An educator wants to evaluate four different methods aimed at reducing the time children spend "off task" in the classroom. To test these methods, she implements one method in each of four similar classrooms and records the time spent off task (in minutes) in each classroom. The results are given in the table.
Classroom Method | |||
---|---|---|---|
A | B | C | D |
1 | 1 | 0 | 3 |
1 | 5 | 0 | 2 |
4 | 3 | 4 | 4 |
4 | 4 | 3 | 5 |
0 | 2 | 3 | 4 |
5 | 2 | 5 | 4 |
2 | 4 | 5 | 2 |
4 | 0 | 6 | 2 |
3 | 1 | 5 | 4 |
(a) Complete the F-table. (Round your answers to two decimal places.)
Source of Variation |
SS | df | MS | F |
---|---|---|---|---|
Between groups |
||||
Within groups (error) |
||||
Total |
(b) Is it necessary to compute a post hoc test? Explain. (Assume
alpha equal to 0.05.)
Yes, post hoc analyses are appropriate because the ANOVA is significant.No, post hoc analyses are not appropriate because the ANOVA is not significant.
Sol:
Install analysis tool pak in excel.
Go to Data>Data analysis>ANOVA single factor
You will get
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
A | 9 | 24 | 2.66666667 | 3 | ||
B | 9 | 22 | 2.44444444 | 2.777778 | ||
C | 9 | 31 | 3.44444444 | 4.777778 | ||
D | 9 | 30 | 3.33333333 | 1.25 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 6.527777778 | 3 | 2.17592593 | 0.737255 | 0.537588 | 2.90112 |
Within Groups | 94.44444444 | 32 | 2.95138889 | |||
Total | 100.9722222 | 35 |
ANSWER;
ANOVA | ||||
Source of Variation | SS | df | MS | F |
Between Groups | 6.53 | 3.00 | 2.18 | 0.74 |
Within Groups | 94.44 | 32.00 | 2.95 | |
Total | 100.97 | 35.00 |
(rounded to 2 decimals as specified
Solutionb:
here p=0.53758
p>0.05
Fail tot reject Ho.
Meaning
All 4 group means are equal.
No need to conduct post hoc test.
No, post hoc analyses are not appropriate because the ANOVA is not significant.
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