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Suppose I have two sets of Pearson correlation coefficients -- call them set A & set B, and they are of the same size. How do I systematically compare the correlations in A against B? E.g., I want to test the hypothesis that A is less than B -- something similar to a two-sample t-test, but the problem here is the two samples are two samples of correlation coefficients.

I've done some research and found out that there is a Fisher's Z-transform for this purpose. But it only tests the difference between one correlation $\rho_1$ against another $\rho_2.$ I couldn't find a way to systematically test the difference between to sets of correlations. Is there such a way? Or can I only compare each pair of correlations using Fisher's method and somehow derive the difference, if any?

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  • $\begingroup$ You can probably look at this problem differently (as a regression problem, possibly with a multilevel model). What are you trying to find out? How did you define the groups? Are the coefficients somehow matched? What do they represent? $\endgroup$ – Gala Jun 11 '13 at 15:53
  • $\begingroup$ Are set A and B two correlation matrices of the same elements? $\endgroup$ – ReliableResearch Jun 12 '13 at 5:46
  • $\begingroup$ Yes, they are of the same elements. In the sense, that $\rho_{1i}$ and $\rho_{2i}$ are comparable using the Fisher's method for all pairs $i$ of set A(1) and B(2). $\endgroup$ – Joe Jun 12 '13 at 6:50
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Simply do a t-test of the transformed correlations, exactly as you would test any two sets of data to compare their means. The test technically is a comparison of the mean transformed correlations, but for most purposes that's not a problem. (How meaningful would an arithmetic mean of correlations be in the first place? Arguably, the transformed correlation coefficients are the meaningful quantities!)


The whole point to the Fisher Z transformation $$\rho\to (\log(1+\rho)-\log(1-\rho))/2$$ is to make comparisons legitimate. When $n$ bivariate data are independently sampled from a near-bivariate Normal distribution with given correlation $\rho,$ the Fisher Z- transformed sample correlation coefficient will have close to a Normal distribution, with mean equal to the transformed value of $\rho$ and variance $1/(n-3)$--regardless of the value of $\rho.$ This is just what is needed to justify applying the Student t test (with equal variances in each group) or Analysis of Variance.

To demonstrate, I simulated samples of size $n=50$ from various bivariate Normal distributions having a range of correlations $\rho,$ repeating this $50,000$ times to obtain $50,000$ sample correlation coefficients for each $\rho$. To make these results comparable, I subtracted the Fisher Z transformation of $\rho$ from each transformed sample correlation coefficient, calling the result "$Z,$" so as to produce distributions that ought to be approximately Normal, all of zero mean, and all with the same standard deviation of $\sqrt{1/(50-3)} \approx 0.15.$ For comparison I have overplotted the density function of that Normal distribution on each histogram.

Figure

You can see that across this wide range of underlying correlations (as extreme as $-0.95$), the Fisher-transformed sample correlations indeed look like they have nearly Normal distributions, as promised.

For those who might be worried about extreme cases, I extended the simulations out to $\rho=0.9999$ (with $\rho=0$ shown as a reference at the left). The transformed distributions are still Normal and still have the promised variances:

Figure 2

Finally, the picture doesn't change much with small sample sizes. Here's the same simulation with samples of just $n=8$ bivariate Normal values:

Figure 3

A tiny bit of skewness towards less extreme values is apparent, and the standard deviations seem a little smaller than expected, but these variations are so small as to be of no concern.

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    $\begingroup$ could we have the source code of this analysis (r). thanks. $\endgroup$ – Maximilian Sep 22 '18 at 0:37
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The problem with using a 2 sample t-test is, presumably, that the correlations are not normally distributed. So, you can use a non-parametric test such as Wilcoxon. Or you could do a permutation test.

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  • $\begingroup$ Since theory demonstrates the Z-transformed correlations are Normal and homoscedastic, the usual reasons for choosing the Wilcoxon don't apply unless the underlying data that produced the correlations are decidedly non-Normal themselves. This is one of those relatively rare cases where one might automatically prefer the classical tests because they will be a little more powerful. $\endgroup$ – whuber Sep 21 '18 at 22:00

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