Worked Example: Re-Estimation For A Binary Endpoint

Why Re-Estimate a Binary-Endpoint Sample Size?

For a binary endpoint, each participant either experiences the event or does not. The sample size needed to compare two arms depends not only on the treatment difference the trial aims to detect, but also on the event rates: the variability of a binary outcome changes with its probability. If an interim pooled event rate differs from the planning value, a prespecified rule can use it to update the enrollment target while keeping the target difference and testing objective fixed.

This tutorial works through that calculation for an equally randomized, two-arm superiority trial. The rule is not a way to discover the treatment effect from blinded interim data. Instead, it uses the pooled rate as a nuisance-rate estimate and places the planned treatment contrast around that rate. The resulting target is approximate and its operating characteristics must be checked for the actual adaptive rule.

Definition: Let \(p_C\) be the control-arm event rate and \(p_T\) the treatment-arm event rate. Let \(\delta>0\) be the planned absolute difference, with \(p_C-p_T=\delta\) when a lower event rate is favorable. With equal allocation and \(n\) participants per arm at an interim analysis, define the blinded pooled rate as \(\hat{p}_{\text{pool}}=(X_C+X_T)/(2n)\), where \(X_C\) and \(X_T\) are the event counts. The pooled-rate re-estimation rule uses \(\hat{p}_{\text{pool}}\) to estimate the nuisance event rate and calculates a revised total sample size per arm.

Model, Assumptions, and Calculation

The calculation assumes independent Bernoulli outcomes within and between arms, a common outcome definition and follow-up period, and equal allocation. It also assumes the endpoint is observed for the participants used in the interim count, with no unaccounted missing outcomes, clustering, or other design feature that changes the variance. The planned absolute difference \(\delta\), significance level, and target power are fixed in advance. The normal approximation should be adequate for the rates and sample sizes involved; sparse event counts may require a more careful method.

For a two-sided test at significance level \(\alpha\), the usual normal-approximation sample size for a difference in two independent proportions is:

n= \frac{ \left[ z_{1-\alpha/2}\sqrt{2p_0(1-p_0)} + z_{1-\beta}\sqrt{p_C(1-p_C)+p_T(1-p_T)} \right]^2 }{(p_C-p_T)^2}.

Here \(n\) is the number of participants per arm, \(1-\beta\) is the target power, and \(z_q\) is the \(q\)-quantile of the standard normal distribution. The first variance term is evaluated under the null hypothesis of equal event rates, using their common rate \(p_0\). The second is evaluated under the planned alternative rates. For the pooled-rate rule, substitute \(\hat{p}_{\text{pool}}\) for \(p_0\), and use symmetric planning rates:

\tilde p_C=\hat{p}_{\text{pool}}+\frac{\delta}{2}, \qquad \tilde p_T=\hat{p}_{\text{pool}}-\frac{\delta}{2}.

These rates have midpoint \(\hat{p}_{\text{pool}}\) and difference \(\delta\). The rule therefore updates the variance terms without changing the planned contrast. It is feasible only if both rates lie between 0 and 1.

Why the Formula Has This Form

Under the null, the estimated difference in event rates has approximate standard error \(\sqrt{2p_0(1-p_0)/n}\). Under the alternative, its approximate standard error is \(\sqrt{[p_C(1-p_C)+p_T(1-p_T)]/n}\). For a two-sided test, the positive-side rejection boundary is approximately \(z_{1-\alpha/2}\) null standard errors above zero. To attain power \(1-\beta\) in the planned direction, the alternative mean difference must exceed that boundary by approximately \(z_{1-\beta}\) alternative standard errors. Thus,

p_C-p_T \approx z_{1-\alpha/2}\sqrt{\frac{2p_0(1-p_0)}{n}} + z_{1-\beta}\sqrt{\frac{p_C(1-p_C)+p_T(1-p_T)}{n}}.

Multiplying by \(\sqrt{n}\), then squaring both sides and dividing by \((p_C-p_T)^2\), gives the sample-size formula above. This is a normal-approximation justification: exact power also includes the small probability of rejection in the opposite tail. The approximation is generally suitable for planning when event counts are not sparse, but the final rule should be assessed using the planned test and design.

Worked Example: Updating the Target From Interim Event Counts

Hypothetical trial. A two-arm trial uses equal allocation and targets a 10-percentage-point reduction in the event rate, so \(\delta=0.1000\). At a blinded interim analysis, 120 participants per arm have complete binary outcomes. There are 36 events altogether. The design uses two-sided \(\alpha=0.05\) and 80% power. Assume these outcomes satisfy the independence and follow-up assumptions above.

The pooled rate is:

\hat{p}_{\text{pool}}=\frac{36}{120+120} =\frac{36}{240}=0.1500.

The symmetric planning rates are \(\tilde p_C=0.1500+0.05000=0.2000\) and \(\tilde p_T=0.1500-0.05000=0.1000\). Both are feasible. Using \(z_{0.975}=1.960\) and \(z_{0.80}=0.8416\), the updated sample size per arm is:

\begin{aligned} n &= \frac{ \left[ 1.960\sqrt{2(0.1500)(0.8500)} + 0.8416\sqrt{(0.2000)(0.8000)+(0.1000)(0.9000)} \right]^2 }{(0.1000)^2}\\ &= \frac{ \left[1.960\sqrt{0.2550}+0.8416\sqrt{0.2500}\right]^2 }{0.01000}\\ &= \frac{[1.960(0.50498)+0.8416(0.5000)]^2}{0.01000}\\ &\approx \frac{(1.4106)^2}{0.01000} \approx 198.97. \end{aligned}

Rounding up gives a revised target of 199 participants per arm, or 398 in total. Since 120 per arm have already been randomized and are included in the final analysis, the trial would plan to enroll 79 additional participants per arm, for 158 additional participants overall. The target is a total, not an instruction to enroll 199 more participants in each arm.

Check the Rule Under Alternative Event Rates

A useful stress test holds the planned difference fixed and varies the midpoint event rate. If equal allocation is maintained, the pooled rate estimates the midpoint in expectation: with \(n\) participants per arm, independent event counts \(X_C\sim\text{Binomial}(n,p_C)\) and \(X_T\sim\text{Binomial}(n,p_T)\) give

E(\hat{p}_{\text{pool}}) = \frac{np_C+np_T}{2n} = \frac{p_C+p_T}{2}.

This is an expectation, not a guarantee that a particular interim estimate equals the midpoint. Moreover, if allocation is unequal, the unweighted pooled rate generally does not estimate the midpoint of the arm-specific rates.

Worked Example: Higher Event Rates and the Revised Target

Suppose, hypothetically, that the control and treatment event rates are 0.3000 and 0.2000, respectively. Their midpoint is 0.2500 and the difference remains \(\delta=0.1000\). With a pooled estimate of 0.2500, the rule uses \(p_0=0.2500\), \(\tilde p_C=0.3000\), and \(\tilde p_T=0.2000\). Keep the two-sided 5% significance level and 80% target power.

The calculation is:

\begin{aligned} n &= \frac{ \left[ 1.960\sqrt{2(0.2500)(0.7500)} + 0.8416\sqrt{(0.3000)(0.7000)+(0.2000)(0.8000)} \right]^2 }{(0.1000)^2}\\ &= \frac{[1.960\sqrt{0.3750}+0.8416\sqrt{0.3700}]^2}{0.01000}\\ &= \frac{[1.960(0.61237)+0.8416(0.60828)]^2}{0.01000}\\ &\approx\frac{(1.7122)^2}{0.01000} \approx 293.2. \end{aligned}

The target rounds up to 294 participants per arm. At that sample size, the alternative standard error under these rates is \(\sqrt{0.3700/294}\approx0.03548\), so the standardized difference is \(0.1000/0.03548\approx2.819\). The approximate two-sided power is

\Phi\left(-1.960\sqrt{\frac{0.3750}{0.3700}}-2.819\right) +1-\Phi\left(1.960\sqrt{\frac{0.3750}{0.3700}}-2.819\right) \approx 0.801,

where \(\Phi\) is the standard normal cumulative distribution function. This is about 80.1% power under the stated rates, close to the 80% target. A higher midpoint rate increases the binary-outcome variance here and therefore increases the required sample size.

Worked Example: A Pooled Rate That Makes the Rule Infeasible

Suppose a different hypothetical interim analysis has 10 events among 250 participants in each arm combined, giving \(\hat{p}_{\text{pool}}=10/250=0.04000\). The planned difference is still \(\delta=0.1000\). The symmetric rates would be:

\tilde p_C=0.04000+0.05000=0.09000, \qquad \tilde p_T=0.04000-0.05000=-0.01000.

A negative event probability is impossible, so the rule does not provide a valid sample-size calculation for this interim result. It would be wrong to insert \(-0.01000\) into the variance formula or silently replace it by zero: either action would fail to implement the specified symmetric-rate rule. A design must prespecify what happens at such a boundary, for example by using a different feasible nuisance-rate rule or a defined enrollment cap. That alternative must preserve the planned contrast and be evaluated as part of the complete rule.

Diagnostics, Alternatives, and Limitations

Before accepting a revised target, check the inputs and the interpretation:

Verify the interim denominator. The pooled event count and total number with observed outcomes must refer to the same endpoint definition and follow-up window. If outcomes are missing, this simple rule does not account for them.
Check the allocation. The midpoint argument uses equal allocation. Do not treat an unweighted pooled rate as the midpoint under unequal allocation without justification.
Check feasibility and bounds. Both symmetric rates must be in \([0,1]\). Also compare the unrounded result with any prespecified minimum or maximum sample size.
Check scale and rounding. The formula gives a number per arm. Round up to a whole participant per arm, then calculate the total and the additional enrollment separately.
Reassess operating characteristics. A plug-in estimate is random, and the final sample size is consequently random. The nominal formula alone does not establish type I error control for the full adaptive trial; evaluate the prespecified rule using the planned final test and relevant scenarios.

A fixed sample size is simpler and avoids an interim-driven change in enrollment, but it must be chosen using an assumed event rate that may be inaccurate. A pooled-rate update can respond to a different nuisance rate while keeping the target difference fixed; its limitation is that a single interim pooled estimate is noisy and the rule may be infeasible near a boundary. A design based on external event-rate information avoids using interim outcomes but depends on how relevant that information is. Re-estimation using arm-specific rates is another possible approach, but it uses more interim information and requires its own prespecified rule and operating-characteristic assessment.

The pooled estimate is not an estimate of the treatment contrast: it combines both arms. The symmetric planning rates encode the assumed contrast rather than estimating it from the blinded data. If the true treatment difference differs from the planned one, the achieved power can differ from the target even when the pooled midpoint is estimated accurately. For this reason, stress-test the rule across plausible pairs of event rates, including values near 0 and 1, and specify how infeasible values or sample-size caps are handled.

Key takeaway: A pooled-rate re-estimation rule updates the variance assumptions, not the primary treatment question. Calculate the pooled rate, form feasible symmetric planning rates around it, apply the two-proportion formula, round up per arm, and assess the complete rule under plausible event-rate scenarios.

Check Your Understanding

Answer each question, then compare your work with the solutions below.

In an equal-allocation trial, 30 events are observed among 200 participants at interim. What is the pooled rate, and what symmetric planning rates correspond to a planned absolute difference of 0.0800?
Why does the pooled rate estimate the midpoint of the two event rates in expectation under equal allocation?
If a calculation gives \(n=198.2\) per arm and 100 participants per arm have already been randomized, what are the revised total target and additional enrollment?
For a planned difference of 0.1200 and a pooled rate of 0.0500, are the symmetric planning rates feasible? Explain.

Solutions. (1) \(\hat{p}_{\text{pool}}=30/200=0.1500\). The rates are \(0.1500+0.04000=0.1900\) and \(0.1500-0.04000=0.1100\). (2) With \(n\) participants per arm, \(E(X_C+X_T)=np_C+np_T\), so \(E[(X_C+X_T)/(2n)]=(p_C+p_T)/2\). (3) Round up to 199 per arm, or 398 total. With 100 already randomized per arm, enroll 99 more per arm, 198 more overall. (4) The rates are \(0.0500+0.0600=0.1100\) and \(0.0500-0.0600=-0.0100\). The negative rate is infeasible, so this symmetric-rate rule cannot supply a valid target for that interim estimate.