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I'm writing functions (in R) that allow working with the $\omega^2$ distribution (the effect size for analysis of variance; the less biased alternative to $\eta^2$). Not being a mathematician or statistician, I do this by converting between $\omega^2$ and $F$ and vice versa (only working on the oneway anova scenario so far).

This works well so far (functions implemented in package userfriendlyscience 0.5-1, see ?pomegaSq if anybody runs into this question because they need those functions) but now I want to add the noncentrality parameter.

For the $t$-distribution, the noncentrality parameter is equal to the $t$-value corresponding to the expected effect in the population (I used the same trick for a Cohen's $d$ distribution). This was simple enough for me to could play around with to be pretty sure that's how it works.

However, for the $F$-distribution, I'm not certain enough that here, the noncentrality parameter is indeed also the relevant $F$-value. For example, assuming $N=100$ participants, $k=3$ groups, and an effect size of $\omega^2 = .06$ (i.e. a 'medium' effect), is it true that the noncentrality parameter of the corresponding $F$-distribution (from which the $F$-value of any given sample of $N=100$ is drawn) is equal to:

convert.omegasq.to.f(.06, 3 - 1, 100 - 3);
[1] 4.191489

(This function is again from userfriendlyscience)

Or, in a formula:

$$ \frac{\omega^2 \frac{df2 + 1}{df1} + 1}{1 - \omega^2} = \frac{.06 \frac{97 + 1}{2} + 1}{1 - .06} = \frac{3.94}{.94} = 4.191489$$

(assuming this formula is correct; it was derived by a colleague of mine who is in fact a statistician)

If this is correct, then that would mean that this would work:

domegaSq <- function(x, df1, df2, populationOmegaSq = 0) {
  return(df(convert.omegasq.to.f(x, df1, df2), df1, df2,
            ncp = convert.omegasq.to.f(populationOmegaSq, df1, df2)));
}

And the distribution of omega squared could be drawn, for example for no effect (grey, $\omega^2=0$), a small effect (red, $\omega^2=0.01$), a medium effect (orange, $\omega^2=0.06$) and a large effect (green, $\omega^2=0.14$) and $N=100$ and $k=3$:

The distribution of omega squared

Does anybody know whether the crucial assumption in this procedure (the noncentrality parameter of the $F$-distribution is the expected value of $F$) is correct?

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  • $\begingroup$ I found this explanation by Conrad Carlberg, where he uses: $$\lambda = \frac{\sum_{j=1}^k \beta^2_j}{\frac{\sigma^2}{n}} = \frac{SS_\text{effect}}{MS_\text{error}}$$ Because $$F = \frac{\frac{SS_\text{effect}}{k-1}}{MS_\text{error}} = \frac{1}{k-1}\frac{SS_\text{effect}}{MS_\text{error}}$$ this would mean that $$\lambda = \frac{F}{\frac{1}{k-1}} = (k - 1)F$$ And $$F = \frac{\lambda}{k-1}$$ For situations with equal group sizes. Does this make sense? $\endgroup$ – Matherion Nov 16 '16 at 13:24
  • $\begingroup$ Yes, this is correct. Check Liu and Raudenbush, eq. 4b, journals.sagepub.com/doi/pdf/10.3102/10769986029002251 $\endgroup$ – Denis Cousineau Nov 13 '17 at 21:30
  • $\begingroup$ Great, that paper covers exactly this - how can I have missed this? Thank you very much! $\endgroup$ – Matherion Dec 11 '17 at 12:50

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