Chapter 14, Problem 24CAP

To determine

Complete the F table for the given study.

Answer to Problem 24CAP

The completed F table is,

Source of Variation	SS	df	MS	$F_{\text{obt}}$
Exam	190	2	95	6.33
Student Class	60	3	20	1.33
Exam $\times$ Student Class	540	6	90	6.00
Error	1,620	108	15
Total	2,410	119

Explanation of Solution

Calculation:

From the information given that, the study of grades for each participants are considered in which the students of four categories from local college were freshmen, sophomores, juniors, and seniors. That is, $q=4$, who were in the same statistics class were given one of three tests (recall, recognition, or a mix of both). That is, $p=3$. Also, the sum of squares for student class is 60, the sum of squares for exam $\times$ student class is 540, the total sum of squares is 2,410, the degrees of freedom for exam are 2, and the mean sum of squares for exam is 95.

The formulas for computing the F table are,

Source of Variation	SS	df	MS	$F_{\text{obt}}$
Factor A	$S{S}_{\text{A}}$	$p-1$	$\frac{S{S}_{\text{A}}}{d{f}_{\text{A}}}$	$\frac{M{S}_{\text{A}}}{M{S}_{\text{E}}}$
Factor B	$S{S}_{\text{B}}$	$q-1$	$\frac{S{S}_{\text{B}}}{d{f}_{\text{B}}}$	$\frac{M{S}_{\text{B}}}{M{S}_{\text{E}}}$
$\text{A}\times \text{B}$	$S{S}_{\text{A}\times \text{B}}$	$\left(p-1\right)\left(q-1\right)$	$\frac{S{S}_{\text{A}\times \text{B}}}{d{f}_{\text{A}\times \text{B}}}$	$\frac{M{S}_{\text{A}\times \text{B}}}{M{S}_{\text{E}}}$
Error (within groups)	$S{S}_{\text{E}}$	$pq\left(n-1\right)$	$\frac{S{S}_{\text{E}}}{d{f}_{\text{E}}}$
Total	$S{S}_{\text{T}}$	$npq-1$

Substituting $d{f}_{\text{Exam}}=2,M{S}_{\text{Exam}}=95$ in the sum of squared for main effect exam formula

$\begin{array}{c}S{S}_{\text{Exam}}=95\times 2\\ =190\end{array}$

The sum of squared for main effect exam is 190.

Substituting $q=4$ in the degrees of freedom for main effect student class formula

$\begin{array}{c}d{f}_{\text{S}}=\left(4-1\right)\\ =3\end{array}$

The degrees of freedom for main effect student class are 3.

Substituting $d{f}_{\text{S}}=3,S{S}_{\text{S}}=60$ in the mean square for main effect student class formula

$\begin{array}{c}M{S}_{\text{S}}=\frac{60}{3}\\ =20\end{array}$

The mean square for main effect student class is 20.

Substituting $p=3,q=4$ in the degrees of freedom for exam $\times$ student class formula

$\begin{array}{c}d{f}_{\text{Exam}\times \text{S}}=\left(3-1\right)\left(4-1\right)\\ =2\times 3\\ =6\end{array}$

The degrees of freedom for interaction exam $\times$ student class are 6.

Substituting $d{f}_{\text{Exam}\times \text{S}}=6,S{S}_{\text{Exam}\times \text{S}}=540$ in the mean square for exam $\times$ student class formula

$\begin{array}{c}M{S}_{\text{Exam}\times \text{S}}=\frac{540}{6}\\ =90\end{array}$

The mean square for exam $\times$ student class is 90.

Substituting $p=3,q=4,n=10$ in the degrees of freedom for error formula

$\begin{array}{c}d{f}_{\text{Error}}=3\times 4\times \left(10-1\right)\\ =3\times 4\times 9\\ =108\end{array}$

The degrees of freedom for error are 108.

Substituting $S{S}_{\text{Exam}}=190,S{S}_{\text{S}}=60,S{S}_{\text{Exam}\times \text{S}}=540,S{S}_{\text{T}}=2,410$ in the sum of squared for error formula

$\begin{array}{c}S{S}_{\text{E}}=S{S}_{\text{T}}-\left(S{S}_{\text{Exam}}+S{S}_{\text{S}}+S{S}_{\text{Exam}\times \text{S}}\right)\\ =2,410-\left(190+60+540\right)\\ =2,410-790\\ =1,620\end{array}$

The sum of squared for error is 1,620.

Substituting $d{f}_{\text{E}}=108,S{S}_{\text{E}}=1,620$ in the mean square for error formula

$\begin{array}{c}M{S}_{\text{E}}=\frac{1,620}{108}\\ =15\end{array}$

The mean square for error is 15.

Substituting $d{f}_{\text{Exam}}=2,d{f}_{\text{S}}=3,d{f}_{\text{Exam}\times \text{S}}=6,d{f}_{\text{E}}=108$ in the total degrees of freedom formula

$\begin{array}{c}d{f}_T=d{f}_{\text{Exam}}+d{f}_{\text{S}}+d{f}_{\text{Exam}\times \text{S}}+d{f}_{\text{E}}\\ =2+3+6+108\\ =119\end{array}$

The total degrees of freedom are 119.

Substituting $M{S}_{\text{Exam}}=95,M{S}_{\text{E}}=15$ in the F obtained value for exam formula

$\begin{array}{c}{F}_{\text{Exam}}=\frac{95}{15}\\ =6.33\end{array}$

The F obtained value for exam is 6.33.

Substituting $M{S}_{\text{S}}=20,M{S}_{\text{S}}=15$ in the F obtained value for student class formula

$\begin{array}{c}{F}_{\text{S}}=\frac{20}{15}\\ =1.33\end{array}$

The F obtained value for student class is 1.33.

Substituting $M{S}_{\text{Exam}\times \text{S}}=90,M{S}_{\text{E}}=15$ in the F obtained value for interaction formula

$\begin{array}{c}{F}_{\text{Exam}\times \text{S}}=\frac{90}{15}\\ =6.00\end{array}$

The F obtained value for interaction is 6.00.

The completed F table is,

Source of Variation	SS	df	MS	$F_{\text{obt}}$
Exam	190	2	95	6.33
Student Class	60	3	20	1.33
Exam $\times$ Student Class	540	6	90	6.00
Error	1,620	108	15
Total	2,410	119