A Novel Sample Size Formula for the Weighted Log-Rank Test Under the Proportional Hazards Cure Model

Abstract

The treatment of cancer has progressed dramatically in recent decades, such that it is no longer uncommon to see a cure or log-term survival in a significant proportion of patients with various types of cancer. To adequately account for the cure fraction when designing clinical trials, the cure models should be used. In this article, a sample size formula for the weighted log-rank test is derived under the fixed alternative hypothesis for the proportional hazards cure models. Simulation showed that the proposed sample size formula provides an accurate estimation of sample size for designing clinical trials under the proportional hazards cure models.

1 Introduction

The treatment of cancer has progressed dramatically in recent decades, such that it is no longer uncommon to see a cure or long-term survival in a significant proportion of patients with various types of cancer, e.g., breast cancer, non-Hodgkin lymphoma, leukemia, prostate cancer, melanoma, and head and neck cancer [1]. To adequately account for cured patients in survival data from clinical trials, the cure models are increasingly useful. Various parametric and semiparametric cure models have been proposed by Farewell [2], Peng et al. [3] and Kuk and Chen [4], among others, and a maximum-likelihood EM algorithm for parametric and semiparametric cure models has been proposed by Peng and Dear [5] and Sy and Taylor [6]. A SAS macro PSPMCM developed by Corbiere and Joly [7] is available to fit both parametric and semiparametric cure models.

The traditional methods for designing survival trials may not be appropriate when there is a cure fraction. Sample size calculations have been developed for clinical trial designs under the cure models. For example, Halpern and Brown [8] developed a computer program to calculate the power and sample size for exponential cure models based on Monte Carlo simulation. Ewell and Ibrahim [1] provided a power formula for exponential cure models by considering a general alternative that allows for the effects of treatment on both short- and long-term survival. Recently, Wang et al. [9] considered a proportional hazards (PH) cure model, a special case of the general alternative proposed by Ewell and Ibrahim [1], and derived a sample size formula for the weighted log-rank test under a series of local alternatives. However, Wu [10] has pointed out that Wang’s formula does not provide adequate sample size or power for clinical trial designs. A series of local alternatives is that the alternative need to change along with the sample size such that the difference in parameter to be detected under the alternative need to diminish towards 0 in a rate of $1/\sqrt{n}$ as the sample size n getting large towards infinity. The fixed alternative is that the difference in parameter to be detected under the alternative is fixed and not depending on the sample size. A sample size established under a series of local alternatives works well only when the value of parameter under the alternative is very close to the one under the null hypothesis, whereas no such restriction for the fixed alternative.

In this paper, we derived a novel sample size formula for the weighted log-rank test under the PH cure model. The rest of the paper is organized as follows. The PH cure models are introduced in section 2. The sample size formula is presented in section 3. Simulations are conducted in section 4 to study the performance of the proposed sample size formula and compared with Wang’s formula. Section 5 illustrates clinical trial design using the proposed methods. The conclusion and additional remarks are presented in section 6.

2 Proportional Hazards Cure Models

The failure time, T^*, is assumed to be T^* = vT + (1 − v)∞, where v is an indicator of whether a subject will eventually (v = 1) or never (v = 0) experience treatment failure, and T denotes the failure time if the subject is not cured, with a survival distribution S(t), which is the conditional distribution for patients who will experience failure, and is often called the latency distribution. Thus, the overall survival distribution of T^* is a mixture model of a cure rate π = P(v = 0) and a latency distribution S(t) given by

$$S^{\ast }(t)=\pi +(1-\pi )S(t).$$

For a two-arm randomized survival trial, let $S_j^{\ast }(t)$ denote the overall survival function and let ${\lambda }_j^{\ast }(t)$ denote its corresponding hazard function for group j, where j = 0, 1 represents control group and treatment group, respectively. Similarly, let S_j(t) denote the survival function in uncured patients and let λ_j(t) denote its hazard function. The cure rate in group j is defined by π_j, where 0 ≤ π_j ≤ 1. For the mixture cure model, we have

$$S_j^{\ast }(t)={\pi }_j+(1-{\pi }_j)S_j(t)$$

and

$${\lambda }_j^{\ast }(t)=\frac{(1-{\pi }_j)S_j(t)}{{\pi }_j+(1-{\pi }_j)S_j(t)}{\lambda }_j(t),$$

for j = 0, 1. To derive the sample size calculation, we consider a class of PH cure models [1][9] in which

$${\lambda }_1(t)=e^{\eta }{\lambda }_0(t)\text{and}\text{logit}({\pi }_1)=\text{logit}({\pi }_0)+\gamma ,$$ (1)

where η is the log-hazard ratio of treatment vs. control for uncured patients and γ is the log-odds ratio of the cure rates for the two groups. Note that π₀ = π₁ = 0 corresponds to the standard PH model. However, if either π₀ ≠ 0 or π₁ ≠ 0, the PH cure model does not satisfy the proportional hazards condition. For a survival trial in which a proportion of patients are cured, we are interested in testing the following null hypothesis:

$$H_0:{\pi }_0={\pi }_1\text{and}{S}_0(t)=S_1(t),$$

which is equivalent to H₀: η = γ = 0. Various alternative hypotheses are of interest: H_1a : η ≠ 0, γ ≠ 0, with differences in both the short-term survival and the cure fraction; H_1b : η ≠ 0, γ = 0, with a difference in the short-term survival but not in the cure fraction; and H_1c : η = 0, γ ≠ 0, with difference in the cure fraction but not in the short-term survival.

3 Sample Size Formula

Consider a class of PH cure models as defined by the equations in (1). The sample size calculation is based on testing the null hypothesis

$$H_0:\eta =0\text{and}\gamma =0$$ (2)

against one of the three alternative hypotheses, H_1a, H_1b, and H_1c, defined in the previous section. The weighted or unweighted log-rank test can be used for testing this hypothesis. It is well known that the log-rank test is asymptotically normal distributed. To derive the asymptotic distribution of the log-rank test statistic under the alternatives hypotheses, suppose a survival trial involving n subjects, and let T_i and C_i denote the survival and censoring times of patient i, respectively, and Z_i = 0/1 denote the treatment group indicator (0 for control group and 1 for treatment group). The observed data then consist of {X_i, Δ_i, Z_i, i = 1, …, n}, where X_i = min(T_i, C_i) and Δ_i = I(T_i ≤ C_i). Let N_i(t) = Δ_iI(X_i ≤ t) and Y_i(t) = I(X_i ≥ t) be the failure and at-risk processes for i = 1, …, n. The weighted log-rank test statistic can then be written as

$$L={\displaystyle \sum _{i=1}^n{\displaystyle {\int }_0^{\tau }{w}_i\{Z_i-\overline{Z}(t)\}d{N}_i(t),}}$$

where τ is the study duration and $\overline{Z}(t)={\displaystyle {\sum }_{i=1}^n{Z}_i{Y}_i(t)}/{\displaystyle {\sum }_{i=1}^n{Y}_i(t)}\to \overline{z}(t)=E_{H_0}(Z|X\ge t)$ and w_i = W(t_i) with W(t) as a weight function that converges to w(t). Under the PH cure models (1), the hazard function of patient i is given by

$${\stackrel{\sim }{\lambda }}_i(t)=\frac{\frac{1-{\pi }_0}{1-{\pi }_0+{\pi }_0{e}^{\gamma Z_i}}{\left[S_0(t)\right]}^{e^{\eta Z_i}}}{\frac{{\pi }_0{e}^{\gamma Z_i}}{1-{\pi }_0+{\pi }_0{e}^{\gamma Z_i}}+\frac{1-{\pi }_0}{1-{\pi }_0+{\pi }_0{e}^{\gamma Z_i}}{\left[S_0(t)\right]}^{e^{\eta Z_i}}}{\lambda }_0(t)e^{\eta Z_i}.$$

Given a type I error of α and power of 1−β under a series of local alternatives, the sample size formula derived by Wang et al. [9] is given by

$$n=\frac{{\left(z_{1-\alpha /2}+z_{1-\beta }\right)}^2{\displaystyle {\int }_0^{\infty }{w}^2(t)G(t)S_0(t)d{\mathrm{\Lambda }}_0(t)}}{p(1-p)(1-{\pi }_0){\left\{{\displaystyle {\int }_0^{\infty }w(t)m(t)G(t)S_0(t)d{\mathrm{\Lambda }}_0(t)}\right\}}^2},$$ (3)

where p is the proportion of sample size allocation for the control group, G(t) is the common survival distribution of censoring time of two groups and $m(t)={\pi }_0\{\gamma +\eta {\mathrm{\Lambda }}_0(t)\}/S_0^{\ast }(t)-\eta$ with Λ₀(t) = − log S₀(t) and $S_0^{\ast }(t)={\pi }_0+(1-{\pi }_0)S_0(t)$.

There are two issues relating to the approach of sample size calculation based on a series of local alternatives. The theoretical issue is that the accuracy of the formula derived under the local alternative is not guaranteed when the alternative departures from the null. The practical issue is that the alternative hypothesis in application is always fixed, which does not change as the sample size changes. Thus, it is expected that the formula performances well only when the alternative is close to the null. Our simulation will show (see section 4) that this formula becomes practically unfitting when the alternative departs reasonably away from the null.

To overcome the inaccuracy of the formula (3), here we derive the asymptotic distribution of the log-rank test under the fixed alternative hypothesis by using a novel approach developed by Xiong [11] in which a theoretic development has showed that the log-rank test is not limited to the tests of proportional hazards model but can be used for more general survival models including the PH cure model. The details of the derivation are very complicated and tedious, and out of scope of this paper. Thus, we present the results in here and omit the details. It can be shown that $n^{-1/2}L-\sqrt{n}\xi (\gamma ,\eta )\to N(0,{\sigma }^2)$, as n → ∞, where

$${\sigma }^2=\frac{p(1-p)(1-{\pi }_0)}{(1-{\pi }_0+{\pi }_0{e}^{\gamma })}{\displaystyle {\int }_0^{\tau }{w}^2(t)q_1(t)G(t)S_0(t)d{\mathrm{\Lambda }}_0(t)},$$ (4)

$$\xi (\gamma ,\eta )=p(1-p)(1-{\pi }_0){\displaystyle {\int }_0^{\tau }w(t)q_2(t)G(t)S_0(t)d{\mathrm{\Lambda }}_0(t)},$$ (5)

where q₁(t) and q₂(t) are two functions given by equations (8) and (9) below. Thus, given a two-sided type I error of α, to achieve a power of 1 − β under the alternative, the total sample size n of the two groups must approximately satisfy the following equation:

$$1-\beta =\mathrm{\Phi }\left\{n^{1/2}\right|\xi (\gamma ,\eta )|/\sigma -z_{1-\alpha /2}\}.$$

The total sample size required for the study can then be determined as

$$n=\frac{{\left(z_{1-\alpha /2}+z_{1-\beta }\right)}^2{\sigma }^2}{{\left\{\xi (\gamma ,\eta )\right\}}^2}.$$ (6)

Substituting equations (4) and (5) into (6), the formula of the total sample size is given by

$$n=\frac{{\left(z_{1-\alpha /2}+z_{1-\beta }\right)}^2{\displaystyle {\int }_0^{\tau }{w}^2(t)q_1(t)G(t)S_0(t)d{\mathrm{\Lambda }}_0(t)}}{p(1-p)(1-{\pi }_0)(1-{\pi }_0+{\pi }_0{e}^{\gamma }){\left\{{\displaystyle {\int }_0^{\tau }w(t)q_2(t)G(t)S_0(t)d{\mathrm{\Lambda }}_0(t)}\right\}}^2},$$ (7)

where

$$q_1(t)=\frac{q(t)\{p(1-{\pi }_0+{\pi }_0{e}^{\gamma })+(1-p)\delta {[S_0(t)]}^{\delta -1}\}}{{\left[p+(1-p)q(t)\right]}^2},$$ (8)

$$q_2(t)=\frac{q(t)\{\delta {[S_0(t)]}^{\delta -1}{[q(t)(1-{\pi }_0+{\pi }_0{e}^{\gamma })]}^{-1}-1\}}{p+(1-p)q(t)},$$ (9)

where δ = e^η is the hazard ratio of group 1 vs. group 0 as defined by (1), and

$$q(t)=\frac{{\pi }_0{e}^{\gamma }+(1-{\pi }_0){\left[S_0(t)\right]}^{\delta }}{[{\pi }_0{e}^{\gamma }+(1-{\pi }_0)][{\pi }_0+(1-{\pi }_0)S_0(t)]}.$$

For the sample size calculation, assume subjects are accrued over an accrual period of duration t_a and an additional follow-up time t_f, that gives a total study duration of τ = t_a + t_f. For simplicity, we assume that the only censoring is administrative censoring at time τ, and that there is no loss to follow-up. The censoring distribution G(t) considered in the trial designs can then be uniform (the distribution of enrollment H(t) with density of h(t) = 1/t_a on [0, t_a]). This leads to the censoring distribution G(t) = H(τ − t) = 1 if t ≤ t_f; = (t_a + t_f − t)/t_a if t_f ≤ t ≤ t_a + t_f; = 0 otherwise. Then the integrations in the sample size formula (6) can be calculated by numeric integrations, for example by using the R function integrate.

4 Simulation

We conducted simulation studies to investigate three important issues: 1) whether the proposed sample size formula provides an accurate estimation of sample size under the PH cure model; 2) what is the relative efficiency of the weighted log-rank test vs. the standard log-rank test; and 3) how is the accuracy of the proposed formula compared with the formula derived by Wang et al. [9].

To answer the first question, we assumed a PH cure model with a latency distribution of Weibull $S\left(t\right)=e^{-\mathrm{\lambda }{t}^{\kappa }}$. The hazard rates of uncured patients and the cure rates were varied to examine three scenarios: (1) where there were differences in both hazard rates and cure rates, that is, η ≠ 0, γ ≠ 0; (2) where there were differences only between hazard rates, that is, η ≠ 0, γ ≠ 0; and (3) where there were differences only between cure rates, that is, η = 0, γ ≠ 0. The hazard rate of control was set to λ = 0.1 and shape parameter was set to κ = 0.5, 1 and 2, and the cure rate of control was set to π₀ = 0.1. The hazard ratio δ = e^η under the alternative was set in a range of 1.4⁻¹ to 2⁻¹, and the log-odds ratio of cure rate γ was set in a range of 0 to 1.6. It was assumed in simulations that the treatment group would have a lower hazard rate and a higher cure rate than the control group if there were differences. In addition, we considered a uniform accrual with accrual period t_a of 1 year and a follow-up period t_f of 10 years. Under each scenario, we calculated sample sizes under equal allocation (p = 0.5) for the standard log-rank test (w(t) = 1) under various parameter configurations, and the corresponding empirical powers via simulations with 10,000 replicates. The results (Table 1) showed that the empirical powers were almost identical to the nominal power of 90% for all three scenarios under all parameter configurations. Therefore, we can conclude that the formula (7) provided accurate sample size estimations.

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

To investigate the relative efficiency of the weighted log-rank test vs. the standard log-rank test, we consider a class of Harrington-Fleming ${\mathcal{G}}^{\rho ,v}$ weight functions $W\left(t\right)={\left\{\widehat{S}\left(t^{-}\right)\right\}}^{\rho }{\left\{1-\widehat{S}\left(t^{-}\right)\right\}}^{\nu }$, where $\widehat{S}\left(t^{-}\right)$ is the left-continues version of the Kaplan-Meier estimate computed from the pooled sample of two groups [12]. Sample sizes were calculated for both weighted log-rank test and standard log-rank test under same scenarios as above. The results (Table 2) showed: a) the weighted log-rank test was not as efficient as the standard log-rank test (with weight function ${\mathcal{G}}^{0,0}$) for scenarios 1 and 2; b) the weighted log-rank test with weight function ${\mathcal{G}}^{-1,0}$ was more efficient than the standard log-rank test for scenario 3. In fact, the weight function ${\mathcal{G}}^{-1,0}$ is the optimal weight function in scenario 3 as shown by Gray and Tsiatis [13] and Wu [14].

Table 2.

Comparison efficiency of the weighted log-rank test versus the standard log-rank test based on calculated sample sizes under the exponential cure model with a nominal type I error of 0.05 and power of 90% (two-sided test, uniform accrual and equal allocation).

		Weight function
Design	δ−1/γ	${\mathcal{G}}^{0,0}$	${\mathcal{G}}^{0,1}$	${\mathcal{G}}^{1,0}$	${\mathcal{G}}^{1,1}$	${\mathcal{G}}^{-1,0}$	${\mathcal{G}}^{-1,1}$
Scenario 1: η ≠ 0 γ ≠ 0
π0 = 0.1	1.2/0.4	1385	1810	1460	1620	1446	2101
λ0 = 0.1	1.3/0.5	734	966	769	866	765	1115
ta = 1	1.4/0.6	469	617	490	556	487	708
tf = 10	1.5/0.7	333	438	347	395	345	499
	1.6/0.8	253	332	264	301	261	376
	1.7/0.9	202	263	210	240	207	296
	1.8/1.0	166	216	173	197	170	242
Scenario 2: η ≠ 0 γ = 0
π0 = 0.1	1.4/0.0	801	1130	819	990	858	1337
λ0 = 0.1	1.5/0.0	562	788	574	694	598	925
ta = 1	1.6/0.0	425	594	435	525	451	692
tf = 10	1.7/0.0	340	472	348	419	359	547
	1.8/0.0	282	389	288	347	296	449
	1.9/0.0	240	330	246	295	251	379
	2.0/0.0	209	287	214	257	218	327
Scenario 3: η = 0 γ ≠ 0
π0 = 0.1	1.0/1.0	1489	1554	1720	1495	1414	1687
λ0 = 0.1	1.0/1.1	1148	1202	1321	1156	1092	1304
ta = 1	1.0/1.2	902	948	1033	910	859	1027
tf = 10	1.0/1.3	720	759	821	728	687	822
	1.0/1.4	583	617	662	591	557	667
	1.0/1.5	478	507	540	486	457	548
	1.0/1.6	396	422	446	404	379	455

Finally, the accuracy of the proposed formula (7) was compared to Wang’s formula. We calculated the sample sizes and the empirical powers for both formulae (Table 3) under the exponential cure model with three scenarios similar to the Table 1 for a relative longer follow-up time (similar results were obtained for a relative shorter follow-up time with t_f = 4 and results are not presented in the paper). The proposed formula (7) gave the accurate estimation of sample sizes for all the scenarios, while Wang’s formula could either overestimate or underestimate the sample sizes when the hazard ratio δ departures from 1. Thus, the formula derived by Wang et al. failed to provide the correct sample size estimation and should not be used for the trial designs in practice.

Table 3.

Comparison sample sizes calculated the exponential cure model using Wang’s formula and the new formula (6) with a nominal type I error of 0.05 and power of 90%: (two-sided test, uniform accrual and equal allocation).

		Wang		Formula (6)				Wang		Formula (6)
Design	δ−1/γ	nW	EPW	nX	EPX	Design	δ−1/γ	nW	EPW	nX	EPX
Scenario 1: η ≠ 0 γ ≠ 0
π0 = 0.4	1.2/0.4	927	0.887	972	0.903	π0 = 0.1	1.2/0.4	1377	0.893	1385	0.906
λ0 = 0.1	1.3/0.5	527	0.881	570	0.906	λ0 = 0.1	1.3/0.5	714	0.894	734	0.899
ta = 1	1.4/0.6	345	0.875	383	0.903	ta = 1	1.4/0.6	448	0.884	469	0.899
tf = 10	1.5/0.7	246	0.857	284	0.903	tf = 10	1.5/0.7	313	0.878	333	0.904
	1.6/0.8	186	0.853	221	0.904		1.6/0.8	234	0.886	253	0.903
	1.7/0.9	147	0.849	180	0.908		1.7/0.9	184	0.877	202	0.900
	1.8/1.0	119	0.834	152	0.912		1.8/1.0	150	0.870	166	0.902
Scenario 2: η ≠ 0 γ = 0
π0 = 0.4	1.4/0.0	1657	0.895	1668	0.902	π0 = 0.1	1.4/0.0	745	0.885	801	0.904
λ0 = 0.1	1.5/0.0	1141	0.903	1156	0.904	λ0 = 0.1	1.5/0.0	513	0.875	562	0.900
ta = 1	1.6/0.0	850	0.892	867	0.904	ta = 1	1.6/0.0	382	0.871	425	0.903
tf = 10	1.7/0.0	667	0.894	686	0.904	tf = 10	1.7/0.0	300	0.869	340	0.904
	1.8/0.0	543	0.888	565	0.906		1.8/0.0	245	0.853	282	0.901
	1.9/0.0	456	0.890	478	0.909		1.9/0.0	205	0.854	240	0.900
	2.0/0.0	391	0.889	414	0.910		2.0/0.0	176	0.855	209	0.907
Scenario 3: η = 0 γ ≠ 0
π0 = 0.4	1.0/1.0	420	0.941	360	0.906	π0 = 0.3	1.0/1.0	255	0.952	209	0.904
λ0 = 0.1	1.0/1.1	347	0.943	297	0.902	λ0 = 0.3	1.0/1.1	211	0.953	171	0.904
ta = 1	1.0/1.2	292	0.942	250	0.905	ta = 1	1.0/1.2	177	0.948	143	0.905
tf = 10	1.0/1.3	249	0.942	214	0.910	tf = 10	1.0/1.3	151	0.959	121	0.908
	1.0/1.4	214	0.944	186	0.910		1.0/1.4	130	0.956	105	0.910
	1.0/1.5	187	0.945	164	0.912		1.0/1.5	114	0.956	91	0.910
	1.0/1.6	164	0.940	146	0.912		1.0/1.6	100	0.954	81	0.908

abbreviation n_W and EP_W : sample size and empirical power calculated by formula derived by Wang at al.; abbreviation n_X and EP_X: sample size and empirical power calculated by formula (6).

5 Example

We illustrate the application of the proposed sample size formula with the use of data from the Eastern Cooperative Oncology Group (ECOG) trial e1684 [7]. The ECOG trial e1684 was a randomized two-arm phase III trial comparing an arm treated with high-dose interferon alpha-2b with an observation arm. The primary endpoint was relapse-free survival (RFS), with RFS defined as the time interval from the date of randomization to the date of disease relapse or death. The trial was originally designed to detect a 50% improvement in median RFS from 1.5 to 2.25 years, with an accrual period of 4 years and a follow-up period of 3 years. A sample size of 286 patients was considered, which was expected to yield a power of 83% based on the standard PH model [9].

Let us now design a randomized two-arm phase III trial similar to the ECOG trial by using the treatment arm of the ECOG trial as the preliminary data for the control arm of the new trial design. As shown by Kaplan-Meier estimate of this survival data in Fig. 1A, the relapse events had occurred almost within 4.3 years after the start of treatment, and approximately 35% of patients were cured (in the sense that they survived throughout the study period). We fitted survival data of all patients with the Weibull distribution, as shown in Fig. 1B. The fitted parametric survival model obviously did not match well the survival data, especially as the parametric model failed to present the later plateau of the empirical survival curve. The Kaplan-Meier estimate in Fig. 1A suggests that the patient cohort consisted of two parts: patients who would probably experience relapse within 4.3 years after the start of treatment and patients who would be cured in the sense that they would relapse-free for 7 years after the start of treatment. Thus, a cure model is appropriate. The SAS macro PSPMCM was applied to this data to fit the treatment arm data under the Weibull cure model, with an estimated shape parameter κ = 1.018, scale parameter λ = 0.836, and cure rate of 35%. Then, we have the Weibull cure model $S_0^{\ast }\left(t\right)={\pi }_0+\left(1-{\pi }_0\right)S_0\left(t\right)$ for the control group which matches well with the Kaplan-Meier survival curve, where π₀ = 0.35 and S₀(t) = exp(−0.836t^1.018) (Fig. 1C and 1D).

Figure 1. KM curves (step function) and fitted Weibull models (solid curve) for the treatment arm of ECOG trial e1684 data.

Fig. 1A: KM curve for all patients; Fig. 1B: KM curve and Weibull model for all patients; Fig. 1C: KM curve and Weibull model for failed patients; Fig. 1D: KM curve and Weibull cure model for all patients.

For the trial design under the PH cure model, assume the alternatives for the three scenarios are as follows: 1) π₁ > π₀ and δ < 1: to detect a hazard ratio δ = 1.5⁻¹ (δ = e^η) and a 10% increase in cure rate (i.e., π₁ = 0.45); 2) π₁ = π₀ and δ < 1: to detect a hazard ratio δ = 2.0⁻¹ and identical 35% cure rates for the two groups; and 3) π₁ > π₀ and δ = 1: to detect a 15% increase in cure rate (i.e., π₁ = 0.50) and keep the hazard rates the same for the two groups (i.e., δ = 1). With equal allocation (p = 0.5) and uniform accrual with t_a = 4, t_f = 3, η = log(δ), and $\gamma =\text{log}\left(\frac{{\pi }_1\left(1-{\pi }_0\right)}{\left(1-{\pi }_1\right){\pi }_0}\right)$, we can calculate the sample sizes by using the sample size formula (7). To achieve a power of 90% with a two-sided type I error of 0.05 under the Weibull PH cure model, sample sizes of 468, 762, and 505 are required for scenarios 1, 2, and 3, respectively, and corresponding empirical powers based on 10,000 simulation runs are 0.902, 0.903 and 0.904, respectively. The corresponding sample sizes calculated by Wang’s formula are 473, 1135, and 557 for scenarios 1, 2, and 3, respectively, and corresponding empirical powers based on 10,000 simulation runs are 0.902, 0.977 and 0.926, respectively. The R code for the sample size calculation is given in Appendix.

6 Conclusion

In this paper, we have derived a new sample size formula for the log-rank test under the PH cure model. The simulations have shown that the proposed formula provides an accurate estimation of sample size under the PH cure models and corrects the sample size calculation proposed by Wang et al. [9]. The efficiency of a class of Harrington-Fleming ${\mathcal{G}}^{\rho ,v}$ weighted log-rank test is explored. For both scenarios 1 and 2, the ${\mathcal{G}}^{\rho ,v}$ weighted log-rank test may not be as efficient as the standard log-rank test. However, the weighted log-rank test with ${\mathcal{G}}^{-1,0}$ weight function is more efficient than the standard log-rank test for the scenario 3. It is interested to investigate the optimal weight function for the scenarios 2 and 3. This will be a future research topic.

Appendix: R code for the sample size calculations of the example

Size=function(kappa, lambda0, pi0, pi1, p, ta, tf, HR, alpha, power)
{z0=qnorm(1−alpha/2); z1=qnorm(power)
 tau=ta+tf; delta=1/HR
 gamma=log(pi1/(1−pi1))−log(pi0/(1−pi0))
 q=function(t){num=pi0*exp(gamma)+(1−pi0)*S0(t)^delta
   den=(pi0*exp(gamma)+(1−pi0))*(pi0+(1−pi0)*S0(t))
   ans=num/den; return(ans)}
 S0=function(t){exp(−lambda0*t^kappa)}
 h0=function(t){kappa*lambda0*t^(kappa−1)}
 G=function(t){1−punif(t, tf, tau)}
 q1=function(t){den=(p+(1−p)*q(t))^2
   num=q(t)*(p*(1−pi0+pi0*exp(gamma))+(1−p)*delta*S0(t)^(delta−1))
   ans=num/den; return(ans)}
 q2=function(t){den=p+(1−p)*q(t)
   num=q(t)*(delta*S0(t)^(delta−1)/(q(t)*(1−pi0+pi0*exp(gamma)))−1)
   ans=num/den; return(ans)}
 f1=function(t){q1(t)*G(t)*S0(t)*h0(t)}
 f2=function(t){q2(t)*G(t)*S0(t)*h0(t)}
 A=integrate(f1, 0, tau)$value
 B=integrate(f2, 0, tau)$value
 nX=(z0+z1)^2*A/(p*(1−p)*(1−pi0)*(1−pi0+pi0*exp(gamma))*B^2)
 m=function(t){pi0*(gamma−log(delta)*log(S0(t)))/(pi0+(1−pi0)*S0(t))−log(delta)}
 g1=function(t){G(t)*S0(t)*h0(t)}
 g2=function(t){m(t)*G(t)*S0(t)*h0(t)}
 C=integrate(g1, 0, tau)$value
 D=integrate(g2, 0, tau)$value
 nW=(z0+z1)^2*C/(p*(1−p)*(1−pi0)*D^2)
 ans=ceiling(c(nX, nW)); return(ans)}
Size(kappa=1.018,lambda0=0.836,pi0=0.35,pi1=0.45,p=0.5,ta=4,tf=3, HR=1.5,alpha=0.05,power=0.90)
468 473
Size(kappa=1.018,lambda0=0.836,pi0=0.35,pi1=0.35,p=0.5,ta=4,tf=3, HR=2.0,alpha=0.05,power=0.90)
762 1135
Size(kappa=1.018,lambda0=0.836,pi0=0.35,pi1=0.50,p=0.5,ta=4,tf=3, HR=1.0,alpha=0.05,power=0.90)
505 557