A partial result of the computation for a two-factor factorial experiment is given as follows (assume that the number of experimental units in each group is fixed):
Source of Variation | SS | df | MS | F |
Factor A | 48 | 4 | 12 | 1 |
Factor B | 6 | 5 | ||
Interaction (AB) | 24 | 15 | ||
Error | 105 | - | ||
Total | 139 | - | - |
A) Fill in all the missing results in the ANOVA table.
B) At the 0.05 level of significance, is there an interaction effect? (Show your work: your hypotheses, test statistic, critical value, and decision).
C) At the 0.05 level of significance, is there evidence of an effect due to factor A? (Show your work: your hypotheses, test statistic, critical value, and decision).
Part A
MS A / MS error = F
12/a = 1
a = 12
MS error = 12
SS/df = MS
MS interaction = 15
15 = SS interaction / 24
SS interaction = 15*24 = 360
MS B / MS error = 5
MS B /12 = 5
MS B = 60
Required ANOVA table is given as below:
Source of Variation |
SS |
df |
MS |
F |
Factor A |
48 |
4 |
12 |
1 |
Factor B |
360 |
6 |
60 |
5 |
Interaction (AB) |
360 |
24 |
15 |
1.25 |
Error |
1260 |
105 |
12 |
- |
Total |
2028 |
139 |
- |
- |
Part b
H0: There is no interaction.
Ha: There is an interaction.
α = 0.05
Test statistic = F = MS interaction / MS error = 60/12 = 5
Critical value = 1.621485
Test statistic > Critical value
So, we reject the null hypothesis
There is sufficient evidence to conclude that there is a statistically significant interaction.
Part c
H0: There is no significant effect due to factor A.
Ha: There is a significant effect due to factor A.
α = 0.05
F = 1
Critical value = 2.45821
Test statistic F is less than critical value
So, we do not reject the null hypothesis
There is not sufficient evidence to conclude that there is significant effect due to factor A.
A partial result of the computation for a two-factor factorial experiment is given as follows (assume...
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