Worked Example Of Pooled Sample Size Updating

Why a Blinded Sample Size Update Needs Care

A blinded interim review can use pooled outcome data to reconsider a nuisance quantity such as variance, then recalculate the sample size needed for a prespecified design target. The procedure does not require the adaptation team to see which treatment group has the higher mean. But “blinded” does not mean the pooled estimate contains only within-group variability: if the treatment groups truly have different means, their separation can contribute to the pooled variance.

This tutorial works through a two-arm, equal-allocation comparison of continuous outcomes. The goal is to show the calculation and its interpretation, not to infer an interim treatment effect. The target effect used in the sample-size formula is a design value specified before the interim review, not an effect estimate extracted from the blinded data.

Definition: A blinded pooled sample size update recalculates a trial’s required sample size using pooled interim outcome information, such as an estimate of variance, while treatment identities and arm-specific outcome summaries remain hidden from the adaptation team. The update must use a prespecified rule and design assumptions.

Model, Target, and Sample-Size Formula

Suppose the primary outcome is measured once per participant, and the planned comparison is the difference between two treatment-group means. For the variance calculation, assume independent observations within and between groups, a common within-group variance \(\sigma^2\), and an interim data set with \(n_1\) and \(n_2\) evaluable participants. A normal model is a standard basis for the usual two-sample \(z\)-based planning approximation; in practice, the suitability of that approximation depends on the outcome distribution and sample sizes.

Let \(\delta>0\) be the prespecified difference in means the design is intended to detect, using a two-sided significance level \(\alpha\) and power \(1-\beta\). For equal planned allocation, the normal-approximation target for the total number of evaluable participants is

N_{\text{eval}} \approx \frac{4\sigma^2\left(z_{1-\alpha/2}+z_{1-\beta}\right)^2}{\delta^2}.

Here, \(z_q\) is the \(q\)-quantile of the standard normal distribution. The factor of four gives the total sample size for two equally sized groups: each group has approximately \(2\sigma^2(z_{1-\alpha/2}+z_{1-\beta})^2/\delta^2\) evaluable participants. This is a planning approximation, not an exact guarantee of the achieved power. If a protocol uses a different test, allocation ratio, multiplicity adjustment, or small-sample method, its matching sample-size procedure should be used instead.

In a blinded update, the adaptation team substitutes a prespecified blinded pooled estimate \(s_{\text{pool}}^2\) for \(\sigma^2\). It then rounds the result up to a feasible allocation, often an even total for equal allocation. An evaluable target is not automatically an enrollment target. If a fraction \(r\) of randomized participants is expected not to contribute an evaluable primary outcome, a simple enrollment allowance is

N_{\text{enroll}} \approx \left\lceil \frac{N_{\text{eval}}}{1-r}\right\rceil.

This calculation treats the attrition fraction as a planning estimate. It does not correct bias from outcome-dependent missingness, and it should be applied consistently with the protocol’s definitions of randomized, treated, and evaluable participants.

Worked Example: Updating the Evaluable Target

Hypothetical data. At a blinded interim review, 120 participants have evaluable outcomes. The pooled sum of squared deviations of those outcomes from their overall mean is 15,600 outcome-units squared. The design specifies a two-sided \(\alpha=0.05\), 90% power, and a target difference \(\delta=5\) outcome units. The interim allocation is 60 participants per group, but the adaptation team sees only pooled outcomes.

Check the pooled estimate. The pooled sample variance uses \(n-1\) in its denominator:

s_{\text{pool}}^2=\frac{15{,}600}{120-1} =\frac{15{,}600}{119} \approx 131.1\ \text{outcome-units}^2.

For a two-sided 5% test, \(z_{1-\alpha/2}\approx1.960\); for 90% power, \(z_{1-\beta}\approx1.282\). Substituting the blinded variance estimate into the prespecified planning formula gives

N_{\text{eval}} \approx \frac{4(131.0924)(1.960+1.282)^2}{5^2} = \frac{524.3696(3.242)^2}{25} \approx 220.5.

The raw approximation is about 220.5 evaluable participants. Rounding up to a feasible even total gives a target of 222 evaluable participants, or 111 per group. This calculation revises the design target using pooled variability; it does not say whether the treatment difference is positive, negative, or close to 5 units.

What the Pooled Variance Estimates

With treatment identities hidden, an overall variance around the grand mean is observable. However, it need not estimate the common within-group variance without bias when the group means differ. The distinction follows from the standard decomposition of the total sum of squares into deviations within groups and deviations between group means.

Result: Under independent observations with common within-group variance \(\sigma^2\), let the two group means be \(\mu_1\) and \(\mu_2\), and let \(n=n_1+n_2\). For the total sum of squares around the grand mean, the expected within-group contribution is \((n-2)\sigma^2\), and the expected between-group contribution is \(\sigma^2+\frac{n_1n_2}{n}(\mu_1-\mu_2)^2\). Thus the pooled sample variance using denominator \(n-1\) has expectation \(\sigma^2+\frac{n_1n_2}{n(n-1)}(\mu_1-\mu_2)^2\).

The extra \(\sigma^2\) in the expected between-group contribution is important: the observed group means themselves vary around their population means. It is not correct to assign the whole treatment-separation term to the within-group sum of squares. The derivation below shows how both contributions arise.

Decompose the total sum of squares.
Write the total squared deviations from the grand mean as the sum of squared deviations from each group’s sample mean plus the between-group sum of squares. The cross-products vanish because deviations within each group sum to zero.

Take the expectation within groups.
Each group’s within-group sum of squares has expectation \((n_j-1)\sigma^2\). Adding the two groups gives \((n_1-1+n_2-1)\sigma^2=(n-2)\sigma^2\).

Take the expectation between groups.
The between-group sum of squares is \(\frac{n_1n_2}{n}(\bar{x}_1-\bar{x}_2)^2\). Since \(E[(\bar{x}_1-\bar{x}_2)^2]=(\mu_1-\mu_2)^2+\sigma^2(1/n_1+1/n_2)\), its expectation is \(\frac{n_1n_2}{n}(\mu_1-\mu_2)^2+\sigma^2\).

Combine and divide.
The expected total sum of squares is \((n-1)\sigma^2+\frac{n_1n_2}{n}(\mu_1-\mu_2)^2\). Dividing by \(n-1\) gives the stated expectation of the pooled sample variance.

Therefore, when \(\mu_1\ne\mu_2\), the expected pooled variance exceeds the common within-group variance. A blinded sample-size update that directly uses the pooled variance may consequently be conservative relative to a calculation based on the true within-group variance. The size of this contribution depends on the group sizes and the squared difference between group means. The team cannot estimate that difference from the pooled variance alone.

Worked Example: Quantifying the Mean-Separation Contribution

Hypothetical model illustration. Suppose each group has 60 participants, the common within-group variance is \(100\) outcome-units squared, and the true group means differ by 4 outcome units. These values are used to understand the estimator’s expectation; they are not information available to a blinded adaptation team.

Here \(n=120\), and \(\frac{n_1n_2}{n}=\frac{60(60)}{120}=30\). The expected within-group sum of squares is

(n-2)\sigma^2=(120-2)(100)=11{,}800.

The expected between-group sum of squares is

\sigma^2+\frac{n_1n_2}{n}(\mu_1-\mu_2)^2 =100+30(4^2)=580.

Together these give an expected total sum of squares of \(11{,}800+580=12{,}380\). Dividing by \(119\), the expected pooled sample variance is

E(s_{\text{pool}}^2)=\frac{12{,}380}{119} \approx 104.0\ \text{outcome-units}^2.

Equivalently, \(100+\frac{30(16)}{119}\approx104.0\). The square root of this expectation is about 10.20 outcome units, but it is not generally the exact expected value of \(s_{\text{pool}}\), because taking a square root and taking an expectation do not commute. The calculation demonstrates the direction and approximate scale of the variance contribution; it does not allow the blinded team to identify the treatment difference.

From the Target to the Enrollment Decision

A trial may have participants already randomized when the revision is made. The operational question is then how many additional participants to enroll, not merely what the total target is. The answer depends on the current count of evaluable outcomes, the planned attrition allowance, allocation constraints, and any protocol maximum. A target should be expressed in the same terms throughout: do not compare a target number of evaluable outcomes with a current number of randomized participants.

Worked Example: Accounting for Attrition and Current Enrollment

Hypothetical continuation. Suppose the revised target is 222 evaluable participants, as in the first example. At the interim snapshot, 120 participants have been randomized and 110 have evaluable outcomes. The planning allowance is 8% non-evaluable outcomes among future enrollees. Assume equal allocation and a protocol that permits the target to be met by adding an even number of participants.

The additional evaluable outcomes needed are \(222-110=112\). The projected additional enrollment is

N_{\text{add,enroll}}= \left\lceil\frac{112}{1-0.08}\right\rceil = \left\lceil121.739\right\rceil =122.

Adding 122 participants is compatible with equal allocation, 61 per group, if the randomization plan and current allocation allow it. At an expected evaluability rate of 92%, the added enrollment contributes \(122(0.92)=112.24\) evaluable outcomes on average; together with the 110 already evaluable, that is 222.24. The calculation supports an operational target of 122 additional participants under the stated assumptions. It does not guarantee exactly 112 additional evaluable outcomes.

Worked Example: A Larger Pooled Variance Changes the Target

Hypothetical sensitivity calculation. Suppose a separate blinded review produces \(s_{\text{pool}}=12.0\) outcome units. Keep the design values \(\alpha=0.05\), 90% power, and \(\delta=5\) outcome units. Then \(s_{\text{pool}}^2=144\), so

N_{\text{eval}} \approx \frac{4(144)(1.960+1.282)^2}{5^2} = \frac{576(3.242)^2}{25} \approx 242.2.

Rounding up to an even evaluable target gives 244 participants. If 110 are currently evaluable, 134 more are needed. With an 8% non-evaluable allowance, the additional enrollment is

\left\lceil\frac{134}{0.92}\right\rceil = \left\lceil145.652\right\rceil =146.

If 120 have already been randomized, this rule would imply 146 additional enrollees and a total of 266 randomized participants, subject to allocation and protocol limits. The difference from the earlier example results from the larger variance estimate, not from a new estimate of the treatment effect.

Diagnostics, Alternatives, and Limits

Before accepting a pooled update, the adaptation team should verify that the interim snapshot and calculation match the prespecified procedure. Key checks include:

Data and denominator: Confirm which participants contribute to the variance calculation, that the snapshot is complete as defined, and that the denominator is appropriate for the pooled variance.
Allocation and outcome structure: Check the actual group counts, the planned allocation ratio, repeated or clustered observations, and whether the formula matches the design. The equal-allocation formula above is not a universal formula for every trial.
Variance assumptions: Consider whether a common within-group variance is plausible and whether outliers, skewness, site differences, or changing outcome measurement could make the interim pooled variance unrepresentative of later participants.
Enrollment projection: Check whether the attrition allowance is realistic and whether missing outcomes could depend on outcome values or participant characteristics. A simple inflation factor cannot fix informative missingness.
Operational constraints: Apply the protocol’s maximum sample size, allocation rounding, and any stopping or feasibility rules exactly as written.

The main alternative is to keep the original fixed sample size rather than revise it using interim variance information. That avoids an interim sample-size decision but may leave the trial underpowered if the design variance was too small, or larger than necessary if it was too large. An unblinded approach that uses treatment-specific information is a different adaptation and is not what the pooled calculation does. Each approach needs its own prespecified statistical and operational safeguards.

Most importantly, a blinded variance-based update does not automatically guarantee preservation of the trial’s Type I error rate. Whether error control is maintained depends on the complete design: the update rule, the test used at the end, the timing and use of interim data, and any other adaptations. The procedure should be specified in advance and justified for that design, often using appropriate statistical results or simulation. It is not enough to say that the review was blinded.

Common pitfall: “The adaptation team saw only pooled data, so the pooled variance must be an unbiased estimate of the within-group variance.” This is not generally correct. Under the common-variance model, the expected pooled variance includes \(\frac{n_1n_2}{n(n-1)}(\mu_1-\mu_2)^2\), which is nonnegative and is positive when the true group means differ. The team can use the pooled estimate under a prespecified rule, but cannot interpret it as a treatment-effect estimate or claim it contains only within-group noise.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

A trial uses two-sided \(\alpha=0.05\), 80% power, equal allocation, \(\delta=4\) outcome units, and \(s_{\text{pool}}=8\) outcome units. What is the approximate evaluable target before allocation rounding?
For two groups of 40 participants, common within-group variance \(25\), and a difference in group means of 2 units, what is the expected pooled sample variance?
A target is 180 evaluable participants, 150 are already evaluable, and expected non-evaluability among future enrollees is 10%. How many additional participants should be enrolled by the simple attrition formula?
Name one quantity the adaptation team can infer from a blinded pooled variance update and one it cannot infer from that update alone.

Solutions. (1) Use \(z_{0.975}=1.960\) and \(z_{0.80}\approx0.842\):

N_{\text{eval}}\approx \frac{4(8^2)(1.960+0.842)^2}{4^2} = \frac{256(2.802)^2}{16} \approx125.6.

The raw target is about 125.6 evaluable participants; an equal-allocation design would round up to an even target of 126. (2) Here \(n=80\) and \(\frac{n_1n_2}{n}=\frac{40(40)}{80}=20\). Therefore

E(s_{\text{pool}}^2)=25+\frac{20(2^2)}{79} =25+\frac{80}{79} \approx26.01\ \text{units}^2.

(3) Thirty additional evaluable outcomes are needed. The enrollment calculation is \(\left\lceil30/0.90\right\rceil=\lceil33.33\rceil=34\) participants. (4) The team can use the pooled data to estimate overall variability under its prespecified method and revise the sample-size target. The pooled variance alone does not reveal the direction or magnitude of the treatment difference, nor does it establish treatment benefit.