From Bioequivalence Assumptions to a Sample Size
A sample-size calculation for average bioequivalence is not just a formula applied to a variability estimate. It is a chain of linked assumptions: the treatment effect is assessed on the log scale, the within-subject variability is anticipated, both equivalence limits must be met, and the planned number of evaluable subjects must be large enough to provide the desired power. A mismatch at any step can produce a plausible-looking but inadequate enrollment target.
The worked analyses here use a balanced two-period, two-sequence crossover design and a single log-transformed pharmacokinetic endpoint. The sample size \(n\) means the number of evaluable subjects, not the number initially enrolled. The calculations assume that the primary endpoint is analyzed with the standard crossover treatment contrast and that the usual average bioequivalence limits apply. They are planning calculations, not claims about a real study.
The standard planning target is power under an assumed true value of \(\theta\), often the log of a ratio close to one. Power is the probability, under that assumed value and the planning model, that both one-sided tests reject. A target such as 80% or 90% is a design choice; it is not a guarantee that the realized study will pass.
Model, Variability, and the Approximate Calculation
For the balanced crossover calculation, let \(\sigma_w\) denote the within-subject standard deviation on the log scale for a treatment measurement. Under the usual model with independent, equal within-subject residual variances for test and reference, the standard deviation of a within-subject treatment difference is \(\sigma_D=\sqrt{2}\sigma_w\). With \(n\) evaluable subjects, the standard error of the estimated treatment difference is approximately \(\sigma_D/\sqrt n\). The residual degrees of freedom in the balanced two-period crossover analysis are typically \(df=n-2\).
If prior information is expressed as a within-subject coefficient of variation \(CV_w\), the log-scale standard deviation is obtained as \(\sigma_w=\sqrt{\ln(1+CV_w^2)}\), where the coefficient of variation is entered as a proportion. For example, 20% is entered as 0.20. This conversion presumes a lognormal relationship between the original-scale measurement and its log-scale variability; it is not valid to substitute 20 directly into the formula.
For a first approximation, replace the finite-sample \(t\) critical value by \(z_{0.95}=1.64485\), appropriate to each one-sided test at \(\alpha=0.05\). If \(\theta\) is the assumed true mean, approximate power is the probability that an approximately normal estimate falls between \(L+z_{0.95}SE\) and \(U-z_{0.95}SE\):
This formula includes the contribution of both equivalence limits. If the assumed true value is near the lower limit, the lower-limit constraint may dominate; when it is near the upper limit, the upper-limit constraint may dominate. Omitting one term is justified only when the probability of failing the other limit is negligible. The two one-sided tests are not two independent chances to succeed: both must pass for the study to conclude average bioequivalence.
Why a Finite-Sample Verification Is Needed
The normal calculation is useful for screening candidate sample sizes, but it treats the standard error as if it were known. In practice, the within-study variance is estimated. Under the normal crossover model, the residual variance estimate has a scaled chi-square distribution with \(df=n-2\). The finite-sample calculation therefore uses the \(t\) critical value \(t_{0.95,df}\) and accounts for the randomness of the estimated standard error.
A direct justification follows by conditioning on the estimated standard deviation. Let \(Q\) be the ratio of its estimate to the true \(\sigma_D\), so that \(Q^2\sim\chi^2_{df}/df\). Conditional on \(Q\), the estimate of \(\theta\) is normal with standard deviation \(SE=\sigma_D/\sqrt n\), while the acceptance interval for that estimate is bounded by \(L+t_{0.95,df}SE Q\) and \(U-t_{0.95,df}SE Q\). Thus the exact model-based power is the conditional probability of falling between those bounds, averaged over \(Q\):
This integral can be evaluated numerically. It states the model and uncertainty assumptions explicitly: normal log-scale treatment differences, a common within-subject variance, and the planned crossover degrees of freedom. For a more complex analysis or design, power should be evaluated under that design’s actual model rather than by borrowing this formula.
Worked Example: Choosing an Evaluable Sample Size
Suppose a hypothetical study plans a balanced two-period crossover and uses \(CV_w=20\%\). The assumed true geometric mean ratio is 1.02, and the desired power is at least 80%, with \(\alpha=0.05\) for each one-sided test. Determine a defensible number of evaluable subjects.
Convert the variability. Enter \(CV_w=0.20\):
The assumed log treatment effect is \(\theta=\ln(1.02)=0.01980\). The limits are \(L=\ln(0.80)=-0.2231\) and \(U=\ln(1.25)=0.2231\). For \(n=14\), \(SE=0.2801/\sqrt{14}=0.07487\). The normal approximation gives:
The screening calculation suggests that 14 evaluable subjects might meet the target, but it is close to the 80% threshold. A finite-sample check matters here: with \(n=14\), \(df=12\), so the critical value is larger than the normal critical value, and the standard error is estimated. Numerical evaluation of the finite-sample integral gives power below 80% for 14 subjects. At \(n=16\), \(df=14\), \(SE=0.2801/\sqrt{16}=0.07002\); the normal approximation is about 86.2%, and finite-sample integration is approximately 82%. Thus 16 evaluable subjects is a more defensible minimum under these stated assumptions. In a real protocol, the calculation should be reproduced with validated software and the intended analysis model.
Check the assumptions. This result depends on the assumed 20% within-subject CV, a true ratio of 1.02, complete evaluable data, and the balanced crossover model. If variability is higher, the ratio is farther from one, or the analysis has fewer effective degrees of freedom, power will be lower. A planned sample size should not be justified by the approximate 80.3% alone when finite-sample verification is readily available.
Changing Variability or the Assumed Ratio
Sample size responds to both variability and the distance from the equivalence limits. Larger \(\sigma_w\) increases the standard error, while an assumed true ratio closer to either limit leaves less room for the confidence interval to fit inside the acceptance region. The effect of a ratio shift is not symmetric on the original scale: calculations are made on the log scale, where the limits are symmetric for 80.00% and 125.00%.
Worked Example: Planning for Higher Variability
Consider a second hypothetical crossover study with \(CV_w=30\%\), an assumed true ratio of 1.00, and a target power of 90%. Convert the CV first:
At \(\theta=0\), the normal approximation is symmetric about zero, so the power is \(2\Phi(m/SE-z_{0.95})-1\), where \(m=\ln(1.25)=0.2231\). For \(n=38\), \(SE=0.4152/\sqrt{38}=0.06735\), giving:
The adjacent smaller balanced choice, \(n=36\), gives \(SE=0.06919\) and approximate power \(2\Phi(1.580)-1\approx0.886\), below 90%. Therefore 38 evaluable subjects is the first even candidate meeting the normal-approximation target. As in the first example, the finite-sample calculation must then be checked; if it falls short, increase \(n\) and repeat. The example also shows why a variability estimate cannot be treated as a minor detail: a 30% CV leads to a substantially larger sample than the 20% CV case.
Check the assumptions. This calculation assumes that the true ratio is exactly 1.00 for planning and that the 30% CV adequately represents the endpoint’s within-subject variability. If the anticipated ratio is not centered at one, calculate both terms in the general power expression; the symmetric shortcut no longer applies.
From Evaluable Subjects to Enrollment
Attrition adjustment is performed after determining the required number of evaluable subjects. If the anticipated proportion of enrolled subjects who will be evaluable is \(r\), a simple planning adjustment is \(\lceil n/r\rceil\), where \(\lceil\cdot\rceil\) means round up to the next integer. This gives an expected-retention adjustment, not a guarantee that at least \(n\) subjects will remain.
Worked Example: Adjusting for Attrition and Balance
Suppose the finite-sample power check supports 16 evaluable subjects, and the planning assumption is that 85% of enrolled subjects will be evaluable. The simple attrition adjustment is:
For a balanced two-sequence crossover, enrollment is usually rounded up to an even number so the sequences can be allocated equally. The practical enrollment target is therefore 20 subjects, 10 per sequence. At 85% evaluability, 20 enrolled subjects imply an expected \(20(0.85)=17\) evaluable subjects. That expectation is one subject above the calculated minimum, providing a modest buffer.
If it is important to have a high probability of retaining at least 16 evaluable subjects, the simple division is not enough. One can model each subject’s evaluability as an independent Bernoulli outcome with probability 0.85 and select the smallest enrollment \(N\) satisfying \(P\{\operatorname{Binomial}(N,0.85)\ge16\}\) at the desired retention probability. That is a different planning criterion from merely achieving 16 evaluable subjects on average.
Diagnostics, Alternatives, and Limitations
Before accepting a result, inspect the inputs and the calculation path. Check that the CV was entered as a proportion, that all calculations use log-scale limits and effects, and that the sample size counts evaluable subjects. Confirm that the assumed true ratio is inside the limits and that the calculated standard error decreases in proportion to \(1/\sqrt n\). Then verify the candidate and at least one smaller and larger sample size with finite-sample power. A reported power that falls when \(n\) increases, or that is unexpectedly insensitive to a large increase in variability, signals an implementation or input error.
A frequent mistake is to calculate power using only the nearer equivalence limit and treat the result as the probability of passing the study. The correction is to calculate the probability of meeting both limits. For instance, if the assumed true ratio is shifted toward the upper limit, the upper-limit failure probability can be substantial even when the lower-limit test has very high power. Likewise, a sample size obtained using a normal approximation should not be described as an exact finite-sample result.
A closed-form normal approximation is useful for transparent planning and quick sensitivity checks. Numerical noncentral-\(t\) or chi-square integration is preferable for finite-sample verification under the specified crossover model. Simulation is a flexible alternative when the planned model includes features that make the analytic calculation unsuitable, such as more complex designs or analysis procedures; the simulation must reproduce the design, estimand, decision rule, and variance assumptions. These approaches do not remove uncertainty in the assumed CV or true ratio. Sensitivity analyses over plausible values are often more informative than reporting a single sample size.
Finally, the calculation addresses statistical power under the planning assumptions. It does not establish that the chosen endpoint, design, sampling schedule, or analysis is scientifically appropriate, and it cannot compensate for systematic bias, protocol deviations, or departures from the model. The enrollment target is justified only to the extent that its assumptions and decision rule are justified.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- For \(CV_w=25\%\), write the formula for \(\sigma_w\) and give its approximate value.
- Why must the general approximate power calculation include contributions from both \(L\) and \(U\)?
- A study requires 24 evaluable subjects and anticipates 80% evaluability. What is the simple attrition-adjusted enrollment before considering sequence balance?
- What does a finite-sample verification add beyond using \(z_{0.95}\) and a fixed standard error?
- If the assumed true ratio moves closer to an equivalence limit while variability stays fixed, what generally happens to power at a fixed \(n\)?
Solutions. (1) \(\sigma_w=\sqrt{\ln(1+0.25^2)}=\sqrt{\ln(1.0625)}\approx0.2462\). (2) Average bioequivalence requires both one-sided tests to pass; either limit can be the limiting test, so the acceptance probability must account for both boundaries. (3) \(\lceil24/0.80\rceil=30\) subjects. Any further rounding depends on the design’s allocation requirements. (4) It uses the finite-sample \(t\) critical value and accounts for estimating variability, including the relevant degrees of freedom. (5) Power generally decreases because the true effect is nearer a boundary, leaving less room for estimation uncertainty while still meeting both equivalence criteria.