[R] survival function with a Weibull dist

Anaid Diaz syadp at yahoo.com.mx
Thu Sep 21 12:48:12 CEST 2006


Hi
I am using R to fit a survival function to my data
(with a weibull distribution).

Data: Survival of individuals in relation to 4
treatments ('a','b','c','g') 

syntax:
----  > survreg(Surv(date2)~males2, dist='weibull')

But I have some problems interpreting the outcome and
getting the parameters for each curve.

---------              Value Std. Error      z       
p
---------  (Intercept)  2.788      0.147 19.022
1.13e-80
--------- males2b     -0.107      0.207 -0.519
6.04e-01
--------- males2c     -0.486      0.586 -0.831
4.06e-01
--------- males2g      0.580      0.207  2.798
5.15e-03
--------- Log(scale)  -1.116      0.139 -8.007
1.18e-15
--------- 
--------- Scale= 0.328 


I know from Venables & Ripley (2002) that the
parameters of this function should be two:

lambda = z (presumably "Value" in R for each
treatment)
alpha = k (scale in R)

Survival function (S):

S(t)= exp-(zt)^k

I don't quite understand how to use the intercept.
First I thought adding it to each other treatment
value, for example:

Survival Males2a (t) = exp – ((intercept + 0) * t) ^ k
Survival Males2b (t) = exp – ((intercept + zb)* t) ^ k

But the curves I get using this interpretation are not
similar to my data


Does anyone use this function and could help me to
interpret the results?

many thanks


Anaid



More information about the R-help mailing list