Solution: Log-rank test for the Leukemia data

We'd like to test at significance level \(0.05\) if the relapse rate between the placebo group (index \(P\)) and the 6-MP group (index \(M\)) are different (at any point) thus our null hypothesis is that the hazard rates are equal throughout the entire experiment, i.e. \[H_0: \alpha_P(t) = \alpha_M(t)\] for all \(t \in [0,t_0]\). Note that \(dN(t)\) are the increments of \(N(t)\), and hence equal to the number of events at time \(t\), meaning that the integrals in \((3.52)\) and \((3.55)\) simplify to sums over the event times \(T_j\) weighted by the number of events at those times, \(d_j\). Let \(T_{Pj}\) be the event times for the placebo group and \(d_{Pj}\) the corresponding number of events, and likewise \(T_{Mj}\) and \(d_{Mj}\) for the 6-MP group. With \(L(t) = \frac{Y_P(t)Y_M(t)}{Y(t)}\) we thus get \begin{align*}Z_1(t_0) &= \int_0^{t_0}\frac{L(t)}{Y_P(t)}dN_P(t) - \int_0^{t_0}\frac{L(t)}{Y_M(t)}dN_M(t) \\ &=\int_0^{t_0}\frac{Y_M(t)}{Y(t)}dN_P(t) - \int_0^{t_0}\frac{Y_P(t)}{Y(t)}dN_M(t) \\ &= \sum_{T_{Pj} \leq t_0} \frac{d_{Pj}Y_M(T_{Pj})}{Y(T_{Pj})} - \sum_{T_{Mj} \leq t_0} \frac{d_{Mj}Y_P(T_{Mj})}{Y(T_{Mj})},\end{align*} which with our data yields \(Z_1(t_0) = 10.2505\). Similarly we get \begin{align*}V_{11}(t_0) &= \int_0^{t_0}\frac{L(t)^2}{Y_P(t)Y_M(t)}dN(t) \\ &= \int_0^{t_0}\frac{Y_P(t)Y_M(t)}{Y(t)^2}dN(t) \\ &= \sum_{T_{j} \leq t_0} \frac{d_{j}Y_P(T_{j})Y_M(T_{j})}{Y(T_{j})^2} ,\end{align*} which yields \(V_{11}(t_0) = 6.595682\). The test statistic in \((3.57)\) then becomes: \[X^2(t_0) = \frac{Z_1(t_0)^2}{V_{11}(t_0)} = 10.2505^2/6.595682 = 15.93054,\] which in turn gives a p-value of \(0.000066\) and hence indicates that there is a significant difference in the relapse rate for the two groups.