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I am trying to calculate confidence interval 95% for hazard ratio from flexsurvreg-weibull distribution output. Please provide the formula of the same.

I have formula for HR = exp(-beta)^shape. But could not find formula for 95% CI for HR.

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User1865345
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kamal
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    The dist="weibull" argument puts this in the context of an accelerated failure time model, which complicates thinking about hazard ratios (HR). Try using dist="weibullPH" for what the author of flexsurv calls a "more interpretable" result in terms of HR in a vignette. – EdM Jan 20 '23 at 20:32
  • Got your point. I did that. but estimates differs between wei and weiph. I need CI for HR from weibull-flexsurvreg output.,agree with your point on AFT context. Otherwise, can you please point me to material for interpreting coefficient from AFT flexsurvreg output ? – kamal Jan 20 '23 at 21:04
  • What is estimated will differ between weibull and weibullPH, but the fundamental Weibull model is the same. It's just represented in a different parameterization. What's called "scale" in the AFT form is not the same as what's called "scale" in the PH form, so covariates modeled as affecting "scale" will have different coefficient values in the model types. Predictions at the same covariate values should still be the same for both model types. This page goes into details and explains how to interpret coefficients in both cases. – EdM Jan 20 '23 at 22:13
  • Just curious, can we not all calculate 95CI for HR from weibull fit? Just looking for formula.we have formula for HR as above. – kamal Jan 21 '23 at 01:29
  • Given the different parameterizations of the Weibull, I'm not sure that your formula for the HR holds for the AFT form of the model. Please edit the question to include a reference to where you found that formula. – EdM Jan 21 '23 at 18:33

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