a. Run a traditional analysis of variance on these data.
b. The following sums of squares have been computed on the data using the appropriate
design matrix (a5Gender, b5SES)Compute the summary table for the analysis of variance using these sums of squares.
16.6 Using the SES portion of the design matrix as our predictor, we find that
.
a. Why is this value the same as in the answer to Exercise 16.5?
b. Will this be the case in all analyses of variance?
16.7 When we take the data in Exercise 16.5 and delete the last two low-SES males, the last three
average-SES males, and the last two high-SES females, we obtain the following sums of
squares:Compute the analysis of variance using these sums of squares.
16.8 Using only the SES predictors for the data in Exercise 16.7, we find.
Why is this not the same as in Exercise 16.7?
16.9 For the data in Exercise 16.5, the complete model isa. Show that this model reproduces the treatment and interaction effects as calculated by
the method shown in Table 16.2.
16.10 For the data in Exercise 16.7, the complete model isa. Show that this model reproduces the treatment and interaction effects as calculated in
Table 16.3.
16.11 Using the following data, demonstrate that Method I (the method advocated in this chapter)
really deals with unweighted means.
B 1 B 2
511
39
A 1
14
6
11
9
10 6
11 2
A 2 12
716.12 Draw a Venn diagram representing the sums of squares in Exercise 16.5.
16.13 Draw a Venn diagram representing the sums of squares in Exercise 16.7.1.2306A 12 3.7167B 12 0.3500B 21 0.4778AB 112 0.5444AB 121 13.67501.1667A 12 3.1667B 12 0.1667B 21 0.8333AB 112 0.1667AB 121 13.4167SSSESSSreg(b)=379.3325SSreg(ab)=15.8132SSreg(b)=379.3325SSreg(a,ab)=112.3392 SSreg(a)=95.4511SSreg(a,b)=437.6338 SSreg(b,ab)=398.7135SSY=750.1951 SSreg(a,b,ab)=458.7285SSSESSSreg(b)=338.6667SSreg(a,ab)=84.000SSreg(a,b)=404.0000 SSreg(b,ab)=357.333SSY=777.6667 SSreg(a,b,ab)=422.6667624 Chapter 16 Analyses of Variance and Covariance as General Linear Models