This is a question for a study guide I can't quite wrap my head around. My thinking is that if Eta-squared is equal to 0, the between group deviations must be equal to 0, therefore it is impossible to have any evidence against the null (thus a p value of < .05). Is my thinking correct? Thanks for any help.
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The formula to compute $\eta^2$ is:
$$\eta^2=\frac{SS_{Between}}{SS_{Total}}$$
So the only way for $\eta^2$ to be equal to 0 is for $SS_{Between}$ (the numerator) to be 0. In this case, the F test for this effect is:
$$F=\frac{SS_{Between}/df_{Between}}{SS_{Within}/df_{Within}}=\frac{0/df_{Between}}{SS_{Within}/df_{Within}}=0$$
Ultimately, the F test is necessarily 0, so your reasoning is correct--you can't get a p < .05 if $\eta^2=0$, because it implies that $SS_{Between}=0$, and thus that $F=0<F_{critical}$.
Patrick Coulombe
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3Unless the sample is huge, and eta-squared is 0.00000001, which is printed as 0. – Jeremy Miles Dec 17 '15 at 06:19