R help - Sep 2006 - survival function with a Weibull dist

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

On Thu, 21 Sep 2006, Anaid Diaz wrote:
> 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:
"this function" being which function? As help(survreg.distributions)
says
      The Weibull distribution is not parameterised the
      same way as in 'rweibull'.

In your model the survival time is a variable whose logarithm has 
distribution

    2.788-0.107males2b-0.486males2c+0.580males2g+epsilon*0.328

where epsilon has cdf F=1-e^{-e^t}.

As in the example on help(survreg.distributions) shows, a survreg Weibull 
model with linear predictor M and scale S corresponds to R's weibull 
distribution with scale exp(M) and shape 1/S.

 	-thomas