Chapter 13, Problem 5E

(a)

To determine

The fault in the provided statement.

Answer to Problem 5E

Solution: The fault in the statement is that the null hypothesis cannot be rejected in a two-way ANOVA when the AB F-statistic is small value.

Explanation of Solution

The provided statement is that the null hypothesis is rejected when AB F-statistic is small and there is no interaction in a two-way ANOVA. The statement is wrong because a higher value of AB F-statistics represents that null hypothesis of no interaction should be rejected. Hence, the hypothesis can be considered as a wrong statement.

(b)

To determine

The fault in the provided statement.

Answer to Problem 5E

Solution: The fault in the statement is that the sum of squares is not similar to mean squares divided by the degrees of freedom.

Explanation of Solution

The provided statement is that the sum of squares is similar to mean of squares divided by the degrees of freedom. The statement is wrong. The correct formula is shown below:

$MS=\frac{SS}{DF}$

The correct statement is the mean of squares is similar to sum of squares divided by the degrees of freedom. Hence, the hypothesis can be considered as a wrong statement.

(c)

To determine

The fault in the provided statement.

Answer to Problem 5E

Solution: The fault in the statement is that when the null hypothesis is true in a two-way ANOVA, the test statistics for the main effects does not follow chi-square distribution.

Explanation of Solution

The provided statement is that when the null hypothesis is true in a two-way ANOVA, the test statistics for the main effects follows chi-square distribution. The statement is wrong because when the null hypothesis is true in a two-way ANOVA, the test statistics for the main effects have F distribution. Hence, the hypothesis can be considered as a wrong statement.

(d)

To determine

The fault in the provided statement.

Answer to Problem 5E

Solution: The provided statement is wrong because the sums of squares add up if there is equal number of observations in two-way ANOVA.

Explanation of Solution

The provided statement is that the sums of squares always include two-way ANOVA. The statement is wrong because the sums of squares for equal number of observation in two-way ANOVA partitioned as

$\text{SS}\left(\text{total}\right)\text{=SS}\left(\text{A}\right)\text{+SS}\left(\text{B}\right)\text{+SS}\left(\text{AB}\right)\text{+SSE}$

If the number of observation are not equal, sum of square may not partitioned as above. Hence, the hypothesis can be considered as a wrong statement.