LibraryPirate

350 CHAPTER 8 ANALYSIS OF VARIANCE
Dialog box:
Session command:
Stat ➤ ANOVA ➤ Twoway MTB > TWOWAY C1 C2 C3;
SUBC> MEANS C2 C3.
Type C1 in Response. Type C2 in Row factor and
Check Display means. Type C3 in Column factor and
Check Display means. Click OK.
Output:
Two-way ANOVA: C1 versus C2, C3
Analysis of Variance for C1
Source DF SS MS F P
C2 17 20238 1190 8.20 0.000
C3 3 2396 799 5.50 0.002
Error 51 7404 145
Total 71 30038
FIGURE 8.4.1 MINITAB procedure and output (ANOVA table) for Example 8.4.1.
8. Statistical decision. Since V.R. = 5.50 is greater than 2.80, we are able
to reject the null hypothesis.
9. Conclusion. We conclude that there is a difference in the four population
means.
10. p value. Since 5.50 is greater than 4.98, the F value for a = .005 and
df = 40, the p value is less than .005.
Figure 8.4.2. shows the SAS ® output for the analysis of Example 8.4.1 and Figure 8.4.3
shows the SPSS output for the same example. Note that SPSS provides four potential
tests. The first test is used under an assumption of sphericity and matches the outputs in
Figures 8.4.1 and 8.4.2. The next three tests are modifications if the assumption of
sphericity is violated. Note that SPSS modifies the degrees of freedom for these three
tests, which changes the mean squares and the p values, but not the V. R. Note that the
assumption of sphericity was violated for these data, but that the decision rule did not
change, since all of the p values were less than a = .05.
■
EXERCISES
For Exercises 8.4.1 to 8.4.3 perform the ten-step hypothesis testing procedure. Let a = .05.
8.4.1 One of the purposes of a study by Liu et al. (A-16) was to determine the effects of MRZ 2579
on neurological deficit in Sprague-Dawley rats. In this study, 10 rats were measured at four time