Complete the F table.

Question

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Answer 1

Question

Chapter 12, Problem 22CAP

(a)

To determine

Complete the F table.

(a)

Expert Solution

Answer to Problem 22CAP

The complete F table is,

Source of Variation	SS	df	MS	$Fobt$
Between groups	30	2	15	11.72
Within groups (error)	50	39	1.28
Total	80	41

Explanation of Solution

Calculation:

From the given data, there are three groups highly experienced, moderately experienced and inexperienced athletes. The sample size is 14. That is, $k=3$ and $n=14$ .

The formulas for computing the F table are,

Source of Variation	SS	df	MS	$Fobt$
Between groups	$SSBG$	$k−1$	$SSBGdfBG$	$MSBGMSE$
Within groups (error)	$SSE$	$N−k$	$SSEdfE$
Total	80	$(N−1) or (kn−1)$

Where $SSBG$ indicates the sum of the squares of variation between the groups, $SST$ denotes the Total sum of squares, $SSE$ denotes the error sum of squares, $MSE$ indicates the means squares due to error, $MSBG$ indicates the means squares due to variation between the groups.k indicates the levels of factor. n represents the number of participants in each groups. N represents the total number of participants.

Source of Variation	SS	df	MS	$Fobt$
Between groups	$80−50=30$	$3−1=2$	$302=15$	$151.28=11.72$
Within groups (error)	50	$42−3=39$	$5039=1.28$
Total	80	$(14×3)−1=41$

Thus, the ANOVA table is,

Source of Variation	SS	df	MS	$Fobt$
Between groups	30	2	15	11.72
Within groups (error)	50	39	1.28
Total	80	41

(b)

To determine

Find the value of eta-squared $(η2)$ .

(b)

Expert Solution

Answer to Problem 22CAP

The value of eta-squared $(η2)$ is 0.38 and it represents thelarge effect size.

Explanation of Solution

Calculation:

The ANOVA table suggests that the value of $SSBG$ is 30 and the value of $SST$ is 80.

Eta-squared $(η2)$ :

One of the measure for effect size is Eta-squared $(η2)$ . Eta-squared computed as the sum of squares between groups divided by the total sum of squares.

Formula is given by,

$η2=SSBGSST$ .

Where, $SSBG$ represents the sum of squares between groups and $SST$ represents the total sum of squares.

Description of effect size using eta-squared:

If the eta-squared value is less than 0.01, then the effect size is small.
If the eta-squared value lies between 0.01 and 0.09, then the effect size is medium.
If the eta-squared value is greater than 0.25, then the effect size is larger.

Substitute the value of $SSBG$ is 30 and the value of $SST$ is 80.

The eta-squared value is,

$η2=SSBGSST=3080=0.38$

Justification: The value of eta-squared is 0.38, which is greater than 0.25. Hence, the effect size is larger.

Hence, the value of eta-squared $(η2)$ is 0.38 and it represents the large effect size.

(c)

To determine

Observe whether the decision is to retain or reject the null hypothesis.

(c)

Expert Solution

Answer to Problem 22CAP

The decision is rejecting the null hypothesis.

Explanation of Solution

Calculation:

The given information suggest that the degrees of freedom for between groups is 2 and the degrees of freedom for error is 39.

Rejection Rule:

If the value of the test statistic is greater than the critical value then reject the null hypothesis.

The given ANOVA table suggests that the degrees of freedom for between groups is 2 and the degrees of freedom for error is 39. That is, the degrees of freedom for numerator is 2 because the between group degrees of freedom represents the numerator degrees of freedom and the degrees of freedom for denominator is 39 because the error group degrees of freedom represents the denominator degrees of freedom.

Critical value:

Assume that the significance level is $α=0.05$ .

The numerator degrees of freedom are 2, the denominator degrees of freedom as 39 and the alpha level is 0.05.

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

Locate the value 2 in the numerator degrees of freedom row.
Locate the value 39 in the denominator degrees of freedom column.
Locate the 0.05 in level of significance.
The intersecting value that corresponds to the numerator degrees of freedom 2, the denominator degrees of freedom 39 with level of significance 0.05 is 3.24.

Thus, the critical value for the numerator degrees of freedom 2, the denominator degrees of freedom 39 with level of significance 0.05 is 3.24.

The ANOVA gives that the test statistic value is $Fobt=11.72$ .

Conclusion:

The value of test statistic is 11.72

The critical value is 3.24.

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

The test statistic value falls under critical region.

By the decision rule, the conclusion is rejecting the null hypothesis.

Hence, the decision is rejecting the null hypothesis.

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Answer 2

Question

Chapter 12, Problem 22CAP

(a)

To determine

Complete the F table.

(a)

Expert Solution

Answer to Problem 22CAP

The complete F table is,

Source of Variation	SS	df	MS	$Fobt$
Between groups	30	2	15	11.72
Within groups (error)	50	39	1.28
Total	80	41

Explanation of Solution

Calculation:

From the given data, there are three groups highly experienced, moderately experienced and inexperienced athletes. The sample size is 14. That is, $k=3$ and $n=14$ .

The formulas for computing the F table are,

Source of Variation	SS	df	MS	$Fobt$
Between groups	$SSBG$	$k−1$	$SSBGdfBG$	$MSBGMSE$
Within groups (error)	$SSE$	$N−k$	$SSEdfE$
Total	80	$(N−1) or (kn−1)$

Where $SSBG$ indicates the sum of the squares of variation between the groups, $SST$ denotes the Total sum of squares, $SSE$ denotes the error sum of squares, $MSE$ indicates the means squares due to error, $MSBG$ indicates the means squares due to variation between the groups.k indicates the levels of factor. n represents the number of participants in each groups. N represents the total number of participants.

Source of Variation	SS	df	MS	$Fobt$
Between groups	$80−50=30$	$3−1=2$	$302=15$	$151.28=11.72$
Within groups (error)	50	$42−3=39$	$5039=1.28$
Total	80	$(14×3)−1=41$

Thus, the ANOVA table is,

Source of Variation	SS	df	MS	$Fobt$
Between groups	30	2	15	11.72
Within groups (error)	50	39	1.28
Total	80	41

(b)

To determine

Find the value of eta-squared $(η2)$ .

(b)

Expert Solution

Answer to Problem 22CAP

The value of eta-squared $(η2)$ is 0.38 and it represents thelarge effect size.

Explanation of Solution

Calculation:

The ANOVA table suggests that the value of $SSBG$ is 30 and the value of $SST$ is 80.

Eta-squared $(η2)$ :

One of the measure for effect size is Eta-squared $(η2)$ . Eta-squared computed as the sum of squares between groups divided by the total sum of squares.

Formula is given by,

$η2=SSBGSST$ .

Where, $SSBG$ represents the sum of squares between groups and $SST$ represents the total sum of squares.

Description of effect size using eta-squared:

If the eta-squared value is less than 0.01, then the effect size is small.
If the eta-squared value lies between 0.01 and 0.09, then the effect size is medium.
If the eta-squared value is greater than 0.25, then the effect size is larger.

Substitute the value of $SSBG$ is 30 and the value of $SST$ is 80.

The eta-squared value is,

$η2=SSBGSST=3080=0.38$

Justification: The value of eta-squared is 0.38, which is greater than 0.25. Hence, the effect size is larger.

Hence, the value of eta-squared $(η2)$ is 0.38 and it represents the large effect size.

(c)

To determine

Observe whether the decision is to retain or reject the null hypothesis.

(c)

Expert Solution

Answer to Problem 22CAP

The decision is rejecting the null hypothesis.

Explanation of Solution

Calculation:

The given information suggest that the degrees of freedom for between groups is 2 and the degrees of freedom for error is 39.

Rejection Rule:

If the value of the test statistic is greater than the critical value then reject the null hypothesis.

The given ANOVA table suggests that the degrees of freedom for between groups is 2 and the degrees of freedom for error is 39. That is, the degrees of freedom for numerator is 2 because the between group degrees of freedom represents the numerator degrees of freedom and the degrees of freedom for denominator is 39 because the error group degrees of freedom represents the denominator degrees of freedom.

Critical value:

Assume that the significance level is $α=0.05$ .

The numerator degrees of freedom are 2, the denominator degrees of freedom as 39 and the alpha level is 0.05.

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

Locate the value 2 in the numerator degrees of freedom row.
Locate the value 39 in the denominator degrees of freedom column.
Locate the 0.05 in level of significance.
The intersecting value that corresponds to the numerator degrees of freedom 2, the denominator degrees of freedom 39 with level of significance 0.05 is 3.24.

Thus, the critical value for the numerator degrees of freedom 2, the denominator degrees of freedom 39 with level of significance 0.05 is 3.24.

The ANOVA gives that the test statistic value is $Fobt=11.72$ .

Conclusion:

The value of test statistic is 11.72

The critical value is 3.24.

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

The test statistic value falls under critical region.

By the decision rule, the conclusion is rejecting the null hypothesis.

Hence, the decision is rejecting the null hypothesis.

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Answer 3

Question

Chapter 12, Problem 22CAP

(a)

To determine

Complete the F table.

(a)

Expert Solution

Answer to Problem 22CAP

The complete F table is,

Source of Variation	SS	df	MS	$Fobt$
Between groups	30	2	15	11.72
Within groups (error)	50	39	1.28
Total	80	41

Explanation of Solution

Calculation:

From the given data, there are three groups highly experienced, moderately experienced and inexperienced athletes. The sample size is 14. That is, $k=3$ and $n=14$ .

The formulas for computing the F table are,

Source of Variation	SS	df	MS	$Fobt$
Between groups	$SSBG$	$k−1$	$SSBGdfBG$	$MSBGMSE$
Within groups (error)	$SSE$	$N−k$	$SSEdfE$
Total	80	$(N−1) or (kn−1)$

Where $SSBG$ indicates the sum of the squares of variation between the groups, $SST$ denotes the Total sum of squares, $SSE$ denotes the error sum of squares, $MSE$ indicates the means squares due to error, $MSBG$ indicates the means squares due to variation between the groups.k indicates the levels of factor. n represents the number of participants in each groups. N represents the total number of participants.

Source of Variation	SS	df	MS	$Fobt$
Between groups	$80−50=30$	$3−1=2$	$302=15$	$151.28=11.72$
Within groups (error)	50	$42−3=39$	$5039=1.28$
Total	80	$(14×3)−1=41$

Thus, the ANOVA table is,

Source of Variation	SS	df	MS	$Fobt$
Between groups	30	2	15	11.72
Within groups (error)	50	39	1.28
Total	80	41

(b)

To determine

Find the value of eta-squared $(η2)$ .

(b)

Expert Solution

Answer to Problem 22CAP

The value of eta-squared $(η2)$ is 0.38 and it represents thelarge effect size.

Explanation of Solution

Calculation:

The ANOVA table suggests that the value of $SSBG$ is 30 and the value of $SST$ is 80.

Eta-squared $(η2)$ :

One of the measure for effect size is Eta-squared $(η2)$ . Eta-squared computed as the sum of squares between groups divided by the total sum of squares.

Formula is given by,

$η2=SSBGSST$ .

Where, $SSBG$ represents the sum of squares between groups and $SST$ represents the total sum of squares.

Description of effect size using eta-squared:

If the eta-squared value is less than 0.01, then the effect size is small.
If the eta-squared value lies between 0.01 and 0.09, then the effect size is medium.
If the eta-squared value is greater than 0.25, then the effect size is larger.

Substitute the value of $SSBG$ is 30 and the value of $SST$ is 80.

The eta-squared value is,

$η2=SSBGSST=3080=0.38$

Justification: The value of eta-squared is 0.38, which is greater than 0.25. Hence, the effect size is larger.

Hence, the value of eta-squared $(η2)$ is 0.38 and it represents the large effect size.

(c)

To determine

Observe whether the decision is to retain or reject the null hypothesis.

(c)

Expert Solution

Answer to Problem 22CAP

The decision is rejecting the null hypothesis.

Explanation of Solution

Calculation:

The given information suggest that the degrees of freedom for between groups is 2 and the degrees of freedom for error is 39.

Rejection Rule:

If the value of the test statistic is greater than the critical value then reject the null hypothesis.

The given ANOVA table suggests that the degrees of freedom for between groups is 2 and the degrees of freedom for error is 39. That is, the degrees of freedom for numerator is 2 because the between group degrees of freedom represents the numerator degrees of freedom and the degrees of freedom for denominator is 39 because the error group degrees of freedom represents the denominator degrees of freedom.

Critical value:

Assume that the significance level is $α=0.05$ .

The numerator degrees of freedom are 2, the denominator degrees of freedom as 39 and the alpha level is 0.05.

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

Locate the value 2 in the numerator degrees of freedom row.
Locate the value 39 in the denominator degrees of freedom column.
Locate the 0.05 in level of significance.
The intersecting value that corresponds to the numerator degrees of freedom 2, the denominator degrees of freedom 39 with level of significance 0.05 is 3.24.

Thus, the critical value for the numerator degrees of freedom 2, the denominator degrees of freedom 39 with level of significance 0.05 is 3.24.

The ANOVA gives that the test statistic value is $Fobt=11.72$ .

Conclusion:

The value of test statistic is 11.72

The critical value is 3.24.

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

The test statistic value falls under critical region.

By the decision rule, the conclusion is rejecting the null hypothesis.

Hence, the decision is rejecting the null hypothesis.

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Answer 4

Question

Chapter 12, Problem 22CAP

(a)

To determine

Complete the F table.

(a)

Expert Solution

Answer to Problem 22CAP

The complete F table is,

Source of Variation	SS	df	MS	$Fobt$
Between groups	30	2	15	11.72
Within groups (error)	50	39	1.28
Total	80	41

Explanation of Solution

Calculation:

From the given data, there are three groups highly experienced, moderately experienced and inexperienced athletes. The sample size is 14. That is, $k=3$ and $n=14$ .

The formulas for computing the F table are,

Source of Variation	SS	df	MS	$Fobt$
Between groups	$SSBG$	$k−1$	$SSBGdfBG$	$MSBGMSE$
Within groups (error)	$SSE$	$N−k$	$SSEdfE$
Total	80	$(N−1) or (kn−1)$

Where $SSBG$ indicates the sum of the squares of variation between the groups, $SST$ denotes the Total sum of squares, $SSE$ denotes the error sum of squares, $MSE$ indicates the means squares due to error, $MSBG$ indicates the means squares due to variation between the groups.k indicates the levels of factor. n represents the number of participants in each groups. N represents the total number of participants.

Source of Variation	SS	df	MS	$Fobt$
Between groups	$80−50=30$	$3−1=2$	$302=15$	$151.28=11.72$
Within groups (error)	50	$42−3=39$	$5039=1.28$
Total	80	$(14×3)−1=41$

Thus, the ANOVA table is,

Source of Variation	SS	df	MS	$Fobt$
Between groups	30	2	15	11.72
Within groups (error)	50	39	1.28
Total	80	41

(b)

To determine

Find the value of eta-squared $(η2)$ .

(b)

Expert Solution

Answer to Problem 22CAP

The value of eta-squared $(η2)$ is 0.38 and it represents thelarge effect size.

Explanation of Solution

Calculation:

The ANOVA table suggests that the value of $SSBG$ is 30 and the value of $SST$ is 80.

Eta-squared $(η2)$ :

One of the measure for effect size is Eta-squared $(η2)$ . Eta-squared computed as the sum of squares between groups divided by the total sum of squares.

Formula is given by,

$η2=SSBGSST$ .

Where, $SSBG$ represents the sum of squares between groups and $SST$ represents the total sum of squares.

Description of effect size using eta-squared:

If the eta-squared value is less than 0.01, then the effect size is small.
If the eta-squared value lies between 0.01 and 0.09, then the effect size is medium.
If the eta-squared value is greater than 0.25, then the effect size is larger.

Substitute the value of $SSBG$ is 30 and the value of $SST$ is 80.

The eta-squared value is,

$η2=SSBGSST=3080=0.38$

Justification: The value of eta-squared is 0.38, which is greater than 0.25. Hence, the effect size is larger.

Hence, the value of eta-squared $(η2)$ is 0.38 and it represents the large effect size.

(c)

To determine

Observe whether the decision is to retain or reject the null hypothesis.

(c)

Expert Solution

Answer to Problem 22CAP

The decision is rejecting the null hypothesis.

Explanation of Solution

Calculation:

The given information suggest that the degrees of freedom for between groups is 2 and the degrees of freedom for error is 39.

Rejection Rule:

If the value of the test statistic is greater than the critical value then reject the null hypothesis.

The given ANOVA table suggests that the degrees of freedom for between groups is 2 and the degrees of freedom for error is 39. That is, the degrees of freedom for numerator is 2 because the between group degrees of freedom represents the numerator degrees of freedom and the degrees of freedom for denominator is 39 because the error group degrees of freedom represents the denominator degrees of freedom.

Critical value:

Assume that the significance level is $α=0.05$ .

The numerator degrees of freedom are 2, the denominator degrees of freedom as 39 and the alpha level is 0.05.

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

Locate the value 2 in the numerator degrees of freedom row.
Locate the value 39 in the denominator degrees of freedom column.
Locate the 0.05 in level of significance.
The intersecting value that corresponds to the numerator degrees of freedom 2, the denominator degrees of freedom 39 with level of significance 0.05 is 3.24.

Thus, the critical value for the numerator degrees of freedom 2, the denominator degrees of freedom 39 with level of significance 0.05 is 3.24.

The ANOVA gives that the test statistic value is $Fobt=11.72$ .

Conclusion:

The value of test statistic is 11.72

The critical value is 3.24.

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

The test statistic value falls under critical region.

By the decision rule, the conclusion is rejecting the null hypothesis.

Hence, the decision is rejecting the null hypothesis.

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Answer 5

Chapter 12, Problem 22CAP

(a)

To determine

Complete the F table.

(a)

Expert Solution

Answer to Problem 22CAP

The complete F table is,

Source of Variation	SS	df	MS	$Fobt$
Between groups	30	2	15	11.72
Within groups (error)	50	39	1.28
Total	80	41

Explanation of Solution

Calculation:

From the given data, there are three groups highly experienced, moderately experienced and inexperienced athletes. The sample size is 14. That is, $k=3$ and $n=14$ .

The formulas for computing the F table are,

Source of Variation	SS	df	MS	$Fobt$
Between groups	$SSBG$	$k−1$	$SSBGdfBG$	$MSBGMSE$
Within groups (error)	$SSE$	$N−k$	$SSEdfE$
Total	80	$(N−1) or (kn−1)$

Where $SSBG$ indicates the sum of the squares of variation between the groups, $SST$ denotes the Total sum of squares, $SSE$ denotes the error sum of squares, $MSE$ indicates the means squares due to error, $MSBG$ indicates the means squares due to variation between the groups.k indicates the levels of factor. n represents the number of participants in each groups. N represents the total number of participants.

Source of Variation	SS	df	MS	$Fobt$
Between groups	$80−50=30$	$3−1=2$	$302=15$	$151.28=11.72$
Within groups (error)	50	$42−3=39$	$5039=1.28$
Total	80	$(14×3)−1=41$

Thus, the ANOVA table is,

Source of Variation	SS	df	MS	$Fobt$
Between groups	30	2	15	11.72
Within groups (error)	50	39	1.28
Total	80	41

(b)

To determine

Find the value of eta-squared $(η2)$ .

(b)

Expert Solution

Answer to Problem 22CAP

The value of eta-squared $(η2)$ is 0.38 and it represents thelarge effect size.

Explanation of Solution

Calculation:

The ANOVA table suggests that the value of $SSBG$ is 30 and the value of $SST$ is 80.

Eta-squared $(η2)$ :

One of the measure for effect size is Eta-squared $(η2)$ . Eta-squared computed as the sum of squares between groups divided by the total sum of squares.

Formula is given by,

$η2=SSBGSST$ .

Where, $SSBG$ represents the sum of squares between groups and $SST$ represents the total sum of squares.

Description of effect size using eta-squared:

If the eta-squared value is less than 0.01, then the effect size is small.
If the eta-squared value lies between 0.01 and 0.09, then the effect size is medium.
If the eta-squared value is greater than 0.25, then the effect size is larger.

Substitute the value of $SSBG$ is 30 and the value of $SST$ is 80.

The eta-squared value is,

$η2=SSBGSST=3080=0.38$

Justification: The value of eta-squared is 0.38, which is greater than 0.25. Hence, the effect size is larger.

Hence, the value of eta-squared $(η2)$ is 0.38 and it represents the large effect size.

(c)

To determine

Observe whether the decision is to retain or reject the null hypothesis.

(c)

Expert Solution

Answer to Problem 22CAP

The decision is rejecting the null hypothesis.

Explanation of Solution

Calculation:

The given information suggest that the degrees of freedom for between groups is 2 and the degrees of freedom for error is 39.

Rejection Rule:

If the value of the test statistic is greater than the critical value then reject the null hypothesis.

The given ANOVA table suggests that the degrees of freedom for between groups is 2 and the degrees of freedom for error is 39. That is, the degrees of freedom for numerator is 2 because the between group degrees of freedom represents the numerator degrees of freedom and the degrees of freedom for denominator is 39 because the error group degrees of freedom represents the denominator degrees of freedom.

Critical value:

Assume that the significance level is $α=0.05$ .

The numerator degrees of freedom are 2, the denominator degrees of freedom as 39 and the alpha level is 0.05.

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

Locate the value 2 in the numerator degrees of freedom row.
Locate the value 39 in the denominator degrees of freedom column.
Locate the 0.05 in level of significance.
The intersecting value that corresponds to the numerator degrees of freedom 2, the denominator degrees of freedom 39 with level of significance 0.05 is 3.24.

Thus, the critical value for the numerator degrees of freedom 2, the denominator degrees of freedom 39 with level of significance 0.05 is 3.24.

The ANOVA gives that the test statistic value is $Fobt=11.72$ .

Conclusion:

The value of test statistic is 11.72

The critical value is 3.24.

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

The test statistic value falls under critical region.

By the decision rule, the conclusion is rejecting the null hypothesis.

Hence, the decision is rejecting the null hypothesis.

Answer 6

Chapter 12, Problem 22CAP

(a)

To determine

Complete the F table.

(a)

Expert Solution

Answer to Problem 22CAP

The complete F table is,

Source of Variation	SS	df	MS	$Fobt$
Between groups	30	2	15	11.72
Within groups (error)	50	39	1.28
Total	80	41

Explanation of Solution

Calculation:

From the given data, there are three groups highly experienced, moderately experienced and inexperienced athletes. The sample size is 14. That is, $k=3$ and $n=14$ .

The formulas for computing the F table are,

Source of Variation	SS	df	MS	$Fobt$
Between groups	$SSBG$	$k−1$	$SSBGdfBG$	$MSBGMSE$
Within groups (error)	$SSE$	$N−k$	$SSEdfE$
Total	80	$(N−1) or (kn−1)$

Where $SSBG$ indicates the sum of the squares of variation between the groups, $SST$ denotes the Total sum of squares, $SSE$ denotes the error sum of squares, $MSE$ indicates the means squares due to error, $MSBG$ indicates the means squares due to variation between the groups.k indicates the levels of factor. n represents the number of participants in each groups. N represents the total number of participants.

Source of Variation	SS	df	MS	$Fobt$
Between groups	$80−50=30$	$3−1=2$	$302=15$	$151.28=11.72$
Within groups (error)	50	$42−3=39$	$5039=1.28$
Total	80	$(14×3)−1=41$

Thus, the ANOVA table is,

Source of Variation	SS	df	MS	$Fobt$
Between groups	30	2	15	11.72
Within groups (error)	50	39	1.28
Total	80	41

Answer 7

Chapter 12, Problem 22CAP

(a)

To determine

Complete the F table.

Answer 8

(a)

Expert Solution

Answer to Problem 22CAP

The complete F table is,

Source of Variation	SS	df	MS	$Fobt$
Between groups	30	2	15	11.72
Within groups (error)	50	39	1.28
Total	80	41

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

From the given data, there are three groups highly experienced, moderately experienced and inexperienced athletes. The sample size is 14. That is, $k=3$ and $n=14$ .

The formulas for computing the F table are,

Source of Variation	SS	df	MS	$Fobt$
Between groups	$SSBG$	$k−1$	$SSBGdfBG$	$MSBGMSE$
Within groups (error)	$SSE$	$N−k$	$SSEdfE$
Total	80	$(N−1) or (kn−1)$

Where $SSBG$ indicates the sum of the squares of variation between the groups, $SST$ denotes the Total sum of squares, $SSE$ denotes the error sum of squares, $MSE$ indicates the means squares due to error, $MSBG$ indicates the means squares due to variation between the groups.k indicates the levels of factor. n represents the number of participants in each groups. N represents the total number of participants.

Source of Variation	SS	df	MS	$Fobt$
Between groups	$80−50=30$	$3−1=2$	$302=15$	$151.28=11.72$
Within groups (error)	50	$42−3=39$	$5039=1.28$
Total	80	$(14×3)−1=41$

Thus, the ANOVA table is,

Source of Variation	SS	df	MS	$Fobt$
Between groups	30	2	15	11.72
Within groups (error)	50	39	1.28
Total	80	41

Answer 13

(b)

To determine

Find the value of eta-squared $(η2)$ .

(b)

Expert Solution

Answer to Problem 22CAP

The value of eta-squared $(η2)$ is 0.38 and it represents thelarge effect size.

Explanation of Solution

Calculation:

The ANOVA table suggests that the value of $SSBG$ is 30 and the value of $SST$ is 80.

Eta-squared $(η2)$ :

One of the measure for effect size is Eta-squared $(η2)$ . Eta-squared computed as the sum of squares between groups divided by the total sum of squares.

Formula is given by,

$η2=SSBGSST$ .

Where, $SSBG$ represents the sum of squares between groups and $SST$ represents the total sum of squares.

Description of effect size using eta-squared:

If the eta-squared value is less than 0.01, then the effect size is small.
If the eta-squared value lies between 0.01 and 0.09, then the effect size is medium.
If the eta-squared value is greater than 0.25, then the effect size is larger.

Substitute the value of $SSBG$ is 30 and the value of $SST$ is 80.

The eta-squared value is,

$η2=SSBGSST=3080=0.38$

Justification: The value of eta-squared is 0.38, which is greater than 0.25. Hence, the effect size is larger.

Hence, the value of eta-squared $(η2)$ is 0.38 and it represents the large effect size.

Answer 14

(b)

To determine

Find the value of eta-squared $(η2)$ .

Answer 15

(b)

Expert Solution

Answer to Problem 22CAP

The value of eta-squared $(η2)$ is 0.38 and it represents thelarge effect size.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

The ANOVA table suggests that the value of $SSBG$ is 30 and the value of $SST$ is 80.

Eta-squared $(η2)$ :

One of the measure for effect size is Eta-squared $(η2)$ . Eta-squared computed as the sum of squares between groups divided by the total sum of squares.

Formula is given by,

$η2=SSBGSST$ .

Where, $SSBG$ represents the sum of squares between groups and $SST$ represents the total sum of squares.

Description of effect size using eta-squared:

If the eta-squared value is less than 0.01, then the effect size is small.
If the eta-squared value lies between 0.01 and 0.09, then the effect size is medium.
If the eta-squared value is greater than 0.25, then the effect size is larger.

Substitute the value of $SSBG$ is 30 and the value of $SST$ is 80.

The eta-squared value is,

$η2=SSBGSST=3080=0.38$

Justification: The value of eta-squared is 0.38, which is greater than 0.25. Hence, the effect size is larger.

Hence, the value of eta-squared $(η2)$ is 0.38 and it represents the large effect size.

Answer 20

(c)

To determine

Observe whether the decision is to retain or reject the null hypothesis.

(c)

Expert Solution

Answer to Problem 22CAP

The decision is rejecting the null hypothesis.

Explanation of Solution

Calculation:

The given information suggest that the degrees of freedom for between groups is 2 and the degrees of freedom for error is 39.

Rejection Rule:

If the value of the test statistic is greater than the critical value then reject the null hypothesis.

The given ANOVA table suggests that the degrees of freedom for between groups is 2 and the degrees of freedom for error is 39. That is, the degrees of freedom for numerator is 2 because the between group degrees of freedom represents the numerator degrees of freedom and the degrees of freedom for denominator is 39 because the error group degrees of freedom represents the denominator degrees of freedom.

Critical value:

Assume that the significance level is $α=0.05$ .

The numerator degrees of freedom are 2, the denominator degrees of freedom as 39 and the alpha level is 0.05.

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

Locate the value 2 in the numerator degrees of freedom row.
Locate the value 39 in the denominator degrees of freedom column.
Locate the 0.05 in level of significance.
The intersecting value that corresponds to the numerator degrees of freedom 2, the denominator degrees of freedom 39 with level of significance 0.05 is 3.24.

Thus, the critical value for the numerator degrees of freedom 2, the denominator degrees of freedom 39 with level of significance 0.05 is 3.24.

The ANOVA gives that the test statistic value is $Fobt=11.72$ .

Conclusion:

The value of test statistic is 11.72

The critical value is 3.24.

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

The test statistic value falls under critical region.

By the decision rule, the conclusion is rejecting the null hypothesis.

Hence, the decision is rejecting the null hypothesis.

Answer 21

(c)

To determine

Observe whether the decision is to retain or reject the null hypothesis.

Answer 22

(c)

Expert Solution

Answer to Problem 22CAP

The decision is rejecting the null hypothesis.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

The given information suggest that the degrees of freedom for between groups is 2 and the degrees of freedom for error is 39.

Rejection Rule:

If the value of the test statistic is greater than the critical value then reject the null hypothesis.

The given ANOVA table suggests that the degrees of freedom for between groups is 2 and the degrees of freedom for error is 39. That is, the degrees of freedom for numerator is 2 because the between group degrees of freedom represents the numerator degrees of freedom and the degrees of freedom for denominator is 39 because the error group degrees of freedom represents the denominator degrees of freedom.

Critical value:

Assume that the significance level is $α=0.05$ .

The numerator degrees of freedom are 2, the denominator degrees of freedom as 39 and the alpha level is 0.05.

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

Locate the value 2 in the numerator degrees of freedom row.
Locate the value 39 in the denominator degrees of freedom column.
Locate the 0.05 in level of significance.
The intersecting value that corresponds to the numerator degrees of freedom 2, the denominator degrees of freedom 39 with level of significance 0.05 is 3.24.

Thus, the critical value for the numerator degrees of freedom 2, the denominator degrees of freedom 39 with level of significance 0.05 is 3.24.

The ANOVA gives that the test statistic value is $Fobt=11.72$ .

Conclusion:

The value of test statistic is 11.72

The critical value is 3.24.

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

The test statistic value falls under critical region.

By the decision rule, the conclusion is rejecting the null hypothesis.

Hence, the decision is rejecting the null hypothesis.

Answer 27

Ch. 12.2 - Prob. 1.1LC Ch. 12.2 - Prob. 1.2LC Ch. 12.2 - Prob. 1.3LC Ch. 12.2 - Prob. 1.4LC Ch. 12.4 - Prob. 2.1LC Ch. 12.4 - Prob. 2.2LC Ch. 12.4 - Prob. 2.3LC Ch. 12.4 - Prob. 2.4LC Ch. 12.6 - Prob. 3.1LC Ch. 12.6 - Prob. 3.2LC

Complete the F table.

Videos

Complete the F table.

Answer to Problem 22CAP

Explanation of Solution

Find the value of eta-squared $(η2)$ .

Answer to Problem 22CAP

Explanation of Solution

Observe whether the decision is to retain or reject the null hypothesis.

Answer to Problem 22CAP

Explanation of Solution

Want to see more full solutions like this?

Chapter 12 Solutions

Complete the F table.

Videos

Complete the F table.

Answer to Problem 22CAP

Explanation of Solution

Find the value of eta-squared (η2)<m:math><m:mrow><m:mrow><m:mo>(</m:mo><m:mrow><m:msup><m:mi>η</m:mi><m:mn>2</m:mn></m:msup></m:mrow><m:mo>)</m:mo></m:mrow></m:mrow></m:math>.

Answer to Problem 22CAP

Explanation of Solution

Observe whether the decision is to retain or reject the null hypothesis.

Answer to Problem 22CAP

Explanation of Solution

Want to see more full solutions like this?

Chapter 12 Solutions

Find the value of eta-squared $(η2)$ .