comparing two Spearman's rho coefficients
Chris Maloof
(asking whether correlations between a pair of
non-normally-distributed variables are different between two different
subject groups). I know this is possible with Pearson's r using the
method described here:
http://www.fon.hum.uva.nl/Service/Statistics/Two_Correlations.html
but I'm not sure if there's a problem with extending the method to
Spearman's rho.
Thanks for any advice or pointers.
Chris
Richard Ulrich
wrote:
When you can do it, you are better off testing whether
the *regression* coefficient is the same in two samples,
since that is less dependent on the variances in the
two samples.
For the test: The Spearman rho is computed as a Pearson
on the rank-transformed data. Thus, as long as the Ns are
moderately large, the Spearman is tested the same as a
Pearson, and the test comparing two can be the same.
- If you find a difference, before you can draw any other
conclusion, you probably have to figure out how to say
that the separate variabilities are not unequal.
I would probably try to find some suitable power transformation
and compare regression coefficients.
--
Rich Ulrich, wpi...@pitt.edu
http://www.pitt.edu/~wpilib/index.html
Chris Maloof
<Rich....@comcast.net> wrote:
>
>When you can do it, you are better off testing whether
>the *regression* coefficient is the same in two samples,
>since that is less dependent on the variances in the
>two samples.
>
>For the test: The Spearman rho is computed as a Pearson
>on the rank-transformed data. Thus, as long as the Ns are
>moderately large, the Spearman is tested the same as a
>Pearson, and the test comparing two can be the same.
> - If you find a difference, before you can draw any other
>conclusion, you probably have to figure out how to say
>that the separate variabilities are not unequal.
>
>I would probably try to find some suitable power transformation
>and compare regression coefficients.
Thanks! That's what I was hoping to hear. The data are actually
percentages; the only reason why I didn't compare logistic regression
coefficients is that the basic analysis has already been done and
rather thoroughly written up in prose and tables using Spearman's rho,
and my professor doesn't want to add weeks to the project by making
the long-distance ex-grad-student redo everything. (The two
coefficients in question are pretty clearly not different, but because
one is significant and the other isn't we want to avoid confusion by
giving numbers.) It's not ideal though...
Hope that makes sense. Thanks for the thoughtful response!
Chris
David Jones
There are two questions:
(i) the benefit or otherwise of the Fisher Z transform being used;
(ii) the approximation for the variance (of the transformed
correlation) as
" 1/(n-3) " ...
A reference I found suggests this needs to be " 1.06/(n-3) " for the
rank correlation. See ...
Choi SC (1977) Test of equality of dependent correlations. Biometrika
64 (3) 645-7
The above may not itself be directly useful but it does reference the
result for the variance as coming from ..
Fieller EC et al (1957) Tests for rank correlation coefficients :I.
Biometrika 44, 470-81.
I have not seen this particular reference: it may contain someting
relevant to (i).
David Jones
Chris Maloof
wrote:
>
>There are two questions:
>
>(i) the benefit or otherwise of the Fisher Z transform being used;
>
>(ii) the approximation for the variance (of the transformed
>correlation) as
> " 1/(n-3) " ...
>A reference I found suggests this needs to be " 1.06/(n-3) " for the
>rank correlation. See ...
>
>Choi SC (1977) Test of equality of dependent correlations. Biometrika
>64 (3) 645-7
>
>The above may not itself be directly useful but it does reference the
>result for the variance as coming from ..
>
>Fieller EC et al (1957) Tests for rank correlation coefficients :I.
>Biometrika 44, 470-81.
>
Wow, that's even more relevant. I'm finding the Choi paper hard to
read (partly because it deals with a very general case of lots of
variables, partly because I just don't know that much statistics) but
the Fieller variance result is clear. Much appreciated!
Chris