Biostatistics

340 CHAPTER 8 ANALYSIS OF VARIANCE
3. Hypotheses.
H 0 : t j ¼ 0 j ¼ 1; 2; 3
H A : not all t j ¼ 0
4. Test statistic. The test statistic is V:R: ¼ MSTr=MSE.
5. Distribution of test statistic. When H 0 is true and the assumptions are
met, V.R. follows an F distribution with 2 and 8 degrees of freedom.
6. Decision rule. Let a ¼ :05. Reject the null hypothesis if the computed
V.R. is equal to or greater than the critical F, which we find in Appendix
Table G to be 4.46.
7. Calculation of test statistic. We compute the following sums of
squares:
SST ¼ð7 10:07Þ 2 þð8 10:07Þ 2 þþð14 10:07Þ 2 ¼ 46:9335
SSBI ¼ 3½ð8:67 10:07Þ 2 þð9:00 10:07Þ 2 þþð12:33 10:07Þ 2 Š¼24:855
SSTr ¼ 5½ð9 10:07Þ 2 þð9:6 10:07Þ 2 þð11:6 10:07Þ 2 Š¼18:5335
SSE ¼ 46:9335 24:855 18:5335 ¼ 3:545
The degrees of freedom are total ¼ð3Þð5Þ 1 ¼ 14, blocks ¼
5 1 ¼ 4, treatments ¼ 3 1 ¼ 2, and residual ¼ð5 1Þð3 1Þ ¼
8. The results of the calculations may be displayed in an ANOVA table
as in Table 8.3.4
8. Statistical decision. Since our computed variance ratio, 20.91, is
greater than 4.46, we reject the null hypothesis of no treatment effects
on the assumption that such a large V.R. reflects the fact that the two
sample mean squares are not estimating the same quantity. The only
other explanation for this large V.R. would be that the null hypothesis is
really true, and we have just observed an unusual set of results. We rule
out the second explanation in favor of the first.
TABLE 8.3.4 ANOVA Table for Example 8.3.1
Source SS d.f. MS V.R.
Treatments 18.5335 2 9.26675 20.91
Blocks 24.855 4 6.21375
Residual 3.545 8 .443125
Total 46.9335 14