LibraryPirate

7.6 HYPOTHESIS TESTING: THE DIFFERENCE BETWEEN TWO POPULATION PROPORTIONS 263
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 7 p M or p F - p M 7 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 = .05. The critical value of z is 1.645. Reject
if computed z is greater than 1.645.
7. Calculation of test statistic. From the sample data we compute Np F =
24>44 = .545, Np M = 11>29 = .379, and p = 124 + 112>144 + 292 =
.479. The computed value of the test statistic, then, is
H 0
z =
A
1.47921.5212
44
1.545 - .3792
+ 1.47921.5212
29
= 1.39
8. Statistical decision. Fail to reject since 1.39 6 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 = .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.
H 0