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Financial constraints are a central consideration in clinical trial design. When per-subject costs differ between treatment arms, equal allocation may achieve the desired power but may not minimize the total study cost. This motivates sample size allocation methods that minimize the variable cost while maintaining a prespecified type I error rate and power.
Optimal allocation problems have been studied in related settings. Brittain and Schlesselman discussed optimal allocation for comparing two proportions[1], and Allison et al. considered statistically powerful study designs under financial constraints[2]. Classical optimal allocation ideas also appear in survey sampling[3]. More recent studies have examined cost-constrained or power-maximizing allocation problems for proportions, trimmed means, and other comparative designs[4−6]. However, closed-form cost-minimizing allocation formulas for several common two-proportion testing scenarios remain valuable for practical trial planning.
This paper derives closed-form continuous sample-size allocations for comparing two independent proportions under unequal unit costs. We consider two-sided tests of equality, non-inferiority/superiority tests, and equivalence tests. For each scenario, we minimize total variable cost subject to the corresponding normal-approximation power constraint. We also show that the normal-approximation power constraint fixes the asymptotic variance of the estimated difference between the two proportions. This provides an inverse Fisher information interpretation of the proposed allocation problem.
We first introduce the notation and assumptions. We assume that the two samples are independent and drawn from two populations. Specifically, let m subjects be assigned to the intervention group and n subjects to the control group with the corresponding binary response indicators
$ X_1,\cdots,X_m $ $ Y_1,\ \ldots,\ Y_n $ $ X_1,\cdots,X_m\sim Bernoulli(\pi_1)\mathrm{\ ,\ }Y_1,\cdots,Y_n\sim Bernoulli\left(\pi_2\right). $ (1) Then, we have
$ \sum\nolimits_{i=1}^{m}{X}_{i}\sim Binomial(m,{\pi }_{1})\; {\mathrm{, }} \;\sum\nolimits_{j=1}^{n}{Y}_{j}\sim Binomial(n,{\pi }_{2}) $ (2) under Eq. (1). Our estimates for
$ {\pi }_{1} $ $ {\pi }_{2} $ $ {\hat{\pi }}_{1}=\dfrac{1}{m}\sum\nolimits_{i=1}^{m}{X}_{i} \;{\mathrm{and}}\; {\hat{\pi }}_{2}=\dfrac{1}{n}\sum\nolimits_{j=1}^{n}{Y}_{j} . $ (3) The cost of a clinical trial consists of two parts. The first part is usually fixed and includes expenses for physicians, nurses, researchers, and other staff members involved in conducting and analyzing the trial. The second part depends on the number of study subjects in the trial. For simplicity, we assume that c1 and c2 denote the unit costs in the intervention and control groups, respectively. The variable cost is therefore
$ C=mc_1+nc_2. $ (4) Our goal is to minimize the variable cost in Eq. (4) subject to the power constraint for each hypothesis-testing scenario. Because fixed costs do not affect the allocation problem, we focus on the variable cost in Eq. (4) and refer to it as the total variable cost hereafter. We also assume that our sample sizes are large enough to allow us to use the normal approximation for sample size calculation.
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The following hypotheses are used to test whether the expected response rates in the intervention (π1) group and the control (π2) group are statistically different:
$ {H}_{0}\colon {\pi }_{1}={\pi }_{2} \;{\rm{versus}}\; {H}_{1}\colon {\pi }_{1}\neq {\pi }_{2}. $ (5) The scientific question is whether the response probabilities differ between the two groups. This is commonly referred to as a two-sided test of equality for two proportions[7].
Theorem 1. The sample sizes for hypotheses (Eq. [5]) to achieve the global minimum cost with power
$ 1-\beta $ $ \alpha $ $ m=\dfrac{\left[\pi_1\left(1-\pi_1\right)\right]^{\frac{1}{2}}\left([c_1\pi_1(1-\pi_1)]^{\frac{1}{2}}+[c_2\pi_2(1-\pi_2)]^{\frac{1}{2}}\right)}{c_1^{\frac{1}{2}}\left(\dfrac{\pi_1-\pi_2}{z_{1-\alpha/2}+z_{1-\beta}}\right)^2} $ and
$ n=\dfrac{\left[\pi_2\left(1-\pi_2\right)\right]^{\frac{1}{2}}\left([c_1\pi_1(1-\pi_1)]^{\frac{1}{2}}+[c_2\pi_2(1-\pi_2)]^{\frac{1}{2}}\right)}{c_2^{\frac{1}{2}}\left(\dfrac{\pi_1-\pi_2}{z_{1-\alpha/2}+z_{1-\beta}}\right)^2}. $ Proof. According to Chow et al.[7], the constraint condition for intervention and control groups to achieve the power
$ 1-\beta $ $ \alpha $ $ \dfrac{|\pi_1-\pi_2|}{\sqrt{\dfrac{\pi_1\left(1-\pi_1\right)}{m}+\dfrac{\pi_2(1-\pi_2)}{n}}}-z_{1-\alpha/2}=z_{1-\beta}. $ Equivalently,
$ \dfrac{\pi_1\left(1-\pi_1\right)}{m}+\dfrac{\pi_2\left(1-\pi_2\right)}{n}-\left(\dfrac{\pi_1-\pi_2}{z_{1-\alpha/2}+z_{1-\beta}}\right)^2=0. $ (6) The corresponding Lagrangian for minimizing Eq. (4) subject to constraint Eq. (6) is, according to Bertsekas[8]:
$ L\left(m,n,\lambda\right)=mc_1+nc_2+\lambda\left[\dfrac{\pi_1\left(1-\pi_1\right)}{m}+\dfrac{\pi_2\left(1-\pi_2\right)}{n}-\left(\dfrac{\pi_1-\pi_2}{z_{1-\alpha/2}+z_{1-\beta}}\right)^2\right]. $ (7) Setting
$ {\nabla }_{\left\{m,n\right\}}L\left(m,n,\lambda \right)=0 , $ we obtain
$ m=\sqrt{\dfrac{\lambda\pi_1\left(1-\pi_1\right)}{c_1}}\; \ \ and\ \ \ \; n=\sqrt{\dfrac{\lambda\pi_2\left(1-\pi_2\right)}{c_2}}. $ (8) Plugging Eq. (8) in Eq. (6), we obtain
$ \sqrt{\dfrac{c_1\pi_1\left(1-\pi_1\right)}{\lambda}}+\sqrt{\dfrac{c_2\pi_2\left(1-\pi_2\right)}{\lambda}}-\left(\dfrac{\pi_1-\pi_2}{z_{1-\alpha/2}+z_{1-\beta}}\right)^2=0, $ which implies that
$ \sqrt{\lambda}=\dfrac{\sqrt{c_1\pi_1\left(1-\pi_1\right)}+\sqrt{c_2\pi_2\left(1-\pi_2\right)}}{\left(\dfrac{\pi_1-\pi_2}{z_{1-\alpha/2}+z_{1-\beta}}\right)^2}. $ (9) By plugging Eq. (9) into Eq. (8), we obtain the solutions in Theorem 1. Note that the Hessian matrix for the Lagrangian function Eq. (7) is
$ \nabla^2L\left(m,n,\lambda\right)=\left[\begin{array}{cc}2\lambda\dfrac{\pi_1\left(1-\pi_1\right)}{m^3} & 0 \\ 0 & 2\lambda\dfrac{\pi_2\left(1-\pi_2\right)}{n^3}\end{array}\right]\ \ , $ which is positive definite. The solution in Theorem 1 is the only KKT point of the problem, which together with the positive definite Hessian, implies it is the global minimizer of the cost under the constraint in Eq. (6).
Example 1. Suppose the expected cure rates are 80% in the intervention group and 65% in the control group receiving standard treatment. Then, the sample sizes under equal allocation with 80% power at the 0.05 significance level are[7]
$ \begin{split} m & =n=\dfrac{\left(z_{1-\alpha/2}+z_{1-\beta}\right)^2\left[\pi_1\left(1-\pi_1\right)+\pi_2\left(1-\pi_2\right)\right]}{\left|\pi_1-\pi_2\right|^2} \\ &=\dfrac{(1.96+0.84)^2[0.8\left(1-0.8\right)+0.65(1-0.65)]}{0.15^2}\approx135. \end{split} $ If the per-subject costs are
${\text{\$}}$ ${\text{\$}}$ $ C=135\times800+135\times200=135,000. $ Using our results from Theorem 1, we have
$ \begin{split}m & =\dfrac{\left[\pi_1\left(1-\pi_1\right)\right]^{\frac{1}{2}}\left([c_1\pi_1(1-\pi_1)]^{\frac{1}{2}}+[c_2\pi_2(1-\pi_2)]^{\frac{1}{2}}\right)}{c_1^{\frac{1}{2}}\left(\dfrac{\pi_1-\pi_2}{z_{1-\alpha/2}+z_{1-\beta}}\right)^2} \\ & =\dfrac{\left[0.8\left(1-0.8\right)\right]^{\frac{1}{2}}\left([800\times 0.8(1-0.8)]^{\frac{1}{2}}+[200\times 0.65(1-0.65)]^{\frac{1}{2}}\right)}{800^{\frac{1}{2}}\left(\dfrac{0.8-0.65}{1.96+0.84}\right)^2}\approx89\end{split} $ and
$ \begin{split} n & =\dfrac{\left[\pi_2\left(1-\pi_2\right)\right]^{\frac{1}{2}}\left(\left[c_1\pi_1\left(1-\pi_1\right)\right]^{\frac{1}{2}}+\left[c_2\pi_2\left(1-\pi_2\right)\right]^{\frac{1}{2}}\right)}{c_2^{\frac{1}{2}}\left(\dfrac{\pi_1-\pi_2}{z_{1-\alpha/2}+z_{1-\beta}}\right)^2} \\ &=\dfrac{\left[0.65\left(1-0.65\right)\right]^{\frac{1}{2}}\left(\left[800\times 0.8\left(1-0.8\right)\right]^{\frac{1}{2}}+\left[200\times 0.65\left(1-0.65\right)\right]^{\frac{1}{2}}\right)}{200^{\frac{1}{2}}\left(\dfrac{0.8-0.65}{1.96+0.84}\right)^2} \\ &\approx212\ , \end{split} $ with a total cost of
$ C^{\ast}=89\times800+212\times200=113,600. $ Moreover,
$ \begin{split}\dfrac{C-C^{\ast}}{C}=\dfrac{135,000-113,600}{135,000}\approx0.1585,\end{split} $ which indicates that the proposed allocation reduces the total cost by approximately 15.85% relative to equal allocation.
Cost minimization for non-inferiority and superiority tests
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We use the following hypotheses to test non-inferiority or superiority[7]:
$ {H}_{0}\colon {\pi }_{1}-{\pi }_{2}\leq {\Delta } \;{\mathrm{versus}} \;{H}_{1}\colon {\pi }_{1}-{\pi }_{2} \gt {\Delta } , $ (10) where Δ is defined as a signed margin, with positive values corresponding to a superiority margin and negative values corresponding to a non-inferiority margin. We assume π1 − π2 − Δ > 0, so that the alternative hypothesis is separated from the null boundary by a positive effect size.
Theorem 2. The sample sizes for hypotheses Eq. (10) to achieve the global minimum cost with the power
$ 1-\beta $ $ m=\dfrac{\left[\pi_1\left(1-\pi_1\right)\right]^{\frac{1}{2}}\left([c_1\pi_1(1-\pi_1)]^{\frac{1}{2}}+[c_2\pi_2(1-\pi_2)]^{\frac{1}{2}}\right)}{c_1^{\frac{1}{2}}\left(\dfrac{\pi_1-\pi_2-\Delta}{z_{1-\alpha}+z_{1-\beta}}\right)^2} $ and
$ n=\dfrac{\left[\pi_2\left(1-\pi_2\right)\right]^{\frac{1}{2}}\left([c_1\pi_1(1-\pi_1)]^{\frac{1}{2}}+[c_2\pi_2(1-\pi_2)]^{\frac{1}{2}}\right)}{c_2^{\frac{1}{2}}\left(\dfrac{\pi_1-\pi_2-\Delta}{z_{1-\alpha}+z_{1-\beta}}\right)^2}. $ Proof. The constraint condition for the intervention and control groups to achieve power 1−β at a significance level α[7] is given as:
$ \dfrac{\pi_1-\pi_2-\Delta}{\sqrt{\dfrac{\pi_1\left(1-\pi_1\right)}{m}+\dfrac{\pi_2(1-\pi_2)}{n}}}-z_{1-\alpha}=z_{1-\beta}, $ which is equivalent to
$ \dfrac{\pi_1\left(1-\pi_1\right)}{m}+\dfrac{\pi_2\left(1-\pi_2\right)}{n}-\left(\dfrac{\pi_1-\pi_2-\Delta}{z_{1-\alpha}+z_{1-\beta}}\right)^2=0. $ (11) Therefore, the Lagrangian function for minimizing Eq. (4) subject to constraint Eq. (11) is[8]
$ L\left(m,n,\lambda\right)=mc_1+nc_2+\lambda\left[\dfrac{\pi_1\left(1-\pi_1\right)}{m}+\dfrac{\pi_2\left(1-\pi_2\right)}{n}-\left(\dfrac{\pi_1-\pi_2-\Delta}{z_{1-\alpha}+z_{1-\beta}}\right)^2\right]. $ (12) Setting
$ \nabla_{\left\{m,n\right\}}L\left(m,n,\lambda\right)=0\ \ , $ we obtain
$ m=\sqrt{\dfrac{\lambda\pi_1\left(1-\pi_1\right)}{c_1}}\; \rm{and}\; n=\sqrt{\dfrac{\lambda\pi_2\left(1-\pi_2\right)}{c_2}}, $ (13) which is the same as Eq. (8). Plugging Eq. (13) into Eq. (11), we obtain
$ \sqrt{\dfrac{c_1\pi_1\left(1-\pi_1\right)}{\lambda}}+\sqrt{\dfrac{c_2\pi_2\left(1-\pi_2\right)}{\lambda}}-\left(\dfrac{\pi_1-\pi_2-\Delta}{z_{1-\alpha}+z_{1-\beta}}\right)^2=0, $ which implies that
$ \sqrt{\lambda}=\dfrac{\sqrt{c_1\pi_1\left(1-\pi_1\right)}+\sqrt{c_2\pi_2\left(1-\pi_2\right)}}{\left(\dfrac{\pi_1-\pi_2-\Delta}{z_{1-\alpha}+z_{1-\beta}}\right)^2}. $ (14) By substituting Eq. (14) into Eq. (13), we obtain the solutions in Theorem 2. Note that the Hessian matrix for the Lagrangian function Eq. (12) is exactly the same as that in Theorem 1, and the solution in Theorem 2 is the only KKT point, which again shows that the solution is the global minimizer of the cost Eq. (4) under the constraint Eq. (11).
Example 2. Non-inferiority. Under the signed-margin convention, suppose that the non-inferiority margin is Δ = −10% and the expected cure rates for the disease in the intervention and control groups are 80% and 75%, respectively. Then, the sample sizes for equal allocation with 80% power at the 0.05 significance level are, according to Chow et al.[7],
$ \begin{split} m & =n=\dfrac{\left(z_{1-\alpha}+z_{1-\beta}\right)^2\left[\pi_1\left(1-\pi_1\right)+\pi_2\left(1-\pi_2\right)\right]}{\left(\pi_1-\pi_2-\Delta\right)^2} \\ &=\dfrac{\left(1.64+0.84\right)^2\left[0.8\left(1-0.8\right)+0.75\left(1-0.75\right)\right]}{\left(0.8-0.75-(-0.1)\right)^2}\approx95. \end{split} $ If the per-subject costs in the intervention and control groups are
${\text{\$}} $ ${\text{\$}} $ $ C=95\times100+95\times800=85,500. $ Applying Theorem 2 yields
$ m=\dfrac{{\left[0.8\left(1-0.8\right)\right]}^{\frac{1}{2}}\left({[100\times 0.8(1-0.8)]}^{\frac{1}{2}}+{[800\times 0.75(1-0.75)]}^{\frac{1}{2}}\right)}{{100}^{\frac{1}{2}}{\left(\dfrac{0.8-0.75+0.1}{1.64+0.84}\right)}^{2}}\approx 178, $ $ n=\dfrac{{\left[0.75\left(1-0.75\right)\right]}^{\frac{1}{2}}\left({[100\times 0.8(1-0.8)]}^{\frac{1}{2}}+{[800\times 0.75(1-0.75)]}^{\frac{1}{2}}\right)}{{800}^{\frac{1}{2}}{\left(\dfrac{0.8-0.75+0.1}{1.64+0.84}\right)}^{2}}\approx 68 $ and the corresponding cost is
$ C^{\ast}=178\times100+68\times800=72,200. $ The relative cost reduction is therefore
$ \dfrac{C-C^{\ast}}{C}=\dfrac{85,500-72,200}{85,500}\approx0.1556. $ Therefore, our method reduces the total cost by 15.56% relative to equal allocation.
Example 3. Superiority. Assume that the superiority margin is 5% and the cure rates are 80% and 65% for the intervention and control groups, respectively. The per-subject costs are
${\text{\$}} $ ${\text{\$}} $ $ \begin{split}m & =n=\dfrac{\left(z_{1-\alpha}+z_{1-\beta}\right)^2\left[\pi_1\left(1-\pi_1\right)+\pi_2\left(1-\pi_2\right)\right]}{\left(\pi_1-\pi_2-\Delta\right)^2} \\ & =\dfrac{\left(1.64+0.84\right)^2\left[0.8\left(1-0.8\right)+0.65\left(1-0.65\right)\right]}{\left(0.8-0.65-0.05\right)^2}\approx238\end{split} $ and the total cost is
$ C=238\times800+238\times200=238,000. $ Incorporating the unequal unit costs, Theorem 2 gives
$ m=\dfrac{\left[0.8\left(1-0.8\right)\right]^{\frac{1}{2}}\left([800\times0.8(1-0.8)]^{\frac{1}{2}}+[200\times0.65(1-0.65)]^{\frac{1}{2}}\right)}{800^{\frac{1}{2}}\left(\dfrac{0.8-0.65-0.05}{1.64+0.84}\right)^2}\approx157, $ and
$ \begin{split} n & =\dfrac{\left[0.65\left(1-0.65\right)\right]^{\frac{1}{2}}\left([800\times 0.8(1-0.8)]^{\frac{1}{2}}+[200\times 0.65(1-0.65)]^{\frac{1}{2}}\right)}{200^{\frac{1}{2}}\left(\dfrac{0.8-0.65-0.05}{1.64+0.84}\right)^2} \\ &\approx375, \end{split} $ resulting in a total cost of
$ C^{\ast}=157\times800+375\times200=200,600. $ Therefore,
$ \dfrac{C-C^{\ast}}{C}=\dfrac{238,000-200,600}{238,000}\approx0.1571. $ Hence, our method reduces the total cost by 15.71% relative to equal allocation.
Cost minimization for equivalence tests
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The equivalence hypotheses are formulated as Chow et al.[7]:
$ {H}_{0}\colon |{\pi }_{1}-{\pi }_{2}|\geq {\Delta }\; {\rm{versus}}\; {H}_{1}\colon {|\pi }_{1}-{\pi }_{2}| \lt {\Delta } , $ (15) where ∆ is the equivalence margin, which is usually specified by investigators rather than statisticians.
Theorem 3. The sample sizes for hypotheses (15) to achieve the global minimum cost with the 1 − β at a significance level α are
$ m=\dfrac{\left[\pi_1\left(1-\pi_1\right)\right]^{\frac{1}{2}}\left([c_1\pi_1(1-\pi_1)]^{\frac{1}{2}}+[c_2\pi_2(1-\pi_2)]^{\frac{1}{2}}\right)}{c_1^{\frac{1}{2}}\left(\dfrac{\Delta-|\pi_1-\pi_2|}{z_{1-\alpha}+z_{1-\beta/2}}\right)^2} $ and
$ n=\dfrac{\left[\pi_2\left(1-\pi_2\right)\right]^{\frac{1}{2}}\left([c_1\pi_1(1-\pi_1)]^{\frac{1}{2}}+[c_2\pi_2(1-\pi_2)]^{\frac{1}{2}}\right)}{c_2^{\frac{1}{2}}\left(\dfrac{\Delta-|\pi_1-\pi_2|}{z_{1-\alpha}+z_{1-\beta/2}}\right)^2} $ Proof. The proof is similar to that of Theorem 2 and is therefore omitted.
Example 4. Equivalence. Assume that the cure rates are 75% (π1 = 0.75) for the intervention and 80% for the control group (π2 = 0.80), and the equivalence margin or limit is ∆ = 20%. The intervention is easier to administer and less expensive than the control group. The per-subject costs for the intervention and control groups are
$ {\text{\$}} $ $ {\text{\$}} $ $ \begin{split} m & =n=\dfrac{\left(z_{1-\alpha}+z_{1-\beta/2}\right)^2\left(\pi_1\left(1-\pi_1\right)+\pi_2(1-\pi_2)\right)}{\left(\Delta-|\pi_1-\pi_2|\right)^2} \\ &=\dfrac{\left(1.64+1.28\right)^2\left(0.75\times 0.25+0.8\times 0.2\right)}{\left(0.2-|0.75-0.8|\right)^2}\approx132. \end{split} $ The total cost is
$ C=132\times100+132\times900=132,000. $ Incorporating the unequal unit costs and applying Theorem 3, we obtain
$ m=\dfrac{{\left[0.75\times 0.25\right]}^{\frac{1}{2}}\left({[100\times 0.75\times 0.25]}^{\frac{1}{2}}+{[900\times 0.8\times 0.2]}^{\frac{1}{2}}\right)}{{100}^{\frac{1}{2}}{\left(\dfrac{0.2-|0.75-0.8|}{1.64+1.28}\right)}^{2}}\approx 268 $ and
$ n=\dfrac{\left[0.8\times0.2\right]^{\frac{1}{2}}\left([100\times0.75\times0.25]^{\frac{1}{2}}+[900\times0.8\times0.2]^{\frac{1}{2}}\right)}{900^{\frac{1}{2}}\left(\dfrac{0.2-|0.75-0.8|}{1.64+1.28}\right)^2}\approx83. $ Therefore,
$ C^{\ast}=268\times100+83\times900=101,500 $ and
$ \dfrac{C-C^{\ast}}{C}=\dfrac{132,000-101,500}{132,000}\approx0.2311, $ which implies that our method reduces the total cost by 23.11% relative to equal allocation.
Fisher information
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The power constraints considered above can be interpreted in terms of inverse Fisher information. For the two independent Bernoulli samples, the Fisher information for π1 and π2 is[9]
$ I_1^{ }\left(\pi_1\right)=\dfrac{m}{\pi_1\left(1-\pi_1\right)},\ \ \ \ I_2^{ }\left(\pi_2\right)\dfrac{n}{\pi_2\left(1-\pi_2\right)}. $ Therefore,
$ I_1^{-1}\left(\pi_1\right)=\dfrac{\pi_1\left(1-\pi_1\right)}{m},\ \ \ I_2^{-1}\left(\pi_2\right)=\dfrac{\pi_2\left(1-\pi_2\right)}{n}. $ Thus, the normal-approximation power constraint fixes
$ I_1^{-1}+I_2^{-1} $ Theorem 4.1. The sum of the inverses Fisher information terms for hypotheses Eq. (5) with power
$ 1-\beta $ $ \left(\dfrac{\pi_1-\pi_2}{z_{1-\alpha/2}+z_{1-\beta}}\right)^2. $ Proof. Since Fisher information
$ I_1\left(\pi_1\right)=\dfrac{m}{\pi_1\left(1-\pi_1\right)} $ $ I_2\left(\pi_2\right)=\dfrac{n}{\pi_2\left(1-\pi_2\right)} $ $ I_1^{-1}+I_2^{-1}=\dfrac{\pi_1\left(1-\pi_1\right)}{m}+\dfrac{\pi_2\left(1-\pi_2\right)}{n}=\left(\dfrac{\pi_1-\pi_2}{z_{1-\alpha/2}+z_{1-\beta}}\right)^2. $ Theorem 4.2. The sum of the inverse Fisher information terms for hypotheses Eq. (10) with power
$ 1-\beta $ $ \left(\dfrac{\pi_1-\pi_2-\Delta}{z_{1-\alpha}+z_{1-\beta}}\right)^2. $ Proof. Using Eq. (11), we have
$ I_1^{-1}+I_2^{-1}=\dfrac{\pi_1\left(1-\pi_1\right)}{m}+\dfrac{\pi_2\left(1-\pi_2\right)}{n}=\left(\dfrac{\pi_1-\pi_2-\Delta}{z_{1-\alpha}+z_{1-\beta}}\right)^2. $ Theorem 4.3. The sum of the inverses of Fisher information terms for hypotheses (15) with power
$ 1-\beta $ $ \left(\dfrac{\Delta-|\pi_1-\pi_2|}{z_{1-\alpha}+z_{1-\beta/2}}\right)^2. $ Proof. Similarly, we have
$ I_1^{-1}+I_2^{-1}=\dfrac{\pi_1\left(1-\pi_1\right)}{m}+\dfrac{\pi_2\left(1-\pi_2\right)}{n}=\left(\dfrac{\Delta-|\pi_1-\pi_2|}{z_{1-\alpha}+z_{1-\beta/2}}\right)^2. $ The sum of the inverse Fisher information terms for the two-sided equality, non-inferiority/superiority, and equivalence tests is fixed under the corresponding normal-approximation power constraint.
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We conducted two simulation studies using SAS®. The first simulation corresponds to Example 1. We used rand('BINOMIAL', p, n) to generate 1,000 binomial random samples under n1 = n2 = 135, π1 = 0.8, and π2 = 0.65, and also generated 1,000 replicates using rand('BINOMIAL', 0.80, 89) and rand('BINOMIAL', 0.65, 212). The power difference between the two comparisons was 0.8%. Therefore, the simulation results were consistent with the theoretical calculation.
The second simulation corresponds to Example 4. We generated 1,000 replicates under the equal allocation (m,n) = (132,132) and another 1,000 replicates under the optimized allocation (m,n) = (268,83). For each replicate, we conducted the prespecified equivalence test and estimated empirical power as the proportion of rejections. The empirical power difference was 3.2%; this discrepancy should be further evaluated with a larger number of simulations and by checking the effect of integer rounding.
These preliminary simulations suggest that the optimized allocations can achieve empirical power close to that of equal allocation while reducing cost. However, more extensive simulations with a larger number of Monte Carlo replicates are needed to quantify the effect of integer rounding and normal-approximation error.
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One limitation is that all results are derived under large-sample assumptions, so they may not apply to small samples. The corresponding optimization problems are more difficult for small samples because sample sizes must take integer values. The proposed large-sample formulas may be less suitable for rare-disease trials, where recruitment constraints and integer sample-size effects often play a more prominent role. These formulas may be useful at the planning stage of comparative studies in which the unit costs differ substantially between treatment arms and recruitment is not the primary limiting factor.
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Financial constraints play an important role in clinical trial design. We derived closed-form formulas for continuous sample-size allocations for comparing two independent binary outcomes under fixed power and a fixed significance level. Because unit costs may differ between study arms, equal allocation does not necessarily minimize total variable cost. The formulas provide useful continuous allocations at the design stage. In practical applications, the rounded integer allocations should be checked against the target power. Although this paper focuses on comparing two proportions, similar cost-minimization ideas may be extended to other endpoints, such as means, relative risks, survival outcomes, and correlations.
Thanks are given to Dr. Rongling Wu for his suggestions.
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The authors confirm contribution to the paper as follows: study conception and design: Luo J (www.researchgate.net/profile/Jiangtao-Luo-2), Liu CM, Pant MD, El Moudden I, Wang Z; theorem proof and algorithm: Luo J, Liu CM, Wang Y, Wang Z, Zhang H; draft manuscript preparation: Luo J, Liu CM, Wang Z, Zhang H. All authors reviewed the results and approved the final version of the manuscript.
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No real datasets were analyzed. Simulation code and generated data are available from the corresponding author, Jiangtao Luo, upon reasonable request.
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The authors declare that they have no conflict of interest.
- Copyright: © 2026 by the author(s). Published by Maximum Academic Press, Fayetteville, GA. This article is an open access article distributed under Creative Commons Attribution License (CC BY 4.0), visit https://creativecommons.org/licenses/by/4.0/.
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Cite this article
Luo J, Liu CM, El Moudden I, Pant MD, Wang Y, et al. 2026. Cost-minimizing sample size allocation for comparing two proportions. Statistics Innovation 3: e009 doi: 10.48130/stati-0026-0009 
Cost-minimizing sample size allocation for comparing two proportions
- Received: 18 August 2025
- Revised: 10 January 2026
- Accepted: 09 February 2026
- Published online: 23 June 2026
Abstract: We develop cost-minimizing sample size allocation methods for comparing two proportions in medical studies while controlling the type I error rate and maintaining prespecified power. Closed-form formulas are derived for two-sided tests of equality, non-inferiority/superiority tests, and equivalence tests. We provide an illustrative example for each testing scenario to demonstrate the cost efficiency of the proposed allocation. The results may help investigators design cost-efficient studies when unit costs differ between treatment arms and financial resources are limited. We further show that the corresponding normal-approximation power constraint can be interpreted as fixing the asymptotic variance of the estimated difference between the two proportions, or, equivalently, the sum of the corresponding inverse Fisher information contributions.