their

imprisonment, resulting in PTSD. A. Ehlers et al. studied various

characteristics of political prisoners from the former East Germany

and presented their findings in the paper “Posttraumatic Stress Disorder (PTSD) Following Political Imprisonment: The Role of Mental Defeat, Alienation, and Perceived Permanent Change” (Journal

of Abnormal Psychology, Vol. 109, pp. 45–55). The researchers randomly and independently selected 32 former prisoners diagnosed

with chronic PTSD and 20 former prisoners that were diagnosed with

PTSD after release from prison but had since recovered (remitted).

The ages, in years, at arrest yielded the following summary statistics.

Chronic



Remitted



x¯1 = 25.8

s1 = 9.2

n 1 = 32



x¯ 2 = 22.1

s2 = 5.7

n 2 = 20



At the 10% significance level, is there sufficient evidence to conclude

that a difference exists in the mean age at arrest of East German prisoners with chronic PTSD and remitted PTSD?



489



Dynamic

7

9



5

10



8

7



8

7



6

7



Static

7

7



7

8



6

7



18

14



9

9



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

to conclude that the mean number of acute postoperative days in the

hospital is smaller with the dynamic system than with the static system? (Note: x¯1 = 7.36, s1 = 1.22, x¯2 = 10.50, and s2 = 4.59.)

10.82 Stressed-Out Bus Drivers. An intervention program designed by the Stockholm Transit District was implemented to improve the work conditions of the city’s bus drivers. Improvements

were evaluated by G. Evans et al., who collected physiological and

psychological data for bus drivers who drove on the improved routes

(intervention) and for drivers who were assigned the normal routes

(control). Their findings were published in the article “Hassles on the

Job: A Study of a Job Intervention with Urban Bus Drivers” (Journal of Organizational Behavior, Vol. 20, pp. 199–208). Following are

data, based on the results of the study, for the heart rates, in beats per

minute, of the intervention and control drivers.

Intervention

68

74

69

68

64



66

58

63

73

76



Control

74

77

60

66

63



52

53

77

71

73



67

76

63

66

59



63

54

60

55

68



77

73

68

71

64



57

54

64

84

82



80



a. At the 5% significance level, do the data provide sufficient evidence to conclude that the intervention program reduces mean

heart rate of urban bus drivers in Stockholm? (Note: x¯1 = 67.90,

s1 = 5.49, x¯2 = 66.81, and s2 = 9.04.)

b. Can you provide an explanation for the somewhat surprising results of the study?

c. Is the study a designed experiment or an observational study? Explain your answer.

10.83 Schizophrenia and Dopamine. Previous research has suggested that changes in the activity of dopamine, a neurotransmitter



490



CHAPTER 10 Inferences for Two Population Means



in the brain, may be a causative factor for schizophrenia. In the

paper “Schizophrenia: Dopamine β-Hydroxylase Activity and Treatment Response” (Science, Vol. 216, pp. 1423–1425), D. Sternberg

et al. published the results of their study in which they examined

25 schizophrenic patients who had been classified as either psychotic

or not psychotic by hospital staff. The activity of dopamine was measured in each patient by using the enzyme dopamine β-hydroxylase

to assess differences in dopamine activity between the two groups.

The following are the data, in nanomoles per milliliter-hour per

milligram (nmol/mL-hr/mg).

Psychotic

0.0150

0.0204

0.0306

0.0320

0.0208



Not psychotic



0.0222

0.0275

0.0270

0.0226

0.0245



0.0104

0.0200

0.0210

0.0105

0.0112



0.0230

0.0116

0.0252

0.0130

0.0200



0.0145

0.0180

0.0154

0.0170

0.0156



At the 1% significance level, do the data suggest that dopamine activity is higher, on average, in psychotic patients? (Note: x¯1 = 0.02426,

s1 = 0.00514, x¯2 = 0.01643, and s2 = 0.00470.)

10.84 Wing Length. D. Cristol et al. published results of their studies of two subspecies of dark-eyed juncos in the article “Migratory

Dark-Eyed Juncos, Junco Hyemalis, Have Better Spatial Memory and

Denser Hippocampal Neurons than Nonmigratory Conspecifics” (Animal Behaviour, Vol. 66, pp. 317–328). One of the subspecies migrates each year, and the other does not migrate. Several physical

characteristics of 14 birds of each subspecies were measured, one of

which was wing length. The following data, based on results obtained

by the researchers, provide the wing lengths, in millimeters (mm), for

the samples of two subspecies.

Migratory

84.5

82.8

80.5

80.1

83.0



81.0

84.5

82.1

83.4

79.7



82.6

81.2

82.3

81.7



Nonmigratory

82.1

87.1

86.3

84.2

87.8



82.4

84.6

86.6

84.3

84.1



83.9

85.1

83.9

86.2



a. At the 1% significance level, do the data provide sufficient evidence to conclude that the mean wing lengths for the two subspecies are different? (Note: The mean and standard deviation for

the migratory-bird data are 82.1 mm and 1.501 mm, respectively,

and that for the nonmigratory-bird data are 84.9 mm and 1.698

mm, respectively.)

b. Would it be reasonable to use a pooled t-test here? Explain your

answer.

c. If your answer to part (b) was yes, then perform a pooled t-test to

answer the question in part (a) and compare your results to that

found in part (a) by using a nonpooled t-test.

In Exercises 10.85–10.90, apply Procedure 10.4 on page 484 to obtain

the required confidence interval. Interpret your result in each case.

10.85 Political Prisoners. Refer to Exercise 10.79 and obtain a

90% confidence interval for the difference, μ1 − μ2 , between the

mean ages at arrest of East German prisoners with chronic PTSD and

remitted PTSD.

10.86 Pancreatic Carcinoma. Refer to Exercise 10.80 and determine a 98% confidence interval for the difference, μ1 − μ2 , between

the mean SUV levels in malignant and benign Pancreatic Carcinoma.



10.87 Acute Postoperative Days. Refer to Exercise 10.81 and find

a 90% confidence interval for the difference between the mean numbers of acute postoperative days in the hospital with the dynamic and

static systems.

10.88 Stressed-Out Bus Drivers. Refer to Exercise 10.82 and find

a 90% confidence interval for the difference between the mean heart

rates of urban bus drivers in Stockholm in the two environments.

10.89 Schizophrenia and Dopamine. Refer to Exercise 10.83 and

determine a 98% confidence interval for the difference between the

mean dopamine activities of psychotic and nonpsychotic patients.

10.90 Wing Length. Refer to Exercise 10.84 and find a 99% confidence interval for the difference between the mean wing lengths of

the two subspecies.

10.91 Sleep Apnea. In the article “Sleep Apnea in Adults With

Traumatic Brain Injury: A Preliminary Investigation” (Archives of

Physical Medicine and Rehabilitation, Vol. 82, Issue 3, pp. 316–

321), J. Webster et al. investigated sleep-related breathing disorders in

adults with traumatic brain injuries (TBI). The respiratory disturbance

index (RDI), which is the number of apneic and hypopneic episodes

per hour of sleep, was used as a measure of severity of sleep apnea. An

RDI of 5 or more indicates sleep-related breathing disturbances. The

RDIs for the females and males in the study are as follows.

Female



Male



0.1 0.5 0.3 2.3

2.0 1.4 0.0



2.6

0.0

4.3



19.3 1.4 1.0

2.1

1.1 5.6

7.5 16.5 7.8



0.0 39.2

5.0 7.0

3.3 8.9



4.1

2.3

7.3



Use the technology of your choice to answer the following questions.

Explain your answers.

a. If you had to choose between the use of pooled t-procedures and

nonpooled t-procedures here, which would you choose?

b. Is it reasonable to use the type of procedure that you selected in

part (a)?

10.92 Mandate Perceptions. L. Grossback et al. examined mandate perceptions and their causes in the paper “Comparing Competing

Theories on the Causes of Mandate Perceptions” (American Journal

of Political Science, Vol. 49, Issue 2, pp. 406–419). Following are

data on the percentage of members in each chamber of Congress

who reacted to mandates in various years.

House

30.3

23.9



41.1

15.2



15.6

11.7



Senate

10.1



21

27



38

17



40

25



39

25



27



Use the technology of your choice to answer the following questions.

Explain your answers.

a. If you had to choose between the use of pooled t-procedures and

nonpooled t-procedures here, which would you choose?

b. Is it reasonable to use the type of procedure that you selected in

part (a)?

10.93 Mutual Funds. A mutual fund is a professionally managed

investment that can be sold to the general public. Mutual funds can

be specialized into different categories such as healthcare-related

or technology-related mutual funds. The following table lists the

3-month rates of return, in percent, for samples of health-care related

and technology-related mutual funds, as reported by Morningstar, an

independent investment research company.



10.3 Inferences for Two Population Means: σ s Not Assumed Equal



Healthcare

9.7

6.7

3.7



9.5

6.6

3.3



7.6

4.6

3.0



6.8

3.9

0.8



Technology

6.7

3.9

−9.9



6.5

4.6

2.9



6.4

4.6

2.8



6.4

4.2

2.7



6.2

3.6

2.5



4.9

3.2



Use the technology of your choice to answer the following questions.

Explain your answers.

a. If you had to choose between the use of pooled t-procedures and

nonpooled t-procedures here, which would you choose?

b. Is it reasonable to use the type of procedure that you selected in

part (a)?

10.94 Acute Postoperative Days. In Exercise 10.81, you conducted a nonpooled t-test to decide whether the mean number of

acute postoperative days spent in the hospital is smaller with the dynamic system than with the static system. Use the technology of your

choice to perform the following tasks.

a. Using a pooled t-test, repeat that hypothesis test.

b. Compare your results from the pooled and nonpooled t-tests.

c. Which test do you think is more appropriate, the pooled or nonpooled t-test? Explain your answer.

10.95 Neurosurgery Operative Times. In Example 10.6 on

pages 482–484, we conducted a nonpooled t-test, at the 5% significance level, to decide whether the mean operative time is less with

the dynamic system than with the static system. Use the technology

of your choice to perform the following tasks.

a. Using a pooled t-test, repeat that hypothesis test.

b. Compare your results from the pooled and nonpooled t-tests.

c. Repeat both tests, using a 1% significance level, and compare

your results.

d. Which test do you think is more appropriate, the pooled or nonpooled t-test? Explain your answer.



Working with Large Data Sets



491



off a regular 2-3/4 wooden tee to those hit off a 3 Stinger Competition golf tee. A Callaway Great Big Bertha driver with 10 degrees

of loft was used for the test and a robot swung the club head at approximately 95 miles per hour. Data on ball velocity (in miles per

hour) with each type of tee, based on the test results, are provided

on the WeissStats site. Use the technology of your choice to do the

following.

a. Obtain normal probability plots, boxplots, and the standard deviations for the two samples.

b. Based on your results from part (a), which would you be inclined

to use to compare the population means: a pooled or a nonpooled

t-procedure? Explain your answer.

c. At the 5% significance level, do the data provide sufficient evidence to conclude that, on average, ball velocity is less with the

regular tee than with the Stinger tee? Perform the required hypothesis test by using both the pooled t-test and the nonpooled t-test,

and compare results.

d. Find a 90% confidence interval for the difference between the

mean ball velocities with the regular and Stinger tees. Obtain the

required confidence interval by using both the pooled t-interval

procedure and the nonpooled t-interval procedure. Compare your

results.

10.98 The Etruscans. Anthropologists are still trying to unravel

the mystery of the origins of the Etruscan empire, a highly advanced

Italic civilization formed around the eighth century B.C. in central

Italy. Were they native to the Italian peninsula or, as many aspects of

their civilization suggest, did they migrate from the East by land or

sea? The maximum head breadth, in millimeters, of 70 modern Italian male skulls and 84 preserved Etruscan male skulls was analyzed

to help researchers decide whether the Etruscans were native to Italy.

The resulting data can be found on the WeissStats site. [SOURCE:

N. Barnicot and D. Brothwell, “The Evaluation of Metrical Data in

the Comparison of Ancient and Modern Bones.” In Medical Biology

and Etruscan Origins, G. Wolstenholme and C. O’Connor, eds., Little, Brown & Co., 1959]

a. Obtain normal probability plots, boxplots, and the standard deviations for the two samples.

b. Based on your results from part (a), which would you be inclined

to use to compare the population means: a pooled or a nonpooled

t-procedure? Explain your answer.

c. Do the data provide sufficient evidence to conclude that a difference exists between the mean maximum head breadths of modern

Italian males and Etruscan males? Perform the required hypothesis test at the 5% significance level by using both the pooled t-test

and the nonpooled t-test. Compare your results.

d. Find a 95% confidence interval for the difference between the

mean maximum head breadths of modern Italian males and Etruscan males. Obtain the required confidence interval by using both

the pooled t-interval procedure and the nonpooled t-interval procedure. Compare your results.



10.96 Treating Psychotic Illness. L. Petersen et al. evaluated the

effects of integrated treatment for patients with a first episode of psychotic illness in the paper “A Randomised Multicentre Trial of Integrated Versus Standard Treatment for Patients With a First Episode

of Psychotic Illness” (British Medical Journal, Vol. 331, (7517):602).

Part of the study included a questionnaire that was designed to measure client satisfaction for both the integrated treatment and a standard

treatment. The data on the WeissStats site are based on the results of

the client questionnaire. Use the technology of your choice to do the

following.

a. Obtain normal probability plots, boxplots, and the standard deviations for the two samples.

b. Based on your results from part (a), which would you be inclined

to use to compare the population means: a pooled or a nonpooled

t-procedure? Explain your answer.

c. Do the data provide sufficient evidence to conclude that, on average, clients preferred the integrated treatment? Perform the required hypothesis test at the 1% significance level by using both

the pooled t-test and the nonpooled t-test. Compare your results.

d. Find a 98% confidence interval for the difference between mean

client satisfaction scores for the two treatments. Obtain the

required confidence interval by using both the pooled t-interval

procedure and the nonpooled t-interval procedure. Compare your

results.



10.99 Suppose that the sample sizes, n 1 and n 2 , are equal for independent simple random samples from two populations.

a. Show that the values of the pooled and nonpooled t-statistics will

be identical. (Hint: Refer to Exercise 10.61 on page 479.)

b. Explain why part (a) does not imply that the two t-tests are equivalent (i.e., will necessarily lead to the same conclusion) when the

sample sizes are equal.



10.97 A Better Golf Tee? An independent golf equipment testing

facility compared the difference in the performance of golf balls hit



10.100 Tukey’s Quick Test. In this exercise, we examine an

alternative method, conceived by the late Professor John Tukey, for



Extending the Concepts and Skills



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