6.5 Proportional hazards or AFT model?
The problem of choosing between a proportional hazards and an accelerated failure time model (everything else equal) can be solved by comparing the AIC of the models. Since the numbers of parameters are equal in the two cases, this amounts to comparing the maximized likelihoods. For instance, in the case with old age mortality:
Let us see what happens with the Gompertz AFT model:
Exactly the same procedure as with the Weibull distribution, but we have to
specify the Gompertz distribution in the call (remember, the Weibull
distribution is the default choice, both for phreg and aftreg).
## Call:
## aftreg(formula = Y ~ sex + civ + birthplace, data = om, dist = "gompertz")
##
## Covariate W.mean Coef Time-Accn se(Coef) Wald p
## sex
## male 0.406 0 1 (reference)
## female 0.594 -0.081 0.922 0.020 0.000
## civ
## unmarried 0.080 0 1 (reference)
## married 0.530 -0.152 0.859 0.034 0.000
## widow 0.390 -0.100 0.905 0.031 0.001
## birthplace
## parish 0.574 0 1 (reference)
## region 0.226 0.022 1.022 0.023 0.340
## remote 0.200 0.038 1.039 0.025 0.132
##
## Baseline parameters:
## log(scale) 2.222 0.042 0.000
## log(shape) -1.584 0.075 0.000
## Baseline life expectancy: 13.6
##
## Events 1971
## Total time at risk 37824
## Max. log. likelihood -7280
## LR test statistic 33
## Degrees of freedom 5
## Overall p-value 3.81949e-06Comparing the corresponding result for the proportional hazards and the AFT models with the Gompertz distribution, we find that the maximized log likelihood in the former case is -7279.973, compared to -7273.701 for the latter. This indicates that the proportional hazards model fit is better. Note however that we cannot formally test the proportional hazards hypothesis; the two models are not nested.