# Does transformation of r into Fisher z benefit a meta-analysis?

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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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? –  Wolfgang Apr 16 '12 at 13:17
@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. –  Adam Hafdahl May 9 at 21:59

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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+1, this is really informative & on-point. I wish I could upvote more than once. –  gung Apr 16 '12 at 13:28
@gung Thanks for the positive feedback! –  Wolfgang Apr 16 '12 at 14:40
@Wolfgang Quite interesting. May be better, if the meta-analytic context was taken. r is an unbiased estimate (Hedges and Olkin, 1985). Should we convert it into Fisher's z for a meta-analysis of sample correlations? please explain from this angle. –  subhash c. davar Jul 8 '12 at 11:37
@subhashdavar Actually, the sample product-moment correlation coefficient $r$ is not an unbiased estimate of $\rho$ (see, for example, Hedges & Olkin, 1985, p. 225). Leaving this aside, the point of my answer was essentially: yes, you should transform. –  Wolfgang Jul 11 '12 at 11:37
@ Wolfgang Please read further if the sample-size is moderately large(say over 15, the bias of the sample correlation r is seldom of practical concern. Even the smaller samples .... the bias is ngligble. Moreover, the formulas cited in the earlier response appear to have a purpose that corrects for sampling error. I wish we could talk about measurement error. –  subhash c. davar Jul 13 '12 at 14:51