In this question they ask how to compare Pearson r for two independent groups (such as males vs females). Reply and comments suggested two ways:

  1. Use Fisher's well-known formula using "z-tranformation" of r;
  2. Use comparison of slopes (regression coefficients).

The latter could be easily performed just via a saturated linear model: $Y = a + bX + cG + dXG$, where $X$ and $Y$ are the correlated variables and $G$ is a dummy (0 vs 1) variable indicating the two groups. The magnitude of $d$ (the interaction term coefficient) is exactly the difference in coefficient $b$ after model $Y = a + bX$ conducted in two groups individually, and its ($d$'s) significance is thus the test of difference in slope between the groups.

Now, slope or regression coef. isn't yet a correlation coef. But if we standardize $X$ and $Y$ - separately in two groups - then $d$ will be equal to the difference r in group 1 minus r in group 0 and therefore its significance will be testing the difference between the two correlations: we're testing slopes but it appeares [as if - ?] we're testing correlations.

Is that I've written correct?

If yes, there's left the question which is a better test of correlations - this one described or Fisher's one? For they will yield not identical results. What do you think?

Later Edit: Thanking @Wolfgang for his reply I nevertheless feel I miss understanding why Fisher's test is more correct a test for r than comparison-of-slope-under-standardization approach described above. So, more answers are welcome. Thank you.


1 Answer 1


Everything that you have written is correct. You can always test out things like that with a toy example. Here is an example with R:


rho <- .5  ### the true correlation in both groups

S1 <- matrix(c( 1,   rho,   rho, 1), nrow=2)
S2 <- matrix(c(16, 4*rho, 4*rho, 1), nrow=2)


xy1 <- mvrnorm(1000, mu=c(0,0), Sigma=S1)
xy2 <- mvrnorm(1000, mu=c(0,0), Sigma=S2)

x <- c(xy1[,1], xy2[,1])
y <- c(xy1[,2], xy2[,2])
group <- c(rep(0, 1000), rep(1, 1000))

summary(lm(y ~ x + group + x:group))

What you will find that the interaction is highly significant, even though the true correlation is the same in both groups. Why does that happen? Because the raw regression coefficients in the two groups reflect not only the strength of the correlation, but also the scaling of X (and Y) in the two groups. Since those scalings differ, the interaction is significant. This is an important point, since it is often believed that to test the difference in the correlation, you just need to test the interaction in the model above. Let's continue:

summary(lm(xy2[,2] ~ xy2[,1]))$coef[2] - summary(lm(xy1[,2] ~ xy1[,1]))$coef[2]

This will show you that the difference in the regression coefficients for the model fitted separately in the two groups will give you exactly the same value as the interaction term.

What we are really interested in though is the difference in the correlations:

cor(xy2)[1,2] - cor(xy1)[1,2]

You will find that this difference is essentially zero. Let's standardize X and Y within the two groups and refit the full model:

x <- c(scale(xy1[,1]), scale(xy2[,1]))
y <- c(scale(xy1[,2]), scale(xy2[,2]))
summary(lm(y ~ x + x:group - 1))

Note that I am not including the intercept or the group main effect here, because they are zero by definition. You will find that the coefficient for x is equal to the correlation for group 1 and the coefficient for the interaction is equal to the difference in the correlations for the two groups.

Now, for your question whether it would be better to use this approach versus using the test that makes use of Fisher's r-to-z transformation.


The standard errors of the regression coefficients that are calculated when you standardize the X and Y values within the groups do not take this standardization into consideration. Therefore, they are not correct. Accordingly, the t-test for the interaction does not control the Type I error rate adequately. I conducted a simulation study to examine this. When $\rho_1 = \rho_2 = 0$, then the Type I error is controlled. However, when $\rho_1 = \rho_2 \ne 0$, then the Type I error of the t-test tends to be overly conservative (i.e., it does not reject often enough for a given $\alpha$ value). On the other hand, the test that makes use of Fisher's r-to-z transformation does perform adequately, regardless of the size of the true correlations in both groups (except when the group sizes get very small and the true correlations in the two groups get very close to $\pm1$.

Conclusion: If you want to test for a difference in correlations, use Fisher's r-to-z transformation and test the difference between those values.

  • $\begingroup$ Does the Fisher transformation have any advantages over the other test? $\endgroup$
    – mark999
    Aug 7, 2011 at 9:37
  • $\begingroup$ Turns out I was a bit too quick. See my edits. That should hopefully answer your question. $\endgroup$
    – Wolfgang
    Aug 7, 2011 at 10:04
  • $\begingroup$ So, @Wolfgang, you hold that comparison-of-slope-under-standardization approach is a valid comparison of r. Fisher's famous alternative is actually an approximation of that. Did I understand you right? $\endgroup$
    – ttnphns
    Aug 7, 2011 at 10:06
  • $\begingroup$ See my edits. I was too quick with my initial conclusions. $\endgroup$
    – Wolfgang
    Aug 7, 2011 at 10:27
  • $\begingroup$ @Wolfgang, later added EDIT of your reply states that Fisher is better. Comparison-of-slope-under-standardization approach is inadequate because "standard errors ... when you standardize ... do not take this standardization into consideration". Please explain me how they ought to take standardization into account so that comparison-of-slope-under-standardization approach become as valid as Fisher's test. $\endgroup$
    – ttnphns
    Aug 7, 2011 at 10:33

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