Effect Size – Calculating Eta Squared from F and Degrees of Freedom (df)

anovaeffect-sizespss

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?

Best Answer

  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.

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