Cox model with a delayed treatment effect in a melanoma trial
A randomised trial enrols 480 patients with advanced melanoma; 240 receive immunotherapy and 240 receive standard care. After median follow-up of 3 years there are 210 deaths. A Cox model adjusted for age and PD-L1 status gives a treatment hazard ratio of 0.78 with 95% confidence interval 0.59 to 1.03. Age HR is 1.02 per year (1.01 to 1.04) and PD-L1 HR is 0.65 (0.51 to 0.83). A Schoenfeld residual test for treatment gives p = 0.008, and a plot of the estimated treatment effect over time crosses the null at about 12 months. Kaplan-Meier curves are almost identical for the first 8 months and separate thereafter. The trial statistician must report the treatment effect and decide how to model it. A Weibull accelerated failure time model fitted to the immunotherapy arm gives shape 1.3 and scale 40 months.
1. Interpret the treatment hazard ratio and its confidence interval. Explain why the Schoenfeld test changes the interpretation.
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The adjusted HR of 0.78 suggests that, at any given time after adjustment for age and PD-L1, immunotherapy patients have 22% lower instantaneous hazard of death than standard care. The 95% CI 0.59 to 1.03 includes 1, so at the 5% level the average effect is not statistically significant. Age and PD-L1 are significant: older age raises the hazard slightly, while higher PD-L1 lowers it.
The Schoenfeld test rejects proportional hazards for treatment (p = 0.008). That means the single HR is not constant over follow-up; it is an average of a time-varying effect. The plot crossing 1 at about 12 months, together with Kaplan-Meier curves that separate only after 8 months, suggests a delayed benefit: little or no effect early, then a lower hazard later. Reporting only 0.78 would mislead clinicians about the early period and may understate the later benefit. The average HR is still a useful summary if labelled as such, but the primary analysis should model the time-varying effect.
2. Propose and specify a time-varying coefficient model or an alternative that addresses the non-proportionality. What would you report?
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Fit a Cox model with a treatment by log-time interaction: h(t|x) = h0(t) exp(beta1 trt + beta2 trt log(t) + other covariates). Then HR(t) = exp(beta1 + beta2 log t). If beta2 is negative, the treatment HR falls over time, consistent with a delayed effect. Test beta2 = 0 and check the shape with a spline in log time if the linear interaction is inadequate. An alternative is a piecewise model with separate treatment effects for 0-12 months and after 12 months, or a flexible parametric model with a time-varying treatment coefficient.
Report HR(t) at clinically relevant times, for example 6, 12 and 24 months, with confidence intervals, and state that the proportional hazards assumption was not met. Also report the Kaplan-Meier curves and the restricted mean survival difference, which does not rely on proportional hazards. If the early effect is truly null, the trial may be underpowered for an overall average; the late effect needs careful interpretation because it is estimated from fewer patients at risk.
3. Using the Weibull AFT output, calculate the predicted 24-month survival and median survival for the immunotherapy arm, and explain how you would obtain the corresponding quantities for standard care without assuming proportional hazards.
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For a Weibull AFT model, S(t) = exp(-(t/lambda)^k) with k = 1.3 and lambda = 40 months. At t = 24, (24/40)^1.3 = 0.6^1.3 = exp(1.3 ln 0.6) = exp(-0.664) = 0.515. Hence S(24) = exp(-0.515) = 0.598, about 60%. The median solves (t/40)^1.3 = ln 2 = 0.693, so t = 40(0.693)^(1/1.3) = 40(0.755) = 30.2 months.
For standard care, do not simply apply the Cox HR of 0.78 if proportional hazards has been rejected. Fit a separate Weibull model to the standard-care arm and compare shapes; if the shapes are similar, the AFT assumption may hold within each arm even when the Cox PH assumption fails. If shapes differ, report arm-specific survival and medians. If a common shape is plausible, the treatment HR implies lambda_standard = lambda_treatment times HR^(1/k) = 40 times 0.78^(1/1.3) = 33.0 months, giving S_standard(24) = exp(-(24/33)^1.3) = 0.516. Check this against the separate fit before using it.