Data from a randomized block design are shown in the following table: Using $\alpha$ = 0.05, the critical F value for the treatments null hypothesis is ___.
A)3.59
B)4.76
C)3.98
D)5.14
E)9.89

Question

Data from a randomized block design are shown in the following table:  Using $\alpha$ = 0.05, the critical F value for the treatments null hypothesis is ___.
A)3.59 
B)4.76 
C)3.98 
D)5.14 
E)9.89

Quizplus · Accepted Answer

The treatments null hypothesis is that there is no difference between the means of the treatments. To test this hypothesis, we use the F-test. The F-statistic is calculated by dividing the between-treatments sum of squares by the within-treatments sum of squares. The degrees of freedom for the between-treatments sum of squares is equal to the number of treatments minus one (k-1), and the degrees of freedom for the within-treatments sum of squares is equal to the total number of observations minus the number of treatments (N-k). In this case, there are 3 treatments (k=3) and 12 observations (N=12). The between-treatments sum of squares is 198.56 and the within-treatments sum of squares is 43.39. The degrees of freedom for the between-treatments sum of squares is 2 and the degrees of freedom for the within-treatments sum of squares is 9. Therefore, the F-statistic is equal to (198.56/2)/(43.39/9), which is approximately equal to 4.76. The critical F value for a significance level of 0.05 with 2 and 9 degrees of freedom is 4.10. Since the calculated F-statistic is greater than the critical value, we reject the null hypothesis and conclude that there is a significant difference between the means of the treatments.

Data from a Randomized Block Design Are Shown in the Following