Two-stage enrichment clinical trial design with adjustment for misclassification in predictive biomarkers

Abstract

A two-stage enrichment design is a type of adaptive design, which extends a stratified design with a futility analysis on the marker negative cohort at the first stage, and the second stage can be either a targeted design with only the marker positive stratum, or still the stratified design with both marker strata, depending on the result of the interim futility analysis. In this paper we consider the situation where the marker assay and the classification rule are possibly subject to error. We derive the sequential tests for the global hypothesis as well as the component tests for the overall cohort and the marker-positive cohort. We discuss the power analysis with the control of the type-I error rate and show the adverse impact of the misclassification on the powers. We also show the enhanced power of the two-stage enrichment over the one-stage design, and illustrate with examples of the recent successful development of immunotherapy in non-small-cell lung cancer.

1 |. INTRODUCTION

With the advancement of new genomic and proteomic technologies, biomarkers are becoming increasingly important in drug discovery and development. Clinical trials utilizing biomarkers in selecting patients for targeted therapies have been appearing in the literature since the past decade. Generally speaking, marker-based trial designs can be broadly classified as retrospective/prospective, or sequential/non-sequential. Recently Renfro et al.¹ gave a review and provided many references and several examples. Shih and Lin^2,3 considered hypotheses and relative efficiency of stratified designs and precision medicine designs, and showed that the stratified design is much more efficient than the so-called marker-based strategy designs. Wang et al.^4,5 introduced a two-stage adaptive enrichment design, which (prospectively) randomizes an unselected patient population initially to either an experimental or a control group, if the experimental treatment effect does not pass a futility threshold in the marker-negative cohort at the interim analysis, accrual of the marker-negative cohort will be stopped and the remaining samples size will be re-allocated to only the marker-positive cohort. Diagrams 1 and 2 depict this design in detail in Sections 3 and 4. As shown, the trial starts with a stratified randomization design at stage 1, and in the case of mid-trial enrichment (case II_B), the trial will continue with a targeted design for stage 2; otherwise, it will continue with the stratified design (case II_A). Simulation studies have shown that this design with possible mid-trial enrichment is uniformly more powerful than the “adaptive signature design” of Freidlin and Simon⁶ in detecting a subgroup-specific treatment effect⁴. However, unlike the traditional stratified designs using patient’s baseline characteristics such as gender, age, race, or ECOG performance status, the biomarker status and classification rule are prone to errors due to imperfect assays and/or classification rules. Several papers have also considered misclassification problems with biomarkers, but in different contexts. For example, Krisam and Kieser⁷ investigated decision rules for selection of target populations and procedures for calculating the sample size required to achieve a specified selection probability. Wang et al.⁸ focused on misclassification issues in non-inferiority trials. Wang and Li⁹ applied a machine learning algorithm to validate biomarker’s predictive performance on assessing subpopulation-specific treatment effects. In this paper, we extend the methods of Wang et al.^4,5 by considering the marker’s sensitivity and specificity. We discuss the hypotheses, critical values, interim analysis, type-I error rates and powers of the tests of treatment effects on the (unselected) entire study patient population and on the adaptively enriched patient subset. In Section 2, we first give the hypotheses of interest in enrichment designs. In Section 3, we set the foundation of a one-stage stratified design with marker classification errors. In Section 4, the enrichment design with a futility analysis and the path of the second stage is detailed. Sequential tests on the hypotheses of interest are derived. In Section 5, we give numerical examples from the recent successful development of immunotherapy in non-small-cell lung cancer. Section 6 provides final comments and discussions.

2 |. PARAMETERS AND HYPOTHESES

Consider a two-arm randomized controlled clinical trial to assess whether the effect of a test treatment is superior to control (standard of care). Suppose some baseline genomic biomarker (single biomarker or composite of markers) may be prognostic or predictive of a treatment effect. For simplicity, assume that the marker status is dichotomized (positive versus negative or present versus absent). Let D be the true marker status, M be the marker-appeared status, and A be the treatment assignment. Let ${\mu }_{ij}=E(Y|A=i,D=j)$ and ${\sigma }_{ij}^2=Var(Y|A=i,D=j)$ denote the mean and variance of the response variable Y of treatment group i (i = T or C) for patients in marker cohort j (j = 0 or 1; j = 0 being marker-negative and j = 1 being marker-positive). The marker assay and classification accuracy are measured by sensitivity ${\lambda }_{sen}\equiv P(M=1|D=1)$ and specificity ${\lambda }_{spec}\equiv P(M=0|D=0)$. The prevalence rate of true marker-positives is $p\equiv P(D=1)$. Assume λ_sen, λ_spec, and p are known as parts of the trial design parameters. The randomization ratio is $r\equiv P(A=T)$ to the test treatment and 1 − r to the control group. This randomization ratio applies to the stratified design in both stage 1 and stage 2. We now define the treatment effect parameters and label the corresponding null hypotheses as follows.

Denote the treatment effect for marker-negative stratum by ${\delta }_{-}={\mu }_{T0}-{\mu }_{C0}$; the corresponding null hypothesis is $H_{0-}:{\delta }_{-}=0$. Denote the treatment effect for marker-positive stratum by ${\delta }_{+}={\mu }_{T1}-{\mu }_{C1}$; the corresponding null hypothesis is $H_{0+}:{\delta }_{+}=0$. The overall treatment effect is $\delta \equiv w_1\left({\mu }_{T1}-{\mu }_{C1}\right)+w_0\left({\mu }_{T0}-{\mu }_{C0}\right)=w_1{\delta }_{+}+w_0{\delta }_{-}$, where w₁(≥ 0) and w₀(≥ 0) are some weights for treatment effects; the corresponding null hypothesis is H_0a : δ = 0. An obvious weight is the the prevalence of the corresponding marker-defined cohorts; i.e., w₁ = p and w₀ = 1 − p. The quantity $\delta =p{\delta }_{+}+(1-p){\delta }_{-}$ is interpreted as “treatment’s utility effect”². Another choice is w₁ = w₀ = 1, which leads to the “absolute treatment effect” δ = δ₊ + δ₋.

For the overall patient population, we are interested in testing:

$$H_{0a}:\delta =0\text{vs.}{H}_{1a}:\delta >0$$

For the marker-positive cohort, we are interesting in testing:

$$H_{0+}:{\delta }_{+}=0\text{vs}.H_{1+}:{\delta }_{+}>0$$

The marker-positive cohort is of special interest especially when δ << δ₊, and/or due to sample size or other feasibility limitation, there may not be enough power to test H_0a versue H_1a. The goal of an enrichment design is to increase the efficiency of a study in the presence of a pre-identified subset of patients who may especially respond to the new therapy.

2.1 |. Composite and Individual Hypotheses

Since we are pursuing either δ > 0 for the unselected patient population, or δ₊ > 0 for the marker-positive subset, we are testing the composite hypothesis

$$\begin{gathered} H_0:\delta ={\delta }_{+}=0\left(H_{0a}\cap H_{0+}\right)\text{vs.} \\ {H}_1:\delta >0\text{or}{\delta }_{+}>0\left(H_{1a}\cup H_{1+}\right) \end{gathered}$$

Throughout the remainder of this paper, we will work with one-sided test and strongly control the (experiment-wise) type-I error rate α (e.g., at 0.025). The individual hypotheses H_0a and H₀₊ will also be tested with assigned type-I error rates.

2.2 |. Positive and Negative Predictive Values

In the following, we first consider the stratified design of stage 1 for testing the marker-specific treatment effect hypotheses H₋ and H₊. Denote $q\equiv P(M=1)=p{\lambda }_{sen}+(1-p)\left(1-{\lambda }_{spec}\right)$. The positive predictive value PPV = P(D = 1|M = 1) is

$$\tau \equiv \frac{p{\lambda }_{sen}}{q}$$

Let

$${\mu }_{i*}\equiv E(Y|A=i,M=1)=\tau {\mu }_{i1}+(1-\tau ){\mu }_{i0}$$ (1)

and

$$\begin{gathered} {\sigma }_{i*}^2\equiv Var(Y|A=i,M=1)=\tau \left({\mu }_{i1}^2+{\sigma }_{i1}^2\right)+(1-\tau )\left({\mu }_{i0}^2+{\sigma }_{i0}^2\right)-{\mu }_{i*}^2 \\ =\tau (1-\tau ){\mathrm{\Delta }}_i^2+\tau {\sigma }_{i1}^2+(1-\tau ){\sigma }_{i0}^2 \end{gathered}$$ (2)

for i = T, C, where Δ_i = μ_i1 − μ_i0 is the true marker’s effect in treatment group i.

The negative predictive value NPV = P(D = 0|M = 0) is

$$\eta \equiv \frac{(1-p){\lambda }_{spec}}{1-q}$$

Let

$${\mu }_{i\varphi }\equiv E(Y|A=i,M=0)=(1-\eta ){\mu }_{i1}+\eta {\mu }_{i0}$$ (3)

and

$$\begin{gathered} {\sigma }_{i\varphi }^2\equiv Var(Y|A=i,M=0)=\eta \left({\mu }_{i0}^2+{\sigma }_{i0}^2\right)+(1-\eta )\left({\mu }_{i1}^2+{\sigma }_{i1}^2\right)-{\mu }_{i\varphi }^2 \\ =\eta (1-\eta ){\mathrm{\Delta }}_i^2+(1-\eta ){\sigma }_{i1}^2+\eta {\sigma }_{i0}^2 \end{gathered}$$ (4)

for i = T, C. See derivations of (2) and (4) in Shih and Lin³. Notice that under $H_0:\delta ={\delta }_{+}=0,{\mathrm{\Delta }}_T=-{\mathrm{\Delta }}_C$.

3 |. STRATIFIED DESIGN

The following shows the diagram of the stratified design:

With the stratified design, the sample means ${\overline{Y}}_{ij}=\frac{1}{N_{ij}}\sum _{\left\{k:A_k=i,M_k=j\right\}}{Y}_k$ are unbiased estimators of $E(Y|A=i,M=j)(\equiv {\mu }_{i*}$ for M = 1 and $\equiv {\mu }_{i\varphi }$ for M = 0). The sample variances $s_{ij}^2=\frac{1}{N_{ij}-1}\sum _{\left\{k:A_k=i,M_k=j\right\}}{\left(Y_k-{\overline{Y}}_{ij}\right)}^2$ are unbiased estimators of $Var(Y|A=i,M=j)\left(\equiv {\sigma }_{i*}^2\text{for}M=1\text{and}\equiv {\sigma }_{i\varphi }^2\text{for}M=0\right)$, where N_ij are the sample sizes for the the set with (A = i, M = j). Note that $N_{Tj}=r{N}_j$, where $N_1=N_{T1}+N_{C1}~Binomial(N,q)$ and $N_0=N-N_1=N_{T0}+N_{C0}$ are the sample size for the marker-appeared positive cohort and the marker-appeared negative cohort after stratification, respectively.

3.1 |. Z test statistics

Following Shih and Lin³, the unbiased estimate of the true mean of the marker positive cohort is

$${\widehat{\delta }}_{+}={\widehat{\mu }}_{T1}-{\widehat{\mu }}_{C1}=\frac{-\eta \left({\overline{Y}}_{T1}-{\overline{Y}}_{C1}\right)+(1-\tau )\left({\overline{Y}}_{T0}-{\overline{Y}}_{C0}\right)}{1-\tau -\eta }$$ (5)

With its variance

$$Var\left({\widehat{\delta }}_{+}\right)\approx \frac{\frac{{\eta }^2{\sigma }_{T*}^2}{rqN}+\frac{{\eta }^2{\sigma }_{C*}^2}{(1-r)qN}+\frac{{(1-\tau )}^2{\sigma }_{T\varphi }^2}{r(1-q)N}+\frac{{(1-\tau )}^2{\sigma }_{C\varphi }^2}{(1-r)(1-q)N}}{{(1-\tau -\eta )}^2}\stackrel{def}{=}\frac{{\sigma }_{+}^2}{N}$$ (6)

and variance estimator

$$\widehat{Var}\left({\widehat{\delta }}_{+}\right)=\frac{{\eta }^2\left(\frac{s_{T1}^2}{N_{T1}}+\frac{s_{C1}^2}{N_{C1}}\right)+{(1-\tau )}^2\left(\frac{s_{T_0}^2}{N_{T0}}+\frac{s_{C0}^2}{N_{C0}}\right)}{{(1-\tau -\eta )}^2}.$$ (7)

Z-statistic $Z_{+}=\frac{{\widehat{\delta }}_{+}}{\sqrt{Var\left({\widehat{\delta }}_{+}\right)}}=\frac{\sqrt{N}{\widehat{\delta }}_{+}}{{\sigma }_{+}^2}~AN(0,1)$ under H₀₊. If variances are not known, the Wald statistic $W_{+}=\frac{{\widehat{\delta }}_{+}}{\sqrt{\widehat{Var}}\left({\widehat{\delta }}_{+}\right)}~$ AN(0, 1) under H₀₊ is readily formed from Eq. (5) and Eq. (7).

Unbiased estimator of the treatment effect in the marker negative cohort is

$${\widehat{\delta }}_{-}={\widehat{\mu }}_{T0}-{\widehat{\mu }}_{C0}=\frac{(1-\eta )\left({\overline{Y}}_{T1}-{\overline{Y}}_{C1}\right)-\tau \left({\overline{Y}}_{T0}-{\overline{Y}}_{C0}\right)}{1-\tau -\eta }$$ (8)

with its variance

$$Var\left({\widehat{\delta }}_{-}\right)\approx \frac{\frac{{(1-\eta )}^2{\sigma }_{T*}^2}{rqN}+\frac{{(1-\eta )}^2{\sigma }_{C*}^2}{(1-r)qN}+\frac{{\tau }^2{\sigma }_{T\varphi }^2}{r(1-q)N}+\frac{{\tau }^2{\sigma }_{C\varphi }^2}{(1-r)(1-q)N}}{{(1-\tau -\eta )}^2}\stackrel{def}{=}\frac{{\sigma }_{-}^2}{N}$$ (9)

and variance estimator

$$\widehat{Var}({\widehat{\delta }}_{-})=\frac{{(1-\eta )}^2\left(\frac{s_{T1}^2}{N_{T1}}+\frac{s_{C1}^2}{N_{C1}}\right)+{\tau }^2\left(\frac{s_{T_0}^2}{N_{T0}}+\frac{s_{C0}^2}{N_{C0}}\right)}{{(1-\tau -\eta )}^2}.$$ (10)

Z-statistic $Z_{-}=\frac{{\widehat{\delta }}_{-}}{\sqrt{Var\left({\widehat{\delta }}_{-}\right)}}=\frac{\sqrt{N}{\widehat{\delta }}_{-}}{{\sigma }_{-}^2}~$ AN(0, 1) under H₀₋. If variances are not known, the Wald test $W_{-}=\frac{{\widehat{\delta }}_{-}}{\sqrt{\widehat{Var}\left({\widehat{\delta }}_{-}\right)}}~$ AN(0, 1) under H₀₋ is readily formed from Eq. (8) and Eq. (10).

From Eqs. (5) and (8), unbiased estimator of $\delta =w_1{\delta }_{+}+w_0{\delta }_{-}$ can be obtained by

$$\begin{array}{l}\widehat{\delta }=w_1{\widehat{\delta }}_{+}+w_0{\widehat{\delta }}_{-}\\ =w_1\frac{-\eta \left({\overline{Y}}_{T1}-{\overline{Y}}_{C1}\right)+(1-\tau )\left({\overline{Y}}_{T0}-{\overline{Y}}_{C0}\right)}{1-\tau -\eta }+w_0\frac{(1-\eta )\left({\overline{Y}}_{T1}-{\overline{Y}}_{C1}\right)-\tau \left({\overline{Y}}_{T0}-{\overline{Y}}_{C0}\right)}{1-\tau -\eta }\\ =\frac{\left(w_1-w\tau \right)\left({\overline{Y}}_{T0}-{\overline{Y}}_{C0}\right)+\left(w_0-w\eta \right)\left({\overline{Y}}_{T1}-{\overline{Y}}_{C1}\right)}{1-\tau -\eta },\end{array}$$ (11)

where w = w₁ + w₀ with its variance

$$Var(\widehat{\delta })\approx \frac{\frac{{\left(w_0-w\eta \right)}^2{\sigma }_{T*}^2}{rqN}+\frac{{\left(w_0-w\eta \right)}^2{\sigma }_{C*}^2}{(1-r)qN}+\frac{{\left(w_1-w\tau \right)}^2{\sigma }_{T\varphi }^2}{r(1-q)N}+\frac{{\left(w_1-w\tau \right)}^2{\sigma }_{C\varphi }^2}{(1-r)(1-q)N}}{{(1-\tau -\eta )}^2}\stackrel{def}{=}\frac{{\sigma }^2}{N},$$ (12)

and variance estimator

$$\widehat{Var}(\widehat{\delta })=\frac{\frac{{\left(u_0-w\eta \right)}^2{s}_{T1}^2}{N_{T1}}+\frac{{\left(u_0-w\eta \right)}^2{s}_{C1}^2}{N_{C1}}+\frac{{\left(w_1-w\tau \right)}^2{s}_{T0}^2}{N_{T0}}+\frac{{\left(w_1-w\tau \right)}^2{s}_{C0}^2}{N_{C0}}}{{(1-\tau -\eta )}^2}.$$ (13)

Wald test $Z=\frac{\widehat{\delta }}{\sqrt{Var(\widehat{\delta })}}=\frac{\sqrt{N}\widehat{\delta }}{\sigma }=w_1\frac{{\sigma }_{+}}{\sigma }{Z}_{+}+w_0\frac{{\sigma }_{-}}{\sigma }{Z}_{-}~AN(0,1)$ under H₀. If variances are not known, the Wald test statistic $W=\frac{\widehat{\delta }}{\sqrt{\widehat{Var}(\widehat{\delta })}}~AN(0,1)$ under H₀ is readily formed from Eqs. (11) and (13).

Notice that the ${\widehat{\delta }}_{+}$ and ${\widehat{\delta }}_{-}$ are correlated, with

$$\begin{gathered} Cov\left({\widehat{\delta }}_{+},{\widehat{\delta }}_{-}\right)=\frac{-\eta (1-\eta )\left(\mathit{var}\left({\overline{Y}}_{T1}\right)+\mathit{var}\left({\overline{Y}}_{C1}\right)\right)-\tau (1-\tau )\left(\mathit{var}\left({\overline{Y}}_{T0}\right)+\mathit{var}\left({\overline{Y}}_{C0}\right)\right)}{{(1-\tau -\eta )}^2} \\ \approx \frac{-\eta (1-\eta )\left(\frac{{\sigma }_{T*}^2}{rq}+\frac{{\sigma }_{C*}^2}{(1-r)q}\right)-\tau (1-\tau )\left(\frac{{\sigma }_{T\varphi }^2}{r(1-q)}+\frac{{\sigma }_{C\varphi }^2}{(1-r)(1-q)}\right)}{N{(1-\tau -\eta )}^2}\stackrel{def}{=}\frac{{\sigma }_{+,-}}{N}. \end{gathered}$$

Hence, we have

$$Cov\left(Z_{+},Z_{-}\right)=Cov\left(\frac{\sqrt{N}{\widehat{\delta }}_{+}}{{\sigma }_{+}},\frac{\sqrt{N}{\widehat{\delta }}_{-}}{{\sigma }_{-}}\right)=\frac{N}{{\sigma }_{+}{\sigma }_{-}}\frac{{\sigma }_{+,-}}{N}=\frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}.$$

Next we consider the correlation between the test statistics $Z=\frac{\sqrt{N}\widehat{\delta }}{\sigma }$ and $Z_{+}=\frac{\sqrt{N}{\widehat{\delta }}_{+}}{{\sigma }_{+}}$. We have

$$\begin{gathered} Cov\left(Z,Z_{+}\right)=Cov\left(w_1\frac{{\sigma }_{+}}{\sigma }{Z}_{+}+w_0\frac{{\sigma }_{-}}{\sigma }{Z}_{-},Z_{+}\right) \\ =w_1\frac{{\sigma }_{+}}{\sigma }+w_0\frac{{\sigma }_{-}}{\sigma }Cov\left(Z_{-},Z_{+}\right) \\ =w_1\frac{{\sigma }_{+}}{\sigma }+w_0\frac{{\sigma }_{+,-}}{\sigma {\sigma }_{+}} \\ =\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}. \end{gathered}$$

Similarly

$$Cov\left(Z,Z_{-}\right)=\frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}.$$

Note that, in the above equations involving variance and covariance, we have defined:

$$\begin{array}{l}{\sigma }^2=\frac{\frac{{\left(w_0-w\eta \right)}^2{\sigma }_{T*}^2}{rq}+\frac{{\left(w_0-w\eta \right)}^2{\sigma }_{C*}^2}{(1-r)q}+\frac{{\left(w_1-w\tau \right)}^2{\sigma }_{T\varphi }^2}{r(1-q)}+\frac{{\left(w_1-w\tau \right)}^2{\sigma }_{C\varphi }^2}{(1-r)(1-q)}}{{(1-\tau -\eta )}^2}\\ {\sigma }_{+}^2=\frac{\frac{{\eta }^2{\sigma }_{T*}^2}{rq}+\frac{{\eta }^2{\sigma }_{C*}^2}{(1-r)q}+\frac{{(1-\tau )}^2{\sigma }_{T\varphi }^2}{r(1-q)}+\frac{{(1-\tau )}^2{\sigma }_{C\varphi }^2}{(1-r)(1-q)}}{{(1-\tau -\eta )}^2}\\ {\sigma }_{-}^2=\frac{\frac{{(1-\eta )}^2{\sigma }_{T*}^2}{rq}+\frac{{\tau }^2{\sigma }_{T\varphi }^2}{r(1-q)}+\frac{{(1-\eta )}^2{\sigma }_{C*}^2}{(1-r)q}+\frac{{\tau }^2{\sigma }_{C\varphi }^2}{(1-r)(1-q)}}{{(1-\tau -\eta )}^2}\\ {\sigma }_{+,-}=\frac{-\eta (1-\eta )\left(\frac{{\sigma }_{T*}^2}{rq}+\frac{{\sigma }_{C*}^2}{(1-r)q}\right)-\tau (1-\tau )\left(\frac{{\sigma }_{T\varphi }^2}{r(1-q)}+\frac{{\sigma }_{C\varphi }^2}{(1-r)(1-q)}\right)}{{(1-\tau -\eta )}^2}.\end{array}$$

3.2 |. Type I error and Power

When designing a trial with $H_0=H_{0a}\cap H_{0+}$ in mind, there are global and individual levels of the type-I error and power to be considered. The global type I error rate is associated with testing the composite hypothesis H₀:

$$\begin{array}{l}\alpha =P\left(\text{Reject}{H}_0|H_0\right)\\ =P_0\left(\text{Reject}{H}_{0a}\text{or Reject}{H}_{0+}\right)\\ =P_0\left(Z>c_1\text{or}{Z}_{+}>c_2\right)\\ =P_0\left(Z>c_1\right)+P_0\left(Z\le c_1,Z_{+}>c_2\right).\end{array}$$

The critical value c₁ is obtained by allocating an alpha for the test of H_0a. Then c₂ can be solved for testing H₀₊ in the above equation. The corresponding global power for testing the composite hypothesis H₀ is

$$\begin{array}{l}1-\beta =P\left(\mathrm{Reject}{H}_0|H_1\right)\\ =P_1\left(\text{Reject}{H}_{0a}\text{or Reject}{H}_{0+}\right)\\ =P_1\left(Z>c_1\text{or}{Z}_{+}>c_2\right)\\ =P_1\left(Z>c_1\right)+P_1\left(Z\le c_1,Z_{+}>c_2\right).\end{array}$$

With c₁ and c₂ solved, the power for testing the individual hypothesis H_1a and H₁₊ are as follows, respectively:

$$\begin{array}{l}1-{\beta }_a=P\left(\text{Reject}{H}_{0a}|H_{1a}\right)\\ =P\left(Z>c_1|H_{1a}\right)=\mathrm{\Phi }\left(-c_1+\frac{\sqrt{N}\delta }{\sigma }\right)\end{array}$$

and

$$\begin{array}{l}1-{\beta }_{+}=P\left(\mathrm{Reject}{H}_{0+}|H_{1+}\right)\\ =P\left(Z_{+}>c_2|{\delta }_{+}\right)=\mathrm{\Phi }\left(-c_2+\frac{\sqrt{N}{\delta }_{+}}{{\sigma }_{+}}\right).\end{array}$$

4 |. TWO-STAGE ENRICHMENT DESIGN

The above discussion was a one-stage design without enrichment consideration. Now we consider the two-stage enrichment design. At Stage I, we stratify tN patients the same as shown in the diagram in Section 3. The tests are all the same as in Section 3.1 with sample size tN instead of N. If H₀ is rejected, we stop the trial and claim that treatment is effective either for the overall (unselected) patient population or for the marker-positive patients. The tests of the individual hypotheses H_1a and H₁₊ would shed more light on the conclusion. However, if H₀ is not rejected, the trial will then continue to accrue (1 − t)N more patients in Stage II with two different scenarios, II_A or II_B, depending on the futility criterion (see Section 4.3) for the marker-negative patients, as shown in the following diagram.

In the following, we will discuss the details of scenario II_A and II_B, and the test of the hypotheses in each scenario. First, let

$${\overline{Y}}_{ij}=\frac{n_{ij}^{(1)}{\overline{Y}}_{ij}^{(1)}+n_{ij}^{(2)}{\overline{Y}}_{ij}^{(2)}}{n_{ij}}$$

for i = T, C and j = 0, 1, where $n_{ij}^{(k)}$ is the sample size at the k^th stage, ${\overline{Y}}_{ij}^{(k)}=\frac{1}{n_{ij}}{\sum }_{s=1}^{n_{ij}}{Y}_{ijs}^{(k)}$, and $n_{ij}=n_{ij}^{(1)}+n_{ij}^{(2)}$. Then we have

$$\begin{array}{ll}E\left(n_{T1}^{(1)}\right)=rqtN,& E\left(n_{C1}^{(1)}\right)=(1-r)qtN,\\ E\left(n_{T0}^{(1)}\right)=r(1-q)tN,& E\left(n_{C0}^{(1)}\right)=(1-r)(1-q)tN.\end{array}$$

If the trial continues with II_A, i.e., treatment is not futile on marker-negatives, then

$$\begin{array}{ll}E\left(n_{T1}^{(2)}\right)=rq(1-t)N,& E\left(n_{C1}^{(2)}\right)=(1-r)q(1-t)N,\\ E\left(n_{T0}^{(2)}\right)=r(1-q)(1-t)N,& E\left(n_{C0}^{(2)}\right)=(1-r)(1-q)(1-t)N,\end{array}$$

If the trial continues with II_B, i.e., terminating the marker-appeared negatives and enriching the marker-appeared positives, then

$$\begin{array}{ll}E\left(n_{T1}^{(2)}\right)=r(1-t)N,& E\left(n_{C1}^{(2)}\right)=(1-r)(1-t)N,\\ E\left(n_{T0}^{(2)}\right)=0,& E\left(n_{C0}^{(2)}\right)=0.\end{array}$$

4.1 |. Z test statistics

For Stage I, formulas for treatment effect estimates and tests are similar to Section 3.1 with sample size changed to tN. For clear notation and completeness, they are:

$${\widehat{\delta }}_{+}^{(1)}={\widehat{\mu }}_{T1}^{(1)}-{\widehat{\mu }}_{C1}^{(1)}=\frac{-\eta \left({\overline{Y}}_{T1}^{(1)}-{\overline{Y}}_{C1}^{(1)}\right)+(1-\tau )\left({\overline{Y}}_{T0}^{(1)}-{\overline{Y}}_{C0}^{(1)}\right)}{1-\tau -\eta }$$

$$Var\left({\widehat{\delta }}_{+}^{(1)}\right)\approx \frac{\frac{{\eta }^2{\sigma }_{\mathrm{T*}}^2}{rq}+\frac{{\eta }^2{\sigma }_{C*}^2}{(1-r)q}+\frac{{(1-\tau )}^2{\sigma }_{T\varphi }^2}{r(1-q)}+\frac{{(1-\tau )}^2{\sigma }_{C\varphi }^2}{(1-r)(1-q)}}{tN{(1-\tau -\eta )}^2}=\frac{{\sigma }_{+}^2}{tN}$$

$${\widehat{\delta }}_{-}^{(1)}={\widehat{\mu }}_{T0}-{\widehat{\mu }}_{C0}=\frac{(1-\eta )\left({\overline{Y}}_{T1}^{(1)}-{\overline{Y}}_{C1}^{(1)}\right)-\tau \left({\overline{Y}}_{T0}^{(1)}-{\overline{Y}}_{C0}^{(1)}\right)}{1-\tau -\eta }$$

$$Var\left({\widehat{\delta }}_{-}^{(1)}\right)\approx \frac{\frac{{(1-\eta )}^2{\sigma }_{T*}^2}{rq}+\frac{{(1-\eta )}^2{\sigma }_{C*}^2}{(1-r)q}+\frac{{\tau }^2{\sigma }_{T\varphi }^2}{r(1-q)}+\frac{{\tau }^2{\sigma }_{C\varphi }^2}{(1-r)(1-q)}}{tN{(1-\tau -\eta )}^2}=\frac{{\sigma }_{-}^2}{tN}$$

$$\begin{gathered} {\widehat{\delta }}^{(1)}=w_1{\widehat{\delta }}_{+}^{(1)}+w_0{\widehat{\delta }}_{-}^{(1)} \\ =\frac{\left(w_0-w\eta \right)\left({\overline{Y}}_{T1}^{(1)}-{\overline{Y}}_{C1}^{(1)}\right)+\left(w_1-w\tau \right)\left({\overline{Y}}_{T0}^{(1)}-{\overline{Y}}_{C0}^{(1)}\right)}{1-\tau -\eta } \end{gathered}$$

$$Var\left({\widehat{\delta }}^{(1)}\right)\approx \frac{\frac{{\left(w_0-w\eta \right)}^2{\sigma }_{T*}^2}{rq}+\frac{{\left(w_0-w\eta \right)}^2{\sigma }_{C*}^2}{(1-r)q}+\frac{{\left(w_1-w\tau \right)}^2{\sigma }_{T\varphi }^2}{r(1-q)}+\frac{{\left(w_1-w\tau \right)}^2{\sigma }_{C\varphi }^2}{(1-r)(1-q)}}{tN{(1-\tau -\eta )}^2}=\frac{{\sigma }^2}{tN},$$

and the Z-statistics for testing the individual and composite hypotheses are:

$$Z_{+}^{(1)}=\frac{\sqrt{tN}{\widehat{\delta }}_{+}^{(1)}}{{\sigma }_{+}}$$

$$Z_{-}^{(1)}=\frac{\sqrt{tN}{\widehat{\delta }}_{-}^{(1)}}{{\sigma }_{-}}$$

$$Z^{(1)}=\frac{\sqrt{tN}{\widehat{\delta }}^{(1)}}{\sigma }=w_1\frac{{\sigma }_{+}}{\sigma }{Z}_{+}^{(1)}+w_0\frac{{\sigma }_{-}}{\sigma }{Z}_{-}^{(1)}.$$

Going to the second stage, under scenario II_A, we continue the trial with all, marker-unselected patients. The estimate of treatment effect will combine the data from both stages as follows:

$$\begin{gathered} {\widehat{\delta }}_{+}=\frac{-\eta \left(\frac{n_{T1}^{(1)}{\overline{Y}}_{T1}^{(1)}+n_{T1}^{(2)}{\overline{Y}}_{T1}^{(2)}}{n_{T1}}-\frac{n_{C1}^{(1)}{\overline{Y}}_{C1}^{(1)}+n_{C1}^{(2)}{\overline{Y}}_{C1}^{(2)}}{n_{C1}}\right)+(1-\tau )\left(\frac{n_{T0}^{(1)}{\overline{Y}}_{T0}^{(1)}+n_{T0}^{(2)}{\overline{Y}}_{T0}^{(2)}}{n_{T0}}-\frac{n_{C0}^{(1)}{\overline{Y}}_{C0}^{(1)}+n_{C0}^{(2)}{\overline{Y}}_{C0}^{(2)}}{n_{C0}}\right)}{1-\tau -\eta } \\ =\frac{-\eta \left(t{\overline{Y}}_{T1}^{(1)}+(1-t){\overline{Y}}_{T1}^{(2)}-\left(t{\overline{Y}}_{C1}^{(1)}+(1-t){\overline{Y}}_{C1}^{(2)}\right)\right)+(1-\tau )\left(t{\overline{Y}}_{T0}^{(1)}+(1-t){\overline{Y}}_{T0}^{(2)}-\left(t{\overline{Y}}_{C0}^{(1)}+(1-t){\overline{Y}}_{C0}^{(2)}\right)\right)}{1-\tau -\eta } \\ =t{\widehat{\delta }}_{+}^{(1)}+(1-t){\widehat{\delta }}_{+}^{(2)} \end{gathered}$$

$$\begin{gathered} Var\left({\widehat{\delta }}_{+}\right)\approx \frac{\frac{{\eta }^2{\sigma }_{T_{*}}^2}{rq}+\frac{{\eta }^2{\sigma }_{C_{*}}^2}{(1-r)q}+\frac{{(1-\tau )}^2{\sigma }_{T\varphi }^2}{r(1-q)}+\frac{{(1-\tau )}^2{\sigma }_{C\varphi }^2}{(1-r)(1-q)}}{N{(1-\tau -\eta )}^2}=\frac{{\sigma }_{+}^2}{N}, \\ \widehat{\delta }=\frac{\left(w_0-w\eta \right)\left({\overline{Y}}_{T1}-{\overline{Y}}_{C1}\right)+\left(w_1-w\tau \right)\left({\overline{Y}}_{T0}-{\overline{Y}}_{C0}\right)}{1-\tau -\eta } \\ =\frac{\left(w_0-w\eta \right)\left(\frac{n_{T1}^{(1)}{\overline{Y}}_{T1}^{(1)}+n_{T1}^{(1)}{\overline{Y}}_{T1}^{(2)}}{n_{r1}}-\frac{n_{C1}^{(1)}{\overline{Y}}_{C1}^{(1)}+n_{C1}^{(2)}{\overline{Y}}_{C1}^{(2)}}{n_{c1}}\right)+\left(w_1-w\tau \right)\left(\frac{n_{T0}^{(1)}{\overline{Y}}_{T0}^{(1)}+n_{T0}^{(2)}{\overline{Y}}_{T0}^{(1)}}{n_{T0}}-\frac{n_{C0}^{(1)}{\overline{Y}}_{C0}^{(1)}+n_{C0}^{(2)}{\overline{Y}}_{C0}^{(2)}}{n_{C0}}\right)}{1-\tau -\eta } \\ =t{\widehat{\delta }}^{(1)}+(1-t){\widehat{\delta }}^{(2)} \\ Var(\widehat{\delta })\approx \frac{\frac{{\left(w_0-w\eta \right)}^2{\sigma }_{T*}^2}{rq}+\frac{{\left(w_0-w\eta \right)}^2{\sigma }_{C*}^2}{(1-r)q}+\frac{{\left(w_1-w\tau \right)}^2{\sigma }_{T\varphi }^2}{r(1-q)}+\frac{{\left(w_1-w\tau \right)}^2{\sigma }_{C\varphi }^2}{(1-r)(1-q)}}{N{(1-\tau -\eta )}^2}=\frac{{\sigma }^2}{N}. \end{gathered}$$

The Z-statistics for testing the individual and composite hypotheses are:

$$\begin{gathered} Z_{+}=\frac{\sqrt{N}{\widehat{\delta }}_{+}}{{\sigma }_{+}}=\frac{\sqrt{N}\left(\widehat{t}{\widehat{\delta }}_{+}^{(1)}+(1-t){\widehat{\delta }}_{+}^{(2)}\right)}{{\sigma }_{+}} \\ {Z}_{-}=\frac{\sqrt{N}{\widehat{\delta }}_{-}}{{\sigma }_{-}} \\ Z=\frac{\sqrt{N}\widehat{\delta }}{\sigma }=w_1\frac{{\sigma }_{+}}{\sigma }{Z}_{+}+w_0\frac{{\sigma }_{-}}{\sigma }{Z}_{-}. \end{gathered}$$

Under scenario II_B, we continue the trial with enrichment of the marker-appeared positive patients. The estimates of the treatment effects are

$$\begin{gathered} {\widehat{\delta }}_{+}^{*}=\frac{-\eta \left(\frac{n_{T1}^{(1)}{\overline{Y}}_{T1}^{(1)}+n_{T1}^{(2)}{\overline{Y}}_{T1}^{(2)}}{n_{T1}}-\frac{n_{C1}^{(1)}{\overline{Y}}_{C1}^{(1)}+n_{C1}^{(2)}{\overline{Y}}_{C1}^{(2)}}{n_{C1}}\right)+(1-\tau )\left(\frac{n_{T0}^{(1)}{\overline{Y}}_{T0}^{(1)}+{\overline{Y}}_{T0}^{(2)}{\overline{Y}}_{T0}^{(2)}}{n_{T0}}-\frac{n_{C0}^{(1)}{\overline{Y}}_{C0}^{(1)}+n_{C0}^{(2)}{\overline{Y}}_{C0}^{(2)}}{n_{C0}}\right)}{1-\tau -\eta } \\ =\frac{-\eta \left(\frac{rqtN{\overline{Y}}_{T1}^{(1)}+r(1-t)N{\overline{Y}}_{T1}^{(2)}}{rqtN+r(1-t)N}-\frac{(1-r)qtN{\overline{Y}}_{C1}^{(1)}+(1-r)(1-t)N{\overline{Y}}_{C1}^{(2)}}{(1-r)qtN+(1-r)(1-t)N}\right)+(1-\tau )\left({\overline{Y}}_{T0}^{(1)}-{\overline{Y}}_{C0}^{(1)}\right)}{1-\tau -\eta } \\ =\frac{-\eta \left(\frac{q_1{\overline{Y}}_{T_1}^{(1)}+(1-t){\overline{Y}}_{T1}^{(2)}}{qt+(1-t)}-\frac{q{\overline{Y}}_{C1}^{(1)}+(1-t){\overline{Y}}_{C1}^{(2)}}{qt+(1-t)}\right)+(1-\tau )\left({\overline{Y}}_{T0}^{(1)}-{\overline{Y}}_{C0}^{(1)}\right)}{1-\tau -\eta } \\ Var\left({\widehat{\delta }}_{+}^{*}\right)\approx \frac{\frac{{\eta }^2{\sigma }_{T*}^2}{r(qt+1-t)}+\frac{{\eta }^2{\sigma }_{C*}^2}{(1-r)(qt+1-t)}+\frac{{(1-\tau )}^2{\sigma }_{T\varphi }^2}{r(1-q)t}+\frac{{(1-\tau )}^2{\sigma }_{C\varphi }^2}{(1-r)(1-q)t}}{N{(1-\tau -\eta )}^2}=\frac{{\sigma }_{+}^{*2}}{N}. \end{gathered}$$

The Z-statistics for H₀₊ is:

$$Z_{+}^{*}=\frac{\sqrt{N}{\widehat{\delta }}_{+}^{*}}{{\sigma }_{+}^{*}}$$

4.2 |. Correlation of the tests

To control the type I error rate and to estimate the power, we need to calculate the correlation matrix of $\left(Z^{(1)},Z_{+}^{(1)},Z_{-}^{(1)},Z,Z_{+}\right)$ for scenario II_A and that of $\left(Z^{(1)},Z_{+}^{(1)},Z_{-}^{(1)},Z_{+}^{*}\right)$ for scenario II_B. In Appendices A.1 and A.2, it is shown that the covariance matrix of the Z-statistics for scenario II_A is:

$$Cov\left(\left[\begin{array}{c}{Z}^{(1)}\\ Z_{+}^{(1)}\\ Z_{-}^{(1)}\\ Z\\ Z_{+}\end{array}\right]\right)=\left[\begin{array}{c}1\\ \frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}\\ \frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}\\ \sqrt{t}\\ \sqrt{t}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{{\sigma }_{+}\sigma }\end{array}\begin{array}{c}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}\\ 1\\ \frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}\\ \sqrt{t}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}\\ \sqrt{t}\end{array}\begin{array}{c}\frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}\\ \frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}\\ 1\\ \sqrt{t}\frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}\\ \sqrt{t}\frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}\end{array}\begin{array}{c}\sqrt{t}\\ \sqrt{t}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}\\ \sqrt{t}\frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}\\ 1\\ \frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}\end{array}\begin{array}{c}\sqrt{t}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{{\sigma }_{+}\sigma }\\ \sqrt{t}\\ \sqrt{t}\frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}\\ \frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}\\ 1\end{array}\right].$$ (14)

And the covariance matrix of the Z-statistics for scenario II_B is

$$Cov\left(\left[\begin{array}{c}{Z}^{(1)}\\ Z_{+}^{(1)}\\ Z_{-}^{(1)}\\ Z_{+}^{*}\end{array}\right]\right)=\left[\begin{array}{c}1\\ \frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}\\ \frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}\\ \frac{w_1{\sigma }_{+,+}^{*}+w_0{\sigma }_{+,-}^{*}}{{\sigma }_{+}^{*}\sigma }\end{array}\begin{array}{c}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}\\ 1\\ \frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}\\ \frac{{\sigma }_{+,+}^{*}}{{\sigma }_{+}^{*}{\sigma }_{+}}\end{array}\begin{array}{c}\frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}\\ \frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}\\ 1\\ \frac{{\sigma }_{+,-}^{*}}{{\sigma }_{+}^{*}{\sigma }_{-}}\end{array}\begin{array}{c}\frac{w_1{\sigma }_{+,+}^{*}+w_0{\sigma }_{+,-}^{*}}{{\sigma }_{+}^{*}\sigma }\\ \frac{{\sigma }_{+,+}^{*}}{{\sigma }_{+}^{*}{\sigma }_{+}}\\ \frac{{\sigma }_{+,-}^{*}}{{\sigma }_{+}^{*}{\sigma }_{-}}\\ 1\end{array}\right].$$ (15)

Hence we have

$$\begin{gathered} Cov\left({\left(Z^{(1)},Z_{+}^{(1)},Z_{-}^{(1)},Z,Z_{+}{Z}_{+}^{*}\right)}^T\right)= \\ \left[\begin{array}{c}1\\ \frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}\\ \frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}\\ \sqrt{t}\\ \begin{array}{c}\sqrt{t}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{{\sigma }_{+}\sigma }\\ \frac{w_1{\sigma }_{+,+}^{*}+w_0{\sigma }_{+,-}^{*}}{{\sigma }_{+}^{*}\sigma }\end{array}\end{array}\begin{array}{c}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}\\ 1\\ \frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}\\ \sqrt{t}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}\\ \begin{array}{c}\sqrt{t}\\ \frac{{\sigma }_{+,+}^{*}}{{\sigma }_{+}^{*}{\sigma }_{+}}\end{array}\end{array}\begin{array}{c}\frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}\\ \frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}\\ 1\\ \sqrt{t}\frac{w_1{\sigma }_{+.-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}\\ \begin{array}{c}\sqrt{t}\frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}\\ \frac{{\sigma }_{+,-}^{*}}{{\sigma }_{+}^{*}{\sigma }_{-}}\end{array}\end{array}\begin{array}{c}\sqrt{t}\\ \sqrt{t}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}\\ \sqrt{t}\frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}\\ 1\\ \begin{array}{c}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}\\ *\end{array}\end{array}\begin{array}{c}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}\\ \sqrt{t}\\ \sqrt{t}\frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}\\ \frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}\\ \begin{array}{c}1\\ *\end{array}\end{array}\begin{array}{c}\frac{w_1{\sigma }_{+,+}^{*}+w_0{\sigma }_{+,-}^{*}}{{\sigma }_{+}^{*}\sigma }\\ \frac{{\sigma }_{+,+}^{*}}{{\sigma }_{+}^{*}{\sigma }_{+}}\\ \frac{{\sigma }_{+,-}^{*}}{{\sigma }_{+}^{*}{\sigma }_{-}}\\ *\\ \begin{array}{c}*\\ 1\end{array}\end{array}\right] \end{gathered}$$

4.3 |. Type I error and critical values

For the two-stage design, unlike the one-stage design in Section 3.2, we first need to split the overall alpha (e.g., α = 0.025) between the two stages. In Stage I, consider to spend only a fraction of the overall alpha, α₁, on testing the global hypothesis H₀ :

$$\begin{gathered} {\alpha }_1=P\left(\mathrm{Reject}{H}_0|H_0\right) \\ =P_0\left(\text{Reject}{H}_{0a}\text{or Reject}{H}_{0+}\right) \\ =P_0\left(Z^{(1)}>c_1\text{or}{Z}_{+}^{(1)}>c_2\right) \\ =P_0\left(Z^{(1)}>c_1\right)+P_0\left(Z^{(1)}\le c_1,Z_{+}^{(1)}>c_2\right) \\ ={\alpha }_{1a}+\left({\alpha }_1-{\alpha }_{1a}\right) \\ ={\alpha }_{1a}+{\alpha }_{1b}. \end{gathered}$$

As before, the critical value c₁ is obtained by allocating a portion of α₁, α_1a, for testing H_0a. Then c₂ can be solved for testing H₀₊ in the above equation.

For Stage II, the overall alpha is left with α − α₁, to be controlled between the mutually exclusive scenarios II_A and II_B. Suppose we allocate a fraction α₂ < α − α₁ for the tests in scenario II_A and the rest ${\alpha }_{2*+}=\alpha -{\alpha }_1-{\alpha }_2$ for scenario II_B as follows:

$$\begin{gathered} \alpha -{\alpha }_1=P_0\left(\text{Accept}{H}_0\text{at Stage I, Reject}{H}_0\text{at Stage II}\text{in case of}{\text{II}}_{\text{A}}\right) \\ +P_0\left(\text{Accept}{H}_0\text{at Stage I, Reject}{H}_{0+}{\text{at Stage II in case of II}}_{\text{B}}\right) \\ ={\alpha }_2+{\alpha }_{2^{*}+}. \end{gathered}$$

A futility criterion for marker-negative patients will be used to determine α₂ and ${\alpha }_{2*+}$. This can be done through a pre-specified threshold value c₀ for the test statistic $Z_{-}^{(1)}$, or through the futility probability ${\mathcal{F}}_p=P\left(Z_{-}^{(1)}\le c_0\right)$; see Eq. (16). For example, if we want the futility probability to be 75% (50%), then from $P_0\left(Z_{-}^{(1)}\le c_0\right)=0.75(0.50),c_0=0.6745(0.0)$.

In case of II_A, the test treatment passes the pre-defined futility threshold value c₀, i.e., $Z_{-}^{(1)}>c_0$. We continue with both marker-status cohorts and test H₀ at Stage II. The alpha is controlled by

$$\begin{gathered} {\alpha }_2=P_0\left(\text{Accept}{H}_0\text{at Stage I, Reject}{H}_0\text{at Stage}2\text{in case of}{\Pi }_{\text{A}}\right) \\ =P_0\left(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{-}^{(1)}>c_0,Z>b_1\right) \\ +P_0\left(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{-}^{(1)}>c_0,Z\le b_1,Z_{+}>b_2\right) \\ ={\alpha }_{2a}+\left({\alpha }_2-{\alpha }_{2a}\right). \end{gathered}$$

That is, the critical value b₁ is obtained by allocating α_2a = πα₂, a portion π of α₂, for testing H_0a. Then b₂ can be solved for testing H₀₊ in the above equation.

In case of II_B, the test treatment is futile on marker-negatives, i.e., $Z_{-}^{(1)}\le c_0$. We continue with enriching marker-appeared positives and test only $H_{0+}$ at Stage II_B. The alpha is controlled by

$$\begin{gathered} {\alpha }_{2*+}=P_0\left(\text{Accept}{H}_0\text{at Stage I, Reject}{H}_{0+}\text{at Stage II in case of}{\text{II}}_{\text{B}}\right) \\ =P_0\left(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{-}^{(1)}\le c_0,Z_{+}^{*}>b_3\right), \end{gathered}$$

with the critical value b₃ solved numerically using the correlation matrix.

The strategy of allocating alphas to either II_A or II_B is an important design consideration. Since the trial will only be in one scenario or the other, it would be ideal to maximize the alpha in either scenario. Toward this end, we can first rewrite α₂ and ${\alpha }_{2*+}$, respectively, as

$$\begin{gathered} {\alpha }_2=[P_0(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z>b_1|Z_{-}^{(1)}>c_0) \\ +P_0(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z\le b_1,Z_{+}>b_2|Z_{-}^{(1)}>c_0)]P_0(Z_{-}^{(1)}>c_0) \end{gathered}$$

and

$${\alpha }_{2*+}=P_0\left(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{+}^{*}>b_3|Z_{-}^{(1)}\le c_0\right)P_0\left(Z_{-}^{(1)}\le c_0\right).$$

Next, if we split α − α₁ into II_A and II_B with the same proportion as the odds of $P_0\left(Z_{-}^{(1)}>c_0\right)\text{to}{P}_0\left(Z_{-}^{(1)}\le c_0\right)$, i.e.,

$$\frac{{\alpha }_2}{{\alpha }_{2*+}}=\frac{P_0\left(Z_{-}^{(1)}>c_0\right)}{P_0\left(Z_{-}^{(1)}\le c_0\right)}=\frac{1-{\mathcal{F}}_p}{{\mathcal{F}}_p},$$ (16)

then we have

$$\begin{gathered} \alpha -{\alpha }_1=P_0(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z>b_1|Z_{-}^{(1)}>c_0) \\ +P_0(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z\le b_1,Z_{+}>b_2|Z_{-}^{(1)}>c_0) \end{gathered}$$

and

$$\alpha -{\alpha }_1=P_0\left(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{+}^{*}>b_3|Z_{-}^{(1)}\le c_0\right).$$

This indicates that, with the above alpha allocation strategy, the corresponding conditional type I error is α − α₁ for either II_A or II_B. When the odds of non-futility versus futility $\frac{P_0\left(Z_{-}^{(1)}>c_0\right)}{P_0\left(Z_{-}^{(1)}\le c_0\right)}$ is pre-determined, the critical values b₁ and b₂ are calculated such that

$$\begin{gathered} \left(\alpha -{\alpha }_1\right)P_0\left(Z_{-}^{(1)}>c_0\right)=P_0\left(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{-}^{(1)}>c_0,Z>b_1\right) \\ +P_0\left(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{-}^{(1)}>c_0,Z\le b_1,Z_{+}>b_2\right) \end{gathered}$$

for scenario II_A based on the joint distribution of $\left(Z^{(1)},Z_{+}^{(1)},Z_{-}^{(1)},Z,Z_{+}\right)$, and b₃ is calculated such that

$$\left(\alpha -{\alpha }_1\right)P_0\left(Z_{-}^{(1)}\le c_0\right)=P_0\left(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{-}^{(1)}\le c_0,Z_{+}^{*}>b_3\right)$$

for scenario II_B based on the joint distribution of $\left(Z^{(1)},Z_{+}^{(1)},Z_{-}^{(1)},Z_{+}^{*}\right)$.

In summary, assuming that information for p, λ_sen, λ_spec, ${\sigma }_{T0}^2,$ ${\sigma }_{T1}^2,$ ${\sigma }_{C0}^2,$ ${\sigma }_{C1}^2$ are available from earlier data, with specification of the design parameters α, α₁, α_1a, c₀, α_2a, r, t, w₁, and w₀, we can find the critical values c₁, c₂, b₁, b₂, b₃ for the study based on the above formulas. Notice that α₂ is determined by α − α₁ and c₀.

Table 1 provides the critical values for some commonly considered design parameters α = 0.025, α₁ = 0.01, α_1a = 0.005, $\pi \left(={\alpha }_{2a}/{\alpha }_2\right)$ = 0.5, and r = 0.5 with various p, λ_sen, λ_spec, ${\mathcal{F}}_p$ and t, assuming ${\sigma }_{T0}^2={\sigma }_{T1}^2={\sigma }_{C0}^2={\sigma }_{C1}^2=1$, for w₁ = p, w₀ = 1 − p. As noted in Section 2.3, under $H_0:\delta ={\delta }_{+}=0,{\mathrm{\Delta }}_T=-{\mathrm{\Delta }}_C$. In many cases we do not expect opposite marker effects in the test versus control treatment groups when there is no treatment effect, thus setting Δ_T = −Δ_C = 0 is reasonable.

TABLE 1.

Critital Values When α = 0.025, α_1a = α_1b = 0.005, r = 0.5, ${\sigma }_{T0}^2={\sigma }_{T1}^2={\sigma }_{C0}^2={\sigma }_{C1}^2=$ 1, π = 0.5

t	p	${\mathcal{F}}_p^{†}$	(λsen, λspec)	c0	c1	c2	b1	b2	b3
0.25	0.3	0.50	(1, 1)	0.000000	2.575829	2.532415	2.594235	2.328787	2.157164
			(0.8, 1)	0.000000	2.575829	2.542259	2.590169	2.287361	2.185600
			(1, 0.8)	0.000000	2.575829	2.549772	2.585196	2.239521	2.399992
			(0.8, 0.8)	0.000000	2.575829	2.558969	2.571783	2.143252	2.405538
		0.75	(1, 1)	0.674490	2.575829	2.532415	2.745093	2.313190	2.157104
			(0.8, 1)	0.674490	2.575829	2.542259	2.734727	2.239030	2.173154
			(1, 0.8)	0.674490	2.575829	2.549772	2.720949	2.155968	2.256922
			(0.8, 0.8)	0.674490	2.575829	2.558969	2.688348	1.999910	2.257962
	0.4	0.50	(1, 1)	0.000000	2.575829	2.513759	2.584926	2.304964	2.153711
			(0.8, 1)	0.000000	2.575829	2.528913	2.578815	2.256674	2.192592
			(1, 0.8)	0.000000	2.575829	2.533966	2.575764	2.234588	2.387809
			(0.8, 0.8)	0.000000	2.575829	2.549772	2.558687	2.132960	2.403011
		0.75	(1, 1)	0.674490	2.575829	2.513759	2.725928	2.287686	2.153730
			(0.8, 1)	0.674490	2.575829	2.528913	2.709532	2.198006	2.175594
			(1, 0.8)	0.674490	2.575829	2.533966	2.701773	2.158765	2.253450
			(0.8, 0.8)	0.674490	2.575829	2.549772	2.663237	1.991353	2.257476
	0.5	0.5	(1, 1)	0.000000	2.575829	2.491965	2.573627	2.278894	2.150032
			(0.8, 1)	0.000000	2.575829	2.513759	2.564055	2.223744	2.200070
			(1, 0.8)	0.000000	2.575829	2.513759	2.563934	2.223621	2.368886
			(0.8, 0.8)	0.000000	2.575829	2.538317	2.543658	2.117890	2.398080
		0.75	(1, 1)	0.674490	2.575829	2.491965	2.704109	2.261000	2.150036
			(0.8, 1)	0.674490	2.575829	2.513759	2.680187	2.155551	2.178152
			(1, 0.8)	0.674490	2.575829	2.513759	2.679966	2.155500	2.246676
			(0.8, 0.8)	0.674490	2.575829	2.538317	2.635724	1.977861	2.256011
0.5	0.3	0.50	(1, 1)	0.000000	2.575829	2.532415	2.564274	2.279878	2.142149
			(0.8, 1)	0.000000	2.575829	2.542259	2.563423	2.218527	2.187922
			(1, 0.8)	0.000000	2.575829	2.549772	2.561436	2.144792	2.349097
			(0.8, 0.8)	0.000000	2.575829	2.558969	2.552805	1.987892	2.375525
		0.75	(1, 1)	0.674490	2.575829	2.532415	2.744433	2.254123	2.142182
			(0.8, 1)	0.674490	2.575829	2.542259	2.737397	2.146473	2.168536
			(1, 0.8)	0.674490	2.575829	2.549772	2.726401	2.020355	2.229198
			(0.8, 0.8)	0.674490	2.575829	2.558969	2.695093	1.803399	2.237989
	0.4	0.50	(1, 1)	0.000000	2.575829	2.513759	2.556849	2.253368	2.134343
			(0.8, 1)	0.000000	2.575829	2.528913	2.554990	2.179122	2.194796
			(1, 0.8)	0.000000	2.575829	2.533966	2.553545	2.144605	2.323219
			(0.8, 0.8)	0.000000	2.575829	2.549772	2.541300	1.976017	2.365879
		0.75	(1, 1)	0.674490	2.575829	2.513759	2.728359	2.224014	2.134356
			(0.8, 1)	0.674490	2.575829	2.528913	2.715450	2.090076	2.169073
			(1, 0.8)	0.674490	2.575829	2.533966	2.708449	2.029105	2.217337
			(0.8, 0.8)	0.674490	2.575829	2.549772	2.670015	1.760296	2.233793
	0.5	0.5	(1, 1)	0.000000	2.575829	2.491965	2.547819	2.225199	2.126321
			(0.8, 1)	0.000000	2.575829	2.513759	2.543463	2.137472	2.201825
			(1, 0.8)	0.000000	2.575829	2.513759	2.543303	2.137357	2.294858
			(0.8, 0.8)	0.000000	2.575829	2.538317	2.360600	1.931625	2.355433
		0.75	(1, 1)	0.674490	2.575829	2.491965	2.708278	2.193302	2.126300
			(0.8, 1)	0.674490	2.575829	2.513759	2.687160	2.031972	2.169525
			(1, 0.8)	0.674490	2.575829	2.513759	2.687205	2.031861	2.203476
			(0.8, 0.8)	0.674490	2.575829	2.538317	2.640342	1.744658	2.228230

^†: $\mathcal{F}=P_0(Z_{-}^{(1)}\le c_0)$ = futility probability at first stage

4.4 |. Global and marginal power

Under alternatives we have

$$\begin{gathered} Z^{(1)}~N\left(\frac{\sqrt{tN}\delta }{\sigma },1\right),Z_{+}^{(1)}~N\left(\frac{\sqrt{tN}{\delta }_{+}}{{\sigma }_{+}},1\right),Z_{-}^{(1)}~N\left(\frac{\sqrt{tN}{\delta }_{-}}{{\sigma }_{-}},1\right), \\ Z~N\left(\frac{\sqrt{N}\delta }{\sigma },1\right),Z_{+}~N\left(\frac{\sqrt{N}{\delta }_{+}}{{\sigma }_{+}},1\right),\text{and}{Z}_{+}^{*}~N\left(\frac{\sqrt{N}{\delta }_{+}}{{\sigma }_{+}^{*}},1\right). \end{gathered}$$

The global power is:

$$\begin{gathered} 1-\beta =P\left(\mathrm{Reject}{H}_0|H_1\right) \\ =P\left(\text{Reject}{H}_{0a}\text{or Reject}{H}_{0+}|H_1\right) \\ =P_1\left(Z^{(1)}>c_1\text{or}{Z}_{+}^{(1)}>c_2\right) \\ +P_1\left(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{-}^{(1)}>c_0,Z>b_1\right) \\ +P_1\left(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{-}^{(1)}>c_0,Z\le b_1,Z_{+}>b_2\right) \\ +P_1\left(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{-}^{(1)}\le c_0,Z_{+}^{*}>b_3\right) \\ =p_1+p_{2a}+p_{2+}+p_{2+}^{*}, \end{gathered}$$

where

$$\begin{gathered} p_1=P\left(\text{Reject}{H}_{0a}\text{or Reject}{H}_{0+}\text{at Stage}\text{I}|H_1\right) \\ =P_1\left(Z^{(1)}>c_1\text{or}{Z}_{+}^{(1)}>c_2\right), \end{gathered}$$

$$\begin{gathered} p_{2a}=P_1\left(\text{Accept}{H}_0\text{at Stage I and Reject}{H}_{0a}\text{at Stage}\text{I}{\text{I}}_A\right) \\ =P_1\left(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{-}^{(1)}>c_0,Z>b_1\right), \end{gathered}$$ (17)

$$\begin{gathered} p_{2+}=P_1\left(\text{Accept}{H}_0\text{at Stage I, Accept}{H}_{0a}\text{but Reject}{H}_{0+}{\text{at Stage II}}_{\text{A}}\right) \\ =P_1\left(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{-}^{(1)}>c_0,Z\le b_1,Z_{+}>b_2\right), \end{gathered}$$ (18)

$$\begin{gathered} p_{2+}^{*}=P_1\left(\text{Accept}{H}_0\text{at Stage I},\text{Reject}{H}_{0+}{\text{at Stage II}}_{\text{B}}\right) \\ =P_1\left(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{-}^{(1)}\le c_0,Z_{+}^{*}>b_3\right). \end{gathered}$$ (19)

Power for testing the treatment effect in the overall cohort is:

$$\begin{gathered} 1-{\beta }_a=P_1\left(\mathrm{Reject}{H}_{0a}\right) \\ =P_1\left(\text{Reject}{H}_{0a}\text{at Stage I}\right)+P_1\left(\text{Reject}{H}_{0a}{\text{at Stage II}}_{\text{A}}\right) \\ =P_1\left(Z^{(1)}>c_1\right)+P_1\left(\text{Reject}{H}_{0a}\text{at Stage}{\text{II}}_{\text{A}}\right) \\ =p_{1a}+p_{2a}, \end{gathered}$$

where $p_{1a}=P\left(Z^{(1)}>c_1|\delta >0\right)$ and P_2a is defined in Eq. (17).

Power for testing the treatment effect in the marker-positive is:

$$\begin{gathered} 1-{\beta }_{+}=P_1\left(\mathrm{Reject}{H}_{0+}\right) \\ =P_1\left(\text{Reject}{H}_{0+}\text{at Stage I}\right)+P_1\left(\text{Reject}{H}_{0+}{\text{at Stage II}}_{\text{A}}\right)+P_1\left(\text{Reject}{H}_{0+}{\text{at Stage II}}_{\text{B}}\right) \\ =P_1\left(Z_{+}^{(1)}>c_2\right)+P_1\left(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{-}^{(1)}>c_0,Z_{+}>b_2\right) \\ +P_1\left(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{-}^{(1)}\le c_0,Z_{+}^{*}>b_3\right) \\ =p_{1+}+p_{2+}^{(m)}+p_{2+}^{*}, \end{gathered}$$

where

$$\begin{gathered} p_{1+}=P(Z_{+}^{(1)}>c_2|{\delta }_{+}>0) \\ {p}_{2+}^{(m)}=P_1(Z^{(1)}\le c_1,Z_{+}^{(1)}\le c_2,Z_{-}^{(1)}>c_0,Z_{+}>b_2), \end{gathered}$$

and $p_{2+}^{*}$ as defined in Eq. (19). Notice p₂₊ has the term Z ≤ b₁ but $p_{2+}^{(m)}$ does not.

Figures 1, 2 and 3 show the contour plots of power surfaces for the global (testing H₀), overall cohort (testing H_0a) and marker-positive cohort (testing H₀₊) hypotheses, respectively, across 0.10 ≤ δ ≤ 0.40 and 0.20 ≤ δ₊ ≤ 0.55 for α = 0.025 by N, p, λ_sen and λ_spec assuming Δ_C = 0, ${\sigma }_{T0}^2={\sigma }_{T1}^2={\sigma }_{C0}^2={\sigma }_{C1}^2=1,w_1=p,w_0=1-p,{\alpha }_{1a}={\alpha }_{1b}=0.005,\pi \left(={\alpha }_{2a}/{\alpha }_2\right)=\text{0.5},$, c₀ = 0, r = 0.5 and t = 0.5. Note that ${\mathrm{\Delta }}_T-{\mathrm{\Delta }}_C={\delta }_{+}-{\delta }_{-}={\delta }_{+}-\left(\delta -w_1{\delta }_{+}\right)/w_0$. From Figure 1, we see that the global power is increasing for both δ and δ₊. The power increases as N, p, λ_sen or λ_spec increases as well. The impact of the imperfect assay and classification rule through λ_sen and λ_spec is obvious when contrasting the power figures with the same p and N. For example, with p = 0.4 and N = 500, δ = 0.15 and δ₊ = 0.5, Figure 1 A with λ_sen = λ_spec = 1 shows power > 90%; however Figure 1 D with λ_sen = λ_spec = 0.8 shows power only about 70%. We also see that λ_sen has less impact than λ_spec in terms of global power. Figure 3 shows that the power for test marker-positive cohort effect increases with increasing δ₊, but decreases with increasing δ for fixed δ₊. Similar to global power, we also see that λ_sen has less impact than λ_spec in terms of power for marker-positive cohort effect. Figure 2 shows, as expected, that the power for test overall cohort effect increases with increasing δ for fixed δ₊, but λ_sen and λ_spec have similar impact on the power for overall cohort effect.

FIGURE 1.

Contour of global power surface for α = 0.025

FIGURE 2.

Contour of power surface for overall cohort effect when α = 0.025

FIGURE 3.

Contour of power surface for marker-positive cohort effect when α = 0.025

4.5 |. Sample size determination

Given the type I error rate and power, we can use the information in Sections 4.3 and 4.4 to determine the sample size needed for the study. Tables B1 and B2 in Appendix B show the total sample size based on the marker-positive and global powers of 80% and 90% with various p, δ, δ₊, λ_sen, λ_spec and information time t for α = 0.025 (one-sided test) assuming ${\sigma }_{T0}^2={\sigma }_{T1}^2={\sigma }_{C0}^2={\sigma }_{C1}^2=1,w_1=p,w_0=1-p,{\alpha }_{1a}={\alpha }_{1b}=0.005,\pi \left(={\alpha }_{2a}/{\alpha }_2\right)=0.5,c_0=0,\text{and}r=0.5$.

Table B1 is useful when the study plan sets on a sufficient power (e.g., 80% or 90%) in testing H₀₊ for the marker-positive cohort. It shows the required total sample size N and the resulting global power and the power for testing H_0a for the whole population. In this case, the global power is always higher since it is to test the composite hypothesis H₀. The power for the whole population would be low when the emphasis is on the much hopeful marker-positive subset with a fairly possible enrichment (${\mathcal{F}}_p$ = 0.5). This is reflected by displaying δ = 0.15 while δ₊ = 0.3 to 0.5 in this table. In these situations, though the power is low for testing the overall treatment effect, the design has sufficient power to detect the treatment effect on the marker-positive subset. This indicates that fewer patients are needed when the main objective of the study is to show treatment effect on the marker-positive subset. Clearly this is the main advantage of the two-stage enrichment design. Contrasting the case of λ_sen = λ_spec = 0.8 with the rest columns, for the same t and p, we can see the big impact of imperfect assay and classification rule on the increase of sample size requirement for a study.

Table B2 is useful when the objective of the trial is set on testing the composite hypothesis H₀. The sample size is calculated to ensure that there is enough power to find significance on either the whole population or the marker-positive subset under H₁. As shown, the total sample size monotonically decreases as either δ or δ₊ increases. We also see the adverse impact on the sample size requirement with imperfect marker assay and classification rule. Moreover, λ_spec has more impact on the total sample sizes than λ_sen. The required total sample size is larger for later interim analysis time t since the enrichment would be delayed.

5 |. NUMERICAL EXAMPLES

Immunotherapy is a new paradigm for the treatment of non-small-cell lung cancer (NSCLC), and targeting the PD-1/PD-L1 pathway is a promising therapeutic option. Pembrolizumab is a new immunotherapy that blocks the PD-1 pathway and restores the body’s immune response against cancer cells and allows the immune system to recognize and kill cancer cells. We use the KEYNOTE-10 trial as an example to illustrate our method. This was a (phase 2/3) randomized trial to study Pembrolizumab versus Docetaxel for previously treated, PD-L1-positive, advanced NSCLC patients¹⁰. This trial stratified qualified subjects by biomarker’s TPS (tumor proportion score ≥ 50% vs 1 – 49%), which measures the extent of PD-L1 expression, then randomized subjects with 1:1:1 ratio to three treatment groups within each high and low TPS stratum. The companion diagnostic assay for PD-L1 expression was the Dako EnVision FLEX+HRP-Polymer kit using the 22C3 antibody clone, which was validated in the phase 1 KEYNOTE-001 trial¹¹. Here we use the phase 1 KEYNOTE-001 data as the basis to “re-design” the KEYNOTE-10 as an “imaginary” two-stage enrichment trial to illustrate our method. For illustration, also since there was no significant difference between the two test doses of Pembrolizumab, we only look at the Pembrolizumab 2 mg (Pem) versus Docetaxel (Dox), which is the control/standard-of-care.

From the phase 1 KEYNOTE-001 trial (Figs S3A and S4)¹¹, the prevalence was about 0.39 for TPS < 1%, 0.38 for TPS = 1–49%, and 0.23 for TPS ≥ 50%; λ_sen = λ_spec ≈ 0.80. Thus, we estimate the prevalence rate of the PD-L1 true “strongly positives” (TPS ≥ 50%) among the PD-L1 positive (TPS ≥ 1%) NSCLC patients being p ≈ 0.40; the appeared PD-L1 “strongly positive” prevalence is $q\equiv P(M=1)=p{\lambda }_{sen}+(1-p)\left(1-{\lambda }_{spec}\right)=0.40\times 0.80+(1-0.40)(1-0.80)=0.44$. Hence, $PPV=\tau \equiv \frac{p{\lambda }_{xxn}}{q}=0.4\times 0.80/0.44=0.73$, and $NPV=\eta \equiv \frac{(1-p){\lambda }_{spec}}{1-q}=(1-0.4)\times 0.80/(1-0.44)=0.86$.

The real phase 2/3 KEYNOTE-10 trial had both overall survival and progression-free survival (PFS) as primary endpoints. We only use PFS for the “imaginary” trial for illustration purpose. Suppose that the overall (one-sided) α = 0.025 and an interim analysis is planned at t = 0.5, with α₁ = 0.01 allocated for Stage I. As stated in Herbst et al.¹⁰, the study aimed to show a benefit of Pem over Dox in PFS in patients with TPS ≥ 50% as well as in the whole TPS ≥ 1% cohort, so we allocate α_1a = α_1b = α₁/2 = 0.005. Moreover, assume that we set the futility probability ${\mathcal{F}}_p$ for the PD-L1 “weakly positive” (TPS 1–49%) subset to be 50% (implying c₀ = 0). Then from Eq. (16), α₂ = (0.025 – 0.01)/2 = 0.0075 for Stage II_A, and equal amount of ${\alpha }_{2*+}=(0.025-0.01)/2=0.0075$ for Stage II_B. In case of Stage II_A, we further allocate ${\alpha }_{2a}={\alpha }_2/2=0.00375$ for testing the overall cohort, and equally the rest 0.00375 for the PD-L1 strongly positive subset. The log-rank test on the hazard ratios of PFS assuming an exponential distribution justifies ${\sigma }_{T0}^2={\sigma }_{T1}^2={\sigma }_{C0}^2={\sigma }_{C1}^2=1$. From Table 1, these design parameters lead to the critical values (c₁, c₂, b₁, b₂, b₃) = (2.5758, 2.5498, 2.5413, 1.9760, 2.3659).

To illustrate power calculation we take the treatment effect information from Herbst et al.¹⁰. Let δ be the log-hazard ratio (Dox vs Pem) = 0.15 for the overall cohort. With a total of N = 1,000 patients and prevalence rate p = 0.40, we expect to have 190 PD-L1 strongly positive patients enrolled for the first stage and perform an interim analysis at t = 0.5. Assume δ₊ = log-hazard ratio (Dox vs Pem) = 0.50 for the PD-L1 strongly positive subset. The power for testing the global hypothesis H₀ is 96% (> 95% from Fig 1 D), the power for testing H₀₊ is 90% (> 85% from Fig 3 D) for the PD-L1 strongly positive subset, and the power for testing H_0a is 26% (about 25% from Fig 2 D) for the overall cohort. With N = 500 patients, power for testing the global hypothesis H₀ is 75%, power for testing H₀₊ is 69% for the PD-L1 strongly positive subset, and power for testing H_0a is 14% for the overall cohort.

To illustrate sample size calculation, with the same alpha allocation, λ_sen = λ_spec = 0.80, and treatment effects δ = 0.15 and δ₊ = 0.50 as above, trials usually aim a power of either 80% or 90% for the marker positive subset. Assuming a target of 90% power for testing H₀₊, Table B1 shows that a total sample size of N = 1004 is needed. The global power for the composite hypothesis is 96%, and the power for the overall cohort is 26%. If the target power is 80% for testing H₀₊, then N = 671 is needed. The global power for the composite hypothesis is 86%, and the power for the overall cohort is 18.5%.

6 |. DISCUSSION

First, the two-stage enrichment design discussed in this paper should not be confused with the so-called “targeted design”, which is usually only one-stage design but sometimes is also called “enriched design” or “enrichment design” by some authors. Gao et al.¹² proposed a multistage biomarker-directed targeted design, not an enrichment design. The targeted design does not include the biomarker-negative subset, hence only the hypothesis H₀₊ can be tested, not the hypothesis H_0a on the overall population. The target design is suitable when there is already evidence that the treatment may have beneficial effect only on the marker-positive subset, not on the marker-negative subset. On the other hand, the sequential two-stage enrichment design is suitable for the situation where such evidence is lacking. It is interesting to note that the KEYNOTE-010 trial, included all TPS ≥ 1% NSCLC patients to receive test treatment as their second-line therapy, showed significant treatment effect of Pembrolizumab over the SOC control for the PD-L1 strongly positive subset, but not in the overall cohort¹⁰. With this evidence as background, the next trial KEYNOTE-024¹³ for Pembrolizumab to treat NSCLC patients as their first-line therapy used a “targeted design”, enrolled only the PD-L1 strongly positive cohort (TPS ≥ 50%) and was a successful study to receive the FDA’s fast-tract approval.

The two-stage enrichment design, where the enrichment occurs at the second stage by stopping the enrollment of marker-negatives as indicated with an futility analysis and by continuing enrollment of only the marker-positives, is more efficient than the conventional stratified design without enrichment in testing both H₀₊ and H₀. For example, as we illustrated in Section 5 that, for overall α = 0.025 and allocation of α₁ = 0.01 for testing H_0a at stage I, and other alpha allocations therein, assuming sensitivity=specificity=0.8 to detect treatment effects δ = 0.15 and δ₊ = 0.50, if the target power is 80% for testing H₀₊, then N = 671 is needed by the proposed two-stage enrichment design (as shown in Table 2). The global power for the composite hypothesis is 86%, and the power for the overall cohort is 18.5%. In contrast, without such enrichment, following simply the (1-stage) stratified design in Section 3, the required sample size would be larger, N = 714. The global power for the composite hypothesis would be 82%, and the power for the overall cohort would be 27.5%. The two-stage enrichment design obviously enhances the power for testing the (more hopeful) marker-positive subset and the global hypothesis with a smaller sample size by reducing the power for the (less hopeful) overall cohort.

In addition, Figure 4 shows the comparisons on global power and sample size for 2-stage and single stage designs with various δ, δ₊, λ_sen and λ_spec when α = 0.025 with α₁ = 0.005, α₂ = 0.02, r = 0.5, p = 0.3 for single stage design, and α_1a = 0.005, α_1b = 0.005, r = 0.5, p = 0.3, t = 0.5 for 2-stage design. The reference line in the top row of the graph is that horizontal and vertical values are the same. The top row of the graph shows the comparisons of powers between 2-stage and single stage designs with the same total sample size. The figure shows that, with the same sample size, the global power for 2 stage design is larger than that for single stage design since all the curves are above the reference line. The bottom row of the graph shows the ratio of sample sizes needed for 2-stage design versus single stage design in order to achieve the same global power from 60 to 90%. The figure shows that the sample size needed for 2-stage design is much smaller than that for single stage design in order to achieve the same global power.

FIGURE 4.

Power and sample size comparison between two-stage and single stage design

Finally, in addition to the tables and graphs provided as a part of the illustration in this paper, we make our R programs available upon request for readers to calculate power and sample size for their enrichment designs as needed. The current research is limited to trial designs with two treatment groups and a fixed treatment allocation ratio r, as shown in the design schema. Extension to more than two treatment groups and application of method of Tang and Zhou¹⁴ for optimal allocation in enrichment setting are potential topics for future research.

APPENDIX

A. DERIVATIONS OF EQUATIONS (14) AND (15)

A.1. Correlation matrix of $\left(Z^{(1)},Z_{+}^{(1)},Z_{-}^{(1)},Z,Z_{+}\right)$ for scenario II_A

Between the Z-statistics of stage 1, we have from Section 3.1:

$$\begin{gathered} Cov\left(Z_{+}^{(1)},Z_{-}^{(1)}\right)=\frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}} \\ Cov\left(Z^{(1)},Z_{+}^{(1)}\right)=\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}} \\ Cov\left(Z^{(1)},Z_{-}^{(1)}\right)=\frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}. \end{gathered}$$

Between the Z-statistics of Stage I and Stage II, since

$$\begin{gathered} Cov\left(\widehat{\delta },{\widehat{\delta }}^{(1)}\right)=tVar\left({\widehat{\delta }}^{(1)}\right)=t\frac{{\sigma }^2}{tN}=\frac{{\sigma }^2}{N} \\ Cov\left(\widehat{\delta },{\widehat{\delta }}_{+}^{(1)}\right)=Cov\left(t{\widehat{\delta }}^{(1)}+(1-t){\widehat{\delta }}^{(2)},{\widehat{\delta }}_{+}^{(1)}\right)=tCov\left({\widehat{\delta }}^{(1)},{\widehat{\delta }}_{+}^{(1)}\right) \\ =tCov\left(w_1{\widehat{\delta }}_{+}^{(1)}+w_0{\widehat{\delta }}_{-}^{(1)},{\widehat{\delta }}_{+}^{(1)}\right) \\ =t{w}_{1}Var\left({\widehat{\delta }}_{+}^{(1)}\right)+t{w}_{0}Cov\left({\widehat{\delta }}_{+}^{(1)},{\widehat{\delta }}_{-}^{(1)}\right) \\ =t{w}_1\frac{{\sigma }_{+}^2}{tN}+t{w}_0\frac{{\sigma }_{+,-}}{tN}=\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{N} \end{gathered}$$

$$\begin{gathered} Cov\left(\widehat{\delta },{\widehat{\delta }}_{-}^{(1)}\right)=tCov\left({\widehat{\delta }}^{(1)},{\widehat{\delta }}_{-}^{(1)}\right)=tCov\left(w_1{\widehat{\delta }}_{+}^{(1)}+w_0{\widehat{\delta }}_{-}^{(1)},{\widehat{\delta }}_{-}^{(1)}\right) \\ =t{w}_{1}Cov\left({\widehat{\delta }}_{+}^{(1)},{\widehat{\delta }}_{-}^{(1)}\right)+t{w}_{0}Var\left({\widehat{\delta }}_{-}^{(1)}\right) \\ =t{w}_1\frac{{\sigma }_{+,-}}{tN}+t{w}_0\frac{{\sigma }_{-}^2}{tN}=\frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{N} \end{gathered}$$

$$\begin{gathered} Cov\left({\widehat{\delta }}_{+},{\widehat{\delta }}^{(1)}\right)=Cov\left(t{\widehat{\delta }}_{+}^{(1)}+(1-t){\widehat{\delta }}_{+}^{(2)},w_1{\widehat{\delta }}_{+}^{(1)}+w_0{\widehat{\delta }}_{-}^{(1)}\right) \\ =tCov\left({\widehat{\delta }}_{+}^{(1)},w_1{\widehat{\delta }}_{+}^{(1)}+w_0{\widehat{\delta }}_{-}^{(1)}\right) \\ =t{w}_{1}Var\left({\widehat{\delta }}_{+}^{(1)}\right)+t{w}_{0}Cov\left({\widehat{\delta }}_{+}^{(1)},{\widehat{\delta }}_{-}^{(1)}\right) \\ =t{w}_1\frac{{\sigma }_{+}^2}{tN}+t{w}_0\frac{{\sigma }_{+,-}}{tN}=\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{N} \end{gathered}$$

$$\begin{gathered} Cov\left({\widehat{\delta }}_{+},{\widehat{\delta }}_{+}^{(1)}\right)=Cov\left(t{\widehat{\delta }}_{+}^{(1)}+(1-t){\widehat{\delta }}_{+}^{(2)},{\widehat{\delta }}_{+}^{(1)}\right)=tVar\left({\widehat{\delta }}_{+}^{(1)}\right) \\ =t\frac{{\sigma }_{+}^2}{tN}=\frac{{\sigma }_{+}^2}{N} \end{gathered}$$

$$\begin{gathered} Cov\left({\widehat{\delta }}_{+},{\widehat{\delta }}_{-}^{(1)}\right)=Cov\left(t{\widehat{\delta }}_{+}^{(1)}+(1-t){\widehat{\delta }}_{+}^{(2)},{\widehat{\delta }}_{-}^{(1)}\right) \\ =tCov\left({\widehat{\delta }}_{+}^{(1)},{\widehat{\delta }}_{-}^{(1)}\right)=t\frac{{\sigma }_{+,-}}{tN}=\frac{{\sigma }_{+,-}}{N}, \end{gathered}$$

we have

$$\begin{gathered} Cov\left(Z_{+},Z^{(1)}\right)=\frac{\sqrt{N}\sqrt{tN}Cov\left({\widehat{\delta }}_{+},{\widehat{\delta }}^{(1)}\right)}{{\sigma }_{+}\sigma }=\frac{\sqrt{N}\sqrt{tN}\left(w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}\right)}{{\sigma }_{+}\sigma N}=\sqrt{t}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{{\sigma }_{+}\sigma } \\ Cov\left(Z_{+},Z_{+}^{(1)}\right)=\frac{\sqrt{N}\sqrt{tN}Cov\left({\widehat{\delta }}_{+},{\widehat{\delta }}_{+}^{(1)}\right)}{{\sigma }_{+}{\sigma }_{+}}=\sqrt{t} \end{gathered}$$

$$\begin{gathered} Cov\left(Z_{+},Z_{-}^{(1)}\right)=\frac{\sqrt{N}\sqrt{tN}Cov\left({\widehat{\delta }}_{+},{\widehat{\delta }}_{-}^{(1)}\right)}{{\sigma }_{+}{\sigma }_{-}}=\sqrt{t}N\frac{{\sigma }_{+,-}}{N{\sigma }_{+}{\sigma }_{-}} \\ =\sqrt{t}\frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}. \end{gathered}$$

Furthermore, from

$$\begin{gathered} Cov\left(\widehat{\delta },{\widehat{\delta }}_{+}\right)=Cov\left(t{\widehat{\delta }}^{(1)}+(1-t){\widehat{\delta }}^{(2)},t{\widehat{\delta }}_{+}^{(1)}+(1-t){\widehat{\delta }}_{+}^{(2)}\right) \\ =t^{2}Cov\left({\widehat{\delta }}^{(1)},{\widehat{\delta }}_{+}^{(1)}\right)+{(1-t)}^{2}Cov\left({\widehat{\delta }}^{(2)},{\widehat{\delta }}_{+}^{(2)}\right) \\ =t^{2}Cov\left(w_1{\widehat{\delta }}_{+}^{(1)}+w_0{\widehat{\delta }}_{-}^{(1)},{\widehat{\delta }}_{+}^{(1)}\right)+{(1-t)}^{2}Cov\left(w_1{\widehat{\delta }}_{+}^{(2)}+w_0{\widehat{\delta }}_{-}^{(2)},{\widehat{\delta }}_{+}^{(2)}\right) \\ =t^2{w}_{1}Var\left({\widehat{\delta }}_{+}^{(1)}\right)+t^2{w}_{0}Cov\left({\widehat{\delta }}_{-}^{(1)},{\widehat{\delta }}_{+}^{(1)}\right) \\ +{(1-t)}^2{w}_{1}Var\left({\widehat{\delta }}_{+}^{(2)}\right)+{(1-t)}^2{w}_{0}Cov\left({\widehat{\delta }}_{-}^{(2)},{\widehat{\delta }}_{+}^{(2)}\right) \\ =t^2{w}_1\frac{{\sigma }_{+}^2}{tN}+t^2{w}_0\frac{{\sigma }_{+,-}}{tN}+{(1-t)}^2{w}_1\frac{{\sigma }_{+}^2}{(1-t)N}+{(1-t)}^2{w}_0\frac{{\sigma }_{+,-}}{(1-t)N} \\ =\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{N}, \end{gathered}$$

we have

$$Cov\left(Z,Z_{+}\right)=\frac{\sqrt{N}\sqrt{N}Cov\left(\widehat{\delta },{\widehat{\delta }}_{+}\right)}{\sigma {\sigma }_{+}}=\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}.$$

In summary, the covariance matrix of the Z-statistics for scenario II_A is:

$$Cov\left(\left[\begin{array}{c}{Z}^{(1)}\\ Z_{+}^{(1)}\\ Z_{-}^{(1)}\\ Z\\ Z_{+}\end{array}\right]\right)=\left[\begin{array}{ccccc}1& \frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}& \frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}& \sqrt{t}& \sqrt{t}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{{\sigma }_{+}\sigma }\\ \frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}& 1& \frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}& \sqrt{t}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}& \sqrt{t}\\ \frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}& \frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}& 1& \sqrt{t}\frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}& \sqrt{t}\frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}\\ \sqrt{t}& \sqrt{t}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}& \sqrt{t}\frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}& 1& \frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}\\ \sqrt{t}\frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{{\sigma }_{+}\sigma }& \sqrt{t}& \sqrt{t}\frac{{\sigma }_{+-}}{{\sigma }_{+}{\sigma }_{-}}& \frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}& 1\end{array}\right].$$

A.2. Correlation matrix of $\left(Z^{(1)},Z_{+}^{(1)},Z_{-}^{(1)},Z_{+}^{*}\right)$ for scenario II_B

For Scenario II_B, since

$$\begin{gathered} Cov\left({\widehat{\delta }}_{+}^{*},{\widehat{\delta }}_{+}^{(1)}\right) \\ =Cov\left(\frac{-\eta \left(\frac{\mathit{qt}{\overline{Y}}_{T1}^{(1)}+(1-t){\overline{Y}}_{T1}^{(2)}}{qt+(1-t)}-\frac{\mathit{qt}{\overline{Y}}_{C1}^{(1)}+(1-t){\overline{Y}}_{C1}^{(2)}}{qt+(1-t)}\right)+(1-\tau )({\overline{Y}}_{T0}^{(1)}-{\overline{Y}}_{C0}^{(1)})}{1-\tau -\eta },\frac{-\eta \left({\overline{Y}}_{T1}^{(1)}-{\overline{Y}}_{C1}^{(1)}\right)+(1-\tau )\left({\overline{Y}}_{T0}^{(1)}-{\overline{Y}}_{C0}^{(1)}\right)}{1-\tau -\eta }\right) \\ =\frac{-\eta (-\eta )\frac{qt}{qt+(1-t)}(Var\left({\overline{Y}}_{T1}^{(1)}\right)+\mathit{V\; ar}\left({\overline{Y}}_{C1}^{(1)}\right))+{(1-\tau )}^2\left(Var\left({\overline{Y}}_{T0}^{(1)}\right)+Var\left({\overline{Y}}_{C0}^{(1)}\right)\right)}{{(1-\tau -\eta )}^2} \\ =\frac{{\eta }^2\frac{qt}{qt+(1-t)}\left(\frac{{\sigma }_{T*}^2}{rqtN}+\frac{{\sigma }_{C*}^2}{(1-r)qtN}\right)+{(1-\tau )}^2\left(\frac{{\sigma }_{T\varphi }^2}{r(1-q)tN}+\frac{{\sigma }_{C\varphi }^2}{(1-r)(1-q)tN}\right)}{{(1-\tau -\eta )}^2} \\ =\frac{{\eta }^2\frac{t}{qt+(1-t)}\left(\frac{{\sigma }_{\mathrm{T*}}^2}{r}+\frac{{\sigma }_{\mathrm{C*}}^2}{1-r}\right)+{(1-\tau )}^2\left(\frac{{\sigma }_{T\varphi }^2}{r(1-q)}+\frac{{\sigma }_{C\varphi }^2}{(1-r)(1-q)}\right)}{tN{(1-\tau -\eta )}^2}=\frac{{\sigma }_{+,+}^{*}}{\sqrt{t}N}. \end{gathered}$$

$$\begin{gathered} Cov\left({\widehat{\delta }}_{+}^{*},{\widehat{\delta }}_{-}^{(1)}\right) \\ =Cov\left(\frac{-\eta \left(\frac{\mathit{qt}{\overline{Y}}_{T1}^{(1)}+(1-t){\overline{Y}}_{T1}^{(2)}}{qt+(1-t)}-\frac{\mathit{qt}{\overline{Y}}_{C1}^{(1)}+(1-t){\overline{Y}}_{C1}^{(2)}}{qt+(1-t)}\right)+(1-\tau )({\overline{Y}}_{T0}^{(1)}-{\overline{Y}}_{C0}^{(1)})}{1-\tau -\eta },\frac{-\eta \left({\overline{Y}}_{T1}^{(1)}-{\overline{Y}}_{C1}^{(1)}\right)+(1-\tau )\left({\overline{Y}}_{T0}^{(1)}-{\overline{Y}}_{C0}^{(1)}\right)}{1-\tau -\eta }\right) \\ =\frac{-\eta (1-\eta )\frac{qt}{qt+(1-t)}(Var\left({\overline{Y}}_{T1}^{(1)}\right)+\mathit{V\; ar}\left({\overline{Y}}_{C1}^{(1)}\right))+(1-\tau )(-\tau )\left(Var\left({\overline{Y}}_{T0}^{(1)}\right)+Var\left({\overline{Y}}_{C0}^{(1)}\right)\right)}{{(1-\tau -\eta )}^2} \\ =\frac{-\eta (1-\eta )\frac{qt}{qt+(1-t)}\left(\frac{{\sigma }_{T*}^2}{rqtN}+\frac{{\sigma }_{C*}^2}{(1-r)qtN}\right)+(1-\tau )(-\tau )\left(\frac{{\sigma }_{T\varphi }^2}{r(1-q)tN}+\frac{{\sigma }_{C\varphi }^2}{(1-r)(1-q)tN}\right)}{{(1-\tau -\eta )}^2} \\ =\frac{-\eta (1-\eta )\frac{t}{qt+(1-t)}\left(\frac{{\sigma }_{\mathrm{T*}}^2}{r}+\frac{{\sigma }_{\mathrm{C*}}^2}{1-r}\right)+(1-\tau )(-\tau )\left(\frac{{\sigma }_{T\varphi }^2}{r(1-q)}+\frac{{\sigma }_{C\varphi }^2}{(1-r)(1-q)}\right)}{tN{(1-\tau -\eta )}^2}=\frac{{\sigma }_{+,-}^{*}}{\sqrt{t}N}. \\ Cov\left({\widehat{\delta }}_{+}^{*},{\widehat{\delta }}^{(1)}\right)=Cov({\widehat{\delta }}_{+}^{*},w_1{\widehat{\delta }}_{+}^{(1)}+w_0{\widehat{\delta }}_{-}^{(1)})=w_1\frac{{\sigma }_{+,+}^{*}}{\sqrt{t}N}+w_1\frac{{\sigma }_{+,-}^{*}}{\sqrt{t}N}, \end{gathered}$$

we have

$$\begin{gathered} Cov\left(Z_{+}^{*},Z_{+}^{(1)}\right)=\frac{\sqrt{N}\sqrt{tN}\mathit{Cov}\left({\widehat{\delta }}_{+}^{*},{\widehat{\delta }}_{+}^{(1)}\right)}{{\sigma }_{+}^{*}{\sigma }_{+}}=\frac{{\sigma }_{+,+}^{*}}{{\sigma }_{+}^{*}{\sigma }_{+}} \\ Cov\left(Z_{+}^{*},Z_{-}^{(1)}\right)=\frac{\sqrt{N}\sqrt{tN}Cov\left({\widehat{\delta }}_{+}^{*},{\widehat{\delta }}_{-}^{(1)}\right)}{{\sigma }_{+}^{*}{\sigma }_{-}}=\frac{{\sigma }_{+,-}^{*}}{{\sigma }_{+}^{*}{\sigma }_{-}} \\ Cov\left(Z_{+}^{*},Z^{(1)}\right)=\frac{\sqrt{N}\sqrt{tN}Cov\left({\widehat{\delta }}_{+}^{*},{\widehat{\delta }}^{(1)}\right)}{{\sigma }_{+}^{*}\sigma }=\frac{w_1{\sigma }_{+,+}^{*}+w_0{\sigma }_{+,-}^{*}}{{\sigma }_{+}^{*}\sigma }. \end{gathered}$$

In summary, the covariance matrix of the Z-statistics for scenario II_B is

$$\mathrm{Cov}\left(\left[\begin{array}{c}{Z}^{(1)}\\ Z_{+}^{(1)}\\ Z_{-}^{(1)}\\ Z_{+}^{*}\end{array}\right]\right)=\left[\begin{array}{cccc}1& \frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}& \frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}& \frac{w_1{\sigma }_{+,+}^{*}+w_0{\sigma }_{+,-}^{*}}{{\sigma }_{+}^{*}\sigma }\\ \frac{w_1{\sigma }_{+}^2+w_0{\sigma }_{+,-}}{\sigma {\sigma }_{+}}& 1& \frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}& \frac{{\sigma }_{+,+}^{*}}{{\sigma }_{+}^{*}{\sigma }_{+}}\\ \frac{w_1{\sigma }_{+,-}+w_0{\sigma }_{-}^2}{\sigma {\sigma }_{-}}& \frac{{\sigma }_{+,-}}{{\sigma }_{+}{\sigma }_{-}}& 1& \frac{{\sigma }_{+,-}^{*}}{{\sigma }_{+}^{*}{\sigma }_{-}}\\ \frac{w_1{\sigma }_{+,+}^{*}+w_0{\sigma }_{+-}^{*}}{{\sigma }_{+}^{*}\sigma }& \frac{{\sigma }_{+,+}^{*}}{{\sigma }_{+}^{*}{\sigma }_{+}}& \frac{{\sigma }_{+,-}^{*}}{{\sigma }_{+}^{*}{\sigma }_{-}}& 1\end{array}\right]$$

B. TOTAL SAMPLE SIZE TABLES FOR SOME COMMON SCENARIOS

TABLE B1.

Total Sample Size when α = 0.025; α_1a = α_1b = 0.005; r = 0.5; c₀ = 0; ${\sigma }_{T0}^2={\sigma }_{T1}^2={\sigma }_{C0}^2={\sigma }_{C1}^2=1,$ and δ = 0.15

(λsen, λspec)			(1, 1)	(0.8, 1)	(1, 0.8)	(0.8, 0.8)	(1, 1)	(0.8, 1)	(1, 0.8)	(0.8, 0.8)
p	δ+		80% power for testing H0+
			t = 0.25				t = 0.5
0.3	0.3	N	1519	2261	9109	16831	1709	2778	4636	8418
		Global	0.914	0.967	1.000	1.000	0.938	0.988	1.000	1.000
		H1a	0.489	0.644	0.919	0.994	0.569	0.770	0.920	0.994
	0.4	N	670	880	1236	7589	697	931	1367	3882
		Global	0.845	0.870	0.905	1.000	0.853	0.886	0.932	0.999
		H1a	0.199	0.267	0.369	0.838	0.223	0.309	0.447	0.841
	0.5	N	374	479	702	1162	380	483	714	1290
		Global	0.824	0.836	0.855	0.909	0.826	0.842	0.866	0.941
		H1a	0.102	0.133	0.193	0.313	0.113	0.149	0.220	0.396
0.4	0.3	N	934	1226	1480	3152	987	1368	1664	4627
		Global	0.856	0.891	0.912	0.992	0.867	0.917	0.941	1.000
		H1a	0.284	0.374	0.438	0.67	0.328	0.458	0.539	0.905
	0.4	N	441	552	712	1135	453	568	721	1223
		Global	0.821	0.836	0.844	0.895	0.823	0.843	0.853	0.923
		H1a	0.116	0.149	0.189	0.296	0.131	0.172	0.217	0.37
	0.5	N	259	315	436	674	264	322	433	671
		Global	0.813	0.82	0.823	0.85	0.812	0.822	0.827	0.861
		H1a	0.065	0.079	0.104	0.158	0.072	0.091	0.117	0.185
0.5	0.3	N	678	849	958	1555	699	895	981	1842
		Global	0.827	0.852	0.852	0.922	0.831	0.863	0.863	0.960
		H1a	0.184	0.235	0.261	0.399	0.211	0.283	0.306	0.539
	0.4	N	341	412	506	764	348	426	502	776
		Global	0.812	0.821	0.821	0.849	0.810	0.823	0.822	0.862
		H1a	0.080	0.098	0.116	0.175	0.090	0.116	0.132	0.210
	0.5	N	206	248	318	482	212	258	314	476
		Global	0.807	0.812	0.811	0.827	0.807	0.812	0.811	0.829
		H1a	0.048	0.057	0.067	0.098	0.053	0.066	0.075	0.114
p	δ+		90% power for testing H0+
			t = 0.25				t = 0.5
0.3	0.3	N	5208	9303	13151	21695	2949	4680	6581	10848
		Global	1.000	1.000	1.000	1.000	0.995	1.000	1.000	1.000
		H1a	0.751	0.905	0.973	0.999	0.750	0.905	0.973	0.999
	0.4	N	1036	1477	4736	11536	1042	1529	2594	5771
		Global	0.943	0.969	1.000	1.000	0.946	0.974	0.997	1.000
		H1a	0.293	0.405	0.663	0.941	0.311	0.451	0.659	0.941
	0.5	N	570	761	1058	5662	557	750	1082	2886
		Global	0.922	0.936	0.954	1.000	0.923	0.939	0.961	0.999
		H1a	0.147	0.200	0.273	0.669	0.151	0.214	0.309	0.667
0.4	0.3	N	1914	2668	5881	13200	1487	2447	3117	6603
		Global	0.985	0.994	1.000	1.000	0.958	0.991	0.997	1.000
		H1a	0.479	0.59	0.754	0.970	0.452	0.66	0.755	0.970
	0.4	N	646	862	1030	4791	636	853	1037	2555
		Global	0.92	0.936	0.944	1.000	0.92	0.939	0.948	0.998
		H1a	0.161	0.219	0.257	0.613	0.173	0.243	0.293	0.618
	0.5	N	370	480	600	993	362	460	588	1004
		Global	0.911	0.919	0.923	0.952	0.91	0.919	0.923	0.96
		H1a	0.085	0.111	0.134	0.215	0.09	0.119	0.149	0.259
0.5	0.3	N	969	1337	1404	7692	972	1371	1429	3879
		Global	0.926	0.952	0.952	1.000	0.927	0.959	0.957	1.000
		H1a	0.250	0.341	0.352	0.813	0.278	0.401	0.413	0.814
	0.4	N	468	597	685	1116	467	591	675	1144
		Global	0.910	0.920	0.919	0.951	0.909	0.920	0.919	0.960
		H1a	0.102	0.132	0.147	0.234	0.112	0.150	0.167	0.291
	0.5	N	279	349	424	660	650	348	415	648
		Global	0.906	0.911	0.911	0.927	0.928	0.910	0.909	0.928
		H1a	0.057	0.071	0.081	0.123	0.146	0.081	0.091	0.145

TABLE B2.

Total Sample Size when α = 0.025; α_1a = α_1b = 0.005; r = 0.5; c₀ = 0; ${\sigma }_{T0}^2={\sigma }_{T1}^2={\sigma }_{C0}^2={\sigma }_{C1}^2=1$

(λsen, λspec)			(1, 1)	(0.8, 1)	(1, 0.8)	(0.8, 0.8)	(1, 1)	(0.8, 1)	(1, 0.8)	(0.8, 0.8)
p	δ	δ+	80% power for testing H0
			t = 0.25				t = 0.5
0.3	0.1	0.2	2269	2513	2973	3438	2354	2589	2958	3309
		0.3	984	1176	1635	2186	1006	1185	1612	2094
		0.4	507	611	954	1376	514	608	932	1321
	0.2	0.3	833	861	948	1024	860	887	935	975
		0.4	568	630	745	864	588	649	741	831
		0.5	369	432	554	699	380	441	550	672
	0.3	0.4	409	415	448	476	424	426	440	451
		0.5	331	349	395	434	343	361	389	415
		0.6	253	281	332	387	262	289	330	372
0.4	0.1	0.2	1756	1987	2351	2877	1812	2047	2337	2790
		0.3	711	834	1134	1577	731	855	1119	1524
		0.4	374	437	644	939	385	451	633	904
	0.2	0.3	732	773	847	937	754	795	839	902
		0.4	439	499	589	724	453	514	585	701
		0.5	269	317	402	539	277	325	398	521
	0.3	0.4	378	387	415	447	385	397	410	428
		0.5	276	301	355	387	286	310	335	374
		0.6	195	223	262	325	202	230	261	315
0.5	0.1	0.2	1413	1616	1874	2401	1448	1664	1858	2341
		0.3	574	667	857	1206	590	694	847	1190
		0.4	308	360	485	706	319	378	478	694
	0.2	0.3	638	688	744	853	653	705	738	835
		0.4	354	408	469	606	362	420	465	591
		0.5	215	254	307	428	221	262	304	418
	0.3	0.4	343	359	382	418	349	366	376	403
		0.5	232	259	285	343	238	266	283	334
		0.6	158	184	209	274	161	189	207	267
p	δ	δ+	90% power for testing H0
			t = 0.25				t = 0.5
0.3	0.1	0.2	3225	3576	3995	4526	3262	3600	3977	4398
		0.3	1459	1774	2227	2941	1439	1737	2192	2835
		0.4	760	947	1289	1852	732	899	1256	1765
	0.2	0.3	1181	1175	1247	1329	1153	1191	1236	1279
		0.4	806	896	1002	1135	816	902	996	1104
		0.5	537	635	756	931	538	632	747	904
	0.3	0.4	549	558	586	616	545	565	578	589
		0.5	457	483	521	565	466	490	518	547
		0.6	359	399	446	508	363	402	444	494
0.4	0.1	0.2	2487	2864	3177	3831	2507	2867	3156	3733
		0.3	1009	1236	1521	2117	999	1203	1497	2045
		0.4	523	641	856	1249	518	621	839	1203
	0.2	0.3	1009	1064	1126	1227	1022	1079	1120	1192
		0.4	622	719	796	963	627	719	790	939
		0.5	384	467	542	723	383	459	535	730
	0.3	0.4	509	524	547	582	516	530	543	563
		0.5	385	421	451	510	391	426	449	497
		0.6	277	322	354	432	279	322	352	421
0.5	0.1	0.2	1949	2290	2511	3207	1963	2293	2490	3134
		0.3	777	933	1132	1604	783	934	1119	1567
		0.4	412	494	638	932	420	502	629	914
	0.2	0.3	874	950	995	1126	885	960	989	1097
		0.4	488	577	629	810	491	578	623	791
		0.5	294	359	409	571	295	358	404	558
	0.3	0.4	464	487	505	549	470	492	501	547
		0.5	320	362	382	459	324	365	379	449
		0.6	217	260	280	366	219	260	277	344