The fault in the provided statement.

Question

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Answer 1

Question

Chapter 13, Problem 5E

(a)

To determine

The fault in the provided statement.

(a)

Expert Solution

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.

(b)

Expert Solution

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=SSDF

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.

(c)

Expert Solution

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.

(d)

Expert Solution

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

SS(total)=SS(A)+SS(B)+SS(AB)+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.

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Answer 2

Question

Chapter 13, Problem 5E

(a)

To determine

The fault in the provided statement.

(a)

Expert Solution

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.

(b)

Expert Solution

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=SSDF

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.

(c)

Expert Solution

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.

(d)

Expert Solution

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

SS(total)=SS(A)+SS(B)+SS(AB)+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.

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Answer 3

Question

Chapter 13, Problem 5E

(a)

To determine

The fault in the provided statement.

(a)

Expert Solution

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.

(b)

Expert Solution

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=SSDF

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.

(c)

Expert Solution

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.

(d)

Expert Solution

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

SS(total)=SS(A)+SS(B)+SS(AB)+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.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 13, Problem 5E

(a)

To determine

The fault in the provided statement.

(a)

Expert Solution

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.

(b)

Expert Solution

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=SSDF

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.

(c)

Expert Solution

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.

(d)

Expert Solution

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

SS(total)=SS(A)+SS(B)+SS(AB)+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.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 13, Problem 5E

(a)

To determine

The fault in the provided statement.

(a)

Expert Solution

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.

(b)

Expert Solution

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=SSDF

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.

(c)

Expert Solution

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.

(d)

Expert Solution

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

SS(total)=SS(A)+SS(B)+SS(AB)+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.

Answer 6

Chapter 13, Problem 5E

(a)

To determine

The fault in the provided statement.

(a)

Expert Solution

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.

Answer 7

Chapter 13, Problem 5E

(a)

To determine

The fault in the provided statement.

Answer 8

(a)

Expert Solution

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.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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.

Answer 13

(b)

To determine

The fault in the provided statement.

(b)

Expert Solution

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=SSDF

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.

Answer 14

(b)

To determine

The fault in the provided statement.

Answer 15

(b)

Expert Solution

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.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

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=SSDF

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.

Answer 20

(c)

To determine

The fault in the provided statement.

(c)

Expert Solution

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.

Answer 21

(c)

To determine

The fault in the provided statement.

Answer 22

(c)

Expert Solution

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.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

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.

Answer 27

(d)

To determine

The fault in the provided statement.

(d)

Expert Solution

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

SS(total)=SS(A)+SS(B)+SS(AB)+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.

Answer 28

(d)

To determine

The fault in the provided statement.

Answer 29

(d)

Expert Solution

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.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

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

SS(total)=SS(A)+SS(B)+SS(AB)+SSE