Chapter 12, Problem 15CAP

1.

To determine

Find the decision for the test at the 0.05 level of significance with 3 and 26 degrees of freedom.

Answer to Problem 15CAP

The decision for the test at the 0.05 level of significance with 3 and 26 degrees of freedom is reject the null hypothesis.

Explanation of Solution

Calculation:

From the information, given that $F_{\left(3,26\right)}=3.00$.

One-way between subjects ANOVA:

It is the statistical procedure that is used to test the hypotheses for the one factor with the two more levels by measuring the variances among two or more means. Moreover it is used when the participants are observed differently at level of the factor with the unknown population variance.

Between group variation:

It is variation attributed to the differences of means between the groups.

Degrees of freedom:$df=k-1$.

Where k indicates the number of groups.

Within group variation:

It is variation attributed to the differences of means within the each group.

Degrees of freedom:$df=N-k$.

Where k indicates the number of groups and N indicates the total participants.

Decision rules

It is the region beyond a critical value in the given hypothesis test. That is,

• If the given value of a test statistic falls beyond the critical value the null hypothesis is rejected.
• If the given value of a test statistic is not greater than the critical value then the null hypothesis is retained.

Critical value:

Critical value is the threshold value defines the limits beyond less than 5% of sample means which can be obtained if the given null hypothesis is true. If the sample mean lies beyond a critical value would lead to the decision of rejecting the null hypothesis.

The given significance level is $\alpha =0.05$.

The test is two tailed, the degrees of freedom are $F_{\left(3,26\right)}$, and the alpha level is 0.05.

From the Appendix C: Table C.3 the F Distribution:

• Locate the value 3 in the numerator degrees of freedom (df) column.
• Locate the value 26 in the denominator degrees of freedom (df) row.
• Locate the 0.05 in the proportion in Two tails combined row.
• The intersecting value that corresponds to the $F_{\left(3,26\right)}$ with level of significance 0.05 is 2.98.

Thus, critical value for the test ANOVA at the 0.05 level of significance with 3 and 26 degrees of freedom is 2.98.

Conclusion:

The value of test statistic is 3.00

The upper critical value is 2.98.

The value of test statistic is greater than the critical value.

The test statistic value falls under critical region.

By the decision rule, the null hypothesis is rejected.

Thus, the decision for the test at the 0.05 level of significance with 3 and 26 degrees of freedom is reject the null hypothesis.

2.

To determine

Find the decision for the test at the 0.05 level of significance with 5 and 20 degrees of freedom.

Answer to Problem 15CAP

The decision for the test at the 0.05 level of significance with 5 and 20 degrees of freedom is retain the null hypothesis.

Explanation of Solution

Calculation:

From the information, given that $F_{\left(5,20\right)}=2.54$.

The given significance level is $\alpha =0.05$.

The test is two tailed, the degrees of freedom are $F_{\left(5,20\right)}=2.54$, and the alpha level is 0.05.

From the Appendix C: Table C.3 the F Distribution:

• Locate the value 5 in the numerator degrees of freedom (df) column.
• Locate the value 20 in the denominator degrees of freedom (df) row.
• Locate the 0.05 in the proportion in Two tails combined row.
• The intersecting value that corresponds to the $F_{\left(5,20\right)}$ with level of significance 0.05 is 2.71.

Thus, critical value for the test ANOVA at the 0.05 level of significance with 5 and 20 degrees of freedom is 2.71.

Conclusion:

The value of test statistic is 2.54.

The upper critical value is 2.71.

The value of test statistic is less than the critical value.

The test statistic value do not falls under critical region.

By the decision rule, the null hypothesis is retained.

Thus, the decision for the test at the 0.05 level of significance with 5 and 20 degrees of freedom is retain the null hypothesis.

3.

To determine

Find the decision for the test at the 0.05 level of significance with 4 and 30 degrees of freedom.

Answer to Problem 15CAP

The decision for the test at the 0.05 level of significance with 4 and 30 degrees of freedom is reject the null hypothesis.

Explanation of Solution

Calculation:

From the information, given that $F_{\left(4,30\right)}=2.72$.

The given significance level is $\alpha =0.05$.

The test is two tailed, the degrees of freedom are $F_{\left(4,30\right)}=2.72$, and the alpha level is 0.05.

From the Appendix C: Table C.3 the F Distribution:

• Locate the value 4 in the numerator degrees of freedom (df) column.
• Locate the value 30 in the denominator degrees of freedom (df) row.
• Locate the 0.05 in the proportion in Two tails combined row.
• The intersecting value that corresponds to the $F_{\left(4,30\right)}$ with level of significance 0.05 is 2.69.

Thus, critical value for the test ANOVA at the 0.05 level of significance with 4 and 30 degrees of freedom is 2.69.

Conclusion:

The value of test statistic is 2.72.

The upper critical value is 2.69.

The value of test statistic is greater than the critical value.

The test statistic value falls under critical region.

By the decision rule, the null hypothesis is rejected.

Thus, the decision for the test at the 0.05 level of significance with 4 and 30 degrees of freedom is reject the null hypothesis.

4.

To determine

Find the decision for the test at the 0.05 level of significance with 2 and 12 degrees of freedom.

Answer to Problem 15CAP

The decision for the test at the 0.05 level of significance with 2 and 12 degrees of freedom is retain the null hypothesis.

Explanation of Solution

Calculation:

From the information, given that $F_{\left(2,12\right)}=3.81$.

The given significance level is $\alpha =0.05$.

The test is two tailed, the degrees of freedom are $F_{\left(2,12\right)}=3.81$, and the alpha level is 0.05.

From the Appendix C: Table C.3 the F Distribution:

• Locate the value 2 in the numerator degrees of freedom (df) column.
• Locate the value 12 in the denominator degrees of freedom (df) row.
• Locate the 0.05 in the proportion in Two tails combined row.
• The intersecting value that corresponds to the $F_{\left(2,12\right)}$ with level of significance 0.05 is 3.89.

Thus, critical value for the test ANOVA at the 0.05 level of significance with 2 and 12 degrees of freedom is 3.89.

Conclusion:

The value of test statistic is 3.81.

The upper critical value is 3.89.

The value of test statistic is less than the critical value.

The test statistic value do not falls under critical region.

By the decision rule, the null hypothesis is retained.

Thus, the decision for the test at the 0.05 level of significance with 2 and 12 degrees of freedom is retain the null hypothesis.