Solution to Exercise 10.24.

  OK, so we have to backfill this ANOVA table. The results we use here appear in the red boxed table on p. 424. The d.f. (``degrees of freedom'') for treatment groups or their sum of squares (``SSTr'') is always I-1 where I is the number of groups. We are told there are

I = 3

groups. The d.f. for ``Error'' is sample size n minus number of treatments I. The sample size here is

$$n \; = \; 21 + 32 + 21 \; = \; 74 .$$

The d.f. for ``Total'' (sum of squares ``SST'') is always the sample size n minus 1. So, we have filled in the ANOVA table to the following extent:

df squares square f

Groups 2 76.09

Error 71

Total 73 1123.14

Now the mean square in each row is the sum of squares over the degrees of freedom, so the sum of squares for groups is

$$76.09 * 2 \; = \; 152.18 .$$

Also, the sum of squares for treatment and error add to the sum of squares for the total, and we get

$$SSE \; = \; SST - SSTr \; = \; 1123.14 - 152.18 \; = \; 970.96 .$$

Hence,

$$MSE \; = \; 970.96/71 \; = \; 13.67549 .$$

Finally, the f statistic is

$$f \; = \; MSTr/MSE \; = \; 76.09/13.67549 \; = \; 5.563969 .$$

Thus, we can fill in the ANOVA table:

df squares square f

Groups 2 152.18 76.09 5.56

Error 71 970.96 13.68

Total 73 1123.14

Is this significant at the $\alpha$ = 0.01 level? Referring to table A.9 in the book, we have

We see that the critical value decreases as the denominator degrees of freedom increases. We want F_.01,2,71, and we can say F_.01,2,71 < 4.98. Since the observed f of 5.56 exceeds this, we do indeed have a significant difference in the means of the groups. (In fact from minitab Calc $\rightarrow$ Probability Distributions we have that the P-value is 1-.9943 = 0.0057.)