Why Exact Power Requires Both Equivalence Boundaries
Average bioequivalence is concluded only when the estimated test-to-reference effect is sufficiently close to zero on the log scale to satisfy both equivalence limits. Power is therefore not simply the chance of rejecting one null hypothesis. It is the probability that the study rejects the lower-boundary null and the upper-boundary null in the same analysis. The exact calculation accounts for uncertainty in both the estimated treatment effect and its estimated standard error.
For the usual 80% to 125% limits, the log-scale bounds are \(L=\ln(0.80)\approx-0.22314\) and \(U=\ln(1.25)\approx0.22314\). More generally, write the limits as \(L<U\), the true mean log test-to-reference ratio as \(\theta\), and the significance level of each one-sided test as \(\alpha\). A common choice is \(\alpha=0.05\).
Model and Exact Power Formula
The calculation here assumes a balanced, complete, two-period, two-sequence crossover analyzed with the standard treatment-effect model. On the log scale, the within-subject error is normal with a common variance across treatments; there is no carryover effect or treatment-by-period interaction affecting the treatment estimate. Subjects are independent, allocation is balanced across the two sequences, and the planned analysis uses the residual variance estimate from the crossover model. Under these assumptions, the treatment estimate is normal, its variance estimate is independent of it, and the scaled residual variance follows a chi-square distribution.
Let \(n\) be the total number of evaluable subjects, \(\sigma_w^2\) the within-subject variance on the log scale, and \(df=n-2\) the residual degrees of freedom for this balanced two-sequence analysis. The standard deviation of the treatment estimate under the planning model is \(v=\sqrt{2\sigma_w^2/n}\). The factor of 2 reflects the variance of a within-subject treatment contrast. The estimated standard error is random; under the model it equals \(v\sqrt{Q/df}\), where \(Q\sim\chi^2_{df}\).
The lower-boundary test rejects when \((\widehat{\theta}-L)/SE>t_{1-\alpha,df}\). The upper-boundary test rejects when \((\widehat{\theta}-U)/SE<-t_{1-\alpha,df}\). Here \(t_{1-\alpha,df}\) is the \(1-\alpha\) quantile of the central \(t\) distribution. The noncentrality parameters for the two one-sided statistics are
Thus, considered separately, the lower test statistic has a noncentral \(t\) distribution with \(df\) degrees of freedom and noncentrality \(\lambda_L\); the upper statistic has the same degrees of freedom and noncentrality \(\lambda_U\). Those distributions give the marginal probabilities of passing each test. However, the two statistics use the same treatment estimate and standard error, so their outcomes are dependent. The exact power is their joint pass probability, not the product of the marginal powers.
One direct expression for that joint probability conditions on \(Q\). Given \(Q\), passing both tests requires the estimate to lie between \(L+t_{1-\alpha,df}v\sqrt{Q/df}\) and \(U-t_{1-\alpha,df}v\sqrt{Q/df}\). If the lower endpoint is not below the upper endpoint, the conditional probability is zero. Otherwise the conditional probability is the difference of two normal probabilities. Averaging over the chi-square distribution gives the exact power:
Here \(\Phi\) is the standard normal cumulative distribution function and \(f_{\chi^2_{df}}\) is the chi-square density. This integral is a convenient exact-power calculation: numerical integration or a validated noncentral-\(t\) power routine can evaluate it. For a calculation based on the two noncentral-\(t\) distributions, the joint probability must still respect their shared random estimate and standard error.
Worked Calculations
Worked Example: Power With a Slightly Higher Expected Test Ratio
Suppose a hypothetical, balanced two-period crossover has \(n=24\) evaluable subjects, an anticipated within-subject CV of 25%, and a true test-to-reference ratio of 1.02. Use the 80% to 125% limits and \(\alpha=0.05\). Assume the model conditions above hold.
First convert the CV to log-scale variance: \(\sigma_w^2=\ln(1+0.25^2)=\ln(1.0625)\approx0.06062\). The degrees of freedom are \(df=24-2=22\), and the standard deviation of the estimated treatment effect is
The expected log effect is \(\theta=\ln(1.02)\approx0.01980\). With \(L=-0.22314\) and \(U=0.22314\), the two noncentrality parameters are
The one-sided critical value is \(t_{0.95,22}\approx1.717\). Substituting these quantities into the exact joint-power integral gives power of approximately 0.82. Thus, if the stated model and planning inputs are correct, repeated studies of this design would have about an 82% probability of passing both equivalence tests. This is not the probability that this particular study will be bioequivalent after its data have been observed.
As a scale check, with \(Q/df\) near 1 the estimate must fall roughly between \(-0.22314+1.717(0.07108)=-0.1011\) and \(0.22314-1.717(0.07108)=0.1011\) to pass both tests. The true value \(0.01980\) is inside that interval, but sampling variability still leaves a meaningful chance of failure.
Worked Example: More Subjects and a Different Expected Ratio
Consider a second hypothetical balanced crossover with \(n=36\), within-subject CV 20%, and expected ratio 1.05. Retain the same equivalence bounds and \(\alpha=0.05\), and assume the same model conditions.
The log-scale variance is \(\sigma_w^2=\ln(1+0.20^2)=\ln(1.04)\approx0.03922\). Therefore \(df=36-2=34\), and
The two boundary noncentralities are
For \(df=34\), \(t_{0.95,34}\approx1.691\). Numerical evaluation of the exact joint-power integral gives power of approximately 0.98. Both noncentralities matter: the expected effect is farther from the upper boundary than from the lower one, but the upper boundary is the more limiting side in this example. The reported value is the chance of passing both tests, not merely the probability of passing the more favorable lower-boundary test.
Worked Example: Centered Ratio Does Not Guarantee High Power
A third hypothetical study has \(n=18\), within-subject CV 30%, and expected ratio 1.00. The true log effect is therefore \(\theta=0\), exactly at the center of the log-scale limits. Assume a balanced complete crossover and the same testing procedure.
The log-scale variance is \(\sigma_w^2=\ln(1+0.30^2)=\ln(1.09)\approx0.08618\). Hence \(df=18-2=16\), and
The critical value is \(t_{0.95,16}\approx1.746\). Exact joint-power evaluation gives approximately 0.41. Although the expected ratio is perfectly centered, the relatively large within-subject variability and modest sample size make the standard error large. The example shows why centering the expected ratio is not a substitute for calculating power.
Calculation Checks, Alternatives, and Limitations
A reliable calculation follows a fixed sequence: specify the equivalence limits and true expected ratio on the log scale; convert the anticipated CV to log-scale variance; determine \(n\), \(df\), and \(v\) under the planned analysis; calculate both boundary noncentralities; and evaluate the joint probability. The following checks help detect errors:
- Check the signs. If \(\theta\) is inside the limits, then \(\lambda_L\) should be positive and \(\lambda_U\) negative. An expected effect near the upper limit makes \(\lambda_U\) close to zero.
- Check the scale. The ratio is logged before it is compared with \(L\) and \(U\). A CV is converted using \(\sigma_w^2=\ln(1+CV^2)\), not treated directly as a log-scale standard deviation.
- Check the design and degrees of freedom. Here \(df=n-2\) follows from the specified balanced two-sequence crossover model. A different design or analysis may have different degrees of freedom and variance structure.
- Check both boundaries. A large probability of passing one test cannot establish high joint power if the other boundary is restrictive.
A common shortcut is to calculate the two marginal noncentral-\(t\) powers and multiply them. That reasoning assumes independence, but both tests use the same estimate and standard error. Another tempting shortcut is to subtract one marginal failure probability from one, or to add the two failure probabilities without considering overlap. Such approximations can be useful in some settings, but they are not the exact joint probability. The conditional integral handles the dependence directly. A normal approximation may also simplify the calculation, but it does not retain the finite-sample \(t\) critical value and random variance estimate; comparing it with the exact result is a separate planning question.
Exactness is conditional on the model and planning inputs. If the CV is underestimated, the expected ratio differs from the assumed value, subjects are lost, or the analysis has a different residual variance structure, the achieved power can depart substantially from the planned value. The calculation also addresses one endpoint and one equivalence requirement at a time; it does not automatically give joint power for multiple pharmacokinetic endpoints. Numerical power software should be checked against the stated bounds, log scale, degrees of freedom, tail directions, and whether it calculates joint or marginal power.
Check Your Understanding
Try each question, then compare your reasoning with the brief solutions.
- For a balanced two-period crossover with 30 evaluable subjects, what residual degrees of freedom are used under the model in this tutorial?
- With \(\theta\) strictly between \(L\) and \(U\), what signs should the lower- and upper-boundary noncentrality parameters have?
- Why is multiplying the two marginal probabilities of passing the boundary tests generally not an exact power calculation?
- If the anticipated within-subject CV increases while the sample size and expected ratio stay fixed, what happens to \(v\), and what is the usual consequence for power?
Solutions. (1) \(df=n-2=30-2=28\). (2) \(\lambda_L=(\theta-L)/v>0\) and \(\lambda_U=(\theta-U)/v<0\). (3) The tests share the same estimated treatment effect and standard error, so their outcomes are dependent; exact power is their joint pass probability. (4) Since \(\sigma_w^2=\ln(1+CV^2)\), a larger CV increases \(\sigma_w^2\) and \(v=\sqrt{2\sigma_w^2/n}\). The wider sampling uncertainty usually lowers the chance that the confidence interval lies entirely within both limits.