Chapter 13, Problem 20CAP

(a)

To determine

Complete the F table.

Make the decision to retain or reject the null hypothesis.

Answer to Problem 20CAP

The complete F table is,

Source of Variation	SS	df	MS	$F_{\text{obt}}$
Between groups	396	2	198	5.50
Between persons	234	9	26
Within groups (error)	648	18	36
Total	1,278	29

The decision is rejecting the null hypothesis.

Explanation of Solution

Calculation:

From the given data, the between groups sum of squares is 234, total sum of squares is 1,278, total degrees of freedom is 29 and the mean square between groups is 198. There are three groups detrimental, safe and helpful.

Degrees of freedom for between groups:

The formula is given by,

$d{f}_{\text{BG}}=k-1$, where k represents the number of groups.

Substitute 3 for k.

$\begin{array}{c}d{f}_{\text{BG}}=k-1\\ =3-1\\ =2\end{array}$

Thus, the value of degrees of freedom for between groups is 2.

The sum of squares between groups is,

$S{S}_{\text{BG}}=M{S}_{\text{BG}}\times d{f}_{\text{BG}}$.

Substitute $M{S}_{\text{BG}}=198$ and $d{f}_{\text{BG}}=2$

$\begin{array}{c}S{S}_{\text{BG}}=M{S}_{\text{BG}}\times d{f}_{\text{BG}}\\ =198\times 2\\ =396\end{array}$

Thus, the sum of squares between groups is 396.

The sum of squares within groups is,

$S{S}_{\text{E}}=S{S}_{\text{T}}-S{S}_{\text{BG}}-S{S}_{\text{BP}}$.

Substitute $S{S}_{\text{T}}=1,278$, $S{S}_{\text{BG}}=396$ and $S{S}_{\text{BP}}=234$,

$\begin{array}{c}S{S}_{\text{E}}=S{S}_{\text{T}}-S{S}_{\text{BG}}-S{S}_{\text{BP}}\\ =1,278-396-234\\ =648\end{array}$

Thus, the sum of squares within groups is 648.

Degrees of freedom for between persons:

The formula is given by,

$d{f}_{\text{BP}}=n-1$, where n represent the number of participants per group.

Substitute 10 for n,

$\begin{array}{c}d{f}_{\text{BP}}=n-1\\ =10-1\\ =9\end{array}$

Thus, theDegrees of freedom for between persons is 9.

Degrees of freedom for within groups:

The formula is given by,

$\begin{array}{c}d{f}_{\text{E}}=\left(k-1\right)\left(n-1\right)\\ =\left(3-1\right)\left(10-1\right)\\ =2\times 9\\ =18\end{array}$.

Mean square between persons:

$M{S}_{\text{BP}}=\frac{S{S}_{\text{BP}}}{d{f}_{\text{BP}}}$

Substitute $S{S}_{\text{BP}}=234$ and $d{f}_{\text{BP}}=9$,

$\begin{array}{c}M{S}_{\text{BP}}=\frac{S{S}_{\text{BP}}}{d{f}_{\text{BP}}}\\ =\frac{234}{9}\\ =26\end{array}$

Thus, the Mean square between persons is 26.

Mean square Error:

$M{S}_{\text{E}}=\frac{S{S}_{\text{E}}}{d{f}_{\text{E}}}$

Substitute $S{S}_{\text{E}}=648$ and $d{f}_{\text{E}}=18$,

$\begin{array}{c}M{S}_{\text{E}}=\frac{S{S}_{\text{E}}}{d{f}_{\text{E}}}\\ =\frac{648}{18}\\ =36\end{array}$

F statistic:

$\begin{array}{c}{F}_{\text{obt}}=\frac{M{S}_{\text{BG}}}{M{S}_{\text{E}}}\\ =\frac{198}{36}\\ =5.5\end{array}$

The ANOVA table is,

Source of Variation	SS	df	MS	$F_{\text{obt}}$
Between groups	396	2	198	5.50
Between persons	234	9	26
Within groups (error)	648	18	36
Total	1,278	29

The data gives that the test statistic value is $F_{\text{obt}}=5.50$, the numerator degrees of freedom is 2 and the denominator degrees of freedom is 9.

Decision rule:

• If the test statistic value is greater than the critical value, then reject the null hypothesis or else retain the null hypothesis.

Critical value:

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

The numerator degrees of freedom is 2, the denominator degrees of freedom as 9 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 9 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 9 with level of significance 0.05 is 4.27.

Thus, the critical value for the numerator degrees of freedom 2, the denominator degrees of freedom 9 with level of significance 0.05 is 4.27.

Conclusion:

The value of test statistic is 5.50

The critical value is 4.27.

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.

(b)

To determine

Find the effect size using partial eta-squared $\left({\omega }^2\right)$.

Answer to Problem 20CAP

The effect size for the test by using ${\eta }_{\text{P}}^2$ is 0.38.

Explanation of Solution

Calculation:

The given ANOVA suggests that the value of $S{S}_{\text{BG}}$ is 396, the value of $S{S}_{\text{T}}$ is 1,278 and $S{S}_{\text{BP}}$ is 234.

Partial Eta-Squared:

The formula is given by,

${\eta }_{\text{P}}^2=\frac{S{S}_{\text{BG}}}{S{S}_{\text{T}}-S{S}_{\text{BP}}}$

Where $S{S}_{\text{BG}}$ represent sum of squares between groups.

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

$S{S}_{\text{BP}}$ represent the sum of squares between persons.

Substitute 396 for $S{S}_{\text{BG}}$, 1,278 for $S{S}_{\text{T}}$ and 234 for $S{S}_{\text{BP}}$ in partial eta-squared formula,

$\begin{array}{c}{\eta }_{\text{P}}^2=\frac{S{S}_{\text{BG}}}{S{S}_{\text{T}}-S{S}_{\text{BP}}}\\ =\frac{396}{1,278-234}\\ =0.38\end{array}$

Hence, the effect size for the test by using ${\eta }_{\text{P}}^2$ is 0.38.