Why Allocation Across Biomarker Strata Matters
When a trial plans to make claims about both a biomarker-positive subgroup and the full target population, total enrollment alone does not determine how informative the study will be. The number enrolled in each stratum controls the precision of its treatment-effect estimate. A small positive stratum, for example, can leave a subgroup estimate imprecise even when the overall trial is large. Conversely, recruiting to the population prevalence may provide more positive participants than necessary for an all-comer estimate, or too few to meet a subgroup goal.
The design problem is to choose stratum sample sizes that meet prespecified goals under stated assumptions about biomarker prevalence, outcome variability, and treatment effects. This tutorial treats prevalence as fixed at a planning value. It does not account for uncertainty in that value.
Planning Model and Precision Formulas
Consider a parallel-group trial with a continuous outcome. Within each biomarker stratum, participants are randomized to treatment or control. Assume independent participant outcomes, independent estimates across strata, and a common outcome variance \(\sigma_s^2\) for treatment and control within stratum \(s\), where \(s\) is \(+\) or \(-\). For planning, treat these variances as specified values. The treatment-effect estimator in each stratum is the difference between its treatment and control sample means.
If the stratum total is split equally between treatment and control, its treatment-effect variance is \(4\sigma_s^2/n_s\). If the arm counts are not equal, the correct variance is \(\sigma_s^2(1/n_{s,T}+1/n_{s,C})\), where \(n_{s,T}\) and \(n_{s,C}\) are the treatment and control counts. Thus the shortcut \(4\sigma_s^2/n_s\) should not be applied to an odd stratum total under a design that requires exact arm balance. If treatment and control variances differ, use \(\sigma_{s,T}^2/n_{s,T}+\sigma_{s,C}^2/n_{s,C}\) instead.
Because the all-comer effect is a weighted sum of the two stratum effects, and the estimates are independent, its variance is the sum of the weighted component variances. With equal allocation to treatment and control in each stratum:
The weights are squared in the variance. The all-comer estimate is \(\hat{\Delta}_{\mathrm{all}}=\pi\hat{\Delta}_+ +(1-\pi)\hat{\Delta}_-\). A normal-approximation confidence interval with confidence level \(1-\alpha\) has half-width \(z_{1-\alpha/2}\sqrt{\operatorname{Var}(\hat{\Delta}_{\mathrm{all}})}\), where \(z_q\) is the \(q\)-quantile of the standard normal distribution. The same calculation with only the relevant stratum variance gives a subgroup interval.
Worked Example: Meeting a Positive-Subgroup Precision Goal
Suppose the assumed prevalence is \(\pi=0.20\), the total enrollment is 400, and the outcome standard deviation is 8 units in both strata. The trial requires equal treatment and control counts within each stratum. The positive-subgroup 95% confidence interval should have a half-width no greater than 3.5 outcome units. Find an allocation that meets this goal and check the all-comer precision.
For the positive subgroup, the planned variance is \(4(8^2)/n_+=256/n_+\). The half-width requirement is \(1.96\sqrt{256/n_+}\leq 3.5\). Squaring both positive sides and rearranging gives \(n_+\geq256(1.96/3.5)^2=80.2816\). The stratum total must be even for exact equal arm sizes, so the smallest eligible total is 82. Set \(n_-=400-82=318\), also even.
The positive-stratum variance is \(256/82=3.122\), so its standard error is \(\sqrt{3.122}=1.767\) units and the 95% half-width is \(1.96(1.767)=3.463\) units. This meets the 3.5-unit goal.
For the all-comer effect, the variance is \(0.20^2(256/82)+0.80^2(256/318)=0.1249+0.5152=0.6401\). Thus its standard error is \(\sqrt{0.6401}=0.8001\) units and its 95% half-width is \(1.96(0.8001)=1.568\) units. This calculation uses the target prevalence weights, despite the enrolled positive fraction being \(82/400=0.205\).
Choosing an Allocation for All-Comer Precision
If the only objective is to minimize all-comer variance for a fixed total \(N=n_++n_-\), the best continuous allocation depends on both prevalence and outcome variability. It is not generally the prevalence-proportional allocation. Define \(a=\pi\sigma_+\) and \(b=(1-\pi)\sigma_-\). Apart from the constant factor 4, the variance to minimize is \(a^2/n_+ + b^2/n_-\).
A short derivation identifies the allocation. Substitute \(n_-=N-n_+\) and differentiate with respect to \(n_+\). At an interior minimum, \(-a^2/n_+^2+b^2/n_-^2=0\). Since sample sizes and standard deviations are positive, this implies \(a/n_+=b/n_-\), or \(n_+/n_-=a/b\). The resulting Neyman allocation is:
This result minimizes the all-comer variance under the model and fixed-total constraint. It does not ensure adequate precision or power in each subgroup. Additional subgroup requirements can force an allocation away from this variance-minimizing ratio.
Worked Example: All-Comer Precision With Unequal Stratum Variability
Assume \(\pi=0.25\), \(\sigma_+=10\) outcome units, \(\sigma_-=8\) outcome units, and \(N=340\). Equal allocation to treatment and control is required within each stratum. Find a near-optimal integer allocation and its all-comer 95% confidence-interval half-width.
The Neyman ratio is \(n_+/n_-=(0.25)(10)/[(0.75)(8)]=2.5/6=5/12\). With \(N=340\), this gives \(n_+=100\) and \(n_-=240\), both even. The stratum treatment-effect variances are \(4(10^2)/100=4.000\) and \(4(8^2)/240=1.067\), in squared outcome units.
The all-comer variance is \((0.25)^2(4.000)+(0.75)^2(1.067)=0.2500+0.6000=0.8500\). The standard error is \(\sqrt{0.8500}=0.9220\) outcome units, giving a 95% half-width of \(1.96(0.9220)=1.807\) outcome units.
For comparison, an even allocation near the prevalence-proportional split is \(n_+=86\), \(n_-=254\). Its all-comer variance is \((0.25)^2(400/86)+(0.75)^2(256/254)=0.2907+0.5669=0.8576\), and its half-width is \(1.96\sqrt{0.8576}=1.815\) units. The difference is modest here, but the Neyman allocation has the smaller modeled variance. A subgroup precision constraint could still make the comparison allocation preferable.
Power Goals and Joint Constraints
Precision goals concern interval width; power goals concern the probability of rejecting a specified null hypothesis when a specified alternative is true. For a two-sided test of a stratum effect against zero, with significance level \(\alpha\), target power \(1-\beta\), equal arm allocation, and known planning standard deviation \(\sigma_s\), the usual normal approximation gives \(n_s=4\sigma_s^2(z_{1-\alpha/2}+z_{1-\beta})^2/\Delta_s^2\). Here \(\Delta_s\) is the effect size used for planning. Round upward to an allowable stratum total. This approximation assumes the stated effect and variance are reasonable planning values; it does not guarantee the achieved power if they are wrong.
Designers can impose several constraints at once: a minimum \(n_+\) for positive-subgroup power or precision, a minimum \(n_-\) for a negative-subgroup goal, and a maximum all-comer variance or interval half-width. With fixed total enrollment, a candidate allocation must satisfy every constraint. If totals must split evenly between arms, both stratum totals must be even. In practice, evaluate integer candidates near the continuous solution and verify each formula using the actual arm counts.
Worked Example: Combining Positive-Subgroup Power and All-Comer Precision
Suppose \(\pi=0.20\), the outcome standard deviation is 10 units in both strata, and a total of 500 participants is planned. The positive-stratum treatment effect of interest is 5 units. Require at least 80% power for a two-sided test at \(\alpha=0.05\), and require the all-comer 95% confidence-interval half-width to be at most 1.8 units. Exact equal treatment and control counts are required within each stratum.
The positive-stratum total required by the normal power approximation is \(4(10^2)(1.96+0.8416)^2/5^2=125.583\). Rounding upward to an even total gives \(n_+=126\), or 63 participants per arm. Allocate the remaining \(n_-=500-126=374\) participants to the negative stratum, or 187 per arm.
The positive-stratum standard error is \(\sqrt{400/126}=1.782\) units. Under the planned 5-unit effect, the standardized mean is \(5/1.782=2.806\). The normal-approximation power is approximately \(P(Z>1.96\mid \text{mean}=2.806)+P(Z<-1.96\mid \text{mean}=2.806)\), which is approximately \(\Phi(0.846)+\Phi(-4.766)=0.801\). Thus the planned positive-subgroup power is about 80.1%.
The all-comer variance is \(0.20^2(400/126)+0.80^2(400/374)=0.1270+0.6845=0.8115\). The standard error is \(\sqrt{0.8115}=0.9008\) units, and the 95% half-width is \(1.96(0.9008)=1.766\) units. Both specified goals are met under the planning assumptions.
Allocation Diagnostics, Alternatives, and Limits
A useful design check is to calculate each stratum’s arm counts, treatment-effect variance, and contribution to all-comer variance, then verify every goal using the integer allocation. A very small stratum count, a large discrepancy between assumed and observed prevalence, or a variance contribution dominated by one stratum signals that the design may not deliver its intended precision. Before enrollment, also check that the assumed prevalence and standard deviations correspond to the target population and planned outcome measurement.
Three allocation approaches answer different questions:
| Approach | How the allocation is chosen | What it prioritizes |
|---|---|---|
| Prevalence-proportional | Allocate in the assumed ratio \(\pi:(1-\pi)\). | Enrollment composition similar to the target population. |
| Neyman | Allocate in the ratio \(\pi\sigma_+:(1-\pi)\sigma_-\). | Minimum all-comer variance for a fixed total under the stated model. |
| Constraint-based | Choose integer counts that satisfy specified subgroup and all-comer goals. | Joint requirements, which may override either simple ratio. |
Prevalence-proportional allocation is not automatically optimal for precision, and Neyman allocation is not automatically adequate for subgroup inference. A constraint-based design is appropriate when subgroup precision or power is a formal objective. If recruitment uses quotas that produce stratum proportions different from \(\pi\), the all-comer estimate must still use the target-population weights for the estimand defined here.
These calculations rely on a specified prevalence, correctly classified biomarker strata, the assumed variance model, and the planned randomization. They use normal approximations and do not account for multiplicity from making several claims, data-dependent population selection, or uncertainty in prevalence and variance. Those considerations can change the design requirements; they are not repaired merely by rounding the allocation.
Common Mistake
A frequent error is to use \(4\sigma_s^2/n_s\) for every stratum total, even when an odd total is split into unequal treatment and control groups. For instance, 73 participants cannot be divided equally between arms. The correction is to either choose an even total, such as 72, when exact balance is required, or calculate the variance using the actual arm counts: \(\sigma_s^2(1/n_{s,T}+1/n_{s,C})\). A second related error is to treat the enrolled positive fraction as the all-comer weight. The estimand’s weight is the assumed target prevalence \(\pi\), unless the target population itself is defined differently.
Check Your Understanding
Answer each question using the stated assumptions. Brief solutions follow the questions.
- With \(\pi=0.30\), equal outcome standard deviations, and fixed total enrollment, what is the continuous Neyman allocation ratio \(n_+/n_-\)?
- Suppose \(\sigma_+=6\), \(n_+=100\), and equal treatment and control allocation. What is the positive-stratum treatment-effect variance?
- With \(\pi=0.10\), \(\sigma_+=12\), and \(\sigma_-=6\), calculate the Neyman ratio \(n_+/n_-\).
- Why can an allocation with many positive participants still yield a poorly estimated all-comer effect if the positive stratum is uncommon?
Solutions. (1) With equal standard deviations, the ratio is \(\pi/(1-\pi)=0.30/0.70=3/7\). (2) The variance is \(4(6^2)/100=144/100=1.440\) squared outcome units. (3) The ratio is \((0.10)(12)/[(0.90)(6)]=1.2/5.4=2/9\). (4) The all-comer variance contribution from the positive stratum is multiplied by \(\pi^2\), so a rare stratum has a small weight in the all-comer estimand; its own subgroup estimate may nevertheless be imprecise if its sample size is inadequate. Allocation goals must therefore be checked for the estimand or subgroup they are intended to support.