# Re: st: probability and z-statistic

 From Deidra Young To Subject Re: st: probability and z-statistic Date Tue, 05 Dec 2006 22:38:51 +0800

```Thanks.

I have a sample size of 315 cases distributed evenly between Australia and
an International database.  55 cases have autism.  All cases have another
genetic condition.  In this case I am looking at the ordered distribution of
the disturbed awake breathing rhythm (thee levels from none, mild & severe).
I am also doing logistic (where autism is the DV and binary) and ologit
(where I use the ordered categorical severity items as DV) so that I can
adjust for age.  However, I wanted to learn how to do tabs without using the
problematic chi squared test which is inappropriate for ordered, categorical
measures and cannot inform which are the cells being compared.

Is this method only appropriate for comparing two groups (df = 1)?

Deidra

On 5/12/06 9:11 PM, "Nick Cox" <n.j.cox@durham.ac.uk>

> 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
> n.j.cox@durham.ac.uk
>
> 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
>>                Cramér'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" <baum@bc.edu>
>>
>>> 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
>>>                 Cramér'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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```