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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


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