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Re: st: Wilcoxon rank-sum (Mann-Whitney) test vs KolmogorovSmirnov test (AWAY)


From   "Arthur Blank" <[email protected]>
To   <[email protected]>
Subject   Re: st: Wilcoxon rank-sum (Mann-Whitney) test vs KolmogorovSmirnov test (AWAY)
Date   Sat, 16 Jul 2005 13:19:14 -0400

e-mails. I will  respond to your e-mail when I return. If this is an
emergency, pl;ease leave a message with Ms. Taneka Banks,
718-920-4484.Thank you. 
  Arthur E. Blank,Ph.D., Co-Director,
  Division of Research
  Department of Family and Social Medicine
  Albert Einstein College of Medicine

>>> "[email protected]" 07/16/05 12:49 >>>

Dear Stata users:

I define two types of firms domestic ones (type 1) and exporting ones
(type 0).
I want to test whether type 0 firms have higher average productivity
of labour (APL) than type 1, which graphical evidence seems to show by
plotting firm-level productivity distributions.
I run two tests:
ranksum l_APL, by(type_firm) porder
ksmirnov l_APL, by(type_firm)

Stata outcomes:

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

    type_firm |      obs    rank sum    expected
-------------+---------------------------------
            0 |      761      376455    364138.5
            1 |      195       80991     93307.5
-------------+---------------------------------
     combined |      956      457446      457446

unadjusted variance    11834501
adjustment for ties  -.16253916
                      ----------
adjusted variance      11834501

Ho: l_APL(type_f~m==0) = l_APL(type_f~m==1)
              z =   3.580
     Prob > |z| =   0.0003

P{l_APL(type_f~m==0) > l_APL(type_f~m==1)} = 0.583



Two-sample Kolmogorov-Smirnov test for equality of distribution
functions:

  Smaller group       D       P-value  Corrected
  ----------------------------------------------
  0:                  0.0185    0.899
  1:                 -0.1755    0.000
  Combined K-S:       0.1755    0.000      0.000

I am confused:
Wilcoxon rank-sum (Mann-Whitney) test implies that type 0 firms have
higher productivity :
P{l_APL(type_f~m==0) > l_APL(type_f~m==1)} = 0.583
Whereas KS test shows that it is the opposite (.899 as a pvalue when
testing that type 0 is smaller).

I guess that there is something I am misunderstanging but I cannot
understand why. Thanks so much for helping!!

Delphine

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