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Re: st: Survival analysis


From   Maarten buis <maartenbuis@yahoo.co.uk>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: Survival analysis
Date   Mon, 31 Aug 2009 09:52:49 +0000 (GMT)

-----------------------------------------
Maarten L. Buis
Institut fuer Soziologie
Universitaet Tuebingen
Wilhelmstrasse 36
72074 Tuebingen
Germany

http://www.maartenbuis.nl
-----------------------------------------


--- moleps islon  wrote:
> > This is the ouput I´m getting using your approach:
> >
> > n=896, failures=292
> > 
> > stcox var,tvc(var) texp((_t>1)_t)
> > 
> > rh
> > 
> > var HR 0.64, p=0.005, CI 0.47-0.87
> > 
> > t
> > var HR 1.01,p=0.001,CI 1.01-1.03
> > 
> > So as far as I understand this the interpretation is
> > that the -var- is protective within the first 24hrs, 
> > but detrimental afterwards ??

--- On Mon, 31/8/09, Maarten buis wrote: 
> No, the coefficient in the t equation is an interaction
> effect. So from t =0 to t=1 the hazard ratio increased
> with 1%. So at t=0 the hazard ratio for var is 
> 0.64/1.01=0.62. In other words, in the first 24hrs var
> was even more protective than afterwards (but only very
> little, so I doubt whether that has any practical
> relevance).

Sorry, I did not see that you turned around the inquality
sign (from < to >). So, in your case you assume that the 
PH assumption holds in the first 24hrs, and that
afterwards the log hazard ratio changes linearly with time.
So, from t=0 to t=1 the hazard ratio of var is .64, and 
after t=1 the hazard ratio increases by 1% every day. At
t=2 the hazard ratio of var is 1.01*.64=.646, at t=3 
1.01^2*.64=.653, at t=4 1.01^3*.64=.659, etc.

To get the interpretation I gave in my previous post you
have to replace 
stcox var,tvc(var) texp((_t>1)_t)

with 
stcox var,tvc(var) texp((_t<1)_t)

Hope this helps,
Maarten



      

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