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In article <[EMAIL PROTECTED]>,
        Brian Sandle <[EMAIL PROTECTED]> writes:
...
>How can he say that Spearman's rank order correlation given by
>
>rho = 1 - [6sigmaD^2]/[N(N^2-1)]
>
> is just Pearson's product moment correlation given by
>
>r =[NsigmaXY -(sigmaX)(sigmaY)]/SQRT{[NsigmaX^2-(sigmaX)^2][
>        NsigmaY^2-(sigmaY)^2]}
>
> using ranks instead of scores?

Quite easily.  If X & Y are ranks (with no ties) then (e.g.)
sigmaX^2 involves SUM_(i=1)^N i^2, which is just N(N+1)(2N+1)/6.
The formula for r then simplifies to give rho.

Note that if there are tied ranks (are there ever not?)
then Pearson's r on the ranks is different from Spearman's rho 
with the "standard" correction for ties (whatever that may be).

If you really really want to calculate a nonparametric correlation,
then I'd suggest using Kendall's tau, which has a natural probabilistic
interpretation and also a smoother distribution than Spearman's rho.
-- 
J.E.H.Shaw   [Ewart Shaw]        [EMAIL PROTECTED]     TEL: +44 2476 523069
  Department of Statistics,  University of Warwick,  Coventry CV4 7AL,  U.K.
  http://www.warwick.ac.uk/statsdept/Staff/JEHS/
3  ((4&({*.(=+/))++/=3:)@([:,/0&,^:(i.3)@|:"2^:2))&.>@]^:(i.@[)  <#:3 6 2
.
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