Hazard Ratio from AFT model

[Dan Murphy]

Dear Statalist community,

I am using the mixed effects Weibull PH and AFT regression, and using the Kidney dataset as an example for my question.

use http://www.stata-press.com/data/r15/catheter
stset

mestreg age female || patient:, distribution(weibull)

failure _d: infect
analysis time _t: time

Fitting fixed-effects model:

Iteration 0: log likelihood = -1700989.9
Iteration 1: log likelihood = -440.1998
Iteration 2: log likelihood = -336.62162
Iteration 3: log likelihood = -334.64937
Iteration 4: log likelihood = -334.57959
Iteration 5: log likelihood = -334.57944
Iteration 6: log likelihood = -334.57944

Refining starting values:

Grid node 0: log likelihood = -336.03604

Fitting full model:

Iteration 0: log likelihood = -336.03604 (not concave)
Iteration 1: log likelihood = -333.14043
Iteration 2: log likelihood = -330.40952
Iteration 3: log likelihood = -329.89242
Iteration 4: log likelihood = -329.87847
Iteration 5: log likelihood = -329.87832
Iteration 6: log likelihood = -329.87832

Mixed-effects Weibull PH regression Number of obs = 76
Group variable: patient Number of groups = 38

Obs per group:
min = 2
avg = 2.0
max = 2

Integration method: mvaghermite Integration pts. = 7

Wald chi2(2) = 10.12
Log likelihood = -329.87832 Prob > chi2 = 0.0063
------------------------------------------------------------------------------
_t | Haz. Ratio Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
age | 1.007348 .013788 0.53 0.593 .9806828 1.034737
female | .1904727 .099992 -3.16 0.002 .0680737 .5329493
_cons | .0072901 .0072274 -4.96 0.000 .0010444 .0508881
-------------+----------------------------------------------------------------
/ln_p | .2243233 .1402795 -.0506195 .4992661
-------------+----------------------------------------------------------------
patient |
var(_cons)| .8308495 .4978425 .256735 2.688808
------------------------------------------------------------------------------
Note: Estimates are transformed only in the first equation.
Note: _cons estimates baseline hazard (conditional on zero random effects).
LR test vs. Weibull model: chibar2(01) = 9.40 Prob >= chibar2 = 0.0011

mestreg age female || patient:, distribution(weibull) time

failure _d: infect
analysis time _t: time

Fitting fixed-effects model:

Iteration 0: log likelihood = -346.46486
Iteration 1: log likelihood = -343.29515
Iteration 2: log likelihood = -335.0513
Iteration 3: log likelihood = -334.58308
Iteration 4: log likelihood = -334.57944
Iteration 5: log likelihood = -334.57944

Refining starting values:

Grid node 0: log likelihood = -335.10428

Fitting full model:

Iteration 0: log likelihood = -335.10428
Iteration 1: log likelihood = -332.13546
Iteration 2: log likelihood = -330.01623
Iteration 3: log likelihood = -329.88013
Iteration 4: log likelihood = -329.87832
Iteration 5: log likelihood = -329.87832

Mixed-effects Weibull AFT regression Number of obs = 76
Group variable: patient Number of groups = 38

Obs per group:
min = 2
avg = 2.0
max = 2

Integration method: mvaghermite Integration pts. = 7

Wald chi2(2) = 13.00
Log likelihood = -329.87832 Prob > chi2 = 0.0015
------------------------------------------------------------------------------
_t | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
age | -.0058496 .010872 -0.54 0.591 -.0271585 .0154592
female | 1.325034 .3719102 3.56 0.000 .596103 2.053964
_cons | 3.932346 .5663757 6.94 0.000 2.82227 5.042422
-------------+----------------------------------------------------------------
/ln_p | .2243237 .1402794 -.0506189 .4992663
-------------+----------------------------------------------------------------
patient |
var(_cons)| .5304902 .2343675 .2231626 1.261053
------------------------------------------------------------------------------
LR test vs. Weibull model: chibar2(01) = 9.40 Prob >= chibar2 = 0.0011


Using the AFT model, the estimate for female in hazard ratio can be calculated as: exp(-exp(.2243237) * 1.325034) = 0.19
I am looking for advice on how to calculate the confidence interval in hazard ratio for female.

Many thanks.

[Maarten Buis]

A Weibull model is special in that you can display the exact same model in terms of hazard ratios and time ratios. So if you want your results in hazard ratios you just remove the time option.

[Dan Murphy]

Dear Maarten,

Thank you for your response.

We can indeed remove the time option in the command. I wanted to get a better understanding of how the result can be estimated in Hazard Ratio when using the AFT model.

Many thanks.