Sample Size for Parallel-Group Bioequivalence

Planning a Parallel-Group Equivalence Study

In a parallel-group bioequivalence study, each participant receives either the test product or the reference product, not both. The treatment comparison therefore uses two independent groups. This distinction matters for planning: the uncertainty of the estimated treatment difference depends on the variance in each group and on the number assigned to each group. A sample size copied from a crossover calculation will generally be inappropriate because it relies on a different variance structure.

The planning target is the probability that the study will conclude average bioequivalence when a specified true treatment difference is assumed. As in other average bioequivalence calculations, exposures are analyzed on the logarithmic scale and equivalence is assessed against prespecified lower and upper limits. The introductory material on average bioequivalence and the earlier material on power targets, equivalence limits, and variance estimation provide the context for selecting these inputs.

Definition: Let \(Y_{Ti}\) and \(Y_{Rj}\) be log-transformed exposure measurements for participants in the test and reference groups. Under the usual common-variance planning model, \(Y_{Ti}\sim N(\mu_T,\sigma^2)\) and \(Y_{Rj}\sim N(\mu_R,\sigma^2)\), independently across participants and groups. The parameter of interest is \(\delta=\mu_T-\mu_R\), and its estimator is \(D=\bar{Y}_T-\bar{Y}_R\). If the group sizes are \(n_T\) and \(n_R\), then \(SE(D)=\sigma\sqrt{1/n_T+1/n_R}\).

The model assumes independent participants, normally distributed log measurements within each group, and a common within-group variance \(\sigma^2\). These are planning assumptions, not guarantees. A model that accounts for covariates or other design features can lead to a different variance and degrees of freedom. The common-variance model is useful here because it makes the role of allocation and variance transparent.

From Equivalence Limits to a Sample-Size Formula

Let \(L\) and \(U\) be the lower and upper equivalence limits on the log scale, with \(L<0<U\). For the usual ratio limits of 0.80 and 1.25, \(L=\ln(0.80)\) and \(U=\ln(1.25)\). The two one-sided tests at level \(\alpha\) establish equivalence when the corresponding \(100(1-2\alpha)\%\) confidence interval for \(\delta\) lies entirely inside \((L,U)\).

For a normal approximation, treat the standard error as known and use the standard normal critical value \(z_{1-\alpha}\). The estimated difference must then fall between \(L+z_{1-\alpha}SE\) and \(U-z_{1-\alpha}SE\). Power is the probability that \(D\) falls in this interval, assuming \(D\) is normally distributed with mean equal to the planned true difference \(\delta\) and standard deviation \(SE\).

\operatorname{Power}_{N} = \Phi\!\left(\frac{U-z_{1-\alpha}SE-\delta}{SE}\right) - \Phi\!\left(\frac{L+z_{1-\alpha}SE-\delta}{SE}\right), \qquad SE=\sigma\sqrt{\frac{1}{n_T}+\frac{1}{n_R}}.

Here \(\Phi\) is the standard normal cumulative distribution function. If either adjusted boundary leaves no interval, so \(L+z_{1-\alpha}SE\ge U-z_{1-\alpha}SE\), the approximate power is zero. This equation also shows why power depends on the expected treatment difference: a difference away from the center of the equivalence region leaves less room on one side.

For symmetric limits \(L=-m\), \(U=m\), and a centered expected difference \(\delta=0\), the approximate power simplifies to \(2\Phi(m/SE-z_{1-\alpha})-1\). Solving for the standard error that gives target power \(1-\beta\), then substituting the independent-groups standard error, gives the equal-allocation formula below.

SE_{\max}=\frac{m}{z_{1-\alpha}+z_{1-\beta/2}}, \qquad n_T=n_R=n \quad\Longrightarrow\quad n\approx \frac{2\sigma^2(z_{1-\alpha}+z_{1-\beta/2})^2}{m^2}.

The result is an approximate number of evaluable participants per group. The derivation uses the normal critical value and does not account for the extra uncertainty from estimating \(\sigma^2\). A finite-sample calculation uses a \(t\) critical value and the sampling distribution of the pooled variance; it can require a larger integer sample size.

Worked Example: Equal-Allocation Planning at a Centered Ratio

Suppose a hypothetical parallel study uses ratio limits 0.80 and 1.25, \(\alpha=0.05\), and target power 80% when the true ratio is 1.00. Planning data suggest a log-scale coefficient of variation of 30%. Find the normal-approximation sample size per group.

For a lognormal exposure measure, the log-scale variance corresponding to coefficient of variation \(CV\) is \(\sigma^2=\ln(1+CV^2)\). Thus \(\sigma^2=\ln(1+0.30^2)=\ln(1.09)=0.08618\). The symmetric log margin is \(m=\ln(1.25)=0.22314\), and the standard normal quantiles are \(z_{0.95}=1.6449\) and \(z_{0.80}=0.8416\) for the power expression's \(z_{1-\beta}\) only if power were 80% one-sided. Here \(1-\beta=0.80\), so \(z_{1-\beta}=z_{0.80}=0.8416\) is not the correct quantile for the centered two-sided acceptance probability derivation, which uses \(z_{1-\beta/2}\). Instead, setting \(2\Phi(m/SE-z_{1-\alpha})-1=0.80\) gives \(\Phi(m/SE-z_{1-\alpha})=0.90\), hence the required quantile is \(z_{0.90}=1.2816\).

Therefore the centered symmetric formula for this two-sided power target is \(n\approx2\sigma^2(z_{1-\alpha}+z_{(1+\text{power})/2})^2/m^2\):

n\approx \frac{2(0.08618)(1.6449+1.2816)^2}{(0.22314)^2} = \frac{0.17236(2.9265)^2}{0.04979} \approx 29.65.

Rounding up gives 30 evaluable participants per group. At \(n=30\), \(SE=\sqrt{0.08618(1/30+1/30)}=0.07580\). The approximate power is \(2\Phi(0.22314/0.07580-1.6449)-1=2\Phi(1.299)-1\approx0.806\). This is just above 80%, as expected from rounding the continuous calculation upward. The calculation assumes equal group variances, independent log measurements, and a true ratio of 1.00.

Allocation Between Groups

If the group sizes differ, use \(SE=\sigma\sqrt{1/n_T+1/n_R}\) directly in the power equation. For a centered expected difference and symmetric limits, the same maximum standard error applies, so the approximate planning condition is

\frac{1}{n_T}+\frac{1}{n_R} \le \frac{1}{\sigma^2} \left(\frac{m}{z_{1-\alpha}+z_{(1+\text{power})/2}}\right)^2.

When the two groups have the same variance and recruitment costs, equal allocation minimizes the variance of the difference for a fixed total sample size. If the group variances differ, minimizing \(\sigma_T^2/n_T+\sigma_R^2/n_R\) at a fixed total gives \(n_T/n_R=\sigma_T/\sigma_R\). In practice, allocation can also reflect recruitment feasibility or other design constraints; the resulting precision should be checked rather than assuming that a balanced design is mandatory.

Worked Example: The Precision Cost of Unequal Allocation

Use the variance and planning limits in the preceding example, but suppose a hypothetical study plans 16 participants receiving test and 32 receiving reference. Compare its approximate power with a balanced allocation of 24 and 24 participants, keeping the total at 48.

For allocation 16:32, \(SE=\sqrt{0.08618(1/16+1/32)}=\sqrt{0.008079}=0.08988\). Its centered approximate power is \(2\Phi(0.22314/0.08988-1.6449)-1=2\Phi(0.838)-1\approx0.598\).

For allocation 24:24, \(SE=\sqrt{0.08618(1/24+1/24)}=\sqrt{0.007182}=0.08475\). The approximate power is \(2\Phi(0.22314/0.08475-1.6449)-1=2\Phi(0.988)-1\approx0.677\).

Both calculations use the same variance in both groups and the same total number of participants. The balanced allocation has the smaller standard error and higher approximate power. The unequal allocation may still be chosen for practical reasons, but its precision cost should be made explicit.

Expected Ratio and Exact Power

A centered expected ratio is a convenient planning assumption, not a universal default. For a planned test-to-reference ratio \(r\), the expected log difference is \(\delta=\ln(r)\). When \(\delta\ne0\), calculate power from both normal tail areas in the general equation; do not use the centered simplification. A modest shift toward one equivalence boundary can meaningfully reduce power even if the expected ratio remains within the limits.

The finite-sample pooled-\(t\) calculation replaces the normal approximation with the pooled standard deviation \(S_p\), where \(S_p^2\) is the weighted average of the two sample variances and \(df=n_T+n_R-2\). Under the common-variance normal model, \(S_p\) is independent of \(D\), and \(dfS_p^2/\sigma^2\sim\chi^2_{df}\). Equivalence is concluded when \(D-t_{1-\alpha,df}S_pq>L\) and \(D+t_{1-\alpha,df}S_pq<U\), where \(q=\sqrt{1/n_T+1/n_R}\).

These facts give a direct exact-power calculation under the model: integrate the conditional probability of that interval over the chi-square distribution of the pooled variance. For each candidate integer pair \((n_T,n_R)\), use the \(t\) critical value for its degrees of freedom and evaluate this probability numerically. Increase the group sizes until the target power is reached. This procedure incorporates both the finite-sample critical value and random variance estimation; the normal formula is a transparent starting approximation, not a substitute when an exact target is required.

Worked Example: Checking Power When the Expected Ratio Is 0.95

Continue with the hypothetical variance \(\sigma^2=0.08618\), limits \(\pm0.22314\), \(\alpha=0.05\), and equal groups. Suppose the planning ratio is 0.95 and examine 40 participants per group using the normal approximation.

The expected log difference is \(\delta=\ln(0.95)=-0.05129\). The standard error is \(SE=\sqrt{0.08618(1/40+1/40)}=0.06564\). The approximate acceptance interval for the observed difference is from \(-0.22314+1.6449(0.06564)\) to \(0.22314-1.6449(0.06564)\), or approximately \((-0.11517,0.11517)\).

Under the planning alternative, \(D\) is approximately normal with mean \(-0.05129\) and standard deviation 0.06564. Thus power is \( \Phi((0.11517+0.05129)/0.06564)-\Phi((-0.11517+0.05129)/0.06564) =\Phi(2.536)-\Phi(-0.973)\approx0.829.\) The candidate exceeds 80% approximate power. This does not prove that 40 per group is the smallest integer sample size, nor does it account for the finite-sample \(t\) calculation; those questions require checking candidate sample sizes, preferably with exact power.

Diagnostics, Alternatives, and Limits

Before relying on a result, check that the inputs and model match the intended study. Confirm that the endpoint, log transformation, equivalence limits, significance level, target power, and expected ratio are specified consistently. Verify that the variance estimate is on the log scale and comes from a population and design relevant to the planned parallel study. A crossover within-subject variance is not automatically the between-participant variance needed here. The earlier material on evaluating variance estimates for sample-size planning is especially relevant to this choice.

A diagnostic review should also consider whether the assumed common variance is credible and whether the groups are genuinely independent. If the expected variances differ, use the unequal-variance standard error \(\sqrt{\sigma_T^2/n_T+\sigma_R^2/n_R}\) for approximate planning and use an analysis and power calculation that accommodate the variance structure. If participants are clustered, repeated, or adjusted through a covariate model, the simple independent-groups formula may not represent the actual standard error.

The main alternative to the normal approximation is exact pooled-\(t\) power under the normal, common-variance model. It is preferable when sample sizes are modest or the requested power threshold must be met closely. The approximation is useful for understanding effects and making an initial calculation, but can be optimistic because it treats the variance as known and uses a normal critical value. Neither method protects against a poor variance estimate, model misspecification, or an expected treatment difference that was chosen unrealistically.

This calculation gives the number of evaluable participants required by the specified model and assumptions. It does not itself account for dropout or nonevaluable profiles. Those adjustments should be handled separately rather than silently folded into the variance or equivalence limits.

Common Mistakes

Using a crossover variance without justification. The parallel-group standard error depends on independent between-participant group variances; crossover variability has a different interpretation.
Using the one-sided power quantile for a centered two-sided power calculation. With symmetric limits and centered truth, the formula uses \(z_{(1+\text{power})/2}\), because the acceptance probability is split across two tails.
Assuming unequal allocation has no precision cost. With equal variances and a fixed total sample size, imbalance increases \(1/n_T+1/n_R\), and therefore increases the standard error.
Treating the normal approximation as an exact finite-sample answer. The pooled-\(t\) calculation accounts for variance estimation and may require more participants.
Ignoring the planned treatment difference. Power is not generally maximized at every value inside the equivalence limits; a shifted expected difference reduces the margin on one side.

Key takeaway: For a parallel-group study, begin with the independent-groups standard error \(\sigma\sqrt{1/n_T+1/n_R}\), calculate power using both equivalence boundaries, and treat the resulting normal-approximation sample size as a planning benchmark. Check the integer design with a finite-sample calculation when the target must be met reliably.

Check Your Understanding

Answer each question, then compare with the brief solutions below.

With equal group sizes and common variance, what happens to the standard error if both group sizes are doubled?
For symmetric log limits \((-m,m)\) and a centered expected difference, why does the power formula use \(z_{(1+\text{power})/2}\)?
With equal variances and a fixed total sample size, which allocation minimizes the variance of the difference?
What is the planning concern if a study uses a crossover within-subject variance for a parallel-group calculation?

Solutions. (1) Since \(SE=\sigma\sqrt{2/n}\), doubling each group size multiplies the standard error by \(1/\sqrt{2}\). (2) The centered acceptance interval has equal tails; setting its probability to the target power requires each upper cumulative probability to be \((1+\text{power})/2\). (3) Equal allocation minimizes \(1/n_T+1/n_R\) when group variances and costs are equal. (4) The crossover variance describes within-subject variation and may not estimate the independent between-participant variance that determines parallel-group precision.