Chapter 12, Problem 23CAP

1.

To determine

Complete the F table.

Answer to Problem 23CAP

The complete F table is,

Source of Variation	SS	df	MS	$F_{\text{obt}}$
Between groups	48	3	16	7.21
Within groups (error)	80	36	2.22
Total	128	39

Explanation of Solution

Calculation:

From the given data, there are four groups like unemployed, retired, part-time and full-time employees. That is, $k=4$. Moreover, the total sum of squares is 128, the degrees of freedom for error is 36 and the mean sum of squares for between groups is 16.

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	$\left(N-1\right)\text{ or }\left(kn-1\right)$

The degrees of freedom for between groups is,

$\begin{array}{c}d{f}_{\text{BG}}=k-1\\ =4-1\\ =3\end{array}$

The total degrees of freedom is,

$\begin{array}{c}d{f}_T=d{f}_{BG}+d{f}_E\\ =3+36\\ =39\end{array}$

The sum of squares for between groups is,

$\begin{array}{c}S{S}_{\text{BG}}=d{f}_{\text{BG}}\times M{S}_{\text{BG}}\\ =3\times 16\\ =48\end{array}$

The sum of squares for error is,

$\begin{array}{c}S{S}_{\text{E}}=S{S}_{\text{T}}-S{S}_{\text{BG}}\\ =128-48\\ =80\end{array}$

The mean sum of squares for error is,

$\begin{array}{c}M{S}_{\text{E}}=\frac{S{S}_{\text{E}}}{d{f}_E}\\ =\frac{80}{36}\\ =2.22\end{array}$

The F statistic is,

$\begin{array}{c}{F}_{\text{obt}}=\frac{M{S}_{\text{BG}}}{M{S}_{\text{E}}}\\ =\frac{16}{2.22}\\ =7.21\end{array}$

The final ANOVA table is,

Source of Variation	SS	df	MS	$F_{\text{obt}}$
Between groups	48	3	16	7.21
Within groups (error)	80	36	2.22
Total	128	39

2.

To determine

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

Answer to Problem 23CAP

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

Explanation of Solution

Calculation:

The given data suggests that the value of $S{S}_{\text{BG}}$ is 48, $d{f}_{\text{BG}}$ is 3, $M{S}_E$ is 2.22 and the value of $S{S}_{\text{T}}$ is 128.

Omega-squared $\left({\omega }^2\right)$:

Omega-squared $\left({\omega }^2\right)$ measures the variability explained by the levels of a factor, divided by the total variability observed.

Formula is given by,

${\omega }^2=\frac{S{S}_{\text{BG}}-d{f}_{\text{BG}}\left(M{S}_{\text{E}}\right)}{S{S}_{\text{T}}+M{S}_{\text{E}}}$.

Where, $S{S}_{\text{BG}}$ represents the sum of squares between groups

$d{f}_{BG}$ represents the degrees of freedom for between groups.

$M{S}_{\text{E}}$ represents the mean square between groups.

$S{S}_{\text{T}}$ represents the total sum of squares.

$M{S}_{\text{E}}$ represents the mean square error.

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 0.06, then the effect size is medium.
• If the eta-squared value is greater than 0.14, then the effect size is larger.

The ANOVA table suggests that the value of $S{S}_{\text{BG}}$ is 48, $d{f}_{\text{BG}}$ is 3, $M{S}_E$ is 2.22 and the value of $S{S}_{\text{T}}$ is 128.

The omega-squared $\left({\omega }^2\right)$ value is,

$\begin{array}{c}{\omega }^2=\frac{S{S}_{\text{BG}}-d{f}_{\text{BG}}\left(M{S}_{\text{E}}\right)}{S{S}_{\text{T}}+M{S}_{\text{E}}}\\ =\frac{48-\left(3\times 2.22\right)}{2.22+128}\\ =\frac{41.34}{130.22}\\ =0.32\end{array}$

Justification: The value of omega-squared is 0.32, which is greater than 0.14. Hence, the effect size is larger.

Hence, the value of omega-squared $\left({\omega }^2\right)$ 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 23CAP

The decision is rejecting the null hypothesis.

Explanation of Solution

Calculation:

The given information suggests that the degrees of freedom for between groups is 3 and the degrees of freedom for error is 36

Rejection Rule:

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

Critical value:

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

The numerator degrees of freedom is 3, the denominator degrees of freedom as 36 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 row.
• Locate the value 36 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 3, the denominator degrees of freedom 36 with level of significance 0.05 is 2.86.

Thus, the critical value for the numerator degrees of freedom 3, the denominator degrees of freedom 36 with level of significance 0.05 is 2.86.

The ANOVA gives that the test statistic value is $F_{\text{obt}}=7.21$.

Conclusion:

The value of test statistic is 7.21.

The critical value is 2.86.

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.