Biostatistics

262 CHAPTER 7 HYPOTHESIS TESTING
Solution:
1. Data. The data consist of information regarding the height status of
Noonan syndrome males and females as described in the statement of
the example.
2. Assumptions. We assume that the patients in the study constitute
independent simple random samples from populations of males and
females with Noonan syndrome.
3. Hypotheses.
H 0 : p F p M or p F p M 0
H A : p F > p M or p F p M > 0
where p F is the proportion of females below the third percentile of
female adult height and p M is the proportion of males below the third
percentile of male adult height.
4. Test statistic. The test statistic is given by Equation 7.6.2.
5. Distribution of test statistic. If the null hypothesis is true, the test
statistic is distributed approximately as the standard normal.
6. Decision rule. Let a ¼ :05. The critical value of z is 1.645. Reject H 0 if
computed z is greater than 1.645.
7. Calculation of test statistic. From the sample data we compute
^p F ¼ 24=44 ¼ :545; ^p M ¼ 11=29 ¼ :379, and p ¼ ð24 þ 11Þ= ð44 þ 29Þ ¼
:479. The computed value of the test statistic, then, is
ð:545 :379Þ
z ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð:479Þð:521Þ
þ ð:479Þð:521Þ
¼ 1:39
44
29
8. Statistical decision. Fail to reject H 0 since 1:39 < 1:645.
9. Conclusion. In the general population of adults with Noonan syndrome
there may be no difference in the proportion of males and females who
have heights below the third percentile of adult height.
10. p value. For this test p ¼ :0823.
&
Tests involving two proportions, using the data from Example 7.6.1, can be carried
out with a variety of computer programs. Outputs from MINITAB and NCSS are shown in
Figure 7.6.1. Again, it should be noted that, because of rounding errors, the results will vary
slightly if calculations are done by hand.