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Biostatistics

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7.8 HYPOTHESIS TESTING: THE RATIO OF TWO POPULATION VARIANCES 269<br />

.05<br />

0 5.05<br />

F (5, 5)<br />

Nonrejection region<br />

Rejection region<br />

FIGURE 7.8.1 Rejection and nonrejection regions,<br />

Example 7.8.1.<br />

3. Hypotheses.<br />

4. Test statistic.<br />

H 0 : s 2 1 s2 2<br />

H A : s 2 1 > s2 2<br />

V:R: ¼ s2 1<br />

s 2 2<br />

(7.8.1)<br />

5. Distribution of test statistic. When the null hypothesis is true, the test<br />

statistic is distributed as F with n 1 1 numerator and n 2 1 denominator<br />

degrees of freedom.<br />

6. Decision rule. Let a ¼ :05. The critical value of F, from Appendix<br />

Table G, is 5.05. Note that if Table G does not contain an entry for the<br />

given numerator degrees of freedom, we use the column closest in value<br />

to the given numerator degrees of freedom. Reject H 0 if V:R: 5:05.<br />

The rejection and nonrejection regions are shown in Figure 7.8.1.<br />

7. Calculation of test statistic.<br />

V:R: ¼ ð30:62Þ2<br />

ð11:37Þ 2 ¼ 7:25<br />

8. Statistical decision. We reject H 0 , since 7:25 > 5:05; that is, the<br />

computed ratio falls in the rejection region.<br />

9. Conclusion. The failure load variability is higher when using the FasT-<br />

FIX method than the vertical suture method.<br />

10. p value. Because the computed V.R. of 7.25 is greater than 5.05, the p<br />

value for this test is less than 0.05. Excel calculates this p value to be<br />

.0243.<br />

&<br />

Several computer programs can be used to test the equality of two variances. Outputs<br />

from these programs will differ depending on the test that is used. We saw in Figure 7.3.3,

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