I know the confidence interval of an estimated correlation can be calculated after performing Fisher transformation.

But I never figured out why. I also couldn't find the explanation online. Can someone point me to the right direction?

  • 1
    $\begingroup$ Sorry, but I find your question not very clearly expressed. What exactly do you want explained? $\endgroup$
    – Glen_b
    Apr 2, 2015 at 0:09

1 Answer 1


Nick Cox has a really good explanation in a 2008 Speaking Stata column from SJ that I will excerpt here:

pwcorr [DVM: Stata's pairwise correlation command] offers only a $p$-value from the standard $t$ test of whether the population correlation $\rho$ is really zero. .... The latter $t$ test cannot be converted into useful confidence intervals, because once the population correlation is really not zero, the sampling distribution of estimates $r$ is substantially skewed, even for large sample sizes. That should not seem surprising, given that correlations are constrained to fall within $[−1, 1]$. Getting at the form of the distribution is a messy problem with, for the most part, only complicated exact solutions, but for practical data analysts an attractive way forward is offered by a suggestion of Ronald Aylmer Fisher (1890–1962): even though $r$ usually has an awkward skewed distribution, the inverse hyperbolic tangent, or atanh, of $r$ is much better behaved. This transform, within statistics often labeled Fisher’s $z$, is said to be normally distributed to a good approximation. The basic tactic is now evident: apply standard normal-based technique on the $z$ scale, and then back-transform using its inverse transform, the hyperbolic tangent, or tanh.

Bootstrapping is a more modern alternative to the classical transformation approach.


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