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:
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
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 $SS_W = MS_W \times (N-k)$.
We also know that:
$$ \eta^2 = \frac{SS_B}{SS_B + SS_W} $$
Thus, if we substitute (1) in (2):
$$ \eta^2 = \frac{F \times MS_W \times (k-1)}{F \times MS_W \times (k-1) + MS_W \times (N-k)} $$
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.