Chapter 12, Problem 22CAP

1.

To determine

Complete the F table.

Answer to Problem 22CAP

The complete F table is,

Source of Variation	SS	df	MS	$F_{\text{obt}}$
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	$F_{\text{obt}}$
Between groups	$S{S}_{\text{BG}}$	$k-1$	$\frac{S{S}_{\text{BG}}}{d{f}_{\text{BG}}}$	$\frac{M{S}_{\text{BG}}}{M{S}_{\text{E}}}$
Within groups (error)	$S{S}_{\text{E}}$	$N-k$	$\frac{S{S}_{\text{E}}}{d{f}_{\text{E}}}$
Total	80	$\left(N-1\right)\text{ or }\left(kn-1\right)$

Source of Variation	SS	df	MS	$F_{\text{obt}}$
Between groups	$80-50=30$	$3-1=2$	$\frac{30}{2}=15$	$\frac{15}{1.28}=11.72$
Within groups (error)	50	$42-3=39$	$\frac{50}{39}=1.28$
Total	80	$\left(14\times 3\right)-1=41$

Thus, the ANOVA table is,

Source of Variation	SS	df	MS	$F_{\text{obt}}$
Between groups	30	2	15	11.72
Within groups (error)	50	39	1.28
Total	80	41

2.

To determine

Find the value of eta-squared $\left({\eta }^2\right)$.

Answer to Problem 22CAP

The value of eta-squared $\left({\eta }^2\right)$ is 0.38 and it represents the large effect size.

Explanation of Solution

Calculation:

The ANOVA table suggests that the value of $S{S}_{\text{BG}}$ is 30 and the value of $S{S}_{\text{T}}$ is 80.

Eta-squared $\left({\eta }^2\right)$:

One of the measure for effect size is Eta-squared $\left({\eta }^2\right)$. Eta-squared computed as the sum of squares between groups divided by the total sum of squares.

Formula is given by,

${\eta }^2=\frac{S{S}_{\text{BG}}}{S{S}_{\text{T}}}$.

Where, $S{S}_{\text{BG}}$ represents the sum of squares between groups and $S{S}_{\text{T}}$ 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 $S{S}_{\text{BG}}$ is 30 and the value of $S{S}_{\text{T}}$ is 80.

The eta-squared value is,

$\begin{array}{c}{\eta }^2=\frac{S{S}_{\text{BG}}}{S{S}_{\text{T}}}\\ =\frac{30}{80}\\ =0.38\end{array}$

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 $\left({\eta }^2\right)$ is 0.38 and it represents the large effect size.

3.

To determine

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

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:

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

The numerator degrees of freedom is 2, the denominator degrees of freedom as 39 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 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 $F_{\text{obt}}=11.72$.

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.