I'm trying to compute ANOVA effect sizes from papers that provide an F value without other information. If I understand correctly, the effect size for a single-factor ANOVA is $$ \eta {2} = \frac{ss_{between}}{ss_{between} + ss_{error}} $$

And the F value is: $$ F = \frac{(N-k)ss_{between}}{(k-1)(ss_{between} + ss_{error})} $$ UPDATE: Nope! the denominator is just [(k-1)*SSerror]. Thus, everything that follows is invalid. Back to first-years stats for me.

Where N = number of observations and k = number of groups.

Question 1: Does it follow that you can calculate eta squared as: $$ \eta {2} = \frac{k-1}{N-k}F $$

Question 2: I tried checking this in some output from SPSS. Here's an example with k=4 and N=158:

SPSS output with relevant values described below

I'm aware that SPSS gives partial eta squared, but for a single-factor ANOVA that should be the same as eta squared, right? And indeed, the ratio of the sums of squares is $\frac{342.872}{(342.872+6133.519)} = .05294$. But using F, we get $2.870*3/154 = .05591$, which is off by much more than rounding error.

Is SPSS subtly adjusting F somehow, or am I confused about how to calculate eta squared?

  • $\begingroup$ @PeterFlom Thank you for the correction. That could have been confusing (: $\endgroup$ – octern Nov 4 '12 at 17:56
  • $\begingroup$ It's got me confused, even without the typo! $\endgroup$ – Peter Flom Nov 4 '12 at 18:12
  1. We know that:

    $$ F = \frac{MS_B} {MS_W} = \frac{SS_B/(k-1)} {SS_W/(N-k)}. $$

    Thus $SS_B = F \times MS_W \times (k-1)$, and $SS_W = MS_W \times (N-k)$.

  2. We also know that:

    $$ \eta^2 = \frac{SS_B}{SS_B + SS_W} $$

  3. Thus, if we substitute (1) in (2):

    $$ \eta^2 = \frac{F \times MS_W \times (k-1)}{F \times MS_W \times (k-1) + MS_W \times (N-k)} $$

  4. The $MS_W$ terms in both numerator and denominator can be removed (simplified), leaving:

    $$ \eta^2 = \frac{F (k-1)}{F (k-1) + (N-k)} = \frac{F (df_B)}{F (df_B) + (df_W)} $$

So, it's possible to compute eta-squared using only F and degrees of freedom.

  • $\begingroup$ Supporting this derivation, it is also discussed by Richardson (2011, Educational Research Review), attributing it to Cohen (1965 and 1973). @Hans Ivers, do you have any input on this related question of mine: stats.stackexchange.com/questions/96151/… $\endgroup$ – jona May 17 '14 at 12:16

This question was based on a huge and very basic error. F is not $$ F = \frac{(N-k)ss_{between}}{(k-1)(ss_{between} + ss_{error})} $$

But rather $$ F = \frac{(N-k)ss_{between}}{(k-1)ss_{error}} $$

With this correction, everything makes sense. Unfortunately, I think it also means that there is no way to calculate etasq if all you know is F and df.

Back to first-year stats for me!


At this IBM/SPSS help page we find:

enter image description here

Terms are defined elsewhere.

enter image description here

It's beyond me, but maybe others can make heads or tails of it.

  • $\begingroup$ This page may be the one you want for explanation of the notation. $\endgroup$ – octern Nov 4 '12 at 20:22

This article by Daniel Lakens explains how eta squared can be calculated from only F and degrees of freedom, but only in cases of one-way ANOVA. This is the example:

For example, for an $F(1, 38) = 7.21$, $η2p = 7.21 \cdot 1/(7.21 \cdot 1 + 38) = 0.16$

  • Lakens, D. (2013). Calculating and reporting effect sizes to facilitate cumulative science: a practical primer for t-tests and ANOVAs. Frontiers in Psychology, 4.

Forgive unearthing an old story, but...

The main reason for confusion in this thread is that SPSS calculates the partial eta-squared instead of the normal eta-squared (and in some versions even incorrectly names it). Formulas which you used are correct, the calculations also, but you read an incorrect result in SPSS, and the problem is broadly described here:

Levine, T. R., & Hullett, C. R. (2002). Eta squared, partial eta squared, and misreporting of effect size in communication research. Human Communication Research, 28(4), 612-625.



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