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)?
Best Answer
Many (including me) get confused by the different ways to define the parameters of a Weibull distribution, particularly since the standard R Weibull-related functions in the
stats
package and thesurvreg()
parametric fitting function in thesurvival
package use different parameterizations.The manual page for the R Weibull-related functions in
stats
says:That's called the "standard parameterization" on the Wikipedia page (where they use $k$ for shape and $\lambda$ for scale).
The
survreg()
function uses a different parameterization, with differences explained on its manual page:The
WeibullReg()
function effectively takes the result fromsurvreg()
and expresses the results in terms of the "standard parameterization."There is a potential confusion, however, as the
$summary
of the object produced byWeibullReg
is "the summary table from the original survreg model." (Emphasis added.) So what you have displayed in the question includes results for both parameterizations.That dual representation of the results helps explain what's going on.
Starting from the bottom, the
survreg
value ofscale
is the reciprocal of the "standard parameterization" value ofshape
. The "standard" shape parameter is calledgamma
in theWeibullReg
$formula
output near the top of your output. The value forgamma
is 0.98434, with a reciprocal of 1.0159, rounding to the value of 1.02 shown asScale
in the last line of your output. The natural logarithm of 1.0159 is 0.01578, shown asLog(scale)
in the next-to-last line. Those last lines of your output, remember, are based on thesurvreg
definition ofscale
.The p-value for that
Log(scale)
is indeed very high. But that just means that the value ofLog(scale)
is not significantly different from 0, or that thescale
itself (as defined insurvreg
) is not different from 1. That has nothing to do with the hazard ratios and so forth for the covariates. It just means that the baseline survival curve of your Weibull model can't be statistically distinguished from a simple exponential survival curve, which would have exactly a value of 1 forsurvreg
scale
or "standard"shape
and a constant baseline hazard over time. So there is nothing to distrust about your results on that basis.