1. An ANOVA was carried out on a set of data with 4 groups of 4 subjects each. The responses are presented in the following table:
Subject |
A1 |
A2 |
A3 |
A4 |
S1 |
4 |
5 |
9 |
8 |
S2 |
5 |
5 |
10 |
8 |
S3 |
6 |
6 |
7 |
6 |
S4 |
5 |
4 |
6 |
6 |
Source |
SS |
df |
MS |
F |
|||
Factor A |
__ |
__ |
_ |
___ |
|||
Subject |
__ |
__ |
___ |
||||
Within/error |
__ |
__ |
___ |
||||
Total |
__ |
___ |
using excel>data>data analysis >two way anova without replication
we have
Anova: Two-Factor Without Replication | ||||||
SUMMARY | Count | Sum | Average | Variance | ||
S1 | 4 | 26 | 6.5 | 5.666667 | ||
S2 | 4 | 28 | 7 | 6 | ||
S3 | 4 | 25 | 6.25 | 0.25 | ||
S4 | 4 | 21 | 5.25 | 0.916667 | ||
A1 | 4 | 20 | 5 | 0.666667 | ||
A2 | 4 | 20 | 5 | 0.666667 | ||
A3 | 4 | 32 | 8 | 3.333333 | ||
A4 | 4 | 28 | 7 | 1.333333 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Factor A | 6.5 | 3 | 2.166667 | 1.695652 | 0.236872 | 3.862548 |
Subject | 27 | 3 | 9 | |||
Error | 11.5 | 9 | 1.277778 | |||
Total | 45 | 15 |
the table is
Source of Variation | SS | df | MS | F | P-value |
Factor A | 6.5 | 3 | 2.166667 | 1.695652 | 0.236872 |
Subject | 27 | 3 | 9 | ||
Error | 11.5 | 9 | 1.277778 | ||
Total | 45 | 15 |
F = 1.69
critical value F(3,9) = 3.86
since F<3.86 so we do not reject Ho and so we cannot perform post hoc test
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