Comparing spearman correlations

[Maguin, Eugene]

All,

I'd like to compare spearman correlations (from crosstabs) from independent
samples. Can anyone comment on whether using standard formula for pearson
correlations from Cohen (as well as others) would be incorrect?

Thanks, Gene Maguin

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[Bruce Weaver]

Gene Maguin wrote

All,

I'd like to compare spearman correlations (from crosstabs) from independent
samples. Can anyone comment on whether using standard formula for pearson
correlations from Cohen (as well as others) would be incorrect?

Thanks, Gene Maguin

Hi Gene.  Good question.  I would argue that you can go ahead.  Here is my thought process.

1. The Spearman correlation = Pearson r computed on ranked data.

2. The test of significance for Pearson r (H0: population correlation = 0) requires/assumes that at least one of the two variables is normally distributed.  But for Spearman's rank correlation, normality is not required.  (Source:  Campbell, Machin & Walters, 2007, Medical Statistics, 4th Ed, p. 156).

3. If the test of significance for Spearman's rank correlation is valid, then the test for the difference between two independent Spearman rank correlations should be too.

I may not be right, but I hope I've at least offered what Robert P. Abelson would call a "principled argument".  ;-)

[Alexander J. Shackman-2]

there's a pretty extensive literature on rank transforms that supports bruce's basic argument.

see eg work by conover and iman, zimmermann and others (http://psyphz.psych.wisc.edu/~shackman/mediation_moderation_resources.htm#DataTransforms)

hth,
alex shackman

On Fri, Jul 30, 2010 at 9:29 AM, Bruce Weaver <[hidden email]> wrote:

Gene Maguin wrote:
>
> All,
>
> I'd like to compare spearman correlations (from crosstabs) from
> independent
> samples. Can anyone comment on whether using standard formula for pearson
> correlations from Cohen (as well as others) would be incorrect?
>
> Thanks, Gene Maguin
>
>

Hi Gene.  Good question.  I would argue that you can go ahead.  Here is my
thought process.

1. The Spearman correlation = Pearson r computed on ranked data.

2. The test of significance for Pearson r (H0: population correlation = 0)
requires/assumes that at least one of the two variables is normally
distributed.  But for Spearman's rank correlation, normality is not
required.  (Source:  Campbell, Machin & Walters, 2007, Medical Statistics,
4th Ed, p. 156).

3. If the test of significance for Spearman's rank correlation is valid,
then the test for the difference between two independent Spearman rank
correlations should be too.

I may not be right, but I hope I've at least offered what Robert P. Abelson
would call a "principled argument".  ;-)


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Alexander J. Shackman, Ph.D.
Wisconsin Psychiatric Institute & Clinics and
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