On Sample Size Calculation for Comparing Survival Curves under General Hypothesis Testing

SUMMARY

The log-rank test is commonly used to test the equivalence of two survival distributions under right censoring. Jung et al. (2005) proposed a modified log-rank test for noninferiority trials and its corresponding sample size calculation. In this paper, we extend the use of the modified log-rank test for clinical trials with various types of non-conventional study objectives and propose its sample size calculation under general null and alternative hypotheses. The proposed formula is so flexible that we can specify any survival distributions and accrual pattern. The proposed methods are illustrated with designing real clinical trials. Through simulations, the modified log-rank test and the derived formula for sample size calculation are shown to have satisfactory small sample performance.

1 INTRODUCTION

In clinical trials, it is often of interest to compare the survival distribution of an experimental therapy, denoted by arm 2, with that of a standard (or control) therapy, denoted by arm 1. In practice, a proportional hazards model is assumed. Let Λ_k(t) denote the cumulative hazard function of arm k(= 1, 2) and Δ = Λ₂(t)/Λ₁(t) the hazard ratio. By the standard log-rank test (Peto and Peto, 1972), we usually test a null hypothesis that two arms have identical survival distributions against an alternative hypothesis that the experimental arm has a longer survival, i.e. H₀ : Δ = 1 against H₁ : Δ < 1 using the standard log-rank test. We will refer to such trials as superiority trials. Sample size for a superiority trial can be calculated under the alternative hypothesis that H₁ : Δ = Δ₁ for a pre-specified clinically significant hazard ratio Δ₁(< 1) (see e.g., George and Desu, 1973; Schoenfeld, 1983; Lakatos, 1988; Yateman and Skene, 1992).

In many cases, the experimental therapy may be less extensive, less toxic or less expensive than the standard therapy. In this case, the former may be acceptable so long as there is evidence that it is not worse than the latter. Thus, we may want to prove that an experimental therapy is stochastically equal or not inferior to the standard therapy. That is, with a given non-inferiority margin (the maximal hazard ratio of clinical insignificance) Δ₀(> 1), H₀ : Δ ≥ Δ₀ against H₁ : Δ < Δ₀ using the non-inferiority log-rank test, e.g., Jung et al. (2005). In a non-inferiority trial, the experimental therapy can be hardly as efficacious as the control, so that we often want to show that the survival distributions are identical between two arms. Thus, we calculate the sample size under H₁ : Δ = 1, see e.g., Chow, Shao and Wang (2003) and Jung et al. (2005).

Generalizing superiority and non-inferiority trials, we consider calculating the sample size for H₀ : Δ = Δ₀ against H₁ : Δ = Δ₁ with Δ₀ ≥ Δ₁. We have Δ₀ = 1 for superiority trials and Δ₁ = 1 for non-inferiority trials. In general, a non-inferiority margin is so small (e.g., Δ₀ ≈ 1.1 contrary to a usual superiority margin Δ₁ ≈ 1.5), so that the required sample size becomes very big. However, possibly due to lower adverse event, a moderate chemotherapy may be able to extend the survival of cancer patients compared to a standard aggressive chemotherapy. In this case, by using Δ₀ larger than 1 and Δ₁ smaller than 1, we can lower the sample size, and the effect size Δ₀/Δ₁ may be close to that of a superiority trial.

For illustration, we consider an example concerning a randomized phase II trial for patients with locally advanced nasopharyngeal carcinoma. A high dose (arm 1) chemotherapy A concurrent with radiotherapy is a standard treatment for the patient population. However, the high dose chemotherapy is not well tolerated by most patients. So, we want to investigate a low dose (arm 2) concurrent chemotherapy A administered more frequently (weekly vs. every 3 weeks). It is known that the high dose chemotherapy has a 3-year progression-free survival (PFS) of S₁(3) = 0.75, corresponding to a hazard rate of λ₁ = 0.096 under an exponential model. In the new randomized phase II trial, we will not be interested in the low dose chemotherapy if its 3-year PFS is S₂(3) = 0.65 or lower, corresponding to Δ₀ = 1.4974, and will be very much interested in it if its 3-year PFS is S₂(3) = 0.8 or higher, corresponding to Δ₁ = 0.7757 under an exponential model. It is believed that the low expected adverse events and more frequent administration will possibly extend the PFS for arm 2 slightly beyond that of arm 1.

Note that the cumulative hazard function for the control arm Λ₁(t) is identical under both H₀ and H₁. Assuming the proportional hazards model, the cumulative hazard function for the experimental arm is Λ₂(t) = Δ₀Λ₁(t) under H₀ and Λ₂(t) = Δ₁Λ₁(t) under H₁. Hence, the survival distributions for a sample size calculation can be specified by (Λ₁(t), Δ₀, Δ₁).

In the next section, a generalized log-rank test is derived under general null and alternative hypotheses. A formula for sample size calculation is proposed in Section 3. The practical application of the proposed sample size calculation method is presented in Section 4. A brief discussion is given in the last section.

2 THE GENERALIZED LOG-RANK TEST

Let n_k denote the sample size in arm k and T_ki the survival time for subject i in arm k (1 ≤ i ≤ n_k; k = 1, 2). Then we usually observe (X_ki, δ_ki), where X_ki is the minimum of T_ki and the censoring time and δ_ki is an event indicator taking 1 if the subject had an event and 0 otherwise.

For arm k, T_k1, …, T_k,nk are IID with hazard function λ_k(t). Under the proportional hazards assumption, Δ = λ₂(t)λ₁(t) denotes the hazard ratio. By Cox (1972), the partial score function W (Δ) and the information function ${\sigma }_n^2\left(\Delta \right)$ are given as

$$W\left(\Delta \right)={\int }_0^{\infty }\frac{Y_1\left(t\right)Y_2\left(t\right)}{Y_1\left(t\right)+\Delta Y_2\left(t\right)}\{\Delta d{\widehat{\Lambda }}_1\left(t\right)-d{\widehat{\Lambda }}_2\left(t\right)\},$$

and

$${\sigma }_n^2\left(\Delta \right)=\Delta {\int }_0^{\infty }\frac{Y_1\left(t\right)Y_2\left(t\right)}{{\{Y_1\left(t\right)+\Delta Y_2\left(t\right)\}}^2}dN\left(t\right),$$

where ${\widehat{\Lambda }}_k\left(t\right)={\int }_0^t{Y}_k^{-1}\left(t\right)d{N}_k\left(t\right)$ is the Aalen-Nelson estimator (Aalen, 1978; Nelson, 1969) for the cumulative hazard function ${\Lambda }_k\left(t\right)={\int }_0^t{\lambda }_k\left(s\right)dS$, $Y_k\left(t\right)={\sum }_{i=1}^{n_k}I(X_{ki}\ge t)$ and $N_k\left(t\right)={\sum }_{i=1}^{n_k}{\delta }_{ki}I(X_{ki}\le t)$ are the at-risk process and the event process for group k, respectively, N(t) = N₁(t) + N₂(t), and I(·) is the indicator function. Note that W (1) is the standard log-rank test statistic (Peto and Peto, 1972). We call W (Δ) the generalized log-rank test in this paper.

As n → ∞and n_k/n → p_k ∈ (0, 1), W (Δ₀)/σ(Δ₀) converges to the standard normal distribution under H₀ : Δ = Δ₀, see e.g. Fleming and Harrington (1991). Let z_1−α denote the 100(1 − α) percentile for the standard normal distribution. Then we reject H₀, in favor of H₁ : Δ < Δ₀, if W (Δ₀)/σ_n(Δ₀) > z_1−α with one-sided type I error rate α.

The partial MLE $\widehat{\Delta }$ is obtained by solving W (Δ) = 0. The MLE is a consistent estimator of the true hazard ratio. Furthermore, by the asymptotic linearity, the score-type test statistic W (Δ₀)/σ_n(Δ₀) is asymptotically equivalent to the Wald-type test statistic ${\sigma }_n\left({\Delta }_0\right)(\widehat{\Delta }-{\Delta }_0)$, so that our sample size formula derived in the following section is valid for both types of test statistics.

3 SAMPLE SIZE FORMULA

We want to estimate the sample size n under a specific alternative hypothesis that H₁ : Δ = Δ₁(<Δ₀) with a desired power. Jung et al. (2005) proposed a sample size formula with Δ₀ > 1 and Δ₁ = 1 for designing non-inferiority trials. This paper is to extend their formula for general Δ₀ and Δ₁ with Δ₁ < Δ₀. Let p_k = n_k/n denote the allocation proportion for arm k and n = n₁ + n₂.

The asymptotic results in this section are derived under H₁. Let S_k(t) and f_k(t) = −∂S_k(t)∂t denote the survival and probability density functions, respectively, for arm k under H₁. Note that S₂(t) = S₁(t)^Δ₁ and f₂(t) = Δ₁f₁(t)S₁(t)^Δ₁−1 under H₁. For a censoring variable C, let G(t) = P (C ≥ t) denote the survivor function of the censoring distribution which is common in two arms. By Jung et al. (2005), ${\sigma }_n^2\left(\Delta \right)$ is asymptotically equivalent to nσ²(Δ), where

$${\sigma }^2\left(\Delta \right)=\Delta p_1{p}_2{\int }_0^{\infty }\frac{G\left(t\right)S_1\left(t\right)S_2\left(t\right)\{p_1{f}_1\left(t\right)+p_2{f}_2\left(t\right)\}}{{\{p_1{S}_1\left(t\right)+\Delta p_2{S}_2\left(t\right)\}}^2}dt.$$ (1)

By the definition of W (Δ), we have

$$W\left({\Delta }_0\right)-W\left({\Delta }_1\right)=n^{-1∕2}({\Delta }_0-{\Delta }_1){\int }_0^{\infty }\frac{Y_1\left(t\right)Y_2\left(t\right)}{\{Y_1\left(t\right)+{\Delta }_0{Y}_2\left(t\right)\}\{Y_1\left(t\right)+{\Delta }_1{Y}_2\left(t\right)\}}dN\left(t\right),$$

which is asymptotically equivalent to nω, where

$$\omega =({\Delta }_0-{\Delta }_1)p_1{p}_2{\int }_0^{\infty }\frac{G\left(t\right)S_1\left(t\right)S_2\left(t\right)\{p_1{f}_1\left(t\right)+p_2{f}_2\left(t\right)\}}{\{p_1{S}_1\left(t\right)+{\Delta }_0{p}_2{S}_2\left(t\right)\}\{p_1{S}_1\left(t\right)+{\Delta }_1{p}_2{S}_2\left(t\right)\}}dt.$$ (2)

The integrals in (1) and (2) are calculated using a numerical method.

Under H₁, the generalized log-rank test statistic can be expressed as

$$\frac{W\left({\Delta }_0\right)}{{\sigma }_n\left({\Delta }_0\right)}=\frac{W\left({\Delta }_1\right)}{{\sigma }_n\left({\Delta }_1\right)}\times \frac{{\sigma }_n\left({\Delta }_1\right)}{{\sigma }_n\left({\Delta }_0\right)}+\frac{W\left({\Delta }_0\right)-W\left({\Delta }_1\right)}{{\sigma }_n\left({\Delta }_0\right)},$$

which, from (1) and (2), can approximated by

$$\frac{W\left({\Delta }_1\right)}{{\sigma }_1}\times \frac{{\sigma }_1}{{\sigma }_0}+\frac{\omega \sqrt{n}}{{\sigma }_0},$$

where ${\sigma }_l^2={\sigma }^2\left({\Delta }_l\right)$ for l = 0, 1.

Suppose that we want to estimate the sample size for detecting H₁ : Δ = Δ₁ with a power of 1 − β by the generalized log-rank test with a one-sided α at H₀ : Δ = Δ₀, i.e.

$$1-\beta =P(\frac{W\left({\Delta }_0\right)}{{\sigma }_n\left({\Delta }_0\right)}>z_{1-\alpha }\mid H_1)\approx P(\frac{W\left({\Delta }_1\right)}{{\sigma }_1}\times \frac{{\sigma }_1}{{\sigma }_0}+\frac{\omega \sqrt{n}}{{\sigma }_0}>z_{1-\alpha }\mid H_1).$$

Since W (Δ₁)/σ₁ is approximately N(0, 1) under H₁, we have

$$-z_{1-\beta }=\left(z_{1-\alpha }-\frac{\omega \sqrt{n}}{{\sigma }_0}\right)\frac{{\sigma }_0}{{\sigma }_1}.$$

Hence the required sample size is obtained as

$$n=\frac{{({\sigma }_0{z}_{1-\alpha }+{\sigma }_1{z}_{1-\beta })}^2}{{\omega }^2}.$$ (3)

Note that ω, ${\sigma }_0^2$ and ${\sigma }_1^2$ are functions of the survival distributions S₁(t) and S₂(t) under H₁, and the common censoring distribution G(t). They are calculated using numerical methods.

The power of the generalized log-rank test roughly depends on the number of events, rather than the number of patients, so that one may want to calculate the expected number of events at the final data analysis. The number of events D is calculated as in the standard log-rank test, i.e. D = n(p₁d₁ + p₂d₂), where $d_k=1+{\int }_0^{\infty }{S}_k\left(t\right)dG\left(t\right)$, see e.g. Schoenfeld (1983). Note that if the survival distributions are shorter, then the required sample size is smaller.

3.1 SOME PRACTICAL INPUT PARAMETER SETTINGS

In this section, we will illustrate the application of the proposed methods when designing a clinical trial for comparing survival distributions with right censoring as follows.

• (A)An exponential distribution can be uniquely specified by a single parameter, such as the hazard rate, the median or the survival probability at a chosen time point. Furthermore, the family of exponential distributions fit real survival data relatively well. So, we often specify the survival distributions using exponential distributions with hazard rates λ_k, S_k(t) = exp(−λ_kt).
• (B)
  When a trial is open, patients are usually uniformly recruited during an accrual period a and additional follow-up period b. In this case, we have

  $$G\left(t\right)=\{\begin{array}{cc}1\hfill & \text{if}t\le b\hfill \\ -t∕b+(a+b)∕a\hfill & \text{if}b<t\le a+b\hfill \\ 0\hfill & \text{if}t>a+b\hfill \end{array}\phantom{\}}$$

Note that these assumptions can be easily extended to non-exponential survival models and a non-uniform censoring (accrual) distribution. Further, we assume that there is no loss to follow-up by (B), but it can be easily extended to account for possible loss to follow-up, see Jung, Kim and Chow (2008).

3.2 WHEN AN ACCRUAL RATE IS SPECIFIED

In Section 3.1, we assume that the accrual period is known. In designing a clinical trial, however, we can estimate the accrual pattern, rather than an accrual period, based on the experience from previous studies on the same patient population. For example, we may assume that patients are expected to be entered to the study at an rate of r during accrual period based on the number of patients treated by the member sites recently. In this case, (B) is replaced by:

• (B′)Patients are accrued following a Poisson distribution with rate r, and are followed for a period b after the completion of accrual.

With (λ₁, Δ₀, Δ₁, α, 1 − β, p₁, b) specified, ω = ω(a) and σ_l = σ_l(a) are functions of a. Hence, under (A) and (B), (3) is expressed as

$$n=\frac{{\{{\sigma }_0\left(a\right)z_{1-\alpha }+{\sigma }_1\left(a\right)z_{1-\beta }\}}^2}{{\omega }^2\left(a\right)}.$$ (4)

On the other hand, under the Poisson accrual distribution (B′), we have

$$n=a\times r.$$ (5)

By equating the right hand sides of (4) and (5), we obtain an equation on a,

$$a\times r=\frac{{\{{\sigma }_0\left(a\right)z_{1-\alpha }+{\sigma }_1\left(a\right)z_{1-\beta }\}}^2}{{\omega }^2\left(a\right)}.$$ (6)

Equation (6) is solved using a numerical method, such as bisection method. Let a* denote the solution to equation (6). Then, given an accrual rate r, instead of an accrual period a, we obtain the sample size by n = a* × r. The procedure for a sample size calculation may be summarized as follows.

• [1]
  Specify the input variables:

  • Type I and II error probabilities, (α, β)
  • Allocation proportions, p₁, p₂
  • Hazard rate λ₁ for the control arm under exponential survival model, and hazard ratios Δ₀ and Δ₁ under H₀ and H₁, respectively
  • Accrual rate r, and follow-up period b
• [2]
  Solve

  $$a\times r=\frac{{\{{\sigma }_0\left(a\right)z_{1-\alpha }+{\sigma }_1\left(a\right)z_{1-\beta }\}}^2}{{\omega }^2\left(a\right)}$$

  with respect to a using the bisection method, where

  $${\sigma }_0^2\left(a\right)={\Delta }_0{\lambda }_1{p}_1{p}_2{\int }_0^{a+b}\frac{G\left(t\right)e^{-{\lambda }_1(1+{\Delta }_1)t}(p_1{e}^{-{\lambda }_{1}t}+{\Delta }_1{p}_2{e}^{-{\Delta }_1{\lambda }_{1}t})}{{(p_1{e}^{-{\lambda }_{1}t}+{\Delta }_0{p}_2{e}^{-{\Delta }_1{\lambda }_{1}t})}^2}dt$$

  $${\sigma }_1^2\left(a\right)={\Delta }_1{\lambda }_1{p}_1{p}_2{\int }_0^{a+b}\frac{G\left(t\right)e^{-{\lambda }_1(1+{\Delta }_1)t}}{p_1{e}^{-{\lambda }_{1}t}+{\Delta }_1{p}_2{e}^{-{\Delta }_1{\lambda }_{1}t}}dt$$

  $$\omega \left(a\right)=({\Delta }_0-{\Delta }_1){\lambda }_1{p}_1{p}_2{\int }_0^{a+b}\frac{G\left(t\right)e^{-{\lambda }_1(1+{\Delta }_1)t}}{p_1{e}^{-{\lambda }_{1}t}+{\Delta }_0{p}_2{e}^{-{\Delta }_1{\lambda }_{1}t}}dt$$

  and

  $$G\left(t\right)=\{\begin{array}{cc}1\hfill & \text{if}t\le b\hfill \\ -t∕b+(a+b)∕a\hfill & \text{if}b<t\le a+b\hfill \\ 0\hfill & \text{if}t>a+b\hfill \end{array}\phantom{\}}$$
• [3]For the solution a = a* to the equation in [2], the required sample size is given as n = a*×r.

3.3 UNDER GENERAL ACCRUAL PATTERNS

In (B′), we assume a constant accrual rate over the whole accrual period. Usually in a multi-center trial, however, it takes a while (e.g. 1 to 2 years) until the study is approved by the institutional review boards of the study centers and the accrual rate is stabilized. Let h(t) for t ≥ 0 be the function representing the pattern of patient accrual over time period. For example, if we expect that the accrual will be linearly increasing for the first a₀ years, called a run-in time, and maintain a constant accrual of r per year after then, then we have

$$h\left(t\right)=\{\begin{array}{cc}rt\hfill & \text{for}0\le t\le a_0\hfill \\ r\hfill & \text{for}t>a_0\hfill \end{array}\phantom{\}}$$ (7)

This is similar to the piecewise linear accrual pattern considered by Yateman and Skene (1992). Given an accrual period of a and an accrual function h(t), (5) is extended to

$$n={\int }_0^{a}h\left(t\right)dt.$$ (8)

The accrual function h(t) is related to the censoring distribution function G(t) as follows. Let E denote the entry time of a patient in the study. The probability density function of E is expressed as

$$\{\begin{array}{cc}h\left(t\right)∕{\int }_0^{a}h\left(s\right)ds\hfill & \text{if}0\le t\le a\hfill \\ 0\hfill & \text{otherwise}\hfill \end{array}\phantom{\}}$$

Since C = a + b − E, the survivor function of C, G(t) = P (C ≥ t), is given as

$$G\left(t\right)=P(E\le a+b-t)=\{\begin{array}{cc}1\hfill & \text{if}t\le b\hfill \\ {\int }_0^{a+b-t}h\left(s\right)ds∕{\int }_0^{a}h\left(s\right)ds\hfill & \text{if}b<t\le a+b\hfill \\ 0\hfill & \text{if}t>a+b\hfill \end{array}\phantom{\}}$$ (9)

By calculating ${\sigma }_0^2\left(a\right)$, ${\sigma }_1^2\left(a\right)$ and ω(a) using this G(t) and equating the right hand sides of (4) and (8), we obtain an equation on a for a general accrual pattern. Using the solution a* to this equation, we obtain the required sample as $n={\int }_0^{a\ast }h\left(t\right)dt$.

Combining (8) and (9), we calculate the probability to observe an event from a patient in arm k by $d_k=1-n^{-1}{\int }_0^{\infty }{S}_k\left(t\right)h(a+b-t)dt$.

For example, for the piecewise linear accrual pattern of (7), we have

$${\int }_0^{a}h\left(s\right)ds=\{\begin{array}{cc}r{a}^2∕2\hfill & \text{if}0\le a\le a_0\hfill \\ r{a}_0^2∕2+r(a-a_0)\hfill & \text{if}a>a_0\hfill \end{array}\phantom{\}}$$

and

$${\int }_0^{a+b-t}h\left(s\right)ds=\{\begin{array}{cc}r{a}_0^2∕2+r(a-a_0)\hfill & \text{if}t<b\hfill \\ r{a}_0^2∕2+r(a+b-a_0-t)\hfill & \text{if}b\le t<a+b-a_0\hfill \\ r{(a+b-t)}^2∕2\hfill & \text{if}a+b-a_0\le t<a+b\hfill \\ 0\hfill & \text{if}t\ge a+b\hfill \end{array}\phantom{\}}$$

The sample size for a general accrual pattern can be summarized as follows. Note that the term ${\int }_0^{a}h\left(s\right)ds$ is cancelled out in the equation for sample size calculation.

• [1]
  Specify the input variables:

  • Type I and II error probabilities, (α, β)
  • Allocation proportions, p₁, p₂
  • Hazard rate λ₁ for the control arm under exponential survival model, and hazard ratios Δ₀ and Δ₁ under H₀ and H₁, respectively
  • Accrual pattern h(t), and follow-up period b
• [2]
  Solve

  $$1=\frac{{\{{\tilde{\sigma }}_0\left(a\right)z_{1-\alpha }+{\tilde{\sigma }}_1\left(a\right)z_{1-\beta }\}}^2}{{\tilde{\omega }}^2\left(a\right)}$$

  with respect to a using the bisection method, where

  $${\tilde{\sigma }}_0^2\left(a\right)={\Delta }_0{\lambda }_1{p}_1{p}_2{\int }_0^{a+b}\frac{\tilde{G}\left(t\right)e^{-{\lambda }_1(1+{\Delta }_1)t}(p_1{e}^{-{\lambda }_{1}t}+{\Delta }_1{p}_2{e}^{-{\Delta }_1{\lambda }_{1}t})}{{(p_1{e}^{-{\lambda }_{1}t}+{\Delta }_0{p}_2{e}^{-{\Delta }_1{\lambda }_{1}t})}^2}dt$$

  $${\sigma }_1^2\left(a\right)={\Delta }_1{\lambda }_1{p}_1{p}_2{\int }_0^{a+b}\frac{\tilde{G}\left(t\right)e^{-{\lambda }_1(1+{\Delta }_1)t}}{p_1{e}^{-{\lambda }_{1}t}+{\Delta }_1{p}_2{e}^{-{\Delta }_1{\lambda }_{1}t}}dt$$

  $$\omega \left(a\right)=({\Delta }_0-{\Delta }_1){\lambda }_1{p}_1{p}_2{\int }_0^{a+b}\frac{\tilde{G}\left(t\right)e^{-{\lambda }_1(1+{\Delta }_1)t}}{p_1{e}^{-{\lambda }_{1}t}+{\Delta }_0{p}_2{e}^{-{\Delta }_1{\lambda }_{1}t}}dt$$

  and

  $$\tilde{G}\left(t\right)={\int }_0^{a+b-t}h\left(s\right)ds$$
• [3]For the solution a = a* to the equation in [2], the required sample size is given as $n={\int }_0^{a\ast }h\left(t\right)dt$.

4 EXAMPLES

We consider some real examples of cancer trials that can be designed using the proposed sample size method.

Example 1

We consider the locally advanced nasopharyngeal carcinoma study that is discussed in the introduction section. Under an equal allocation p₁ = p₂ = 0.5, an annual accrual rate of r = 55 patients and b = 3 years of additional follow-up after completion of accrual, we need n = 138 patients (n_k = 69 per arm) for 1 − β = 80% power for detecting H₁ : Δ₁ = 0.7757 by the generalized log-rank test with one-sided α = 10% for H₀ : Δ₀ = 1.4974. We expect about 42 events (progressions or deaths) at the final data analysis.

In order to validate this sample size, 10,000 samples of size n = 138 are generated from the design setting of (p₁, λ₁, Δ₁, r, b) = (0.5, 0.096, 0.7757, 55, 3) as in the above sample size calculation, and the generalized log-rank test with one-sided α = 0.1 is applied to each simulated sample. The empirical power is obtained as 0.7920, which is very close to the nominal 1 − β = 0.8. We also conduct simulations to check the small sample property of the generalized log-rank test by generating 10,000 samples of size n = 138 under H₀ : Δ₀ = 1.4974. The empirical type I error rate is 0.0923, which is very close to the nominal α = 0.1.

Extending this example, suppose that the accrual is expected to linearly increase for the first year (a₀ = 1) and maintain a constant accrual rate of r = 55 per year from the beginning of the second year. Then, we obtain a* = 3 and n = 136. With the slightly longer accrual period (3 years vs. 138/55 = 2.5 years), this design setting requires a slightly smaller sample size than the setting with a constant accrual rate which was considered above.

Example 2

Chemotherapy B has been a standard regimen for patients with non-bulky stage I and II Hodgkin lymphoma. In a previous study on 6 cycles of B, each patient had a FDG-PET (fluorodeoxyglucose positron-emission tomography) imaging after 2 cycles of B. It was found that the patients with a negative PET image (group 1) and those with a positive PET image (group 2) had a 3-year PFS, defined as the time to disease progression or death, of S₁(3) = 0.86 and S₂(3) = 0.52, respectively, and the hazard ratio λ₂/λ₁, was estimated as Δ₀ = 4.3.

In a new (single-arm) phase II trial, the patients with a negative PET image after 2 cycles of B will be treated by additional 4 cycles of the standard chemotherapy B, whereas those with a positive PET image after 2 cycles of B will be treated by 4 cycles of a more aggressive chemotherapy C. By the PET-guided chemotherapy strategy, it is believed that the 3-year PFS of the PET-positive patients can be increased to S₂(3) = 0.74, from 0.52, resulting in a hazard ratio of Δ₁ = 2. Note that the group 1 patients will receive the same treatment as that of the previous study. Based on an exponential distribution model for PFS, the annual hazard rate for group 1, corresponding to S₁(3) = 0.86, is λ₁ = 0.05. The previous study observed about p₂ = 20% of PET-positivity. Assuming an annual accrual rate of r = 60 patients and b = 3 years of additional follow-up after completion of accrual, we need n = 195 patients for 1 − β = 90% power for detecting H₁ : Δ₁ = 2 by the generalized log-rank test with one-sided α = 10% for H₀ : Δ₀ = 4.3. Under this specific alternative hypothesis, we expect about 47 events (progressions or deaths) at the data analysis. Note that this is not a randomized trial, so that the resulting allocation proportions may be slightly different from the specified (p₁, p₂) = (0.8, 0.2). If the observed allocation proportions are closer to 0.5, the planned sample size has enough power. However, if they are farther from 0.5, we may consider checking the statistical power of the study based on the observed power while other design parameters are fixed at the values used when designing the study. In this power checking, we may use the estimated censoring distribution $\widehat{G}\left(t\right)$ based on the accrual times of patients too.

Simulation studies are conducted to evaluate the calculated sample size under the above design settings of H₀ and H₁, respectively. Using 10,000 simulation samples of size n = 132 under each hypothesis, the empirical type I error rate and power are observed as 0.0984 (to be compared to α = 0.1) and 0.8749 (to be compared to 1 − β = 0.9), respectively.

5 DISCUSSIONS

We have proposed a sample size calculation method of the generalized log-rank test for hypotheses H₀ : Δ = Δ₀ vs. H₀ : Δ = Δ₁ (Δ₀ > Δ₁). The proposed sample size formula is identical to that of the log-rank test by Schoenfeld (1983) and Lakatos (1988) if Δ₀ = 1, and that of the futility log-rank test by Jung et al. (2005) if Δ₁ = 1.

From (1)–(3), we observe that n is complicatedly dependent on the allocation proportions p_k, but the total sample size n is approximately minimized when p₁ = p₂ = 1/2. The sample size increases in λ₁, r and 1 − β, and decreases in Δ₀/Δ₁, b and α.

Our sample size method is so flexible that we can assume any survival distributions for two arms and any accrual pattern during study period. As a typical setting, we specify type I error rate α, power 1 − β, allocation proportion for arm 1 p₁, accrual rate r, follow-up period b, and survival distributions for two arms under H₀ and H₁ (or the survival distribution of the control arm and the hazard ratios Δ₀ and Δ₁ under H₀ and H₁. respectively); and we obtain the required sample size using our formula. However, we can obtain any of these input and output parameters given other parameter values. Furthermore, accrual rate r can be replaced by an expected accrual rate a in these calculations. The general accrual pattern approach will be very useful when checking the statistical power in the middle of a study with the realized accrual times.