Usually $r$ is transformed into Fisher $z$ to test difference between two $r$ values. But, when a meta-analysis is to be performed, why we should take such a step? Does it correct for measurement error or non-sampling error and why should we assume that $r$ is an imperfect estimate of population correlation?

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    $\begingroup$ The last part of your question ("Why should we assume that r is an imperfect estimate of population correlation?") is somewhat unrelated to the previous part. And what do you mean by "imperfect"? Do you mean biased? $\endgroup$
    – Wolfgang
    Apr 16, 2012 at 13:17
  • $\begingroup$ @subhash: Can you state more precisely what you mean by "correct for measurement error or non-sampling error"? Answering your question might be easier if you could define these terms unambiguously, such as be expressing them in terms of things such as random variables, distributions, parameters, or estimators. $\endgroup$ May 9, 2013 at 21:59

1 Answer 1


There is actually quite a bit of a debate in the literature whether one should conduct a meta-analysis with the raw correlation coefficients or with the r-to-z transformed values. However, leaving aside this discussion, there are really two reasons why the transformation is applied:

  1. Many meta-analytic methods assume that the sampling distribution of the observed outcomes is (at least approximately) normal. When $\rho$ (the true correlation) in a particular study is far away from 0 and the sample size is small, then the sampling distribution of the (raw) correlation becomes very skewed and is not at all well approximated by a normal distribution. Fisher's r-to-z transformation happens to be a rather effective normalizing transformation (even though this is not the primary purpose of the transformation -- see below).

  2. Many meta-analytic methods assume that the sampling variances of the observed outcomes are (at least approximately) known. For example, for the raw correlation coefficient, the sampling variance is approximately equal to:

$$\text{Var}[r] = \frac{(1-\rho^2)^2}{n-1}$$

In order to actually calculate $\text{Var}[r]$, we must do something about that unknown value of $\rho$ in that equation. For example, we could just plug the observed correlation (i.e., $r$) into the equation. This will give us an estimate of the sampling variance, but this happens to be a rather inaccurate estimate (especially in smaller samples). On the other hand, the sampling variance of an r-to-z transformed correlation is approximately equal to:

$$\text{Var}[z] = \frac{1}{n-3}$$

Note that this no longer depends on any unknown quantities. This is in fact the variance-stabilizing property of the r-to-z transformation (which is the actual purpose of the transformation).

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    $\begingroup$ +1, this is really informative & on-point. I wish I could upvote more than once. $\endgroup$ Apr 16, 2012 at 13:28
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    $\begingroup$ Yes, I know that the bias is usually negligible (and in practice is never corrected for), but it is not correct to say that $r$ is unbiased. Also, the formulas do not correct for sampling error. They are simply used to calculate the sampling variance, which is then used to compute a weighted average of either the raw of transformed correlations. Measurement error is another issue. Using the attenuation correction, we can also correct a correlation for measurement error. $\endgroup$
    – Wolfgang
    Jul 13, 2012 at 21:09
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    $\begingroup$ @subhash: Can you clarify what you mean by "r is unbiased (for measurement error)"? Are you referring to a notion from classical test theory, perhaps as used by F. Schmidt, J. Hunter, and several of their colleagues and other authors in meta-analytic techniques for validity generalization? As you may know, their methods emphasize estimating the between-studies mean and variance of "true" correlations that have been "corrected" for "artifacts" (e.g., unreliability, range restriction, dichotomization). $\endgroup$ May 9, 2013 at 22:14
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    $\begingroup$ If we take a random-effects view of meta-analysis, where $\rho$ varies randomly (e.g., among studies), we might also consider whether $\rho$ or its Fisher-z counterpart $\zeta = \tanh^{-1} \rho$ better satisfies any meta-analytic assumptions about the effect-size parameter. For example, it's often unclear whether $\rho$ or $\zeta$ is more likely to be normally distributed, which some procedures assume (e.g., certain maximum-likelihood estimators and "credibility" or prediction intervals). $\endgroup$ May 9, 2013 at 22:43
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    $\begingroup$ Could you provide a citation or two arguing the opposite point—that r should be used—since you remarked that there is some debate? $\endgroup$
    – Mark White
    Jun 28, 2017 at 3:18

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