(a) To test for the equality of 4 treatment means for a completely randomized design, we will perform a right-tailed F test. The hypotheses areH_(0):mu_(1)=mu_(2)=mu_(3)=mu_(4)H_(1): Not all the means are equal(b) For a one-factor ANOVA with sample sizes n_(1)=8,n_(2)=5,n_(3)=6, and n_(4)=6, the total number of observations is the sum of the sample sizes for each group:{:[n=n_(1)+n_(2)+n_(3)+n_(4)],[=8+5+6+6],[=25]:}Since there are c=4 groups and n=25 observations, degrees of freedom for the F test are Numerator:{:[df_(1)=c-1],[=4-1],[=3" (between treatments, factor) "]:}Denominator:{:[df_(2)=n-c],[=25-4],[=21quad" (within treatments, error) "]:}(c) For alpha=.05 in this right-tailed F test, Excel's function = F.INV.RT (.05,3,21) yields the 5 percent right-tail critical value F_(3,21)=3.072. So the decision rule i ... See the full answer