Variance Re-Estimation With Unequal Allocation

Why Allocation Ratio Changes a Variance Update

In a two-arm trial with a continuous outcome, an interim variance estimate can change the sample size needed to detect a prespecified treatment difference. With unequal randomization, the update must account for how many participants are assigned to each arm: adding a participant to a smaller arm generally reduces the variance of the treatment-control difference more than adding one to the larger arm. The calculation therefore depends on both the estimated outcome variance and the allocation ratio.

The focus here is a fixed treatment-to-control allocation ratio and a continuous outcome analyzed by comparing the two group means. This is a planning calculation using a normal approximation; it does not replace the protocol’s prespecified analysis or the protections needed for an adaptive trial. The earlier material on uncertainty in interim nuisance-parameter estimates motivates why the variance estimate itself should not be treated as exact.

Definition: Let \(n_T\) and \(n_C\) be the treatment and control sample sizes, and define the fixed allocation ratio as \(r=n_T/n_C\). The total sample size is \(N=n_T+n_C\). Let \(\delta>0\) be the treatment difference the design aims to detect, in outcome units, and let \(\sigma^2\) be the common outcome variance in the two groups. A variance-based re-estimation substitutes an interim estimate of \(\sigma^2\) into the sample-size formula while retaining the prespecified \(\delta\), type I error rate, power target, and allocation ratio.

Deriving the Unequal-Allocation Formula

Assume independent participants, normally distributed outcomes within each arm, a common variance \(\sigma^2\), and a fixed ratio \(r\). These assumptions imply that the variance of the difference between the treatment and control sample means is \(\sigma^2/n_T+\sigma^2/n_C\). For a two-sided test at level \(\alpha\), the usual normal-approximation planning equation for power \(1-\beta\) is

\delta=(z_{1-\alpha/2}+z_{1-\beta})\sqrt{\sigma^2\left(\frac{1}{n_T}+\frac{1}{n_C}\right)}.

Here \(z_q\) is the \(q\) quantile of the standard normal distribution. This equation approximates the separation between the null and the design alternative, measured in standard errors, needed to attain the target power. Because \(n_T=rn_C\), and \(N=(1+r)n_C\), the variance term can be written using total enrollment:

\sigma^2\left(\frac{1}{n_T}+\frac{1}{n_C}\right) =\sigma^2\left(\frac{1}{rn_C}+\frac{1}{n_C}\right) =\frac{\sigma^2(1+r)^2}{rN}.

Squaring the planning equation and solving for \(N\) gives the required total sample size. This derivation also shows why simply multiplying an equal-allocation sample size by an arbitrary factor is not generally correct: the ratio enters through the variance of the mean difference.

N_{\text{req}} =\frac{(z_{1-\alpha/2}+z_{1-\beta})^2\sigma^2}{\delta^2} \frac{(1+r)^2}{r}.

In an interim re-estimation, the same expression is used with the interim variance estimate \(s^2\) in place of \(\sigma^2\). If the target effect and design error rates do not change, the total-sample-size update is proportional to the variance update:

\frac{N_{\text{new}}}{N_{\text{old}}} =\frac{s_{\text{new}}^2}{s_{\text{old}}^2},

provided both calculations use the same allocation ratio, effect target, and normal quantiles. This proportionality applies before integer rounding, floors, or caps.

Worked Example: Initial Design With a 2:1 Ratio

Suppose a hypothetical continuous-outcome trial plans two treatment participants for each control participant. The design targets a mean difference of 4 outcome units, assumes a common standard deviation of 10 units, uses a two-sided \(\alpha=0.05\), and aims for 90% power. Use \(z_{0.975}=1.960\) and \(z_{0.90}=1.282\).

Here \(r=2\), \(\sigma^2=100\) square units, and \(\delta=4\) units. Substituting the displayed quantiles into the formula gives

N_{\text{req}} =\frac{(1.960+1.282)^2(100)}{4^2}\frac{(1+2)^2}{2} =\frac{(3.242)^2(100)}{16}\frac{9}{2} \approx 295.6.

The calculation uses the rounded quantiles shown, so the reported unrounded result is approximately 295.6, not a result obtained from more precise quantiles. Since a 2:1 allocation is easiest to implement in blocks of three, round the target up to 297 participants: 198 assigned to treatment and 99 to control. Rounding up avoids falling below the calculated total and preserves the planned ratio.

Estimating the Variance and Updating Enrollment

If a common variance is assumed, a standard interim estimate is the pooled within-arm sample variance. For interim arm sizes \(n_T\) and \(n_C\), with arm-specific sample variances \(s_T^2\) and \(s_C^2\), it is

s_p^2= \frac{(n_T-1)s_T^2+(n_C-1)s_C^2}{n_T+n_C-2}.

The degrees-of-freedom weights matter. Taking the unweighted average of the two arm variances is generally not the pooled estimate when the interim arm sizes differ. Under independent normal outcomes with a common variance, \(s_p^2\) estimates that common variance; it remains uncertain, especially with a small interim sample. The formula assumes complete outcome data for the participants included in the estimate and that the common-variance model is appropriate.

Worked Example: Updating From a Larger Interim Variance

Continue the hypothetical design above. Suppose the interim estimate of the common standard deviation is 12 outcome units, rather than the planning value of 10. Keep the same 2:1 ratio, target difference, \(\alpha\), and power.

The new variance estimate is \(12^2=144\) square units. Using the same displayed quantiles,

N_{\text{new}} =\frac{(1.960+1.282)^2(144)}{4^2}\frac{9}{2} =(3.242)^2\frac{144}{16}\frac{9}{2} \approx 425.7.

Equivalently, the unrounded target increases by the variance ratio \(144/100=1.44\): \(295.6096\times1.44\approx425.6778\). The first calculation, using the displayed rounded quantiles, gives approximately 425.7 as well. Rounding the target up to a multiple of three gives 426 participants, or 284 treatment and 142 control. If 60 participants have already been enrolled in the intended 2:1 proportions, the remaining target is \(426-60=366\). The update reflects increased uncertainty in the outcome scale, not evidence that the treatment effect is larger or smaller.

Worked Example: Pooling Unequal Interim Arms

In another hypothetical trial with the same design targets, an interim analysis has 30 treatment participants with \(s_T=11\) units and 15 control participants with \(s_C=9\) units. The enrollment ratio is 2:1. Under the common-variance assumption, calculate the pooled variance and the updated total target.

The pooled variance uses 29 and 14 degrees of freedom as weights:

s_p^2 =\frac{(30-1)(11^2)+(15-1)(9^2)}{30+15-2} =\frac{29(121)+14(81)}{43} =\frac{4643}{43} \approx107.98\text{ square units}.

Using \(r=2\), \(\delta=4\), and the same rounded quantiles as before, the unrounded total target is

N_{\text{new}} =\frac{(3.242)^2(107.9767)}{16}\frac{9}{2} \approx319.2.

Rounding up to a multiple of three yields 321 total participants, allocated as 214 to treatment and 107 to control. The arm-specific standard deviations differ in this interim sample, but that fact alone does not establish unequal population variances; this calculation explicitly uses the common-variance model. If that assumption is not scientifically or empirically tenable, the variance calculation and analysis plan need reconsideration rather than automatic pooling.

How the Ratio Affects Precision and Enrollment

For a fixed total \(N\) and common variance, the allocation-dependent factor in the variance is \(g(r)=(1+r)^2/r=r+2+1/r\). Its minimum is 4 at \(r=1\): with equal per-participant costs and a common variance, equal allocation gives the smallest variance for a fixed total. Unequal allocation may still be appropriate for clinical, ethical, logistical, or cost reasons, but it has a precision cost under these assumptions.

Worked Example: Comparing Precision at a Fixed Total

Suppose a hypothetical trial can enroll 300 participants and the common outcome variance is 100 square units. Compare equal allocation with a 2:1 treatment-to-control ratio.

With equal allocation, \(n_T=n_C=150\), so the variance of the mean difference is \(100(1/150+1/150)=1.333\) square units and its standard error is \(\sqrt{1.333}\approx1.155\) units. With a 2:1 ratio, \(n_T=200\) and \(n_C=100\), giving variance \(100(1/200+1/100)=1.500\) square units and standard error \(\sqrt{1.500}\approx1.225\) units. Thus the unequal allocation has 12.5% greater variance and about 6.1% greater standard error at this fixed total. These are precision comparisons, not changes in the treatment effect or in the nominal type I error rate.

A different comparison is needed when the ratio itself is being chosen rather than held fixed. If the two arms truly have the same variance and participant costs are equal, the calculation above favors equal allocation for precision. When outcome variances or costs differ, the best ratio can differ; that is a design-choice problem, not a reason to change the ratio silently during a variance-only update. The update derived here holds \(r\) fixed so that the effect of changing the estimated variance is interpretable.

Operational Checks, Alternatives, and Limitations

A protocol should specify the variance estimator, the interim time at which it is calculated, the effect size and error-rate inputs, the allocation ratio, and the rounding rule. It should also define any minimum or maximum total sample size in advance. These rules make the adaptation reproducible and prevent a discretionary choice of estimator or cap after seeing the interim results. In an adaptive design, the final testing procedure must remain consistent with the design’s error-control plan; this sample-size equation alone does not establish that property.

Before applying the update, check that the assumptions match the data and design:

Independence: observations must be independent across participants for the stated variance formula. Dependence requires a different variance calculation.
Common variance: pooling assumes a common population variance in both arms. Strong evidence of different variances calls for a justified alternative model and a compatible analysis.
Stable target: \(\delta\), \(\alpha\), power, and the allocation ratio are held fixed in this update. Changing them at the same time makes the source of the sample-size change harder to interpret.
Usable interim data: check arm counts, missing outcomes, implausible values, and data errors. An unusual variance estimate can arise from data problems as well as genuine variability.
Feasible target: confirm that the rounded target respects the planned ratio and that the number remaining is nonnegative. A cap below the calculated requirement means the stated power target is no longer achieved under the estimated variance and other assumptions.

This normal-approximation calculation is most defensible when the variance estimate is reasonably stable and the planned analysis is compatible with a normal model. It does not automatically account for small-sample degrees of freedom, losses to follow-up, nonadherence, clustering, or repeated measurements. Those features change the effective variance or the enrollment needed and require their own design calculations. Likewise, this tutorial concerns continuous outcomes; a binary outcome has a different variance structure.

A common mistake is to use the equal-allocation formula for a 2:1 design, or to multiply an old sample size by the ratio of standard deviations. The correction is to use the variance ratio \(s_{\text{new}}^2/s_{\text{old}}^2\) and the allocation factor \((1+r)^2/r\), keeping the target difference and other design inputs explicit. Another mistake is to treat a cap as if it preserved the original power: if the capped total is smaller than the recalculated requirement, the planned power is not guaranteed.

Key takeaway: Under a common-variance, independent normal-outcome model with fixed allocation ratio \(r\), total enrollment is proportional to the variance estimate and to \((1+r)^2/r\). Estimate the variance appropriately, update the target transparently, and interpret any cap or allocation constraint as a design trade-off.

Check Your Understanding

Answer each question, then check the short solutions below.

For \(r=3\), what is the allocation factor \((1+r)^2/r\), and what does it represent in the variance formula?
A variance estimate changes from 64 to 100 square units while all other design inputs remain fixed. By what factor does the unrounded sample-size target change?
With 20 treatment participants and 10 control participants, the arm-specific variances are 25 and 36 square units, respectively. Under a common-variance model, calculate the pooled estimate.
Why does a 2:1 allocation have a larger standard error than equal allocation when total enrollment and common variance are held fixed?

Solutions. (1) The factor is \((1+3)^2/3=16/3\approx5.333\); it multiplies the common-variance contribution to the difference-of-means variance when expressed at a fixed total \(N\). (2) The ratio is \(100/64=1.5625\), so the unrounded target increases by a factor of 1.5625. (3) The pooled estimate is \(\{(20-1)(25)+(10-1)(36)\}/(20+10-2)=(475+324)/28=28.54\) square units. (4) Unequal allocation places fewer participants in the control arm; because its contribution \(\sigma^2/n_C\) is then larger, the sum \(\sigma^2(1/n_T+1/n_C)\) increases.