Chapter 13, Problem 23CAP

1.

To determine

Complete the F table.

Make the decision to retain or reject the null hypothesis.

Answer to Problem 23CAP

The complete F table is,

Source of Variation	SS	df	MS	$F_{\text{obt}}$
Between groups	24	2	12	4.80
Between persons	33	11	3
Within groups (error)	55	22	2.5
Total	112	35

The decision is rejecting the null hypothesis.

Explanation of Solution

Calculation:

From the given data, the between groups mean squares is 12, the between persons mean squares is 3, sum of squares within groups is 55. There are three groups and the sample size is 12.

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.

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 12 for n,

$\begin{array}{c}d{f}_{\text{BP}}=n-1\\ =12-1\\ =11\end{array}$

Thus, the degrees of freedom for between persons is 11.

Degrees of freedom for within groups:

The formula is given by,

$\begin{array}{c}d{f}_{\text{E}}=\left(k-1\right)\left(n-1\right)\\ =\left(3-1\right)\left(12-1\right)\\ =2\times 11\\ =22\end{array}$

Thus, the degrees of freedom for within groups is 22.

Total Degrees of freedom:

The formula is given by,

$\begin{array}{c}d{f}_{\text{T}}=kn-1\\ =\left(3\times 12\right)-1\\ =36-1\\ =35\end{array}$

Thus, the total degrees of freedom is 35.

The sum of squares between groups is,

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

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

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

Substitute $M{S}_{\text{BG}}=12$ and $d{f}_{\text{BG}}=2$

$\begin{array}{c}S{S}_{\text{BG}}=M{S}_{\text{BG}}\times d{f}_{\text{BG}}\\ =12\times 2\\ =24\end{array}$

Thus, the sum of squares between groups is 24.

The sum of squares between persons is,

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

Where, $S{S}_{\text{BP}}$ represents the between persons sum of squares.

$d{f}_{\text{BP}}$ represents the between persons degrees of freedom.

$M{S}_{\text{BP}}$ represents the between persons mean sum of squares.

Substitute $M{S}_{\text{BP}}=3$ and $d{f}_{\text{BP}}=11$

$\begin{array}{c}S{S}_{\text{BP}}=M{S}_{\text{BP}}\times d{f}_{\text{BP}}\\ =3\times 11\\ =33\end{array}$

Thus, the sum of squares between persons is 33.

The total sum of squares is,

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

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

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

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

$S{S}_{\text{E}}$ represents the error sum of squares.

Substitute $S{S}_{\text{E}}=55$, $S{S}_{\text{BG}}=24$ and $S{S}_{\text{BP}}=33$,

$\begin{array}{c}S{S}_{\text{T}}=S{S}_{\text{E}}+S{S}_{\text{BG}}+S{S}_{\text{BP}}\\ =55+24+33\\ =112\end{array}$

Thus, the total sum of squares is 112.

Mean square Error:

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

Where, $S{S}_{\text{E}}$ represents the error sum of squares.

$d{f}_{\text{E}}$ represents the error degrees of freedom.

Substitute $S{S}_{\text{E}}=55$ and $d{f}_{\text{E}}=22$,

$\begin{array}{c}M{S}_{\text{E}}=\frac{S{S}_{\text{E}}}{d{f}_{\text{E}}}\\ =\frac{55}{22}\\ =2.5\end{array}$

F statistic:

$\begin{array}{c}{F}_{\text{obt}}=\frac{M{S}_{\text{BG}}}{M{S}_{\text{E}}}\\ =\frac{12}{2.5}\\ =4.8\end{array}$

The ANOVA table is,

Source of Variation	SS	df	MS	$F_{\text{obt}}$
Between groups	24	2	12	4.80
Between persons	33	11	3
Within groups (error)	55	22	2.5
Total	112	35

The data gives that the test statistic value is $F_{\text{obt}}=4.80$, the numerator degrees of freedom is 2 and the denominator degrees of freedom is 11.

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 11 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 11 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 11 with level of significance 0.05 is 3.98.

Thus, the critical value for the numerator degrees of freedom 2, the denominator degrees of freedom 11 with level of significance 0.05 is 3.98.

Conclusion:

The value of test statistic is 4.48.

The critical value is 3.98.

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

The test statistic value falls under critical region.

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

2.

To determine

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

Answer to Problem 23CAP

The effect size for the test by using $\left({\omega }_{\text{P}}^2\right)$ is 0.22.

Explanation of Solution

Calculation:

The given ANOVA suggests that the value of $S{S}_{\text{BG}}$ is 24, the value of $S{S}_{\text{T}}$ is 112, the value of $d{f}_{\text{BG}}$ is 2, the value of $M{S}_{\text{E}}$ is 2.5 and $S{S}_{\text{BP}}$ is 33.

Partial Omega-Squared:

The formula is given by,

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

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.

$d{f}_{\text{BG}}$ represent the degrees of freedom for between groups.

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

Substitute 24 for $S{S}_{\text{BG}}$, 112 for $S{S}_{\text{T}}$, 2 for $d{f}_{\text{BG}}$, 2.5 for $M{S}_{\text{E}}$ and 33 for $S{S}_{\text{BP}}$ in partial omega-squared formula,

$\begin{array}{c}{\omega }_{\text{P}}^2=\frac{S{S}_{\text{BG}}-d{f}_{\text{BG}}\left(M{S}_{\text{E}}\right)}{\left(S{S}_{\text{T}}-S{S}_{\text{BP}}\right)+M{S}_{\text{E}}}\\ =\frac{24-\left(2\times 2.5\right)}{\left(112-33\right)+2.5}\\ =\frac{19}{81.5}\\ =0.22\end{array}$

Hence, the effect size for the test by using ${\omega }_{\text{P}}^2$ is 0.22.