180 CHAPTER 8/HYPOTHESIS TESTING: TWO SAMPLE INFERENCE
between the control and the high NAPAP groups, but not between the control and the low NAPAP
groups. We will perform two-sample t-tests accordingly.
Test for Equal Variances: Change_1, Change_3
95% Bonferroni confidence intervals for standard deviations
N Lower StDev Upper
Test for Equal Variances: Change_2, Change_3
95% Bonferroni confidence intervals for standard deviations
N Lower StDev Upper
Two-Sample T-Test and CI: Change_1, Change_3
Two-sample T for Change_1 vs Change_3
N Mean StDev SE Mean
Two-Sample T-Test and CI: Change_2, Change_3
Two-sample T for Change_2 vs Change_3
N Mean StDev SE Mean
8.118 If the standard deviation (
V
) of change in blood pressure (bp) is the same in each group and is known
without error, then the 97.5% CI for is given by
P
P
12
CHAPTER 8/HYPOTHESIS TESTING: TWO SAMPLE INFERENCE 181
8.119 The 95% CI must be narrower than the 97.5% CI. Since both intervals are centered at the same value
8.122 We have the test statistic
d
8.123 We wish to test the hypothesis 01 2 11 2
:vs.:HH
P
PPP
z
. We will first compare the variance of the
8.124 We have the test statistic
182 CHAPTER 8/HYPOTHESIS TESTING: TWO SAMPLE INFERENCE
12
22
12
12
xx
t
ss
nn
8.125 The 95% CI
1,.975n
dt se
r . Thus, the width of the
1,.975
CI 2 0.42 0.17 0.25
n
tse
.
8.126 A 2-sided 90% CI for the mean change score for retired women
8.127 We wish to use a 2 sample t test for independent samples to test the hypothesis 01 2
:vs.H
P
P
CHAPTER 8/HYPOTHESIS TESTING: TWO SAMPLE INFERENCE 183
8.128 The test statistic is
8.130 We calculate the differences in BMI for heavy smoking women as follows:
ID
Change in BMI for heavy
smoking women who quit
smoking
11 + 5.5
12 + 3.2
184 CHAPTER 8/HYPOTHESIS TESTING: TWO SAMPLE INFERENCE
8.131 Let
1
P
underlying mean BMI change among ex-smokers and 2
P
underlying mean BMI change
ID
Change in BMI for never
smoking women
1 + 2.8
2 – 0.9
8.132 We wish to test the hypothesis 22 22
01 21 2 11 21 2
:, ,vs.:,HH
P
PV V P PV V
z
. We have the test
statistic
CHAPTER 8/HYPOTHESIS TESTING: TWO SAMPLE INFERENCE 185
8.133 We use the power formula:
8.135 We wish to test the hypothesis:
H
0
:
P
1
P
2
vs. H
0
:
P
1
z
P
2
.
186 CHAPTER 8/HYPOTHESIS TESTING: TWO SAMPLE INFERENCE
8.137 We have the pooled variance estimate
We then use the t statistic given by
8.138 We use the sample size formula
CHAPTER 8/HYPOTHESIS TESTING: TWO SAMPLE INFERENCE 187
8.139-141
To address these questions, we will need to create many new variables. Using the original data, we need
to create a BMI variable, noting that the height recorded in the data must be divided by 100 to convert to
Two-Sample T-Test and CI: WtChange, Good A1C
Good
A1C N Mean StDev SE Mean
Two-Sample T-Test and CI: HtChange, Good A1C
Good
Two-Sample T-Test and CI: BMIChange, Good A1C
Good
A1C N Mean StDev SE Mean
0 53 1.322 0.670 0.092
188 CHAPTER 8/HYPOTHESIS TESTING: TWO SAMPLE INFERENCE
8.142 Using the relationship that the upper bound of a 95% CI = 1.96 /
x
n
V
, we get
8.144 Since the estimate of the standard deviation within the 4th quartile is identical to that calculated above,
8.147 First, we need the pooled variance estimate
22
216.8 9388 17.2 3166 285.7
12554
s
.
8.148 For this problem, we need to use the sample size formula given in Eq. 8.27, using k=2. We will compute
the sample size required for n2.
2,n
8.149 Here, we use the power formula:
CHAPTER 8/HYPOTHESIS TESTING: TWO SAMPLE INFERENCE 189
8.154 Let x = Change = (5 min heart rate – Baseline heart rate). A 95% CI for mean change in heart rate will be
8.155 Comparing baseline heart rate in 1996 to baseline heart rate in 2006, we have the following data:
190 CHAPTER 8/HYPOTHESIS TESTING: TWO SAMPLE INFERENCE
8.157 From 8.155, we estimate the mean change in heart rate to be 9.4, with a standard error of 1.946. Since we
8.159 Based on the results in (a) we will use a two sample t-test with equal variances. We obtain the pooled
variance estimate as follows:
8.161 We will use the one-sample power formula, with
8.162 Sample size is given by
CHAPTER 8/HYPOTHESIS TESTING: TWO SAMPLE INFERENCE 191
8.165 First, we must calculate the pain difference scores:
Subject Pain E Pain C Difference
11.38.8 –7.5
27.31.3 6
300.8 –0.8
8.167 We find that X=15 in our data set, since only one patient experienced greater pain in the experimental
192 CHAPTER 8/HYPOTHESIS TESTING: TWO SAMPLE INFERENCE
8.171 The test statistic t= 22 2 2
1 2 1122
( ) / / / (92.5 57.0) / 50.4 / 306 26.3 /17
xx snsn
8.174 We have the F statistic: F = 38.22/20.92 = 3.34~F18,10 under H0.
8.176 The pooled variance estimate is: