2
$\begingroup$

We use the Fisher $z$-transformation for our correlation matrices, s.t. we arrive at approximately normally distributed data.

As you might now, the Fisher $z$-transformation is equivalent to the function

$z = 0.5 \log\left(\frac{1+x}{1-x}\right)$ , $x$ being Pearson's correlation coefficient, ranging between -1 and 1.

Now I have been told that Fisher's $z$ has a domain between -4 and 4 - something which does not make sense to me, since when approximating 1 and -1 for the $x$ values, we arrive at much bigger or smaller values respectively.

Have I been misinformed or am I missing out on something?

$\endgroup$
1
  • 2
    $\begingroup$ Fisher's $z$ is indeterminate for perfect correlations $r$ of $+1$ or $-1$. That should make sense too. Any uncertainty about whether the result is credible has to be answered otherwise. (Notation $x$ for a correlation seems a little perverse here.) $\endgroup$
    – Nick Cox
    Mar 13, 2016 at 6:52

1 Answer 1

4
$\begingroup$

I suspect that what you heard was meant as a "rule of thumb": under the null hypothesis the Fisher's $z$ should follow a standard normal sampling distribution and values less than $-4$ and more than $4$ are then extremely unlikely.

$\endgroup$
1
  • 1
    $\begingroup$ Ah..thought it must be something "pragmatic"..thank you very much!! $\endgroup$
    – Pugl
    Sep 7, 2015 at 10:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.