Why the Assumed Ratio Matters
A sample-size calculation for average bioequivalence needs an assumed true test-to-reference ratio, not just equivalence limits and a variability estimate. The ratio affects how far the assumed treatment effect lies from the nearest limit. When the assumed ratio is one, the effect is at the centre of symmetric log-scale limits. As the ratio moves away from one, it approaches one boundary, leaving less room for sampling error while still meeting both equivalence criteria. Under reciprocal limits, as the assumed effect moves away from zero toward either boundary, power falls and the sample size needed for a chosen power generally rises.
The effect is most naturally expressed on the log scale. Let \(\theta\) be the true mean test-minus-reference difference in log exposure. The geometric mean ratio (GMR) is \(\exp(\theta)\), so a GMR of one corresponds to \(\theta=0\). This tutorial holds the equivalence limits and within-subject variability fixed while examining the effect of changing \(\theta\).
Power Under Fixed Limits
Assume a standard two-period crossover analysis of log-transformed exposure. For planning, let \(\hat{\theta}\) estimate the population mean log contrast and suppose, initially, that its standard error \(SE\) is fixed and known. Under the normal approximation, \(\hat{\theta}\sim N(\theta,SE^2)\). The approximation treats the estimated contrast as unbiased and assumes independent subjects, a common within-subject variance, and a correctly specified crossover analysis. It also assumes that the planned sample is evaluable and that the equivalence limits and variance are fixed.
With a one-sided significance level \(\alpha\), the two one-sided tests (TOST) declare equivalence when the corresponding \(100(1-2\alpha)\%\) confidence interval for \(\theta\) lies wholly inside \((L,U)\). Equivalently, under the fixed-\(SE\) normal approximation, \(\hat{\theta}\) must lie between \(L+z_{1-\alpha}SE\) and \(U-z_{1-\alpha}SE\), where \(z_{1-\alpha}\) is the \(1-\alpha\) standard-normal quantile. Thus the power at assumed effect \(\theta\) is:
The CDF-difference expression applies when \(U-L-2z_{1-\alpha}SE>0\). If this quantity is zero or negative, the acceptance interval is empty or has zero width, so the power is 0.
Here \(\Phi\) is the standard-normal cumulative distribution function. The first term is the probability of meeting the upper-limit test; subtracting the second removes the probability of falling below the lower-limit test. The formula gives the probability of meeting both criteria, not the power of either test alone.
Worked Example: Power at a GMR of 1.10
Suppose the limits are 0.80 and 1.25, \(\alpha=0.05\), and the planned standard error is \(SE=0.057735\). Assume these limits are exact reciprocals, so their log limits are \(L=\ln(0.80)=-0.2231436\) and \(U=\ln(1.25)=0.2231436\). For a hypothetical assumed GMR of 1.10, \(\theta=\ln(1.10)=0.0953102\). The standard-normal critical value is \(z_{0.95}=1.644854\).
The upper standardized bound is \((0.2231436-0.0953102)/0.057735-1.644854\approx0.5693\). The lower standardized bound is \((-0.2231436-0.0953102)/0.057735+1.644854\approx-3.8709\), or approximately \(-3.871\). Therefore,
Under the stated assumptions, the probability of concluding equivalence is about 71.5%. This is below an 80% target. The assumed GMR is still inside the limits, but it is closer on the log scale to the upper limit than to the centre.
Why Power Falls Away from the Centre
For reciprocal limits, \(L=-c\) and \(U=c\) for some \(c>0\). In the fixed-\(SE\) formula, let \(a=c/SE-z_{1-\alpha}\) and \(t=\theta/SE\). When the acceptance interval has positive width, the power becomes \(\Phi(a-t)-\Phi(-a-t)\). This expression makes the symmetry around \(\theta=0\) visible: the same absolute log-scale departure in either direction gives the same power.
The decline for positive \(\theta\) can be justified directly. Differentiating with respect to \(\theta\) gives \((\phi(-a-t)-\phi(a-t))/SE\), where \(\phi\) is the standard-normal density. For \(\theta>0\), the point \(-a-t\) is farther from zero than \(a-t\), so \(\phi(-a-t)<\phi(a-t)\). The derivative is therefore negative. By symmetry, power also declines as \(\theta\) moves below zero. If the limits are not symmetric on the log scale, the maximum need not occur at zero; the relevant centre is the midpoint \((L+U)/2\). For the usual reciprocal limits, that midpoint is zero, corresponding to GMR one.
This argument concerns the fixed-\(SE\) normal approximation. In an exact calculation, the variance is estimated, so the critical value and standard error are random. Under the same symmetric-limit and normal-model assumptions, power can be evaluated by integrating over the variance estimate as well as the treatment-contrast estimate. Conditional on a given estimated standard error, the acceptance interval remains centred at zero; moving the assumed effect away from zero reduces the probability that the estimate falls inside it. Exact power therefore reflects the same basic loss of margin, though its numerical value can differ from the normal approximation.
Worked Example: A Smaller Departure from One
Keep the limits, \(\alpha\), and \(SE=0.057735\) from the previous example, but assume a hypothetical GMR of 1.05. Then \(\theta=\ln(1.05)=0.0487902\). The upper and lower standardized bounds are:
Consequently, the approximate power is \(\Phi(1.375)-\Phi(-3.065)\approx0.914\). This exceeds the approximately 71.5% power at a GMR of 1.10 because 1.05 is closer to one, the centre of these reciprocal limits. The calculation does not say that the true ratio will be near 1.05; it reports power conditional on that planning assumption.
From Reduced Power to a Larger Sample
For a fixed design and within-subject variance, increasing the sample size reduces \(SE\). That makes the confidence interval narrower and can restore power lost when the assumed effect moves toward a limit. The standard-error relationship depends on the design and its variance parameterization; it must be taken from the applicable sample-size model rather than guessed. The following illustration uses a hypothetical subject-level contrast standard deviation \(\tau=0.20\), with \(SE=\tau/\sqrt{n}\), and uses the fixed-\(SE\) normal approximation. Here \(n\) is the number of evaluable subjects under this illustrative model.
For a target power of 80%, the sample size can be checked by substituting \(SE=0.20/\sqrt{n}\) into the two-bound power formula. At the centre, symmetry simplifies the calculation: the power is \(2\Phi(c/SE-z_{0.95})-1\). For a noncentral assumed effect, both bounds should be evaluated; a one-sided margin calculation alone is not generally sufficient.
Worked Example: Comparing Sample Sizes for an 80% Target
Use \(L=-0.2231436\), \(U=0.2231436\), \(\alpha=0.05\), \(\tau=0.20\), and a target power of 80%. First assume a GMR of one, so \(\theta=0\). At \(n=7\), \(SE=0.20/\sqrt{7}=0.07559\). The common positive standardized bound is \(0.2231436/0.07559-1.644854\approx1.307\), giving power \(2\Phi(1.307)-1\approx0.809\). At \(n=6\), \(SE=0.08165\), the bound is approximately \(1.089\), and power is approximately \(2\Phi(1.089)-1=0.724\). Thus \(n=7\) is the smallest integer meeting the 80% target in this approximation at the centre.
Now assume a GMR of 1.10, so \(\theta=0.0953102\). At \(n=15\), \(SE=0.20/\sqrt{15}=0.05164\). The upper standardized bound is approximately \((0.2231436-0.0953102)/0.05164-1.644854=0.830\); the lower bound is about \(-4.52\). Hence power is approximately \(\Phi(0.830)-\Phi(-4.52)=0.797\), just below target. At \(n=16\), \(SE=0.05000\), the upper bound is \(0.9118\), and the lower bound is approximately \(-4.725\). Power is then approximately \(\Phi(0.9118)-\Phi(-4.725)=0.819\). In this illustrative normal calculation, 16 subjects are needed at the assumed GMR of 1.10, compared with 7 at a GMR of one.
The exact integer requirements would be obtained using the appropriate exact power calculation when the variance is estimated. The illustration isolates the effect of changing the assumed ratio: all other inputs are held fixed.
Planning Choices, Diagnostics, and Limitations
The main alternative to the fixed-\(SE\) approximation is an exact power calculation that accounts for estimation of the within-subject variance and uses the relevant degrees of freedom. It is preferable when the planned sample is small or when the approximation is not sufficiently accurate for the decision. Simulation can also check a more complicated model, provided it reproduces the planned design and analysis. Whichever method is used, compute the joint TOST success probability rather than treating the two one-sided tests as independent or reporting only the less favourable marginal power.
A sound planning calculation should be checked for sensible behaviour. Under reciprocal limits and fixed variability, power at assumed GMRs equally distant from one on the log scale should match. Power should be highest at one and decline as the assumed ratio approaches either limit. Increasing \(n\) should not reduce power under the stated model. A result that violates these patterns may indicate a sign error in \(\theta\), an incorrect log limit, inconsistent units, or a calculation that has used one test’s power in place of the joint probability.
The assumed ratio is a planning input, not a value established by the power calculation. Choosing a value too close to one can produce an optimistic sample size if the actual ratio is farther from one. Conversely, an assumed value nearer a boundary can lead to a larger design. The calculation also does not establish equivalence: that conclusion depends on the observed estimate and its uncertainty after data collection. Finally, this tutorial holds within-subject variability fixed; the effect of changing that input is a separate planning question.
- Mistake: Using the ratio itself as the effect in a log-scale power formula. Correction: Convert it to \(\theta=\ln(\mathrm{GMR})\) and compare it with log-scale limits.
- Mistake: Concluding that a true GMR inside the limits guarantees high power. Correction: Power also depends on distance to the nearer limit relative to the standard error.
- Mistake: Using the power of the upper-limit test as the overall TOST power. Correction: Calculate the probability that both one-sided criteria are met.
Check Your Understanding
Answer each question using the log-scale effect and the assumptions stated in the question.
- What is \(\theta\) when the assumed GMR is 0.90, and how does its power compare with that for a GMR of \(1/0.90\) under reciprocal limits?
- With reciprocal limits 0.80 and 1.25, why is a GMR of 1.10 expected to have lower power than a GMR of one at the same standard error?
- Why is the probability of meeting both TOST criteria not generally equal to either one-sided test’s power?
- Under the illustrative model \(SE=\tau/\sqrt{n}\), what change to \(n\) can compensate for a departure from a GMR of one?
Solutions. (1) \(\theta=\ln(0.90)=-0.10536\). Since \(1/0.90\approx1.1111\) has log effect \(+0.10536\), the two powers match under symmetric log limits and the same standard error. (2) The log effect \(\ln(1.10)=0.09531\) is closer to the upper limit \(0.22314\) than zero is, so the estimate has less room to vary while its confidence interval remains inside both limits. (3) Equivalence requires both tests to reject their respective null hypotheses; the joint probability accounts for both conditions, which are based on the same estimated contrast. (4) Increase \(n\), which decreases \(SE\) under the stated model and narrows the confidence interval; recalculate power to find the integer sample size meeting the target.