Biostatistics

7.7 HYPOTHESIS TESTING: A SINGLE POPULATION VARIANCE 265
Solution:
1. Data. See statement in the example.
2. Assumptions. The study sample constitutes a simple random sample
from a population of similar children. The IGF-I levels are normally
distributed.
3. Hypotheses.
H 0 : s 2 ¼ 600
H A : s 2 6¼ 600
4. Test statistic. The test statistic is given by Equation 7.7.1.
5. Distribution of test statistic. When the null hypothesis is true, the test
statistic is distributed as x 2 with n 1 degrees of freedom.
6. Decision rule. Let a ¼ :05. Critical values of x 2 are 6.262 and 27.488.
Reject H 0 unless the computed value of the test statistic is between
6.262 and 27.488. The rejection and nonrejection regions are shown in
Figure 7.7.1.
7. Calculation of test statistic.
x 2 ¼ 15ð670:81Þ
600
¼ 16:77
8. Statistical decision. Do not reject H 0 since 6:262 < 16:77 < 27:488.
9. Conclusion. Based on these data we are unable to conclude that the
population variance is not 600.
10. p value. The determination of the p value for this test is complicated by
the fact that we have a two-sided test and an asymmetric sampling
distribution. When we have a two-sided test and a symmetric sampling
distribution such as the standard normal or t, we may, as we have
seen, double the one-sided p value. Problems arise when we attempt to
.025
.025
0
6.262
27.488
x 2 15
Rejection region
Nonrejection region
Rejection region
FIGURE 7.7.1 Rejection and nonrejection regions for Example 7.7.1.