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ST 520 Statistical Principles of Clinical Trials - NCSU Statistics ...

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CHAPTER 9 <strong>ST</strong> <strong>520</strong>, A. TSIATIS and D. Zhang<br />

• γ > 0 implies that individuals on treatment 1 have worse survival (i.e. die faster)<br />

• γ = 0 implies the null hypothesis<br />

• γ < 0 implies that individuals on treatment 1 have better survival (i.e. live longer)<br />

If the proportional hazards assumption were true; that is,<br />

then this would imply that<br />

or<br />

Consequently,<br />

and<br />

d log S1(t)<br />

−<br />

dt<br />

λ1(t) = λ0(t) exp(γ),<br />

d log S0(t)<br />

= − exp(γ),<br />

dt<br />

− log S1(t) = − log S0(t) exp(γ).<br />

S1(t) = {S0(t)} exp(γ) ,<br />

log{− log S1(t)} = log{− log S0(t)} + γ.<br />

This last relationship can be useful if we want to assess whether a proportional hazards as-<br />

sumption is a reasonable representation <strong>of</strong> the data. By plotting the two treatment-specific<br />

Kaplan-Meier curves on a log{− log} scale we can visually inspect whether these two curves<br />

differ from each other by a constant over time.<br />

Also, in the special case where we feel comfortable in assuming that the survival distributions<br />

follow an exponential distribution; i.e. constant hazards, the proportional hazards assumption is<br />

guaranteed to hold. That is,<br />

λ1(t)<br />

λ0(t)<br />

λ1<br />

= .<br />

λ0<br />

In section 9.1 we showed that the median survival time for an exponential distribution with<br />

hazard λ is equal to m = log(2)/λ. Therefore, the ratio <strong>of</strong> the median survival times for two<br />

treatments whose survival distributions are exponentially distributed with hazard rates λ1 and<br />

λ0 is<br />

m1<br />

m0<br />

= {log(2)/λ1}<br />

{log(2)/λ0}<br />

PAGE 150<br />

λ0<br />

= .<br />

λ1

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