Worked Example: Variance Re-Estimation For A Continuous Endpoint

Why Re-Estimate Variance?

For a continuous endpoint, the required sample size depends on the variability of the outcome as well as the treatment difference the trial is designed to detect. If the true standard deviation is larger than the value used at planning, a design based on that value may have less power than intended. A blinded internal pilot can provide an interim estimate of variability from participants already enrolled; a prespecified rule then updates the total sample size without discarding those participants.

The calculation is useful only when the endpoint, target treatment difference, significance level, allocation ratio, estimator, and allowable sample-size range are specified in advance. The example below uses a two-arm, equal-allocation trial with one continuous outcome per participant. It illustrates a normal-approximation planning calculation; an actual protocol should use the planned final test and verify the adaptive design's operating characteristics under its assumptions.

Definition: A blinded internal-pilot variance re-estimation uses outcome data from an interim group of participants already enrolled in the trial, with treatment assignments concealed, to estimate outcome variability and update the planned total sample size. The internal-pilot participants remain part of the final analysis. The re-estimation algorithm, including its sample-size bounds, must be prespecified.

Model and Sample-Size Rule

Assume independent participants in two treatment groups, equal allocation, and normally distributed outcomes with a common within-group variance \(\sigma^2\). Let \(\Delta\) be the absolute treatment difference the design aims to detect. The hypotheses are two-sided, \(H_0\): the treatment means are equal, and \(H_a\): they differ. The design targets type I error \(\alpha\) and power \(1-\beta\). For planning, use the large-sample normal approximation to the two-sample comparison.

If \(n\) is the final number of participants per group, the standard error of the difference in sample means under the planning model is \(\sqrt{2\sigma^2/n}\). The noncentrality under a treatment difference of magnitude \(\Delta\) is therefore \(\lambda=\Delta\sqrt{n/(2\sigma^2)}\). A normal-approximation sample size for a two-sided test is obtained by setting this noncentrality near the sum of the critical value and the power quantile:

n_{\mathrm{req}} = \frac{2\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 calculation gives a per-group sample size; round it upward to an integer. If the protocol specifies a minimum \(n_{\min}\) and maximum \(n_{\max}\) per group, the implemented size is

n_{\mathrm{final}} = \min\!\left(n_{\max}, \max\!\left(n_{\min},\lceil n_{\mathrm{req}}\rceil\right)\right).

At the interim, the variance estimate replaces \(\sigma^2\) in the planning formula. For a blinded internal pilot with \(N_1\) observed outcomes \(Y_1,\ldots,Y_{N_1}\), a basic pooled blinded sample variance is

s_{\mathrm{blind}}^2 = \frac{1}{N_1-1} \sum_{i=1}^{N_1}(Y_i-\bar{Y})^2, \qquad \bar{Y}=\frac{1}{N_1}\sum_{i=1}^{N_1}Y_i.

The exact estimator and its handling of interim data must be prespecified. A blinded pooled variance is not automatically identical to the within-group variance under every treatment-effect configuration; the design's assumptions and estimator therefore need scrutiny. The numerical examples use the stipulated \(s_{\mathrm{blind}}^2\) as the variance input to the stated rule.

Why the Formula Has This Form

The derivation follows from the rejection threshold and the desired power. Under the planning alternative, the standardized difference in means is approximately normal with variance 1 and mean \(\lambda\). A two-sided test rejects when the absolute standardized statistic exceeds \(z_{1-\alpha/2}\). Thus the approximate power at a proposed \(n\) is

\operatorname{Power}(n) \approx \Phi\!\left(-z_{1-\alpha/2}-\lambda\right) + \Phi\!\left(\lambda-z_{1-\alpha/2}\right), \qquad \lambda=\Delta\sqrt{\frac{n}{2\sigma^2}},

where \(\Phi\) is the standard normal cumulative distribution function. For a small \(\alpha\) and a target power above one-half, the probability of crossing the lower rejection boundary under the positive alternative is usually negligible. Requiring the upper-tail probability to be approximately \(1-\beta\) gives \(\lambda-z_{1-\alpha/2}\approx z_{1-\beta}\). Substituting \(\lambda=\Delta\sqrt{n/(2\sigma^2)}\) and squaring both sides yields the sample-size formula. This is a planning approximation; a final analysis using a \(t\) test, small samples, or a non-normal outcome may require a more exact calculation.

Worked Application

Worked Example: Updating the Sample Size From a Blinded Pilot

Hypothetical design. A two-arm trial plans a two-sided comparison with \(\alpha=0.05\), 90% power, and a clinically relevant difference of \(\Delta=10\) outcome units. The original planning standard deviation was 15 units, giving a rounded target of 48 per group by the normal approximation; the protocol chose 50 per group. After 50 participants per group have outcomes, treatment assignments remain concealed and the prespecified blinded variance calculation gives \(s_{\mathrm{blind}}^2=400\) units squared. The allowed final sample size is 70 to 120 per group.

Step 1: Calculate the unbounded updated requirement. For these design targets, \(z_{0.975}=1.960\) and \(z_{0.90}=1.2816\). Substituting the blinded variance estimate gives

n_{\mathrm{req}} = \frac{2(400)(1.960+1.2816)^2}{10^2} = \frac{800(10.50797)}{100} = 84.06.

Rounding upward gives 85 participants per group. Since 85 is between the prespecified bounds of 70 and 120, the final target is \(n_{\mathrm{final}}=85\) per group, or 170 participants total.

Step 2: Count additional enrollment correctly. The 50 participants per group already enrolled are included in the final target. The trial therefore needs \(85-50=35\) additional participants per group, or 70 additional participants in total, assuming the pilot participants contribute usable primary outcomes.

Step 3: Interpret the planned power. At the updated variance estimate, the noncentrality is

\lambda = \frac{10}{\sqrt{2(400)/85}} = 3.260.

The approximate two-sided power is \(\Phi(-1.960-3.260)+\Phi(3.260-1.960)\), or approximately 0.903. Thus the rounded target gives about 90.3% normal-approximation power for a true difference of 10 units under the stated model. This is a design calculation, not a claim that the trial has a 90.3% probability of producing a significant result regardless of the true effect.

Worked Example: What the Sample-Size Bounds Do

Keep the same hypothetical design targets, but consider two possible blinded variance estimates. The examples show why the bounds are part of the rule, rather than optional adjustments made after seeing the result.

Low estimated variance. If \(s_{\mathrm{blind}}^2=250\), then

n_{\mathrm{req}} = \frac{2(250)(10.50797)}{100} = 52.54.

Rounding upward would give 53 per group, below the minimum of 70. The rule therefore selects 70 per group. Its approximate noncentrality is \(10/\sqrt{2(250)/70}=3.742\), giving power \(\Phi(-1.960-3.742)+\Phi(3.742-1.960)\approx0.963\). The lower bound protects against an overly small trial even when the variance estimate is favorable.

High estimated variance. If \(s_{\mathrm{blind}}^2=700\), then

n_{\mathrm{req}} = \frac{2(700)(10.50797)}{100} = 147.1.

The rounded requirement exceeds the maximum, so the rule selects 120 per group. Its approximate noncentrality is \(10/\sqrt{2(700)/120}=2.927\), giving power \(\Phi(-1.960-2.927)+\Phi(2.927-1.960)\approx0.833\). The cap means the requested 90% power is not attained under this variance estimate and the model. It would be incorrect to report the target power as achieved simply because the maximum sample size was used.

Expected Enrollment Under the Rule

A bounded update produces a random final sample size because the interim variance estimate is random. Expected enrollment is calculated by averaging the final enrollment over possible re-estimation outcomes and their probabilities. It is not generally the enrollment obtained by inserting an average variance into the sample-size formula: rounding and the lower and upper bounds make that shortcut unreliable.

Worked Example: Averaging Enrollment Across Outcomes

Suppose, for planning illustration, the team assigns probabilities 0.25, 0.50, and 0.25 to three possible re-estimation outcomes. They lead respectively to final sizes of 70, 85, and 120 participants per group. The internal pilot has already enrolled 50 per group in every scenario.

Scenario	Probability	Final per-group size	Total final enrollment
Low variance	0.25	70	140
Intermediate variance	0.50	85	170
High variance	0.25	120	240

The expected final size per group is \(0.25(70)+0.50(85)+0.25(120)=90\). Expected total enrollment is therefore \(2(90)=180\) participants. Subtracting the 100 already enrolled in the pilot gives expected additional enrollment of 80 participants. These probabilities are hypothetical inputs to the planning calculation; they must not be treated as universal properties of the re-estimation rule.

Checks, Alternatives, and Limitations

Before implementing a variance update, verify that the interim outcomes are available at the planned look, the variance is calculated from the prespecified eligible observations, and the rule uses the correct per-group or total sample-size convention. Confirm that the pilot participants remain in the final analysis and that every scenario is subject to the same rounding and bounds. If the primary endpoint has missing observations, repeated measurements, clustering, or a distribution poorly represented by the normal model, a simple independent-outcome variance formula may not represent the actual design.

A fixed sample-size design is simpler: it does not use interim variance information, so its enrollment target is not changed by the pilot. Its weakness is that a misspecified planning variance can leave the trial underpowered or unnecessarily large. An external-pilot estimate avoids using trial interim outcomes, but may not describe the population or measurement process in the main trial. The internal-pilot approach uses data from the trial itself, but requires prespecification and careful evaluation of the full adaptive procedure.

The calculation above is a normal-approximation illustration. With small samples, a final test based on a \(t\) distribution, or additional design features, calculate power using the intended final analysis rather than assuming the normal approximation is exact. The adaptive rule should also be evaluated for its operating characteristics under plausible variance and treatment-effect scenarios; a variance update alone does not establish that the complete design controls its intended error rate.

Common mistake: “The blinded pilot estimate produced a required size of 147 per group, so using the allowed maximum of 120 per group still gives the planned 90% power.” The calculation in the high-variance example instead gives approximately 83.3% power. A cap limits enrollment; it does not preserve the power target. Report the capped size and its resulting power separately, and state when the target is not reached.

Check Your Understanding

Use the normal-approximation rule and the design targets from the examples: \(\alpha=0.05\), 90% power, \(\Delta=10\), equal allocation, and bounds of 70 to 120 per group. Solutions follow each question.

If the blinded variance estimate is 500 units squared, what is the unbounded sample-size requirement, and what final size does the rule select?
With a variance estimate of 500 and the selected final size, how many additional participants are needed if 50 per group are already enrolled?
Why is the expected total enrollment in the three-scenario example 180 rather than 100, the number enrolled at the interim?
What should be reported if the variance estimate implies a requirement above the maximum sample size?

Solutions. (1) \(n_{\mathrm{req}}=2(500)(10.50797)/100=105.1\), so the rounded and bounded final size is 106 per group. (2) The additional enrollment is \(2(106-50)=112\) participants total. (3) The 100 participants are the interim enrollment common to all scenarios, not the final enrollment. Averaging the scenario-specific final totals gives \(0.25(140)+0.50(170)+0.25(240)=180\). (4) Report the capped final size and calculate power at that size under the stated assumptions; do not claim that the target power was achieved unless the calculation supports that claim.