# st: RE: metan command & binary variables

 From Leonelo Bautista <[email protected]> To [email protected] Subject st: RE: metan command & binary variables Date Wed, 21 Apr 2004 07:45:29 -0500

```Odds ratios and relative risks are calculated different.
(4/5)/(4/51)=10.2 this is the ratio of two odds
(4/(5+4))/(4/(51+4))=6.1 this is the ratio of two probabilities (risks)

Leonelo E. Bautista,
University of Wisconsin
Population Health Sciences

-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of
[email protected]
Sent: Tuesday, April 20, 2004 5:22 PM
To: [email protected]
Subject: st: metan command & binary variables

Hello All --
I'm using Stata version 8.2 and I hope someone can clarify a
question I have regarding the -metan- command and it's
estimation of odds ratios versus risk ratios.  Suppose I have
two rates that capture success on a diet where the first rate
represents those who adhered to a certain protocol and the
second rate represents those who dropped from the protocol
after follow-up.  Each rate is captured by the number
successful divided by the number in the respective group
(adhere vs follow-up).  I created four variables:  num1, den1,
num2, & den2 where num1 is the # of successful dieters that
follows for num2 & den2.  When I run the following code

-metan num1 den1 num2 den2, random or label(namevar=Trial)-

I get an 'odds ratio' that works out to equal the ratio of the
rate from the adherers over the rate for the droppers.  If I
specify -rr- instead of -or- the results differ and I am at a
loss as to why.  Consider the following values:
num1=4, den1=5, num2=4, den2=51
where the calculation of (4/5)/(4/51) = 10.2, which is
consistent with -metan..., or- but if I specify -metan..., rr-
I get a 'risk ratio' of 6.111.  Am I not understanding the
vernacular or can someone clue me into how the 6.111 was
calculated?

As always, your input is much appreciated!
thank you -- Clint Thompson
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