Biostatistics

7.3 HYPOTHESIS TESTING: THE DIFFERENCE BETWEEN TWO POPULATION MEANS 241
reduce aortic stiffness. In the 15 patients with hypertension (group 1), the mean aortic
stiffness index was 19.16 with a standard deviation of 5.29. In the 30 control subjects
(group 2), the mean aortic stiffness index was 9.53 with a standard deviation of 2.69. We
wish to determine if the two populations represented by these samples differ with respect to
mean aortic stiffness index.
Solution:
1. Data. The sample sizes, means, and sample standard deviations are:
n 1 ¼ 15; x 1 ¼ 19:16; s 1 ¼ 5:29
n 2 ¼ 30; x 2 ¼ 9:53; s 2 ¼ 2:69
2. Assumptions. The data constitute two independent random samples,
one from a population of subjects with hypertension and the other from a
control population. We assume that aortic stiffness values are approximately
normally distributed in both populations. The population variances
are unknown and unequal.
3. Hypotheses.
H 0 : m 1 m 2 ¼ 0
H A : m 1 m 2 6¼ 0
4. Test statistic. The test statistic is given by Equation 7.3.3.
5. Distribution of test statistic. The statistic given by Equation 7.3.3 does
not follow Student’s t distribution. We, therefore, obtain its critical
values by Equation 7.3.4.
6. Decision rule. Let a ¼ :05. Before computing t 0 we calculate w 1 ¼
ð5:29Þ 2 =15 ¼ 1:8656 and w 2 ¼ ð2:69Þ 2 =30 ¼ :2412. In Appendix Table
E we find that t 1 ¼ 2:1448 and t 2 ¼ 2:0452. By Equation 7.3.4 we
compute
t 0 1:8656ð2:1448Þþ:2412 2:0452
¼ ð Þ ¼ 2:133
1:8656 þ :2412
Our decision rule, then, is reject H 0 if the computed t is either 2:133
or 2:133.
7. Calculation of test statistic. By Equation 7.3.3 we compute
t 0 19:16 9:53
¼ s ð ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Þ 0 ¼ 9:63
ð5:29Þ 2
þ ð2:69
1:4515 ¼ 6:63
Þ2
15 30