# Calculating CIs for $\eta^2$ via Z scores - sample size?

In the thread Confidence Interval for $\eta^2$ it was proposed that if only limited statistics are available (in my case, F, df1, df2, means), one could calculate the 95% CI for $\eta^2$ by:

• transforming $\eta^2$, which is equivalent to $R^2$, into r
• transforming r into a Z score (artanh)
• calculating the CI of the Z score (as +/- 1.96*SE)
• back-transforming all values (tanh) and squaring them to get to $R^2$/$\eta^2$

1. Is this general approach sound?
2. The SE of the Z score is given as $\frac{1}{\sqrt{N-3}}$. What does N correspond to here? For example, my $\eta^2$ comes from a repeated-measures ANOVA (one factor, 4 levels). Should I use the total number of samples, the N, or df2, or ..? And possibly: why?

In case you are still interested in this topic, I would recommend you to take a look at the papers, referenced in my answer, especially the first one (by Lakens). Also, check MBESS R package: see home page and JSS paper (note that the software's current version most likely contains additional features and improvements, not described in the referenced original JSS paper).