Chapter 12, Problem 20E

To determine

Complete the two-way ANOVA table.

Find whether there is a significant difference in Factor A means.

Find whether there is a significant difference in Factor B means.

Find whether there is a significant interaction means.

Answer to Problem 20E

The two-way ANOVA table is given below:

Source of variation	Sum of Squares	Degrees of freedom	Mean Squares	F-statistic
Factor A	75	3	25	2.25
Factor B	25	2	12.25	1.10
Interaction	300	6	50	4.08
Error	600	48	12.25
Total	1,000	59

There is no significant difference among Factor A means.

There is no significant difference among Factor B means.

There is a significant interaction between Factor A and Factor B.

Explanation of Solution

Degrees of freedom for Factor A and Factor B:

The degrees of freedom for Factor A and Factor B can be obtained as follows:

There are four levels of Factor A and three levels of Factor B.

$\begin{array}{c}df\left(A\right)=k-1\\ =4-1\\ =3\end{array}$

The degrees of freedom for Factor A is 3.

$\begin{array}{c}df\left(B\right)=b-1\\ =3-1\\ =2\end{array}$

Thus, degrees of freedom for Factor B is 2.

Degrees of freedom for interaction:

The degrees of freedom for interaction are calculated below:

$\begin{array}{c}df\left(AB\right)=\left(k-1\right)\left(b-1\right)\\ =3\times 2\\ =6\end{array}$

Thus, the degrees of freedom for interaction is 6.

Degrees of freedom for total:

It is mentioned that the number of replications per cell is 5. There are 4 levels, in which Factor A has 4 levels and Factor B has 3 levels. Thus, the total number of observations is 60.

$\begin{array}{c}df\left(Total\right)=n-1\\ =60-1\\ =59\end{array}$

Therefore, the total degrees of freedom is 59.

Degrees of freedom for Error:

The error degrees of freedom can be obtained as follows:

$\begin{array}{c}df\left(Error\right)=n-kb\\ =60-4\times 3\\ =60-12\\ =48\end{array}$

Thus, the error degrees of freedom is 48.

Mean square of Factor A:

From the given output, it is found that the sum of squares of Factor A (SSA) is 75.

The mean square of Factor A is calculated below:

$\begin{array}{c}MS\left(A\right)=\frac{SS\left(A\right)}{df}\\ =\frac{75}{3}\\ =25\end{array}$

Therefore, the mean square of Factor A is 25.

Mean square of Factor B:

The mean square of Factor B is given below:

$\begin{array}{c}MS\left(B\right)=\frac{SS\left(B\right)}{df}\\ =\frac{25}{2}\\ =12.5\end{array}$

Thus, the mean square of Factor B is 12.5.

Mean square for interaction:

$\begin{array}{c}MS\left(AB\right)=\frac{SS\left(AB\right)}{df}\\ =\frac{300}{6}\\ =50\end{array}$

Thus, the mean square for interaction is 50.

Mean square error:

The sum of squares due to error is 600.

$\begin{array}{c}MSE=\frac{SSE}{df}\\ =\frac{600}{48}\\ =12.25\end{array}$

Thus, the mean square error is 12.25.

F-Statistic for Factor A:

$\begin{array}{c}F=\frac{MS\left(A\right)}{MSE}\\ =\frac{25}{12.25}\\ =2.04\end{array}$

Thus, the F statistic for Factor A is 2.04.

F-Statistic for Factor B:

$\begin{array}{c}F=\frac{MS\left(B\right)}{MSE}\\ =\frac{12.25}{12.25}\\ =1.00\end{array}$

The F statistic for Factor B is 1.00.

F-Statistic for AB:

$\begin{array}{c}F=\frac{MS\left(AB\right)}{MSE}\\ =\frac{50}{12.25}\\ =4.08\end{array}$

Thus, the F statistic for AB is 4.08.

The ANOVA table is given below:

Source of variation	Sum of Squares	Degrees of freedom	Mean Squares	F-statistic
Factor A	75	3	25	2.25
Factor B	25	2	12.25	1.10
Interaction	300	6	50	4.08
Error	600	48	12.25
Total	1,000	59

The null and alternative hypotheses are stated below:

For Factor A:

$H_0:$ There is no significant difference among Factor A means.

$H_1:$ There is a significant difference among Factor A means.

For Factor B:

$H_0:$ There is no significant difference among Factor B means.

$H_1:$ There is a significant difference among Factor B means.

For interaction:

$H_0:$ There is no significant interaction between Factor A and Factor B.

$H_1:$ There is a significant interaction between Factor A and Factor B.

Decision Rule:

Reject the null hypothesis, if the computed F test statistic value is greater than the critical value at the 0.05 significance level. Otherwise, fail to reject the null hypothesis.

Conclusion:

 Factor A:

From Appendix B.6A, at the 0.05 significance level, the critical value of Factor A is 2.80.

Since the F test statistic is less than the critical value of Factor A. Hence, one is failed to reject the null hypothesis at the 0.05 significance level.

Therefore, there is no significant difference among Factor A means.

Factor B:

The critical value of Factor B is 3.19 and the F test statistic for Factor B is 1.10.

Since the F test statistic is less than the critical value of Factor B. Hence, one is failed to reject the null hypothesis at the 0.05 significance level.

Thus, there is no significant difference among Factor B means.

Interaction AB:

The critical value for interaction is 2.29 and the F test statistic is 4.08.

Since the F test statistic is greater than the critical value. Hence, one can reject the null hypothesis at the 0.05 significance level.

Therefore, there is a significant interaction between Factor A and Factor B.