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Sample Size and Power · Tutorial 133 of 399

Sample Size for Replicate Crossover Average Bioequivalence

Calculate and interpret approximate sample sizes for replicate crossover bioequivalence studies, using the correct within-subject variance for each design.

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What You'll Learn

  • Derive the variance of the test-minus-reference contrast for fully and partially replicated designs.
  • Distinguish replicate-design variance inputs from the two-period crossover variance.
  • Calculate approximate sample sizes for centered and noncentered expected log ratios.
  • Account for equivalence limits, significance level, target power, and sequence balance.
  • Identify assumptions and diagnostics that affect the reliability of the calculation.

Why Replication Changes the Sample-Size Calculation

A replicate crossover gives participants more than one exposure to at least one treatment. That repetition can make a study more informative about within-subject variability, but it also changes the variance of the estimated test-to-reference contrast. A sample-size calculation that uses the two-period crossover variance without adjustment can therefore be too large or too small for the planned replicate layout.

The planning target is usually the log-scale mean treatment difference for a pharmacokinetic endpoint such as AUC or Cmax. Average bioequivalence is assessed by testing whether this difference lies within prespecified log-scale limits. Sample size depends on how far the expected difference is from those limits, the target power and significance level, and the variance of the contrast under the particular crossover design.

Definition: Let \(\theta\) be the population mean log test-to-reference difference, and let \(L\) and \(U\) be the lower and upper log-equivalence limits. Average bioequivalence requires the \(100(1-2\alpha)\%\) confidence interval for \(\theta\) to lie within \((L,U)\), equivalently the two one-sided tests of \(H_0:\theta\leq L\) and \(H_0:\theta\geq U\) to reject at level \(\alpha\). A planning variance \(v_D\) is the variance of an individual subject’s treatment contrast under the specified design; for \(N\) independent subjects, the approximate standard error of the estimated mean contrast is \(\sqrt{v_D/N}\).

For the common limits corresponding to ratios 0.80 and 1.25, \(L=\ln(0.80)\approx-0.22314\) and \(U=\ln(1.25)\approx0.22314\). If the expected test-to-reference ratio is \(GMR\), the planning value of the log difference is \(\theta=\ln(GMR)\). The calculation below assumes that the limit, expected ratio, and variability refer to the same log-transformed endpoint.

Deriving the Replicate-Design Variance

Consider a simple within-subject error model on the log scale. Each observed value consists of a treatment mean, any modeled period contribution, and a within-subject error. Assume the within-subject errors from distinct periods are independent, have mean zero, and share variance \(\sigma_w^2\). The treatment contrast uses averages of a subject’s observations under test and reference, so its error variance is the sum of the variances of those two averages. This is the key design-specific step.

$$ \operatorname{Var}(\bar{e}_T-\bar{e}_R) = \frac{\sigma_w^2}{r_T} + \frac{\sigma_w^2}{r_R} $$

Here \(r_T\) and \(r_R\) are the number of test and reference observations contributing to the subject-level contrast. The calculation assumes equal within-subject variances for the two treatments. If the design or planning model allows treatment-specific variances or a different covariance structure, use the variance of the actual contrast under that model instead.

In a fully replicated four-period design, such as TRTR/RTRT, each subject receives test twice and reference twice. Thus \(r_T=r_R=2\), giving \(v_D=\sigma_w^2\). In a partially replicated three-period design, such as TRR/RTR/RRT, test is observed once and reference twice. Thus \(v_D=\sigma_w^2+\sigma_w^2/2=1.5\sigma_w^2\). By contrast, in a two-period crossover each treatment is observed once, so the corresponding variance is \(2\sigma_w^2\). These factors are not interchangeable.

Planning result: Under the equal-variance, independent-period error model, use \(v_D=\sigma_w^2\) for a fully replicated design with two observations of each treatment, and \(v_D=1.5\sigma_w^2\) for a partially replicated design with one test and two reference observations. For a two-period crossover, the corresponding variance is \(2\sigma_w^2\). These are variances of the subject-level contrast, not variances of a single log observation.

When variability is supplied as a within-subject coefficient of variation \(CV_w\), convert it to log-scale variance using \(\sigma_w^2=\ln(1+CV_w^2)\), with \(CV_w\) expressed as a proportion. Then apply the design-specific multiplier. This distinction matters: using \(\sigma_w^2\) directly as the contrast variance in a two-period crossover would omit a factor of two, while applying that same factor to a fully replicated contrast would overstate its variance under the stated model.

Approximate Sample-Size Calculation

For a planning approximation, treat the estimated log treatment difference as normally distributed with mean \(\theta\) and standard error \(SE=\sqrt{v_D/N}\). At significance level \(\alpha\), average bioequivalence is declared when the estimate is sufficiently far from both limits. Under this approximation, the probability of passing both one-sided tests is

$$ \operatorname{Power}(N) \approx \max\left\{0,\, \Phi\left(\frac{U-\theta}{SE}-z_{1-\alpha}\right) - \Phi\left(\frac{L-\theta}{SE}+z_{1-\alpha}\right) \right\}, \qquad SE=\sqrt{\frac{v_D}{N}}. $$

Here \(\Phi\) is the standard normal cumulative distribution function, and \(z_q\) is its \(q\) quantile. The first term is the probability of satisfying the upper-bound test; subtracting the second term removes the probability of failing the lower-bound test. For a centered expectation \(\theta=0\) and symmetric limits \((- \Delta,\Delta)\), this simplifies to \(2\Phi(\Delta/SE-z_{1-\alpha})-1\). Solving for the approximate sample size at target power \(1-\beta\) gives

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

The \(z_{1-\beta/2}\) quantile appears because power is the probability of passing both one-sided tests: at the centered value, each tail of the acceptance event receives half of the total non-equivalence probability. The expression is an initial normal approximation, not a guarantee that the final integer sample size will attain the target under a finite-sample analysis. In practice, evaluate candidate integer sample sizes using the planned model and an appropriate finite-sample power calculation; also respect the design’s sequence allocation.

Worked Example: Centered Expectation in a Fully Replicated Design

Suppose a hypothetical four-period fully replicated study targets 80% power at \(\alpha=0.05\), expects a test-to-reference ratio of 1.00, and uses limits 0.80 and 1.25. Assume \(CV_w=30\%\). Find the normal-approximation sample size before sequence-balance adjustment.

First convert the coefficient of variation to log-scale variance: \(\sigma_w^2=\ln(1+0.30^2)=\ln(1.09)\approx0.08618\). There are two observations per treatment, so the fully replicated contrast variance is \(v_D=\sigma_w^2\approx0.08618\). The symmetric log margin is \(\Delta=\ln(1.25)\approx0.22314\).

For 80% target power, \(z_{1-\beta/2}=z_{0.90}\approx1.2816\), while \(z_{1-\alpha}=z_{0.95}\approx1.6449\). Therefore \[ N\geq\frac{0.08618}{0.22314^2}(1.6449+1.2816)^2 =\frac{0.08618}{0.04979}(2.9265)^2 \approx14.82. \]

Rounding up gives 15 subjects as a normal-approximation starting value. The two sequences in this design are usually balanced, so an allocation constrained to equal sequence sizes would require checking the next suitable even total, 16. This calculation does not establish that 16 achieves 80% power under a finite-sample model; that must be checked using the planned analysis and its degrees of freedom.

Worked Example: Partial Replication and Sequence Balance

A hypothetical three-period partially replicated study has the same limits and expected ratio, but targets 90% power. Assume \(CV_w=30\%\), with one test and two reference observations per subject. Calculate the approximate total and account for three-sequence balance.

The log-scale within-subject variance remains \(\sigma_w^2=\ln(1.09)\approx0.08618\). The contrast variance is not this value alone: \(v_D=1.5\sigma_w^2=1.5(0.08618)\approx0.12927\). For 90% power, \(z_{1-\beta/2}=z_{0.95}\approx1.6449\). Thus \[ N\geq\frac{0.12927}{0.22314^2}(1.6449+1.6449)^2 =\frac{0.12927}{0.04979}(3.2898)^2 \approx28.10. \]

The normal approximation therefore gives 29 subjects before considering allocation. With three sequences, equal allocation requires a total divisible by three, so the next suitable total is 30, or 10 subjects per sequence. As in the first example, the sequence-balanced total is a candidate for the finite-sample power calculation, not a substitute for it.

When the Expected Ratio Is Not One

A nonzero expected log difference reduces the distance to one equivalence limit and increases the required sample size. The centered closed-form expression no longer applies directly; use the two-term power expression and examine integer candidate sizes. The following example illustrates why sequence balance can affect the final planned total even when the normal approximation is used.

Worked Example: Noncentered Expectation in a Fully Replicated Design

Suppose a hypothetical fully replicated study expects a ratio of 1.05, has \(CV_w=25\%\), uses limits 0.80 and 1.25, and targets 80% power at \(\alpha=0.05\). Calculate approximate power for candidate totals of 12 and 14 subjects.

Here \(\theta=\ln(1.05)\approx0.04879\), and \(\sigma_w^2=\ln(1+0.25^2)=\ln(1.0625)\approx0.06062\). For this fully replicated design, \(v_D=0.06062\). At \(N=12\), \(SE=\sqrt{0.06062/12}\approx0.07108\). The two arguments of \(\Phi\) are \[ \frac{0.22314-0.04879}{0.07108}-1.6449\approx0.808, \qquad \frac{-0.22314-0.04879}{0.07108}+1.6449\approx-2.181. \] Thus power is approximately \(\Phi(0.808)-\Phi(-2.181)\approx0.7905-0.0146=0.776\), below 80%.

At \(N=14\), \(SE=\sqrt{0.06062/14}\approx0.06580\). The corresponding arguments are approximately \(1.005\) and \(-2.488\), so power is \(\Phi(1.005)-\Phi(-2.488)\approx0.8425-0.0064=0.836\). Four-period fully replicated designs use two sequences; 14 is balanced if divided equally. On this approximation, 14 is the first of these two checked balanced candidates to exceed 80%. A complete search would also evaluate any smaller balanced candidate under the chosen power method.

Assumptions, Checks, and Alternatives

These calculations are only as credible as their assumptions and inputs. The design must be specified correctly, including which treatments are repeated and how sequences are allocated. The within-subject variability must be relevant to the endpoint, population, and study conditions being planned. The simple variance derivation assumes independent period errors with a common variance; period and sequence effects must be handled by the analysis model rather than confused with within-subject error. The expected log ratio and equivalence limits must also be stated on the same scale.

A practical workflow is to identify the treatment contrast and its replication pattern, derive or obtain its variance under the planned model, convert any coefficient of variation to log-scale variance, and calculate power across feasible sequence-balanced totals. Then verify the selected total using a finite-sample method consistent with the planned analysis. Degrees of freedom, variance estimation, and the test statistic can make exact or model-based power differ from the normal approximation, especially at small sample sizes.

Diagnostics for a planning calculation are mostly checks on inputs and structure. Verify that the CV is entered as a proportion, that the logarithm is natural, and that the contrast-variance multiplier matches the replicate layout. Check whether rounding for sequence balance changes the candidate total, and inspect power at the selected total and the preceding feasible total. If the required sample size changes sharply under modest changes in CV or expected ratio, report that sensitivity rather than presenting one planning number as certain.

The two-period crossover is the main alternative when replication is not required: its contrast variance is \(2\sigma_w^2\) under equal within-subject variances. A fully replicated design has contrast variance \(\sigma_w^2\), while the stated partially replicated design has \(1.5\sigma_w^2\). Replication can therefore improve precision for the treatment contrast under these assumptions, while also requiring additional periods and a design-specific allocation and analysis. These variance comparisons alone do not establish which design is preferable; feasibility, burden, and the study’s objectives also matter.

A common mistake is to take a CV-derived \(\sigma_w^2\) and use \(2\sigma_w^2\) for every crossover, or to use \(\sigma_w^2\) as though it were always the contrast variance. The correction is to count the observations contributing to each treatment average and calculate the contrast variance under the planned covariance model. Another mistake is to round a normal-approximation result upward and call it the final required sample size. It is only a candidate: sequence balance and finite-sample power still need checking.

Key takeaway: Replicate-crossover sample size depends on the variance of the subject-level test-minus-reference contrast. Derive that variance from the actual replication pattern before applying an average bioequivalence power calculation.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. Under the equal-variance independent-period model, what is the contrast variance for a fully replicated design with two observations per treatment?
  2. A partially replicated design observes test once and reference twice. If \(\sigma_w^2=0.04\), what is the contrast variance?
  3. Why does the centered normal-approximation sample-size formula use \(z_{1-\beta/2}\) for target power \(1-\beta\)?
  4. What is the first design-specific check to make if a planned replicate study’s sample size appears implausibly small?

Solutions. (1) It is \(\sigma_w^2/2+\sigma_w^2/2=\sigma_w^2\). (2) It is \(0.04+0.04/2=0.060\). (3) For a centered expectation, power is \(2\Phi(\Delta/SE-z_{1-\alpha})-1\); setting this equal to \(1-\beta\) gives \(\Delta/SE-z_{1-\alpha}=z_{1-\beta/2}\). (4) Verify that the variance multiplier matches the replication pattern and that the variance is the contrast variance, not merely the variance of one log observation.