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RE: st: probability and z-statistic


From   "Nick Cox" <[email protected]>
To   <[email protected]>
Subject   RE: st: probability and z-statistic
Date   Tue, 5 Dec 2006 13:11:35 -0000

Kit's suggestion was identical to mine. 

This looks correctly calculated assuming that
your z genuinely does relate to N(0,1). 

In this example, the question of significance is 
at best cosmetic. Even if a much bigger sample size 
led to this being declared significant, a rank
correlation of 0.03 is still clear-cut! 

Nick 
[email protected] 

Deidra Young
 
> I just tried Kit's suggestion.  One variable has two 
> categories and the
> other has three. My test is to determine if there is a 
> difference between
> cases with and without Autism Diagnosis (2) using an ordered severity
> measure with three levels (none, mild or severe).  Using your 
> suggestion, I
> obtained the following result.  Does this look like the p value is
> calculated correctly?
 
> . tab autism_dx kerr_breath_score, all
> 
>                     |  Kerr Score for Disturbed Awake
> Initially Diagnosed |         Breathing Rhythm
>         with Autism |      None       Mild     Severe |     Total
> --------------------+---------------------------------+----------
> No Autism Diagnosis |        75        109         76 |       260
>    Autism Diagnosis |        11         29         15 |        55
> --------------------+---------------------------------+----------
>               Total |        86        138         91 |       315
> 
>           Pearson chi2(2) =   2.5711   Pr = 0.277
>  likelihood-ratio chi2(2) =   2.6204   Pr = 0.270
>                Cram�r's V =   0.0903
>                     gamma =   0.0759  ASE = 0.116
>           Kendall's tau-b =   0.0324  ASE = 0.050
> 
> . ret li
> 
> scalars:
>                  r(N) =  315
>                  r(r) =  2
>                  r(c) =  3
>               r(chi2) =  2.571055140705914
>                  r(p) =  .2765046694792958
>            r(chi2_lr) =  2.620364965212246
>               r(p_lr) =  .2697708234098308
>           r(CramersV) =  .0903442295432575
>              r(gamma) =  .0758665794637018
>            r(ase_gam) =  .1163419665428843
>               r(taub) =  .0324087958656579
>           r(ase_taub) =  .0497834038046439
> 
> . di r(taub) / r(ase_taub)
> .65099598
> 
> . scalar z=r(taub)/r(ase_taub)
> 
> . display "z= " z " with p-value " normden(1-z)
> z= .65099598 with p-value .37537099
> 
> 
> 
> 
> 
> 
> 
> 
> On 5/12/06 8:22 PM, "Kit Baum" <[email protected]>
> 
> > Deidra wants to retrieve the p-value and Kendall tau-b score from a
> > tab of two variables.
> > 
> > 
> > . tab anxiety depress,all
> > 
> >             |             DEPRESS
> >     ANXIETY |         1          2          3 |     Total
> > -----------+---------------------------------+----------
> >           1 |         8          1          0 |         9
> >           2 |        14         42          3 |        59
> >           3 |         3         21         11 |        35
> >           4 |         0          1          3 |         4
> > -----------+---------------------------------+----------
> >       Total |        25         65         17 |       107
> > 
> >            Pearson chi2(6) =  46.1892   Pr = 0.000
> > likelihood-ratio chi2(6) =  40.9054   Pr = 0.000
> >                 Cram�r's V =   0.4646
> >                      gamma =   0.7778  ASE = 0.087
> >            Kendall's tau-b =   0.4951  ASE = 0.070
> > 
> > . return list
> > 
> > scalars:
> >                   r(N) =  107
> >                   r(r) =  4
> >                   r(c) =  3
> >                r(chi2) =  46.18917309609633
> >                   r(p) =  2.71459405298e-08
> >             r(chi2_lr) =  40.90537799483023
> >                r(p_lr) =  3.02263811177e-07
> >            r(CramersV) =  .4645828854557538
> >               r(gamma) =  .7777777777777778
> >             r(ase_gam) =  .087050646901156
> >                r(taub) =  .4950723482288552
> >            r(ase_taub) =  .0704364215558283
> > 
> > If she wants the p-value of Kendall's tau-b, it can be easily
> > calculated (if indeed taub is ~N(0,1)) from what is available in the
> > return list:
> > 
> > . scalar z=r(taub)/r(ase_taub)
> > 
> > . display "z= " z " with p-value " normden(1-z)
> > z= 7.0286414 with p-value 5.114e-09

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