Date:         Thu, 17 Feb 2005 11:27:36 +0100
Reply-To:     Marta García-Granero
              <biostatistics@terra.es>
Sender:       "SPSSX(r) Discussion" <SPSSX-L@LISTSERV.UGA.EDU>
From:         Marta García-Granero
              <biostatistics@terra.es>
Organization: Asesoría Bioestadística
Subject:      Re: Questions on Odds Ratios
In-Reply-To:  <136e4d060502170038113510ee@mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-15

Hi

See the following example:

DATA LIST LIST / smoker coffee problem freq (4 F8.0).
BEGIN DATA
1       1       1       11
1       1       2       72
2       1       1       17
2       1       2       73
1       2       1       15
1       2       2       61
2       2       1       47
2       2       2       528
END DATA.
WEIGHT  BY freq .
VALUE LABELS smoker 1 'Yes' 2 'No'
 /coffee 1' Yes' 2 'No' /problem 1'Yes' 2 'No'.

* Assuming problem is the outcome, coffee is the IV and tobacco is the
  adjusting factor *.

PJV> 2 - Mantel-Haenzel statistics, which is available  through the
PJV> crosstabs command. If the Mantel-Haenzel statistic is not significant
PJV> then you don't need to adjust by race and can calculate a simple Odds
PJV> ratio.

CROSSTABS
  /TABLES=problem  BY coffee  BY smoker
  /FORMAT= AVALUE TABLES
  /STATISTIC=CHISQ RISK CMH(1)
  /CELLS= COUNT.

* Last part of the output will give you Mantel-Haenszel OR, the
  adjusted estimator of the effect of coffee comsumption, adjusted by
  smoker status.


PJV> 1 - Try with the logistic regression method, I think that this command
PJV> can help you.

RECODE problem (2=0)  .
EXECUTE .

VALUE LABELS problem 1'Yes' 0 'No'.
LOGISTIC REGRESSION  problem
  /METHOD = ENTER smoker coffee
  /CONTRAST (smoker)=Indicator
  /CONTRAST (coffee)=Indicator
  /PRINT = CI(95)
  /CRITERIA = PIN(.05) POUT(.10) ITERATE(20) CUT(.5) .

--
Regards,
 Marta                            mailto:biostatistics@terra.es