Chapter 13, Problem 26CAP

To determine

Complete the F table.

Make the decision to retain or reject the null hypothesis.

Answer to Problem 26CAP

The complete F table is,

Source of Variation	SS	df	MS	$F_{\text{obt}}$
Between groups	30	1	30	2.00
Between persons	27	9	3
Within groups (error)	135	9	15
Total	192	19

The decision is retaining the null hypothesis. The number of outbursts during did not significantly vary by the type of teacher.

Explanation of Solution

Calculation:

From the given data, the total sum of squares is 192, the total degrees of freedom is 19, the between persons degrees of freedom is 9, the between groups degrees of freedom is 9, the mean square between persons is 3 and the mean square between groups is 30.

Degrees of freedom for between groups:

The formula is given by,

$d{f}_{\text{BG}}=d{f}_{\text{T}}-d{f}_{\text{BP}}-d{f}_{\text{E}}$

Where $d{f}_{\text{T}}$ represents the total degrees of freedom.

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

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

Substitute 19 for $d{f}_{\text{T}}$, 9 for $d{f}_{\text{BP}}$ and 9 for $d{f}_{\text{E}}$,

$\begin{array}{c}d{f}_{\text{BG}}=d{f}_{\text{T}}-d{f}_{\text{BP}}-d{f}_{\text{E}}\\ =19-9-9\\ =1\end{array}$

Thus, the value of degrees of freedom for total is 1.

Sum of squares for between persons:

The formula is given by,

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

Where, $M{S}_{\text{BP}}$ represents the mean square between persons

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

Substitute 3 for $M{S}_{\text{BP}}$ and 9 for $d{f}_{\text{BP}}$,

$\begin{array}{c}S{S}_{\text{BP}}=M{S}_{\text{BP}}\times d{f}_{\text{BP}}\\ =3\times 9\\ =27\end{array}$

Thus, the sum of squares for between persons is 27.

Sum of squares for between groups:

The formula is given by,

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

Where, $M{S}_{\text{BG}}$ represents the mean square between groups

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

Substitute 30 for $M{S}_{\text{BG}}$ and 1 for $d{f}_{\text{BG}}$,

$\begin{array}{c}S{S}_{\text{BG}}=M{S}_{\text{BG}}\times d{f}_{\text{BG}}\\ =30\times 1\\ =30\end{array}$

Thus, the sum of squares for between groups is 30.

The sum of squares within groups is,

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

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 $S{S}_{\text{T}}=192$, $S{S}_{\text{BG}}=30$ and $S{S}_{\text{BP}}=27$,

$\begin{array}{c}S{S}_{\text{E}}=S{S}_{\text{T}}-S{S}_{\text{BG}}-S{S}_{\text{BP}}\\ =192-30-27\\ =135\end{array}$

Thus, the sum of squares total is 135.

Mean square error:

$\begin{array}{c}M{S}_{\text{E}}=\frac{S{S}_{\text{E}}}{d{f}_{\text{E}}}\\ =\frac{135}{9}\\ =15\end{array}$

The ANOVA table is,

Source of Variation	SS	df	MS	$F_{\text{obt}}$
Between groups	30	1	30	2.00
Between persons	27	9	3
Within groups (error)	135	9	15
Total	192	19

Null Hypothesis:

$H_0:$ The number of outbursts during a class did not significantly vary by the type of teacher.

Alternative Hypothesis:

$H_1:$ The number of outbursts during a class significantly vary by the type of teacher.

The data gives that the test statistic value is $F_{\text{obt}}=2.00$, the numerator degrees of freedom is 1 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 1, the denominator degrees of freedom as 9 and the alpha level is 0.05.

From the Appendix C: Table C.3 the F Distribution:

• Locate the value 1 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 5.12.

Thus, the critical value for the numerator degrees of freedom 1, the denominator degrees of freedom 9 with level of significance 0.05 is 5.12.

Conclusion:

The value of test statistic is 2.00.

The critical value is 5.12.

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

The test statistic value does not falls under critical region.

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

Therefore, it can be concluded that the number of outbursts during did not significantly vary by the type of teacher.