I have two correlations (genetic correlations actually), their standard errors, and their z-statistics (but I don't know the sample sizes). Is it possible to do a significance test for the difference between these (genetic) correlations?
1 Answer
Yes, this is possible. The reason is because if you have $r$ and $SE_r$, you do indeed have the sample sizes: $$SE_r = \sqrt{\frac{1-r}{n-2}}$$ or $$n = \frac{1-r}{SE_r^2}+2$$
In your OP, you mention "z-statistics". I believe you are referencing the Fisher's z-transformation (as if you were referring to a test statistic for the correlation, it would be a t-ratio). This is important, because the Fisher transform is needed for the hypothesis test to compare the two correlations: $$H_0 : \rho_1 = \rho_2$$
To conduct the test, transform the correlations to Fisher z: $$z_i = \frac{1}{2} \ln \left(\frac{1+r_i}{1-r_i}\right)$$ Using the standard error for the differences: $$s=\sqrt{\frac{1}{n_1-3} + \frac{1}{n_2-3}}$$ you can use the standard normal distribution to find the P-value for $$\zeta = \frac{z_1-z_2}{s}$$ (This last value is just a z-score, but as we've used z for something else, I called it zeta instead.)