#StackBounty: #self-study #survival #interpretation #weibull Interpretation of Weibull Accelerated Failure Time Model Output

Bounty: 50

In this case study I have to assume a baseline Weibull distribution, and I’m fitting an Accelerated Failure Time model, which will be interpreted by me later on regarding both hazard ratio and survival time.

The data looks like this.

head(data1.1)

TimeSurv IndSurv Treat Age
1     6 days       1     D  27
2    33 days       1     D  43
3   361 days       1     I  36
4   488 days       1     I  54
5   350 days       1     D  49
6   721 days       1     I  49
7  1848 days       0     D  32
8   205 days       1     D  47
9   831 days       1     I  24
10  260 days       1     I  38

I’m fitting a model using the function Weibullreg() in R. The survival function is built reading TimeSurv as the time measures and IndSurv as the indicator of censoring. The covariates considered are Treat and Age.

My issue deals with understanding the output properly:

wei1 = WeibullReg(Surv(TimeSurv, IndSurv) ~ Treat + Age, data=data1.1)
wei1


$formula
Surv(TimeSurv, IndSurv) ~ Treat + Age

$coef
            Estimate           SE
lambda  0.0009219183 0.0006803664
gamma   0.9843411517 0.0931305471
TreatI -0.5042111027 0.2303038312
Age     0.0180225253 0.0089632209

$HR
              HR       LB       UB
TreatI 0.6039819 0.384582 0.948547
Age    1.0181859 1.000455 1.036231

$ETR
             ETR        LB        UB
TreatI 1.6690124 1.0574337 2.6343045
Age    0.9818574 0.9644488 0.9995801

$summary

Call:
survival::survreg(formula = formula, data = data, dist = "weibull")
               Value Std. Error     z      p
(Intercept)  7.10024    0.41283 17.20 <2e-16
TreatI       0.51223    0.23285  2.20  0.028
Age         -0.01831    0.00913 -2.01  0.045
Log(scale)   0.01578    0.09461  0.17  0.868

Scale= 1.02 

Weibull distribution
Loglik(model)= -599.1   Loglik(intercept only)= -604.1
    Chisq= 9.92 on 2 degrees of freedom, p= 0.007 
Number of Newton-Raphson Iterations: 5 
n= 120

I don’t really get how Scale = 1.02 and log(scale) = 0.015, and if the p-value of this log(scale) is a big non-signfificant one, from how the documentation of the function shows some conversions it makes, am I to assume that the values of the alphas are also not to be trusted (considering they were reached using the scale value)?


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