[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
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