There are two tasks so degrees of freedom for task = 2 - 1 = 1
Now, degrees of freedom for interaction term = 79 - 76 - 1 - 1 = 1
MStask = SStask/df = 2.451 = 2.45
MSinteraction = SSinteraction/df = 0.05/1 = 0.05
Ftask = MStask/MSerror = 2.45/0.816 = 3.0024
Finteraction = MSinteraction/MSerror = 0.05/0.816 = 0.06
The p-value for interactio is .807156. The result is not significant at p < .01.
No, even at 5%, Eleanor cannot conclude that interaction is significant.
At 5% both the main effects are significant
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