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st: R: the sign of z statatistic after -ranksum-? (flag: Stata 9/2 SE)


From   "Carlo Lazzaro" <carlo.lazzaro@tin.it>
To   <statalist@hsphsun2.harvard.edu>
Subject   st: R: the sign of z statatistic after -ranksum-? (flag: Stata 9/2 SE)
Date   Sun, 25 Oct 2009 12:18:29 +0100

Dear Rose,
the following examples may shed light on the oddity you came across in
performing -ranksum-.

webuse fuel2,clear
ranksum mpg, by(treat)

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

       treat |      obs    rank sum    expected
-------------+---------------------------------
           0 |       12         128         150
           1 |       12         172         150
-------------+---------------------------------
    combined |       24         300         300

unadjusted variance      300.00
adjustment for ties       -4.04
                     ----------
adjusted variance        295.96

Ho: mpg(treat==0) = mpg(treat==1)
             z =  -1.279
    Prob > |z| =   0.2010

return list

scalars:
               r(N_2) =  12
               r(N_1) =  12
             r(Var_a) =  295.9565217391304
                 r(z) =  -1.278817949868369
           r(sum_exp) =  150
           r(sum_obs) =  128
            r(group1) =  0

As you can see, after -return list-, Stata consider - r(sum_obs) =  128 -.

z can be obtained via the following formula (see also: Pagano M, Gauvreau K.
Principles of Biostatistics. 2nd edition. Brooks/Cole, 2000)

di (128-150)/295.96^.5
-1.2788104

When you decide to invert the samples rank-sum:

ranksum mpg, by(treat2)

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

      treat2 |      obs    rank sum    expected
-------------+---------------------------------
           0 |       12         172         150
           1 |       12         128         150
-------------+---------------------------------
    combined |       24         300         300

unadjusted variance      300.00
adjustment for ties       -4.04
                     ----------
adjusted variance        295.96

Ho: mpg(treat2==0) = mpg(treat2==1)
             z =   1.279
    Prob > |z| =   0.2010

. return list

scalars:
               r(N_2) =  12
               r(N_1) =  12
             r(Var_a) =  295.9565217391304
                 r(z) =  1.278817949868369
           r(sum_exp) =  150
           r(sum_obs) =  172
            r(group1) =  0

As you can see after -return list-, Stata consider - r(sum_obs) =  172 -.

Again, z can be obtained via the following formula (see also: Pagano M,
Gauvreau K. Principles of Biostatistics. 2nd edition. Brooks/Cole, 2000)

di (172-150)/295.96^.5
1.2788104.

As Martin said, usually <the test is against a two-sided alternative, so the
sign hardly matters...>

However, your concern is probably driven by the awareness that some
textbooks on statistics report the following formula for calculating
Wilkoxon rank sum test (see again: Pagano M, Gauvreau K. Principles of
Biostatistics. 2nd edition. Brooks/Cole, 2000):


zw = (W-mw)/sw

Where the z-statistic is obtained by subtracting the mean of the ranks sum
from W (the smallest of the two rank sums)and ; in this way, as you stated
in your thread, <z statistic the result should be the same no matter which
group is compared with which group>.

HTH and Kind Regards,
Carlo
-----Messaggio originale-----
Da: owner-statalist@hsphsun2.harvard.edu
[mailto:owner-statalist@hsphsun2.harvard.edu] Per conto di gjhxmu@sina.com
Inviato: domenica 25 ottobre 2009 4.29
A: statalist
Oggetto: st: the sign of z statatistic after -ranksum-?

Dear statalists,

I use -ranksum- in the stata to do  the Wilcoxon rank-sum test, which is
also known as the Mann-Whitney two-sample statistic.

My typing is as followings,

webuse fuel2,clear
ranksum mpg, by(treat) // the z statistic equals -1.279
replace treat=treat==0
ranksum mpg, by(treat) // the z statistic equals 1.279


My doubt is why the sign of z statistic is opposite. As far as I know, in
the computation of z statistic the result should be the same no matter which
group is compared with which group.


Any help will be appreciated!


Best regards,

Rose.
 

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