Sample Size Calculations for Combination Drugs of 2 Monotherapies With a Single Approved Dose Level

Abstract

Background:

In this article, we study the sample size calculations for the combination drugs of 2 monotherapies with a single approved dose level when the primary endpoints are binary.

Methods:

Two study cases are examined: In the first, each monotherapy has the same indication, while in the second, each monotherapy has a different indication. The sample sizes are calculated by using an asymptotic joint distribution of test statistics and employing unequal allocation for 3 popular measures of 2 proportions: the risk difference, the log relative risk, and the log odds ratio.

Results:

Results show that our proposed method produces smaller total sample sizes compared with the heuristic method.

Conclusions:

The total sample sizes can be reduced by incorporating unequal allocation and dependency between 2 test statistics.

Keywords

unequal allocation power bivariate normal distribution correlation coefficient

Background

Combination drugs are widely used to treat many diseases. including cancer, hypertension, asthma, diabetes, and arthritis. Hence, pharmaceutical companies often seek to develop combination drugs that combine 2 monotherapies. In such cases, the pharmaceutical companies need to prove that the combination drug has greater efficacy than its 2 monotherapies in order to obtain approval from a regulatory agency. In this article, we consider 3-arm parallel designs, which are often used to compare 3 treatments (the combination drug and its 2 monotherapies) for the development of a combination drug.

The calculation of sample sizes is a key issue in planning such clinical trials. Several articles have been devoted to the study of sample size calculations for combination drugs.^1

–8 For cases where the primary endpoint is a continuous type, Kang and Shin⁹ studied the sample size calculations for the combination drugs of 2 monotherapies where each has only 1 approved dose level. The purpose of this article was to extend their results to binary endpoints for 3 popular metrics: the risk difference, the log relative risk, and the log odds ratio.

Specifically, in this article, we modify the method of Laska and Meisner¹ in evaluating the asymptotic power function numerically and using unequal allocation to minimize the total sample sizes. We focus on instances in which each monotherapy of the binary primary endpoint has only 1 approved dose level. Although Laska and Meisner¹ mentioned a binomial model, they did not provide any specific results for binary endpoints. Therefore, in this article, we compare the total sample sizes computed by the proposed method with those calculated using the heuristic method for 2 monotherapies when each monotherapy has the same indication (study case 1) and when they have different indications (study case 2).

Study Case 1: When 2 Monotherapies Have the Same Indication

Notation and Hypotheses

Let A and B denote 2 monotherapies that have only 1 approved dose level for the same disease. When a pharmaceutical company seeks to develop a combination drug AB combining A and B, a 3-arm parallel clinical trial is often planned to compare the 3 treatments (monotherapy A, monotherapy B, and combination drug AB). Let X denote the primary endpoint for the same disease. Because A and B are 2 monotherapies for the same disease, X is the common primary endpoint for the 3 treatment groups. For instance, Jalyn (GlaxoSmithKline, London, UK) is a fixed-dose combination of dutasteride and tamsulosin that is used to reduce the risk of acute urinary retention.¹⁰

Let X_A represent the number of favorable events of interest from the group of treatment A. Similarly, X_B and X_AB can be defined. It is assumed that

X_{A} \sim B (n_{A}, p_{A}), X_{B} \sim B (n_{B}, p_{B}), X_{A B} \sim B (n_{A B}, p_{A B}) .

When 2 binomial proportions are compared, there are 3 popular measures: the risk difference, the log relative risk, and the log odds ratio.^11,12 The hypotheses for the development of the combination drug AB in terms of the risk difference are therefore given as follows:

H_{0} : p_{A B} \leq p_{A} or p_{A B} \leq p_{B} vs H_{1} : p_{A B} > p_{A} and p_{A B} > p_{B} .

Similarly, the hypotheses in terms of the log relative risk are given by

\begin{matrix} H_{0} : log (\frac{p_{A B}}{p_{A}}) \leq 0 or log (\frac{p_{A B}}{p_{B}}) \leq 0 \\ vs H_{1} : log (\frac{p_{A B}}{p_{A}}) > 0 and log (\frac{p_{A B}}{p_{B}}) > 0 \end{matrix}

and the hypotheses in terms of the log odds ratio are given by

H_{0} : log (\frac{p_{A B} / (1 - p_{A B})}{p_{A} / (1 - p_{A})}) \leq 0 or log (\frac{p_{A B} / (1 - p_{A B})}{p_{B} / (1 - p_{B})}) \leq 0

vs H_{1} : log (\frac{p_{A B} / (1 - p_{A B})}{p_{A} / (1 - p_{A})}) > 0 and log (\frac{p_{A B} / (1 - p_{A B})}{p_{B} / (1 - p_{B})}) > 0.

In other words, the pharmaceutical company must show that the combination drug is more effective than each of the 2 monotherapies in relation to a preselected measure such as the risk difference, the log relative risk, or the log odds ratio.

The Heuristic Method

First of all, we would like to emphasize that the heuristic method is incorrect because of the assumption of independence described below. Nevertheless, the heuristic method has often been used by some pharmaceutical companies based on my personal experience. That is why we consider the heuristic method in this study.

For simplicity, we describe the heuristic method in terms of the risk difference. The heuristic method in terms of the log relative risk and the log odds ratio can be obtained similarly. The heuristic method to compute the sample sizes for the hypotheses in equation 1 is to divide them into 2 hypotheses as follows:

H_{0 A} : p_{A B} \leq p_{A} vs H_{1 A} : p_{A B} > p_{A}

H_{0 B} : p_{A B} \leq p_{B} vs H_{1 B} : p_{A B} > p_{B} .

Note that rejecting both H _0A and H _0B in equations 2 and 3 is equivalent to rejecting H ₀ in equation 1. Let E_A and E_B represent the events of rejecting each null hypothesis:

E_{A} = {reject H_{0 A}}, E_{B} = {reject H_{0 B}}

Then, by assuming that E_A and E_B are independent, the power of testing the hypotheses in equation 1 is

\begin{matrix} 1 - β & = & P (reject H_{0} | H_{1}) \\ = & P (E_{A} \cap E_{B} | H_{1}) \\ = & P (E_{A} | H_{1 A}) P (E_{B} | H_{1 B}) \end{matrix} .

Let $n_{s, d (A)}^{h} and n_{s, d (B)}^{h}$ be the sample sizes for testing the hypotheses in equation 2 with $\sqrt{1 - β}$ power based on the asymptotic test when equal allocation is used. The superscript h represents the heuristic method, and the first subscript s and second subscript d denote the same disease and the risk difference, respectively. On this basis, it is well known (eg, Chow et al^13(p87)) that

n_{s, d (A)}^{h} = {(\frac{z_{α} - z_{\sqrt{1 - β}}}{p_{A B} - p_{A}})}^{2} [p_{A B} (1 - p_{A B}) + p_{A} (1 - p_{A})], n_{s, d (A)}^{h} = n_{s, d (A B)}^{h},

where z_α is the 100αth upper percentile of the standard normal distribution. Similarly, $n_{s, d (B)}^{h}$ can be calculated.

By a similar argument, the sample size based on the heuristic method in terms of the log relative risk is

n_{s, r (A)}^{h} = {(\frac{z_{α} - z_{\sqrt{1 - β}}}{log (p_{A B}) - log (p_{A})})}^{2} [\frac{(1 - p_{A B})}{p_{A B}} + \frac{(1 - p_{A})}{p_{A}}], n_{s, r (A)}^{h} = n_{s, r (A B)}^{h}

and the sample size based on the heuristic method in terms of the log odds ratio is

n_{s, o (A)}^{h} = {(\frac{z_{α} - z_{\sqrt{1 - β}}}{log (\frac{p_{A B} / (1 - p_{A B})}{p_{A} (1 - p_{A})})})}^{2} [\frac{1}{p_{A B} (1 - p_{A B})} + \frac{1}{p_{A} (1 - p_{A})}], n_{s, o (A)}^{h} = n_{s, o (A B)}^{h}

Therefore, the total sample sizes based on the heuristic method for the risk difference, the log relative risk, and the log odds ratio are

\begin{matrix} n_{s, d}^{h} = 3 \times max (n_{s, d (A)}^{h}, n_{s, d (B)}^{h}) \\ n_{s, r}^{h} = 3 \times max (n_{s, r (A)}^{h}, n_{s, r (B)}^{h}) \\ n_{s, o}^{h} = 3 \times max (n_{s, o (A)}^{h}, n_{s, o (B)}^{h}) \end{matrix} .

The heuristic method presents 2 problems. The first problem is that, since the group of the combination drug is involved with both E_A and E_B , the 2 events E_A and E_B are not independent. The second problem is that the total sample size in equation 5 might be unnecessarily large because the maximum is taken. The proposed method in study case 2 below solves these problems and reduces the total sample size.

Proposed Method for Sample Size Calculation

Kang and Shin⁹ modified the method of Laska and Meisner¹ in order to calculate sample sizes for a continuous primary endpoint. In this section, we extend their results to binary endpoints. For ease of explanation, we explain the proposed method in terms of the risk difference. The proposed method in terms of the log relative risk and the log odds ratio can be obtained similarly.

Let T_s _,d(A) and T_s _,d(B) define the following test statistics for equations 2 and 3.

T_{s, d (A)} = \frac{{\hat{p}}_{A B} - {\hat{p}}_{A}}{\sqrt{\frac{1}{n_{A B}} {\hat{p}}_{A B} (1 - {\hat{p}}_{A B}) + \frac{1}{n_{A}} {\hat{p}}_{A} (1 - {\hat{p}}_{A})}}

T_{s, d (B)} = \frac{{\hat{p}}_{A B} - {\hat{p}}_{B}}{\sqrt{\frac{1}{n_{A B}} {\hat{p}}_{A B} (1 - {\hat{p}}_{A B}) + \frac{1}{n_{B}} {\hat{p}}_{B} (1 - {\hat{p}}_{B})}}

In this article we suggest that the power of testing the hypotheses in equation 1 be calculated directly from the asymptotic joint cumulative probability distribution of 2 test statistics T_s _,d(A) and T_s _,d(B). In other words,

\begin{matrix} 1 - β & = & P (T_{s, d (A)} > z_{α} and T_{s, d (B)} > z_{α} | H_{1}) \\ ≅ & P (Z_{s, d (A)} > w_{s, d (A)} and Z_{s, d (B)} > w_{s, d (B)} | H_{1}) \end{matrix}

where Z_s _,d(A) and Z_s _,d(B) are the same as T_s _,d(A) and T_s _,d(B) except that ${\hat{p}}_{A B}, {\hat{p}}_{A}, and {\hat{p}}_{B}$ in the denominator of T_s _,d(A) and T_s _,d(B) are replaced with p_AB , p_A , and p_B , respectively, and

w_{s, d (A)} = z_{α} - \frac{p_{A B} - p_{A}}{\sqrt{\frac{1}{n_{A B}} p_{A B} (1 - p_{A B}) + \frac{1}{n_{A}} p_{A} (1 - p_{A})}}

w_{s, d (B)} = z_{α} - \frac{p_{A B} - p_{B}}{\sqrt{\frac{1}{n_{A B}} p_{A B} (1 - p_{A B}) + \frac{1}{n_{B}} p_{B} (1 - p_{B})}}

and

(\begin{matrix} Z_{s, d (A)} \\ Z_{s, d (B)} \end{matrix}) \sim N_{2} ((\begin{matrix} 0 \\ 0 \end{matrix}), (\begin{matrix} 1 & ρ_{s, d} \\ ρ_{s, d} & 1 \end{matrix}))

ρ_{s, d} = \frac{1}{\sqrt{1 + \frac{n_{A B}}{n_{A}} \frac{p_{A} (1 - p_{A})}{p_{A B} (1 - p_{A B})}} \sqrt{1 + \frac{n_{A B}}{n_{B}} \frac{p_{B} (1 - p_{B})}{p_{A B} (1 - p_{A B})}}} .

Note that z_α is used in 2 inequalities in equation 6 because the min test is employed.¹ The min test requires both test statistics T_s _,d(A) and T_s _,d(B) to reject the null hypotheses H _0A and H _0B at the significance level α at most. Thus, the overall type I error rate is controlled under the significance level α regardless of the values of the unknown parameters.^1,14

Therefore, the power can be expressed in terms of the cumulative distribution function of the standardized bivariate normal distribution

1 - β ≅ 1 - [Φ_{2} (w_{s, d (A)}, \infty) + Φ_{2} (\infty, w_{s, d (B)}) - Φ_{2} (w_{s, d (A)}, w_{s, d (B)})],

where Φ₂ denotes the cumulative distribution function of the standardized bivariate normal distribution.

Since ρ_s _,d ≠ 0, the joint probability density functions of the standardized bivariate normal distribution have the shapes of rotated ellipses and the power function in equation 7 may not achieve the maximum power when w_s _,d(A) = w_s _,d(B). In addition, the difference between the probability of interest of the combination drug and those of the 2 monotherapies are not equal in many instances. That is,

p_{A B} - p_{A} \neq p_{A B} - p_{B} .

Therefore, we consider the unequal allocation, n_AB : n_A : n_B = 1 : k ₁ : k ₂. Hence,

ρ_{s, d} = \frac{1}{\sqrt{1 + \frac{1}{k_{1}} \frac{p_{A} (1 - p_{A})}{p_{A B} (1 - p_{A B})}} \sqrt{1 + \frac{1}{k_{2}} \frac{p_{B} (1 - p_{B})}{p_{A B} (1 - p_{A B})}}} .

Let N represent the total sample size: N = n_AB + n_A + n_B . Then,

n_{A B} = \frac{N}{1 + k_{1} + k_{2}}, n_{A} = k_{1} \frac{N}{1 + k_{1} + k_{2}}, n_{B} = k_{2} \frac{N}{1 + k_{1} + k_{2}} .

For given values of p_AB , p_A , p_B , and N, the values of k ₁ and k ₂ are escalated from 0.1 to 3.0 by 0.01 increments to discover the maximum power and the values of k ₁ and k ₂ that maximize the power function in equation 7. For given values of p_AB , p_A , and p_B , we can find the minimal N and the optimal values of k ₁ and k ₂ that achieve 80% power by trying several values of N. Note that the optimal values of k ₁ and k ₂ depend on the values of p_AB , p_A , p_B , and N. Table 1 displays the numerical results. The proposed method produces smaller total sample sizes than the heuristic method in all cases. For example, when p_AB = 0.8, p_A = 0.6, and p_B = 0.7, the total sample sizes produced by the heuristic method and the proposed method are 931 and 575, respectively. The proposed method should allocate 575 patients as follows in order to achieve 80% power.

\begin{matrix} n_{A B} & = & \frac{575}{1 + 0.41 + 1.11} = 228 \\ n_{A} & = & 0.41 \times \frac{575}{1 + 0.41 + 1.11} = 94 \\ n_{B} & = & 1.11 \times \frac{575}{1 + 0.41 + 1.11} = 253 \end{matrix}

Table 1.

Total Sample Sizes (n_AB + n_A + n_B ) of the Same Indication in Terms of Risk Difference From the heuristic Method and the Proposed Method (α = 0.05, β = 0.2, p_AB = 0.8).

p_A	p_B	Heuristic	Proposed	k ₁	k ₂	p_A	p_B	Heuristic	Proposed	k ₁	k ₂
0.2	0.2	23	21	0.76	0.76	0.5	0.2	115	66	1.22	0.30
	0.3	38	29	0.63	0.99		0.3	115	73	1.18	0.48
	0.4	63	41	0.47	1.15		0.4	115	86	1.11	0.70
	0.5	115	66	0.30	1.22		0.5	115	110	0.95	0.95
	0.6	252	133	0.15	1.22		0.6	252	177	0.63	1.13
	0.7	931	467	0.05	1.16		0.7	931	514	0.22	1.14
0.3	0.2	38	29	0.99	0.63	0.6	0.2	252	133	1.22	0.15
	0.3	38	36	0.87	0.87		0.3	252	141	1.22	0.26
	0.4	63	48	0.70	1.06		0.4	252	153	1.18	0.40
	0.5	115	73	0.48	1.18		0.5	252	177	1.13	0.63
	0.6	252	141	0.26	1.22		0.6	252	241	0.93	0.93
	0.7	931	476	0.08	1.14		0.7	931	575	0.41	1.11
0.4	0.2	63	41	1.15	0.47	0.7	0.2	931	467	1.16	0.05
	0.3	63	48	1.06	0.70		0.3	931	476	1.14	0.08
	0.4	63	61	0.93	0.93		0.4	931	489	1.14	0.13
	0.5	115	86	0.70	1.11		0.5	931	514	1.14	0.22
	0.6	252	153	0.40	1.18		0.6	931	575	1.11	0.41
	0.7	931	489	0.13	1.14		0.7	931	885	0.87	0.87

For the log relative risk, the same arguments hold with the following conditions.

T_{s, r (A)} = \frac{log ({\hat{p}}_{A B}) - log ({\hat{p}}_{A})}{\sqrt{\frac{1}{n_{A B}} {\hat{p}}_{A B} (1 - {\hat{p}}_{A B}) + \frac{1}{n_{A}} {\hat{p}}_{A} (1 - {\hat{p}}_{A})}}

T_{s, r (B)} = \frac{log ({\hat{p}}_{A B}) - log ({\hat{p}}_{B})}{\sqrt{\frac{1}{n_{A B}} {\hat{p}}_{A B} (1 - {\hat{p}}_{A B}) + \frac{1}{n_{B}} {\hat{p}}_{B} (1 - {\hat{p}}_{B})}}

w_{s, r (A)} = z_{α} - \frac{log (p_{A B}) - log (p_{A})}{\sqrt{\frac{1}{n_{A B}} p_{A B} (1 - p_{A B}) + \frac{1}{n_{A}} p_{A} (1 - p_{A})}}

w_{s, r (B)} = z_{α} - \frac{log (p_{A B}) - log (p_{B})}{\sqrt{\frac{1}{n_{A B}} p_{A B} (1 - p_{A B}) + \frac{1}{n_{B}} p_{B} (1 - p_{B})}}

ρ_{s, r} = \frac{1}{\sqrt{1 + \frac{n_{A B}}{n_{A}} \frac{(1 - p_{A}) / p_{A}}{(1 - p_{A B}) / p_{A B}}} \sqrt{1 + \frac{n_{A B}}{n_{B}} \frac{(1 - p_{B}) / p_{B}}{(1 - p_{A B}) / p_{A B}}}}

Table 2 shows the total sample sizes produced by the heuristic method and the proposed method for the log relative risk. The total sample sizes from the proposed method are smaller than those from the heuristic method in all cases.

Table 2.

Total Sample Sizes (n_AB + n_A + n_B ) of the Same Indication in Terms of Log Relative Risk From the Heuristic Method and the Proposed Method (α = 0.05, β = 0.2, p_AB = 0.8).

p_A	p_B	Heuristic	Proposed	k ₁	k ₂	p_A	p_B	Heuristic	Proposed	k ₁	k ₂
0.2	0.2	56	48	2.97	2.97	0.5	0.2	143	95	1.97	1.07
	0.3	68	55	2.33	2.65		0.3	143	100	1.92	1.18
	0.4	92	68	1.67	2.31		0.4	143	112	1.79	1.34
	0.5	143	95	1.07	1.97		0.5	143	136	1.50	1.50
	0.6	279	165	0.56	1.63		0.6	279	205	0.97	1.52
	0.7	958	505	0.18	1.31		0.7	958	548	0.34	1.30
0.3	0.2	68	55	2.65	2.33	0.6	0.2	279	165	1.63	0.56
	0.3	68	61	2.28	2.28		0.3	279	171	1.62	0.63
	0.4	92	74	1.76	2.15		0.4	279	182	1.60	0.76
	0.5	143	100	1.18	1.92		0.5	279	205	1.52	0.97
	0.6	279	171	0.63	1.62		0.6	279	270	1.23	1.23
	0.7	958	511	0.20	1.30		0.7	958	610	0.53	1.27
0.4	0.2	92	68	2.31	1.67	0.7	0.2	958	505	1.31	0.18
	0.3	92	74	2.15	1.76		0.3	958	511	1.30	0.20
	0.4	92	86	1.83	1.83		0.4	958	524	1.31	0.25
	0.5	143	112	1.34	1.79		0.5	958	548	1.30	0.34
	0.6	279	182	0.76	1.60		0.6	958	610	1.27	0.53
	0.7	958	524	0.25	1.31		0.7	958	923	0.99	0.99

For the log odds ratio, the same arguments also hold with the following conditions.

T_{s, o (A)} = \frac{log (\frac{{\hat{p}}_{A B} / (1 - {\hat{p}}_{A B})}{{\hat{p}}_{A} / (1 - {\hat{p}}_{A})})}{\sqrt{\frac{1}{n_{A B} {\hat{p}}_{A B} (1 - {\hat{p}}_{A B})} + \frac{1}{n_{A} {\hat{p}}_{A} (1 - {\hat{p}}_{A})}}}

T_{s, o (A)} = \frac{log (\frac{{\hat{p}}_{A B} / (1 - {\hat{p}}_{A B})}{{\hat{p}}_{B} / (1 - {\hat{p}}_{B})})}{\sqrt{\frac{1}{n_{A B} {\hat{p}}_{A B} (1 - {\hat{p}}_{A B})} + \frac{1}{n_{B} {\hat{p}}_{B} (1 - {\hat{p}}_{B})}}}

w_{s, o (A)} = z_{α} - \frac{log (\frac{p_{A B} (1 - p_{A B})}{p_{A} (1 - p_{A})})}{\sqrt{\frac{1}{n_{A B} p_{A B} (1 - p_{A B})} + \frac{1}{n_{A} p_{A} (1 - p_{A})}}}

w_{s, o (B)} = z_{α} - \frac{log (\frac{p_{A B} (1 - p_{A B})}{p_{B} (1 - p_{B})})}{\sqrt{\frac{1}{n_{A B} p_{A B} (1 - p_{A B})} + \frac{1}{n_{B} p_{B} (1 - p_{B})}}}

ρ_{s, o} = \frac{1}{\sqrt{1 + \frac{n_{A B}}{n_{A}} \frac{p_{A B} (1 - p_{A B})}{p_{A} (1 - p_{A})}} \sqrt{1 + \frac{n_{A B}}{n_{B}} \frac{p_{A B} (1 - p_{A B})}{p_{B} (1 - p_{B})}}} .

Table 3 presents the total sample sizes produced by the heuristic method and by the proposed method for the log odds ratio. In all cases, the proposed method produces smaller total sample sizes than the heuristic method.

Table 3.

Total Sample Sizes (n_AB + n_A + n_B ) of the Same Indication in Terms of Log Odds Ratio From the Heuristic Method and the Proposed Method (α = 0.05, β = 0.2, p_AB = 0.8).

p_A	p_B	Heuristic	Proposed	k ₁	k ₂	p_A	p_B	Heuristic	Proposed	k ₁	k ₂
0.2	0.2	41	38	0.76	0.76	0.5	0.2	135	82	0.78	0.34
	0.3	56	45	0.64	0.76		0.3	135	87	0.76	0.39
	0.4	82	57	0.49	0.76		0.4	135	96	0.71	0.48
	0.5	135	82	0.34	0.78		0.5	135	118	0.62	0.62
	0.6	273	150	0.19	0.81		0.6	273	182	0.42	0.75
	0.7	954	486	0.07	0.88		0.7	954	515	0.16	0.87
0.3	0.2	56	45	0.76	0.64	0.6	0.2	273	150	0.81	0.19
	0.3	56	50	0.67	0.67		0.3	273	154	0.80	0.22
	0.4	82	62	0.55	0.72		0.4	273	162	0.79	0.29
	0.5	135	87	0.39	0.76		0.5	273	182	0.75	0.42
	0.6	273	154	0.22	0.80		0.6	273	240	0.63	0.63
	0.7	954	490	0.08	0.88		0.7	954	565	0.29	0.85
0.4	0.2	82	57	0.76	0.49	0.7	0.2	954	486	0.88	0.07
	0.3	82	62	0.72	0.55		0.3	954	490	0.88	0.08
	0.4	82	72	0.63	0.63		0.4	954	497	0.87	0.10
	0.5	135	96	0.48	0.71		0.5	954	515	0.87	0.16
	0.6	273	162	0.29	0.79		0.6	954	565	0.85	0.29
	0.7	954	497	0.10	0.87		0.7	954	857	0.67	0.67

Study Case 2: When 2 Monotherapies Have Different Indications

Notation and Hypotheses

In the previous section, we considered 2 monotherapies for the same indication, while here we deal with 2 monotherapies for different indications used to make a combination drug. For example, Actifed (Johnson & Johnson, New Brunswick, NJ) is a fixed-dose combination of triprolidine and pseudoephedrine that is used to treat seasonal allergic rhinitis (SAR).¹⁵ Triprolidine is an antihistamine used to treat hay fever–like symptoms, while pseudoephedrine is a sympathomimetic used to alleviate nasal congestion.

Let A and B denote 2 monotherapies for 2 different diseases, each with only 1 approved dose level. Suppose that a pharmaceutical company would like to develop a combination drug that combines A and B. Laska and Meisner¹ did not explain how their method might be applied to this situation.

Suppose that the pharmaceutical company plans a 3-arm parallel clinical trial in order to compare the 3 treatments (monotherapy A, monotherapy B, and combination drug AB). Let X and Y be the primary endpoints of the 2 different diseases for the 2 drugs A and B, respectively. It is assumed that

\begin{matrix} X_{A} \sim B (n_{A}, p_{A, X}) X_{B} \sim B (n_{B}, p_{B, X}) X_{A B} \sim B (n_{A B}, p_{A B, X}) \\ Y_{A} \sim B (n_{A}, p_{A, Y}) Y_{B} \sim B (n_{B}, p_{B, Y}) Y_{A B} \sim B (n_{A B}, p_{A B, Y}) \end{matrix}

and

X_{A B} = \sum_{i = 1}^{n_{A B}} X_{A B, i}, Y_{A B} = \sum_{i = 1}^{n_{A B}} Y_{A B, i} .

Furthermore, it is assumed that

C o r r (X_{A B, i}, Y_{A B, i}) = ρ_{A B} for all i

because the 2 primary endpoints (X and Y) are measured from the same patient in the group of the combination drug.

On this basis, we would like to test the following hypotheses for the development of the combination drug AB in terms of the risk difference:

\begin{matrix} H_{0} : p_{A B, X} \leq p_{B, X} or p_{A B, Y} \leq p_{A, Y} \\ vs & H_{1} : p_{A B, X} > p_{B, X} and p_{A B, Y} > p_{A, Y} \end{matrix} .

It is noted that p_A _{, X} and p_B _{, Y} do not appear in the hypotheses in equation 8. The hypothesis H _1A: p_AB _,X > p_B _,X implies that the combination drug AB has significant additional efficacy in treating the first disease by adding the treatment A. Similarly, the hypothesis H _1B: p_AB _,X > p_A _,X implies that the combination drug AB has greater efficacy in treating the second disease compared with monotherapy A. Similarly, the hypotheses in terms of the log relative risk are given by

\begin{matrix} H_{0} : log (\frac{p_{A B, X}}{p_{B, X}}) \leq 0 or log (\frac{p_{A B, Y}}{p_{B, Y}}) \leq 0 \\ vs & H_{1} : log (\frac{p_{A B, X}}{p_{B, X}}) > 0 and log (\frac{p_{A B, Y}}{p_{B, Y}}) > 0 \end{matrix} .

and the hypotheses in terms of the log odds ratio are given by

\begin{matrix} H_{0} : log (\frac{p_{A B, X} / (1 - p_{A B, X})}{p_{B, X} / (1 - p_{B, X})}) \leq 0 or log (\frac{p_{A B, Y} / (1 - p_{A B, Y})}{p_{A, Y} / (1 - p_{A, Y})}) \leq 0 \\ vs & H_{1} : log (\frac{p_{A B, X} / (1 - p_{A B, X})}{p_{B, X} / (1 - p_{B, X})}) > 0 and log (\frac{p_{A B, Y} / (1 - p_{A B, Y})}{p_{A, Y} / (1 - p_{A, Y})}) > 0 \end{matrix} .

The Heuristic Method

As seen in the previous section, we would like to emphasize that the heuristic method is incorrect because of the assumption of independence. The heuristic method for calculating sample sizes for the hypotheses in equation 8 is to divide the hypotheses into 2 hypotheses:

\begin{matrix} H_{0 A} : p_{A B, X} \leq p_{B, X} & vs & H_{1 A} : p_{A B, X} > p_{B, X} \\ H_{0 B} : p_{A B, Y} \leq p_{A, Y} & vs & H_{1 B} : p_{A B, Y} > p_{A, Y} \end{matrix} .

Following similar arguments to those for study case 1, let $n_{d, d (A)}^{h} and n_{d, d (A B)}^{h}$ be the sample sizes for testing equation 9 with $\sqrt{1 - β}$ power based on the asymptotic test when equal allocation is employed, where the first subscript d represents the different disease and the second subscript d denotes the risk difference. As a result, the well-known sample size calculation formula for the risk difference (eg, Chow et al^13(p87)) is

\begin{matrix} n_{d, d (A)}^{h} = & {(\frac{z_{α} - z_{\sqrt{1 - β}}}{p_{A B, X} - p_{B, X}})}^{2} [\frac{1}{p_{A B, X} (1 - p_{A B, X})} + \frac{1}{p_{B, X} (1 - p_{B, X})}] \\ n_{d, d (A)}^{h} = & n_{d, d (A B)}^{h} \end{matrix} .

With a similar method, $n_{d, d (A B)}^{h}$ can be computed. For the log relative risk, a similar reasoning produces

\begin{matrix} n_{d, r (A)}^{h} = & {(\frac{z_{α} - z_{\sqrt{1 - β}}}{log (p_{A B, X}) - log (p_{B, X})})}^{2} [\frac{(1 - p_{A B, X})}{p_{A B, X}} + \frac{(1 - p_{B, X})}{p_{B, X}}] \\ n_{d, r (A)}^{h} = & n_{d, r (A B)}^{h} \end{matrix}

and the sample size based on the heuristic method in terms of the log odds ratio is

\begin{matrix} n_{d, o (A)}^{h} = & {(\frac{z_{α} - z_{\sqrt{1 - β}}}{log (\frac{p_{A B, X} / (1 - p_{A B, X})}{p_{B, X} / (1 - p_{B, X})})})}^{2} [\frac{1}{p_{A B, X} (1 - p_{A B, X})} + \frac{1}{p_{B, X} (1 - p_{B, X})}] \\ n_{d, o (A)}^{h} = & n_{d, o (A B)}^{h} \end{matrix} .

Therefore, the total sample sizes based on the heuristic method for the risk difference, the log relative risk, and the log odds ratio are

\begin{matrix} n_{d, d}^{h} = 3 \times max (n_{d, d (A)}^{h}, n_{d, d (B)}^{h}) \\ n_{d, r}^{h} = 3 \times max (n_{d, r (A)}^{h}, n_{d, r (B)}^{h}) \\ n_{d, o}^{h} = 3 \times max (n_{d, o (A)}^{h}, n_{d, o (B)}^{h}) \end{matrix} .

The heuristic method presents 2 problems, as pointed out in the previous section. For this reason, the total sample size based on the heuristic method might be unnecessarily large.

Proposed Method for Sample Size Calculation

We now present a new sample size calculation method similar to the method presented in the previous section. The hypotheses in equation 8 can be divided into 2 hypotheses as follows.

H_{0 A} : p_{A B, X} \leq p_{B, X} vs H_{1 A} : p_{A B, X} > p_{B, X}

H_{0 B} : p_{A B, Y} \leq p_{A, Y} vs H_{1 B} : p_{A B, Y} > p_{A, Y} .

Let T_d _,d(A) and T_d _,d(B) denote the following test statistics for equations 11 and 12:

T_{d, d (A)} = \frac{{\hat{p}}_{A B, X} - {\hat{p}}_{B, X}}{\sqrt{\frac{1}{n_{A B}} {\hat{p}}_{A B, X} (1 - {\hat{p}}_{A B, X}) + \frac{1}{n_{B}} {\hat{p}}_{B, X} (1 - {\hat{p}}_{B, X})}}

T_{d, d (B)} = \frac{{\hat{p}}_{A B, Y} - {\hat{p}}_{A, Y}}{\sqrt{\frac{1}{n_{A B}} {\hat{p}}_{A B, Y} (1 - {\hat{p}}_{A B, Y}) + \frac{1}{n_{A}} {\hat{p}}_{A, Y} (1 - {\hat{p}}_{A, Y})}}

The power of testing the hypotheses in equation 8 is calculated directly from the asymptotic joint cumulative probability distribution of 2 test statistics, T_d _,d(A) and T_d _,d(B).

\begin{matrix} 1 - β & = & P (T_{d, d (A)} > z_{α} and T_{d, d (B)} > z_{α} | H_{1}) \\ ≅ & P (Z_{d, d (A)} > w_{d, d (A)} and Z_{d, d (B)} > w_{d, d (B)} | H_{1}) \end{matrix}

where Z_d _{, d(A)} and Z_d _{, d(B)} are the same as T_d _,d(A) and T_d _,d(B) except that ${\hat{p}}_{A B}, {\hat{p}}_{A}, and {\hat{p}}_{B}$ in the denominator of T_d _,d(A) and T_d _,d(B) are replaced with p_AB , p_A , and p_B , respectively, and

w_{d, d (A)} = z_{α} - \frac{p_{A B, X} - p_{B, X}}{\sqrt{\frac{1}{n_{A B}} p_{A B, X} (1 - p_{A B, X}) + \frac{1}{n_{B}} p_{B, X} (1 - p_{B, X})}}

w_{d, d (B)} = z_{α} - \frac{p_{A B, Y} - p_{A, Y}}{\sqrt{\frac{1}{n_{A B}} p_{A B, Y} (1 - p_{A B, Y}) + \frac{1}{n_{A}} p_{A, Y} (1 - p_{A, Y})}}

and

[\begin{matrix} Z_{d, d (A)} \\ Z_{d, d (B)} \end{matrix}] \overset{d}{\to} N_{2} (0, (\begin{matrix} 1 & ρ_{d, d} \\ ρ_{d, d} & 1 \end{matrix}))

ρ_{d, d} = \frac{ρ_{A B}}{\sqrt{1 + \frac{n_{A B}}{n_{B}} \frac{p_{B, X} (1 - p_{B, X})}{p_{A B, X} (1 - p_{A B, X})}} \sqrt{1 + \frac{n_{A B}}{n_{A}} \frac{p_{A, Y} (1 - p_{A, Y})}{p_{A B, Y} (1 - p_{A B, Y})}}}

The mathematical derivation of ρ_d _,d is provided in the Appendix. The same approach as that used in study case 1 is used in this section, with n_AB : n_A : n_B = 1 : k ₁ : k ₂. Table 4 displays the total sample sizes computed by the heuristic method and the proposed method. The proposed method produces smaller total sample sizes than the heuristic method in all instances.

Table 4.

Total Sample Sizes (n_AB + n_A + n_B ) of Different Indications in Terms of Risk Difference From the Heuristic Method and the Proposed Method (α = 0.05, β = 0.2, p_AB _,X = 0.8, p_AB _,Y = 0.7, ρ_AB = 0.5).

p_B,X	p_A,Y	Heuristic	Proposed	k ₁	k ₂	p_B,X	p_A,Y	Heuristic	Proposed	k ₁	k ₂
0.2	0.2	37	29	0.75	0.55	0.5	0.2	114	72	0.41	1.13
	0.3	66	43	0.93	0.41		0.3	114	87	0.62	1.01
	0.4	125	72	1.05	0.26		0.4	125	118	0.84	0.83
	0.5	289	153	1.09	0.13		0.5	289	199	1.00	0.53
	0.6	1131	568	1.08	0.04		0.6	1131	617	1.05	0.18
0.3	0.2	37	36	0.68	0.77	0.6	0.2	251	138	0.22	1.20
	0.3	66	51	0.87	0.61		0.3	251	153	0.38	1.15
	0.4	125	80	1.01	0.41		0.4	251	183	0.60	1.04
	0.5	289	161	1.08	0.22		0.5	289	264	0.86	0.81
	0.6	1131	577	1.08	0.07		0.6	1131	681	1.04	0.35
0.4	0.2	62	48	0.57	0.97	0.7	0.2	930	472	0.07	1.16
	0.3	66	63	0.78	0.82		0.3	930	487	0.12	1.13
	0.4	125	93	0.95	0.60		0.4	930	516	0.22	1.12
	0.5	289	174	1.05	0.34		0.5	930	593	0.43	1.06
	0.6	1131	591	1.07	0.11		0.6	1131	1001	0.84	0.75

For the log relative risk, a similar conclusion can be obtained under the following conditions.

T_{d, r (A)} = \frac{log ({\hat{p}}_{A B, X}) - log ({\hat{p}}_{B, X})}{\sqrt{\frac{1}{n_{A B}} \frac{(1 - {\hat{p}}_{A B, X})}{{\hat{p}}_{A B, X}} + \frac{1}{n_{B}} \frac{(1 - {\hat{p}}_{B, X})}{{\hat{p}}_{B, X}}}}

T_{d, r (B)} = \frac{log ({\hat{p}}_{A B, Y}) - log ({\hat{p}}_{A, Y})}{\sqrt{\frac{1}{n_{A B}} \frac{(1 - {\hat{p}}_{A B, Y})}{{\hat{p}}_{A B, Y}} + \frac{1}{n_{A}} \frac{(1 - {\hat{p}}_{A, Y})}{{\hat{p}}_{A, Y}}}}

ρ_{d, r} = \frac{ρ_{A B}}{\sqrt{1 + \frac{(1 - p_{B, X}) / p_{B, X}}{(1 - p_{A B, X}) / p_{A B, X}} \frac{1}{k_{2}}} \sqrt{1 + \frac{(1 - p_{A, Y}) / p_{A, Y}}{(1 - p_{A B, Y}) / p_{A B, Y}} \frac{1}{k_{1}}}}

For the log odds ratio, similar results can also be achieved with the following conditions.

T_{d, O (A)} = \frac{log ({\hat{p}}_{A B, X} / (1 - {\hat{p}}_{A B, X})) - log ({\hat{p}}_{B, X} / (1 - {\hat{p}}_{B, X}))}{\sqrt{\frac{1}{n_{A B} {\hat{p}}_{A B, X} (1 - {\hat{p}}_{A B, X})} + \frac{1}{n_{B} {\hat{p}}_{B, X} (1 - {\hat{p}}_{B, X})}}}

T_{d, O (B)} = \frac{log ({\hat{p}}_{A B, Y} / (1 - {\hat{p}}_{A B, Y})) - log ({\hat{p}}_{A, Y} / (1 - {\hat{p}}_{A, Y}))}{\sqrt{\frac{1}{n_{A B} {\hat{p}}_{A B, Y} (1 - {\hat{p}}_{A B, Y})} + \frac{1}{n_{A} {\hat{p}}_{A, Y} (1 - {\hat{p}}_{A, Y})}}}

ρ_{d, o} = \frac{ρ_{A B}}{\sqrt{1 + \frac{p_{A B, X} (1 - p_{A B, X})}{p_{B, X} (1 - p_{B, X})} \frac{1}{k_{2}}} \sqrt{1 + \frac{p_{A B, Y} (1 - p_{A B, Y})}{p_{A, Y} (1 - p_{A, Y})} \frac{1}{k_{1}}}}

The mathematical derivation of ρ_d _,r and ρ_d _,o can be calculated per the Appendix. Tables 5 and 6 present the total sample sizes for the log relative risk and the log odds ratio. The total sample sizes from the proposed method are smaller than those from the heuristic method in all cases.

Table 5.

Total Sample Sizes (n_AB + n_A + n_B ) of Different Indications in Terms of Log Relative Risk From the Heuristic Method and the Proposed Method (α = 0.05, β = 0.2, p_AB _,X = 0.8, p_AB _,Y = 0.7, ρ_AB = 0.5).

p_B,X	p_A,Y	Heuristic	Proposed	k ₁	k ₂	p_B,X	p_A,Y	Heuristic	Proposed	k ₁	k ₂
0.2	0.2	70	57	2.61	2.23	0.5	0.2	142	103	1.20	1.87
	0.3	96	71	2.18	1.54		0.3	142	115	1.33	1.70
	0.4	154	102	1.84	0.96		0.4	154	145	1.43	1.36
	0.5	317	186	1.51	0.48		0.5	317	228	1.41	0.84
	0.6	1159	607	1.25	0.15		0.6	1159	652	1.25	0.29
0.3	0.2	70	63	2.24	2.17	0.6	0.2	278	173	0.65	1.60
	0.3	96	77	2.03	1.62		0.3	278	185	0.78	1.56
	0.4	154	108	1.78	1.05		0.4	278	213	0.98	1.43
	0.5	317	192	1.50	0.54		0.5	317	294	1.18	1.09
	0.6	1159	614	1.25	0.17		0.6	1159	717	1.20	0.45
0.4	0.2	91	76	1.76	2.06	0.7	0.2	957	514	0.21	1.30
	0.3	96	89	1.76	1.70		0.3	957	527	0.26	1.30
	0.4	154	120	1.67	1.20		0.4	957	555	0.36	1.29
	0.5	317	204	1.49	0.66		0.5	957	632	0.57	1.23
	0.6	1159	627	1.25	0.21		0.6	1159	1041	0.98	0.87

Table 6.

Total Sample Sizes (n_AB + n_A + n_B ) of Different Indications in Terms of Log Odds Ratio From the Heuristic Method and the Proposed Method (α = 0.05, β = 0.2, p_AB _,X = 0.8, p_AB _,Y = 0.7, ρ_AB = 0.5).

p_B,X	p_A,Y	Heuristic	Proposed	k ₁	k ₂	p_B,X	p_A,Y	Heuristic	Proposed	k ₁	k ₂
0.2	0.2	55	47	0.89	0.67	0.5	0.2	134	91	0.46	0.75
	0.3	83	60	0.88	0.52		0.3	134	102	0.55	0.69
	0.4	143	89	0.89	0.35		0.4	143	128	0.67	0.59
	0.5	306	169	0.91	0.19		0.5	306	205	0.81	0.40
	0.6	1150	585	0.94	0.06		0.6	1150	616	0.92	0.14
0.3	0.2	55	53	0.77	0.68	0.6	0.2	272	159	0.27	0.79
	0.3	83	66	0.81	0.56		0.3	272	169	0.35	0.78
	0.4	143	94	0.85	0.40		0.4	272	194	0.48	0.72
	0.5	306	174	0.90	0.22		0.5	306	268	0.68	0.58
	0.6	1150	589	0.94	0.07		0.6	1150	671	0.89	0.26
0.4	0.2	81	66	0.63	0.71	0.7	0.2	953	496	0.10	0.88
	0.3	83	77	0.70	0.62		0.3	953	505	0.13	0.87
	0.4	143	105	0.78	0.48		0.4	953	528	0.19	0.86
	0.5	306	183	0.86	0.28		0.5	953	596	0.35	0.82
	0.6	1150	597	0.93	0.09		0.6	1150	983	0.71	0.60

Discussion

In this article, we calculated the total sample sizes for the development of combination drugs of two monotherapies with reference to the 3 popular measures of binary endpoints: the risk difference, the log relative risk, and the log odds ratio. The results in Tables 1 to 6 show that the risk difference produces the smallest total sample sizes and the log relative risk yields the largest total sample sizes for various values of (p_A , p_B ) (or p_B _,X, p_A _,Y) when p_AB = 0.8 (or p_AB,X = 0.8, p_AB,Y = 0.7, ρ_AB = 0.5). It would be interesting to investigate whether the same results hold for the different values of (p_AB , p_A , p_B ) (or p_AB _,X, p_AB _,Y, ρ_AB , p_B _,X, p_A _,Y).

The largest difference in sample size between the newly proposed method and the heuristic method happens when there is a large difference between p_A and p_B . When one treatment is obviously better than another, it may not be ethical to include the inferior treatment.

Footnotes

Appendix. The Asymptotic Joint Distribution of the Test Statistics

By the multivariate central limit theorem, we have

where

Since

where

we have the following result by continuous mapping theorem

because

and

Declaration of Conflicting Interests

No potential conflicts were declared.

Funding

This research was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF), funded by the Ministry of Education, Science and Technology (2013R1A1A2004920).

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