st: RE: ranksum: p-value?

["Newson, Roger B" <r.newson@imperial.ac.uk>]

Your true P-value is  

Prob > z =   0.0216

The other probability (0.604) is not a P-value, but is the probability
that a randomly-chosen member of Group A has a higher outcome value than
a randomly-chosen member of Group B. This probability is known as
Harrell's c-index. If you want a confidence interval for Harrell's
c-index (or for Somers' D = 2c-1), then you can use the -somersd-
package, downloadable from SSC using the -ssc- command in Stata. The
-somersd- package can also estimate median differences, ratios and
slopes.

I hope this helps.

Roger


Roger Newson
Lecturer in Medical Statistics
Respiratory Epidemiology and Public Health Group
National Heart and Lung Institute
Imperial College London
Royal Brompton campus
Room 33, Emmanuel Kaye Building
1B Manresa Road
London SW3 6LR
UNITED KINGDOM
Tel: +44 (0)20 7352 8121 ext 3381
Fax: +44 (0)20 7351 8322
Email: r.newson@imperial.ac.uk 
www.imperial.ac.uk/nhli/r.newson/

Opinions expressed are those of the author, not of the institution.

-----Original Message-----
From: owner-statalist@hsphsun2.harvard.edu
[mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of Vanessa
Mahlberg
Sent: 12 June 2007 16:54
To: statalist@hsphsun2.harvard.edu
Subject: st: ranksum: p-value?

Hi,

I?ve done a Wilcoxon test for testing the difference between two groups.
Now I have a doubt about the interpretation of the result of this test.
I don?t know which one is my real p-value. The one Prob > z =   0.0216
or 0.604??? The obtained result is:

.ranksum a_firmenbewertung, by (zugeh) porder

Two-sample Wilcoxon rank-sum (Mann-Whitney) test

zugeh                   obs    rank sum    expected

Ehemalige Pr        84      7327.5        6678
Direkteinste          74      5233.5        5883

combined           158       12561       12561

unadjusted variance    82362.00
adjustment for ties    -2423.03
----------
adjusted variance      79938.97

Ho: a_firm~g(zugeh==Ehemalige Praktikanten) =
a_firm~g(zugeh==Direkteinsteiger)
z =   2.297
Prob > z =   0.0216

P{a_firm~g(zugeh==Ehemalige Praktikanten) >
a_firm~g(zugeh==Direkteinsteiger)} = 0.604

Thanks for your support
Vanessa


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