
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 | |
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,
Where
Substitute 19 for
Thus, the value of degrees of freedom for total is 1.
Sum of squares for between persons:
The formula is given by,
Where,
Substitute 3 for
Thus, thesum of squares for between persons is 27.
Sum of squares for between groups:
The formula is given by,
Where,
Substitute 30 for
Thus, thesum of squares for between groups is 30.
The sum of squares within groups is,
Substitute
Thus, the sum of squares total is 135.
Mean square error:
The ANOVA table is,
Source of Variation | SS | df | MS | |
Between groups | 30 | 1 | 30 | 2.00 |
Between persons | 27 | 9 | 3 | |
Within groups (error) | 135 | 9 | 15 | |
Total | 192 | 19 |
Null Hypothesis:
Alternative Hypothesis:
The data gives that the test statistic value is
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
The numerator degrees of freedom are 1, 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 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.
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Chapter 13 Solutions
Statistics for the Behavioral Sciences
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