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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?
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jona
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2 Answers2

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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).

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Aleksandr Blekh
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This is an interesting suggestion. It has one limitation: Fisher's z -- which is the name of this approach -- gives results between -1 and 1 but eta's are only positive. In UniMult, I am using Cox and Hinkley's definition of confidence intervals as the range of possible population values from which the observed value is non-significant. Than a simple loop in a computer program is used to test the range of values. With a bit of patience or the help of a web calculator, this could be done for an occasional eta.

Rich Gorsuch
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