Table 1.

Sample sizes were calculated under the Weibull cure model for the standard log-rank test with a nominal level of 0.05 and power of 90% (two-sided test, uniform accrual and equal allocation). The corresponding empirical type I errors and powers were estimated based on 10,000 simulation runs.

	δ−1/γ	n	$\widehat{\alpha }$	EP	n	$\widehat{\alpha }$	EP	n	$\widehat{\alpha }$	EP
Design		κ = 0.5			κ = 1			κ = 2
Scenario 1: η ≠ 0 γ ≠ 0
π0 = 0.1	1.2/0.4	3445	.050	.903	1385	.050	.902	1075	.047	.897
λ0 = 0.1	1.3/0.5	1818	.047	.902	734	.051	.901	599	.052	.902
ta = 1	1.4/0.6	1165	.052	.904	469	.053	.898	389	.052	.901
tf = 10	1.5/0.7	833	.048	.902	333	.052	.897	276	.049	.905
	1.6/0.8	638	.053	.899	253	.051	.905	208	.048	.897
	1.7/0.9	512	.053	.900	202	.051	.906	164	.050	.900
	1.8/1.0	426	.049	.909	166	.052	.900	133	.053	.903
Scenario 2: η ≠ 0 γ = 0
π0 = 0.1	1.4/0	1783	.049	.900	801	.048	.900	1335	.048	.902
π1 = 0.1	1.5/0	1266	.044	.904	562	.052	.902	927	.052	.900
λ0 = 0.1	1.6/0	970	.049	.903	425	.048	.903	696	.049	.900
ta = 1	1.7/0	783	.048	.907	340	.049	.904	551	.051	.902
tf = 10	1.8/0	655	.053	.904	281	.052	.897	454	.050	.901
	1.9/0	563	.052	.911	240	.053	.904	385	.054	.896
	2.0/0	495	.051	.904	209	.050	.908	333	.052	.905
Scenario 3: η = 0 γ ≠ 0
π0 = 0.1	1/1.0	5627	.049	.900	1489	.050	.899	427	.052	.905
λ0 = 0.1	1/1.1	4306	.049	.904	1148	.050	.904	338	.047	.902
λ1 = 0.1	1/1.2	3356	.047	.897	902	.052	.900	272	.052	.901
ta = 1	1/1.3	2657	.052	.903	720	.049	.897	222	.050	.899
tf = 10	1/1.4	2133	.050	.900	583	.047	.903	184	.051	.908
	1/1.5	1734	.053	.895	478	.050	.902	154	.049	.902
	1/1.6	1425	.046	.908	396	.050	.898	131	.051	.911