I was recently looking at a paper in the journal Psychological Science and came across this:

F(1, 71) = 4.5, p = .037, $\mu^2$ = .06

F(1, 71) = 0.08, p = .78, $\mu^2$ = .001

I was wondering what the $\mu^2$ is in the above. Typically in APA the third thing should be either the MSE or it should be a standardized effect size (or you should have all 4). I'm guessing it's a standardized effect size of some sort but I'm not familiar with it and searching the net has turned up nothing. The actual effect, as near as I can tell from the graph, is about 12 for the first one.

Is this an effect size I haven't heard of yet or a typo in the article?

Farrelly, D., Slater, R., Elliott, H. R., Walden, H. R. and Wetherell, M. A. (2013) Competitors Who Choose to Be Red Have Higher Testosterone Levels. Psychological Science, DOI:10.1177/0956797613482945

Here's a screen shot of the text (p.2)

enter image description here

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I can only think of this referring to $\eta^2$, computed as:

$\eta^2={SS_{effect} \over SS_{total}}$

This is the proportion of variance explained in the dependent variable by the grouping variable (in this case, a binary variable). This would be indeed the same value as the $R^2$ obtained if the difference between the two groups was estimated using simple linear regression:


I can see from the paper that the second F test is actually that of an interaction term, and since it has 1 degree of freedom, I am deducing that the second factor was also a binary variable. In this case, the $\eta^2$'s are partial $\eta^2$'s, which are the proportion of variance explained by the grouping variable (or the interaction term) controlling for the other grouping variable. In this more complex case, the partial $\eta^2$'s are the same as the partial $R^2$'s obtained from the multiple linear regression:

$y_i=\beta_0+\beta_1group_{1i}+\beta_2group_{1i}+\beta_3 \cdot group_{1i} \cdot group_{2i} + \epsilon_i$

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  • $\begingroup$ Given that no one else has come up with an alternative I'll mark this correct. However Patrick it's entirely possible this is an (misapplied) $\omega^2$. The $\eta^2$ better suits a typo I think (upside down $\mu$?). $\endgroup$ – John Aug 14 '13 at 21:25

This is eta-squared and it is a fairly poor measure of effect size (partial eta-squared is often reported in statistical software such as SPSS when calculating ANOVA)

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  • $\begingroup$ If it is $\eta^2$ (and it may well be) then it is indeed a typo, since the question listed it as $\mu^2$ $\endgroup$ – Peter Flom Nov 3 '13 at 13:35

My apologies, I misread Patrick's response above; I now see he also indicated it was $_{partial}\eta^2$. I added the formulas for finding $_{partial}\eta^2$ should any need those.

I am posting on this old thread to clarify should others visit in the future. The value reported is $_{partial}\eta^2$, not $\eta^2$. Since the analysis is a two-way ANOVA, the formula $\frac{(F)(df_a)}{(F)(df_a)+df_e}$ (where $df_a$ is df for the predictor and $df_e$ is the error df) which normally provides $\eta^2$ for a one-way ANOVA, instead produces the $_{partial}\eta^2$ for a multi-way ANOVA.

Thus $_{partial}\eta^2$ = $\frac{(F)(df_a)}{(F)(df_a)+df_e}$ = $\frac{4.5}{4.5+71}$ = .059 ~ .06


$_{partial}\eta^2$ = $\frac{(F)(df_a)}{(F)(df_a)+df_e}$ = $\frac{.08}{.08+71}$ = .001

The authors mislabeled the effect, rather than $\mu^2$ it should be $_{partial}\eta^2$.

It is also possible the authors meant to report Cohen's $f^2$ effect size since the results of these two F-ratios provide both $_{partial}\eta^2$ and $f^2$ that are within rounding error, but I have never seen Cohen $f^2$ reported like this.


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  • $\begingroup$ Nice first answer. Note that you can make your math more readable by using Latex math notation (see here for more details: math.stackexchange.com/help/notation ) $\endgroup$ – Martin Modrák Mar 9 '18 at 19:56
  • $\begingroup$ Thanks Martin, I have updated the response to use the Latex option - I was not aware of that option. $\endgroup$ – Bryan Mar 9 '18 at 20:43

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