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Sample Size Re-Estimation · Tutorial 337 of 799

Re-Estimation For Equivalence Trials

Derive and apply variance-based sample-size updates for equivalence trials, then quantify how uncertainty in the interim variance estimate can affect the planned power.

Advanced 12 min read

What You'll Learn

  • Express equivalence as two one-sided hypotheses and identify the interval in which the estimated treatment difference must fall.
  • Derive a normal-approximation sample-size rule for a centered treatment difference and a prespecified power target.
  • Update the total per-arm sample size from an interim estimate of the common outcome variance.
  • Quantify the sampling variability of the variance estimate and explain why a plug-in update does not guarantee target power.
  • Recognize assumptions, operational safeguards, and diagnostics relevant to equivalence-trial re-estimation.

Why Re-Estimation Is Different for Equivalence

In an equivalence trial, the goal is not to show that a treatment differs from control, nor merely to rule out a difference in one direction. The goal is to show that the treatment difference is sufficiently close to zero: specifically, that it lies within prespecified lower and upper equivalence margins. Because success requires two one-sided tests to pass, the sample size must support both conclusions. Uncertainty in the outcome variance matters directly: if the variance is larger than planned, the confidence interval is wider and equivalence is harder to establish.

This tutorial develops a variance-based re-estimation rule for a continuous outcome with equal allocation. The calculations use a normal approximation and a common variance. They are planning approximations, not a substitute for checking the final test and the full adaptive design. The earlier material on sample size re-estimation for noninferiority trials uses a one-sided margin; equivalence instead requires both one-sided conditions below.

Definition: Let \(\theta=\mu_T-\mu_C\) be the population mean difference between treatment and control, and let \(\Delta>0\) be the prespecified symmetric equivalence margin, in the outcome's units. Equivalence means \(-\Delta<\theta<\Delta\). The two one-sided tests (TOST) use \(H_{0L}:\theta\leq-\Delta\) versus \(H_{aL}:\theta>-\Delta\), and \(H_{0U}:\theta\geq\Delta\) versus \(H_{aU}:\theta<\Delta\). Equivalence is concluded only when both null hypotheses are rejected.

Assume independent participants, a continuous outcome, equal numbers \(n\) in the two groups, and normally distributed outcomes with a common variance \(\sigma^2\). For the planning calculation, treat \(\sigma\) as known or use a large-sample normal approximation. The estimated difference \(D\) then has approximate distribution \(D\sim N(\theta,2\sigma^2/n)\), with \(SE=\sigma\sqrt{2/n}\). At one-sided significance level \(\alpha\), both tests pass exactly when

$$ -\Delta+z_{1-\alpha}SE < D < \Delta-z_{1-\alpha}SE, $$

where \(z_q\) is the \(q\)-quantile of the standard normal distribution. If the lower endpoint is not below the upper endpoint, this acceptance interval is empty and equivalence cannot be declared. The corresponding two-sided confidence interval has confidence level \(1-2\alpha\); for \(\alpha=0.05\), it is a 90% interval.

Power and the Sample-Size Rule

Under a true difference \(\theta\), the probability that both tests pass under this approximation is the probability that \(D\) falls inside the acceptance interval. Subtracting the normal probabilities below its lower endpoint and above its upper endpoint gives

$$ \operatorname{Power}(n,\theta,\sigma) = \max\!\left\{0,\; \Phi\!\left(\frac{\Delta-z_{1-\alpha}SE-\theta}{SE}\right) - \Phi\!\left(\frac{-\Delta+z_{1-\alpha}SE-\theta}{SE}\right) \right\}, \qquad SE=\sigma\sqrt{\frac{2}{n}}, $$

where \(\Phi\) is the standard normal cumulative distribution function. The formula makes the two-sided requirement explicit. For planning at the centered alternative \(\theta=0\), symmetry reduces power to

$$ \operatorname{Power}(n,0,\sigma) = \max\!\left\{0,\;2\Phi\!\left(\frac{\Delta}{SE}-z_{1-\alpha}\right)-1\right\}. $$

To ensure this power is at least \(1-\beta\), it is sufficient to require \(\Delta/SE-z_{1-\alpha}\geq z_{1-\beta/2}\). This allocates no more than \(\beta/2\) probability to failure on either side. Substituting \(SE=\sigma\sqrt{2/n}\) and rearranging gives the familiar conservative planning rule:

$$ n \geq \frac{2\sigma^2\left(z_{1-\alpha}+z_{1-\beta/2}\right)^2}{\Delta^2}. $$

The resulting \(n\) is the number of participants per group, rounded up to an integer. For the centered alternative, the rule gives the exact continuous-\(n\) threshold for the centered normal-approximation power equation; rounding up gives the smallest integer \(n\) meeting the target. It is not an exact solution for a noncentered alternative or for a different final testing procedure. If the anticipated true difference is not zero, use the general power expression and solve for \(n\) numerically. Power generally declines as the anticipated difference moves toward either equivalence margin.

Worked Example: Initial Equivalence-Trial Size

Suppose a hypothetical trial compares mean change in a laboratory measure between two treatments. The outcome standard deviation is assumed to be \(8\) mg/dL, the symmetric equivalence margin is \(\Delta=5\) mg/dL, the one-sided level is \(\alpha=0.05\), and the target power is 90% at \(\theta=0\). Assume independent outcomes, equal allocation, and a common normal variance for this planning calculation.

Here \(z_{0.95}=1.6449\) and \(z_{0.95}=1.6449\) for the \(z_{1-\beta/2}\) term because \(\beta=0.10\). Thus

$$ n\geq \frac{2(8^2)(1.6449+1.6449)^2}{5^2} = \frac{128(3.2898)^2}{25} \approx 55.41. $$

Rounding up gives 56 participants per group. At \(n=55\), \(SE=8\sqrt{2/55}=1.52554\) mg/dL, and the centered-alternative power is approximately \(2\Phi(5/1.52554-1.6449)-1\approx0.897\), below 90%. At \(n=56\), \(SE=8\sqrt{2/56}=1.51186\) mg/dL and power is approximately \(2\Phi(5/1.51186-1.6449)-1\approx0.903\). The 90% target is therefore met at 56 per group under this approximation.

At \(n=56\), the treatment difference must fall between \(-5+1.6449(1.51186)\) and \(5-1.6449(1.51186)\), or approximately \(-2.513\) to \(2.513\) mg/dL, for both tests to pass. These are the boundaries for the estimated difference, not the equivalence margins themselves.

Updating the Size From an Interim Variance

Suppose an interim review estimates the common outcome variance without examining the treatment difference. Let \(s_p^2\) denote the pooled within-group variance estimate. Replacing \(\sigma^2\) in the planning rule by \(s_p^2\) gives a plug-in target total per-group size:

$$ n_{\mathrm{new}} = \max\!\left\{ n_1,\; \left\lceil \frac{2s_p^2\left(z_{1-\alpha}+z_{1-\beta/2}\right)^2}{\Delta^2} \right\rceil \right\}, $$

where \(n_1\) is the number already enrolled per group and \(\lceil\cdot\rceil\) rounds upward. The update is a total sample size per group, not the number of additional participants. If a prespecified minimum or maximum is used, apply it according to the protocol, while ensuring the final total cannot be less than the number already enrolled.

Worked Example: Re-Estimating From a Larger Variance

Continue the hypothetical laboratory-outcome trial. After 25 participants per group have been enrolled, a blinded interim review estimates a pooled standard deviation of \(10.5\) mg/dL. Retain \(\Delta=5\) mg/dL, \(\alpha=0.05\), and 90% target power. Assume the estimate is based on within-group variability and that the design has prespecified this variance-only update.

The revised total per-group size is

$$ n_{\mathrm{new}} = \left\lceil \frac{2(10.5^2)(1.6449+1.6449)^2}{5^2} \right\rceil = \lceil95.46\rceil = 96. $$

The trial therefore targets 96 participants per group in total, or 71 additional participants per group beyond the 25 already enrolled. At the updated size, the planning standard error is \(10.5\sqrt{2/96}=1.5155\) mg/dL. The centered normal-approximation power is approximately \(2\Phi(5/1.5155-1.6449)-1\approx0.902\), consistent with the target after rounding. This calculation uses the interim estimate as if it were the true standard deviation; that qualification is important.

Why Variance Uncertainty Changes the Interpretation

A plug-in update does not make the nuisance parameter known. Under the normal, common-variance model, if the interim estimate pools \(n_1\) independent observations per group, its degrees of freedom are \(\nu=2n_1-2\), and

$$ \frac{\nu s_p^2}{\sigma^2}\sim\chi^2_{\nu}, \qquad \frac{\operatorname{SD}(s_p^2)}{\sigma^2}=\sqrt{\frac{2}{\nu}}. $$

The second expression is the coefficient of variation of the variance estimate. Since the unrounded re-estimated size is proportional to \(s_p^2\), its sampling variability is substantial when the interim sample is small. A realized estimate below \(\sigma^2\) tends to produce a smaller trial than the true variance would require; an estimate above it tends to produce a larger one. Thus, achieving the target power conditional on the observed estimate is not the same as guaranteeing that target over repeated interim samples.

The following example quantifies this point using the same hypothetical design. It treats the variance estimate and the resulting size as random across repeated internal pilots, before any floor, cap, or rounding is applied.

Worked Example: Quantifying Interim Variance Uncertainty

Suppose the interim estimate in the preceding example is based on 20 participants per group, so \(\nu=2(20)-2=38\). If the true standard deviation is \(10.5\) mg/dL, the standard deviation of the pooled variance estimate is

$$ \operatorname{SD}(s_p^2) = \sigma^2\sqrt{\frac{2}{\nu}} = 110.25\sqrt{\frac{2}{38}} \approx25.29\;(\mathrm{mg/dL})^2. $$

Its coefficient of variation is \(\sqrt{2/38}\approx0.2294\), or 22.9%. The unrounded size rule has coefficient \(2(3.2898)^2/25\approx0.8658\), so its standard deviation across repeated interim estimates is approximately \(0.8658(25.29)=21.90\) participants per group. This is a measure of variability in the recalculated size, not a confidence interval for the required sample size. It shows why a single estimate of 10.5 mg/dL should not be treated as certain. Smaller interim samples would have fewer degrees of freedom and still greater relative uncertainty.

Assumptions, Diagnostics, and Safeguards

The rule is most defensible when the interim estimate reflects the variance relevant to the final endpoint, the groups have a common variance, allocation is balanced, and participants contribute independent outcomes. The normal approximation also becomes less reliable with very small samples or markedly nonnormal outcomes. If the final analysis uses a pooled \(t\) statistic rather than a normal statistic, planning should use an appropriate \(t\)-based calculation or simulation, particularly when sample sizes are small. The normal formula is a transparent approximation, not an exact guarantee for a finite-sample \(t\) procedure.

Before applying an update, check that the interim estimate is based on the prespecified variance estimator and that data handling, endpoint definition, and follow-up are consistent with the final analysis. Review group-specific spreads for a conspicuous variance imbalance, and inspect outliers or data errors that could dominate \(s_p^2\). Such diagnostics do not license an informal change of method: if the common-variance assumption is implausible, a preplanned alternative calculation is needed. A floor and cap can control operational consequences but may leave the final design below its target power or require more enrollment than is feasible.

A blinded, variance-only update is different from changing sample size in response to an observed treatment difference. Blinding helps keep the decision from being driven by apparent efficacy, but the adaptation rule and final testing procedure still need to be prespecified and assessed as a design. In particular, do not infer that any adaptive update automatically preserves the nominal type I error. The rule should specify the interim timing, variance estimator, target, rounding, minimum and maximum size, and how the final test uses the accrued data. A fixed sample size avoids interim-estimate variability but may be inefficient if the planned variance is wrong. A plug-in update responds to observed nuisance information, while a more comprehensive predictive or simulation-based assessment can incorporate uncertainty in that information and the details of the final procedure.

Common pitfall: “The interim standard deviation is 10.5 mg/dL, so the updated design has exactly 90% power.” The calculation only gives approximately 90% power if that estimate equals the true standard deviation and the planning assumptions hold. Correctly, it is a plug-in estimate of power; sampling error in the interim variance can make the actual operating power lower or higher.
Key takeaway: For a centered alternative and equal allocation, the approximate per-group size scales with \(\sigma^2/\Delta^2\). Replacing \(\sigma^2\) with an interim estimate gives a useful prespecified update, but the uncertainty of that estimate—and the final test's operating characteristics—must remain part of the design assessment.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. Why must a TOST equivalence trial reject two null hypotheses rather than just one?
  2. For \(\alpha=0.05\), what confidence level corresponds to the two one-sided tests?
  3. If the interim pooled variance estimate is four times the variance used in the original planning calculation, what happens to the unrounded sample-size target under this rule?
  4. With 15 participants per group in a normal-theory internal pilot, what are the pooled variance degrees of freedom and its coefficient of variation?
  5. Does a variance-only plug-in update guarantee that repeated trials will have at least the target power? Explain briefly.

Solutions. (1) Equivalence requires evidence both that \(\theta>-\Delta\) and that \(\theta<\Delta\); failing either one leaves at least one equivalence boundary unexcluded. (2) The corresponding two-sided interval is 90%, since its two tails each have probability \(\alpha=0.05\). (3) The unrounded target is proportional to the variance, so it becomes four times as large, subject to floors, caps, and rounding. (4) \(\nu=2(15)-2=28\), and the coefficient of variation is \(\sqrt{2/28}\approx0.2673\), or 26.7%. (5) No. The interim estimate is random, so the plug-in size can be too small or too large relative to the size required by the true variance; repeated-trial power requires an operating-characteristic assessment.

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