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st: rate ratios from xtlogit using margins - how?

From   SeyiLSHTM <>
Subject   st: rate ratios from xtlogit using margins - how?
Date   Fri, 17 Dec 2010 05:30:19 -0800 (PST)

Dear Listers,
I have what I hope is a simple question.
I am running an xtlogit model (re) to compare stillbirth rates in an
intervention (=1) and a control group (=0) with clustering. Stata outputs
the odds ratio, though I'd rather convert this to show the adjusted rates,
and an adjusted rate ratio or a rate difference (with CIs), as this is more
intuitive when presenting results to mixed audiences.

my model:

xtlogit still intvn, re i(zone) or

I used the margins command to obtain what I understood to be the predicted
probabilities (adjusted rates) of stillbirth in the two groups of intvn (the
binary var described above):

. margins intvn

0	.6546526	.0915509	7.15	0.000	.4752162	.8340891
1	1.028017	.0932528	11.02	0.000	.8452453	1.210789

I see xb (the linear prediction), not the probabilities - is this correct?
In the stata manual it says for margins after logit: "....numbers reported
in the “Margin” column are average predicted probabilities...", but I am
assuming that this is not the same for xtlogit as the above numbers are
clearly not probabilities, so I exponentiate the values; e(0.65) and
e(1.03). This gives me the odds for each group, right? not the probabilities
(as often greater than 1).
So, after transforming xb, I think I now have the odds of stillbirth in
intvn=0 and intvn=1,  as the ratio of these matches exactly the odds ratio
in my xtlogit model. 

So, to get adjusted rates (probabilities) and a rate ratio, I then
transformed these odds to probabilities (0.654/(1+0.654), and
1.028/(1+1.028)) and found the ratio of these, 0.736/0.658 = 1.12 = rate

Is this correct? How can I obtain a single CI for this rate ratio?
Can I use this formula to transform the upper and lower limits of the odds

odds ratio CI/ (1- [CI in intvn=0 group]) + ([CI in intvn=0 group] * odds
ratio CI)
(Obtained using help/findit oddsrisk)

Apologies for length, any advice appreciated.

Have a good Christmas,


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