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I have independent correlation coefficient measures for each subject. I would like to test whether the correlation coefficient of the group is significantly different from 0. My understanding is that the best way to do this would be to use a t-test with an r-value per subject.

It's been recommended to me that I first perform a Fisher's transformation on the r-values. I'm wondering why this is necessary. My understanding is that the Fisher's transform is used because the r's are not normally distributed.

Therefore, it seems that the transform makes sense if one is just comparing a single r-value to 0 (i.e. in lieu of testing against a t-distribution with the test statistic $t=\frac{r*\sqrt{n−2}}{\sqrt{1−r^2}}$). However, in my t-test, I am comparing the sample to the sampling distribution (which I think can be assumed normal even if the underlying distribution is not). If this is the case, does it still make sense to employ the transformation before performing the t-test?

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You are right: it's not necessary to perform Fisher's transform. Do the t-test. It uses an exact null distribution, whereas comparing Fisher z-transform to a normal distribution would be an approximation.

Trying to do both the z-transform and the transformation to t-distribution would be complete nonsense. The formula for a t-statistic that you give is only for Pearson correlation coefficients, not for z-statistics.

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    $\begingroup$ Thank you! "The formula for a t-statistic that you give is only for Pearson correlation coefficients, not for z-statistics." Yes. My understanding is that, if you want to compare a single value then, you can either use the t-statistic formula that I wrote above or transform to z and then reference the z-distribution (the latter controls for the lack of underlying normality). Is this correct? In my case, if I do a t-test then neither is relevant. (Just trying to get a better understanding of the other 2 methods.) $\endgroup$ – Rita Feb 8 '17 at 16:34
  • $\begingroup$ So when drawing a conclusion, is it valid to say that you either perform a t-test on the correlation coefficient or a z-transformation? When do I use the one over the other one? I'm a bit confused at the little and try to separate those tools. $\endgroup$ – Ben Feb 6 '20 at 7:10
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If you analyse the $r$ values directly you are assuming they all have the same precision which is only likely to be true if they are (a) all based on the same $n$ (b) all more or less the same. You could compute the standard errors and then do your analysis weighting each by the inverse of its sampling variance. It would seem easier to transform them to $z$ especially if they are all based on the same $n$ as then you could assume equal variances. If they are not based on the same $n$ then you definitely need to weight them.

If I were doing this I would treat it as a meta-analysis problem because software is readily available for doing this on correlation coefficients and it takes care of the weighting. I would enter the $z$ with their standard errors and get an overall summary $z$ (which I would transform back to $r$ obviously) and more importantly a confidence interval for $z$ (and hence $r$). It would also provide a significance test if you really like significance tests. Meta-analysis software would also give you an estimate of the heterogeneity of the estimated coefficients which would indicate whether in fact summarising them as a single number was a fruitful thing to so.

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    $\begingroup$ Thank you! Is this in lieu of a one-sample t-test? Why would this be preferable? If I understand correctly, the standard-error is contained in the test statistic I wrote above. Do you mean that I should get this test-statistic for each participant, average this across participants, and do NHST on this one-point value? This seems to keep the central tendency of the set of subjects but excludes the variance of the individual subjects. $\endgroup$ – Rita Feb 8 '17 at 16:38
  • $\begingroup$ Added some more as an edit to the answer. Meta-analysis does weight the estimates by the way. $\endgroup$ – mdewey Feb 8 '17 at 16:48
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If I am reading you correctly, you are comparing the mean r values of two groups. In general, even though the t test is robust to violations of normality, you have greater power with normal distributions. Therefore, if some of your r's are high (over .6 or so) it would be a good idea to transform them. Naturally, the t test doesn't care what the numbers are (they are correlations) but only their distribution.

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