[R] AFT model time-dependent with weibull distribution

[JPF]

Göran Broström wrote:
> 
> 
> Good. Do you still need answers to your other questions?
> 
> 

Yes. Could answer the following two questions: 

1- Can I use phreg function to estimate  a model with time-dependent
covariates? In case of a positive answer, how? 
 
2- I could not find any example that clearly explains how to interpret
aftreg output. Specially, refering to the diference between survreg and
aftreg output (intercept and sign of the estimates).

I include below an example of output of a regression with coxph, survreg,
phreg and aftreg and a time-independent variable. I would appreciate if you
could explain it or provide an external example that explains how it works.

n=26
events=25
time at risk=45

a/ 

coxph(Surv(time,s) ~ Z1,  data=data.frame(data))

     coef     exp(coef)  se(coef)    z     p
Z1 0.0249      1.03      0.00907 2.75 0.006


b/

phreg(Surv(time,s) ~ Z1,  data=data.frame(data), dist="weibull")

Covariate          W.mean      Coef Exp(Coef)  se(Coef)    Wald p
Z1                      43.689     0.033     1.033     0.009     0.000 

log(scale)                          0.641     1.899     0.065     0.000 
log(shape)                         1.172     3.230     0.158     0.000 

Max. log. likelihood      -22.135 
LR test statistic         13.1 
Degrees of freedom        1 
Overall p-value           0.000302689


c/

aftreg(Surv(time,s) ~ Z1,  data=data.frame(data), dist="weibull")

Covariate          W.mean      Coef Exp(Coef)  se(Coef)    Wald p
mas                  43.689       0.010    1.010     0.002     0.000 

log(scale)                          1.147     3.149     0.141     0.000 
log(shape)                         1.172     3.230     0.158     0.000 

Max. log. likelihood      -22.135 
LR test statistic         13.1 
Degrees of freedom        1 
Overall p-value         0.000302692


d/

survreg(Surv(time,s) ~ Z1,  data=data.frame(data), dist="weibull")

                      Value     Std. Error     z        p
(Intercept)      1.1476    0.13498     8.50  1.87e-17
mas               -0.0101   0.00232    -4.34  1.45e-05
Log(scale)      -1.1724    0.15787    -7.43 1.11e-13

Scale= 0.310 

Weibull distribution
Loglik(model)= -22.1   Loglik(intercept only)= -28.7
        Chisq= 13.05 on 1 degrees of freedom, p= 3e-04 
Number of Newton-Raphson Iterations: 5 


Thank you very much, 

J







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