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
±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.
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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.)
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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.
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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