From a Population Criterion to a Power Calculation
A population bioequivalence (PBE) criterion combines the difference in population means with a comparison of population variances. Power therefore depends on more than how close the treatment and reference means are: the variance estimates also contribute uncertainty, and the treatment variance affects the criterion itself. A useful power calculation makes these pieces explicit rather than treating PBE as an ordinary test of a mean difference alone.
For illustration, consider a conventional PBE metric on the log scale. Let \(\mu_T\) and \(\mu_R\) be the treatment and reference population means, and let \(\sigma_T^2\) and \(\sigma_R^2\) be their population variances. Let \(\theta\) be the prespecified PBE constant. The criterion is that the treatment-reference mean difference squared, plus the difference in variances, not exceed \(\theta\sigma_R^2\). The calculation below uses independent normal treatment and reference samples. It is a transparent planning approximation, not a claim that every replicate or crossover design has these sampling distributions.
The margin \(h\) and all its terms are on the squared log scale. They are dimensionless when the original exposure measures are logged. The boundary \(h=0\) is the least favourable point of the null hypothesis for the usual one-sided test; if its actual type I error is controlled at \(\alpha\), power at that boundary is approximately \(\alpha\). Power is the probability of passing under specified true parameter values, not the probability that the criterion is true.
Estimator Distribution and Approximate Power
Suppose there are \(n\) independent observations in each group. Within each group, observations are normally distributed, with treatment and reference means \(\mu_T,\mu_R\) and variances \(\sigma_T^2,\sigma_R^2\), respectively. The two groups are independent. Write \(\bar{x}_T,\bar{x}_R\) for the sample means and \(s_T^2,s_R^2\) for the unbiased sample variances. The estimated margin is \(\hat{h}=(\bar{x}_T-\bar{x}_R)^2+s_T^2-(1+\theta)s_R^2\). Under these assumptions, the contrast \(D=\bar{x}_T-\bar{x}_R\) is normal with mean \(\delta=\mu_T-\mu_R\) and variance \(v=(\sigma_T^2+\sigma_R^2)/n\). For normal samples, each sample mean is independent of its sample variance, and the two groups are independent.
The squared contrast is not itself normally distributed. However, its mean and variance are available exactly. If \(D\sim N(\delta,v)\), then \(E(D^2)=\delta^2+v\) and \(\operatorname{Var}(D^2)=4\delta^2v+2v^2\). Also, \((n-1)s_j^2/\sigma_j^2\) has a chi-square distribution with \(n-1\) degrees of freedom, so \(\operatorname{Var}(s_j^2)=2\sigma_j^4/(n-1)\). These facts yield the mean and variance of \(\hat{h}\).
The \(v\) term means that \(\hat{h}\) is upward biased for \(h\): squaring a noisy estimated mean difference adds its variance. A simple correction is \(\hat{h}_c=\hat{h}-v\), which has expectation \(h\) when the planning variances are known. Its variance is the same as the variance of \(\hat{h}\). For the normal approximation below, use this bias-corrected estimator; this makes the power expression correspond directly to the population margin.
The last term includes \((1+\theta)^2\) because the reference sample variance is multiplied by \(-(1+\theta)\) in the margin. The two sample variance contributions are added, not subtracted: independent estimation errors contribute positive variances even though one component is subtracted in \(\hat{h}_c\).
For planning, approximate \(\hat{h}_c\) by a normal distribution with mean \(h\) and variance \(V_h\), using the specified population parameters to calculate \(SE_h\). The resulting approximation is most defensible when the variance estimates have adequate degrees of freedom and the distribution of the complete estimator is not dominated by a highly skewed squared-contrast term. For an upper one-sided confidence bound at level \(1-\alpha\), the approximate pass rule is \(\hat{h}_c+z_{1-\alpha}SE_h<0\), where \(z_{1-\alpha}\) is the corresponding standard normal quantile. Hence:
Here \(\Phi\) is the standard normal cumulative distribution function. This formula uses the true parameter values for planning and treats the standard error as fixed. In an actual analysis, variance components and the standard error are estimated; their randomness, design-specific covariance, and any prescribed confidence-bound procedure can change the achieved power.
Worked Calculations
Worked Example: Power When the Population Criterion Is Met
Suppose a hypothetical planning scenario has \(n=400\) participants per independent group, \(\theta=0.50\), \(\delta=0.04\), \(\sigma_T=0.28\), and \(\sigma_R=0.25\), all on the log scale. Take \(\alpha=0.05\), so \(z_{1-\alpha}=1.645\). First check that the assumed population values meet the criterion:
Thus the criterion is met under these assumed values. The variance of the estimated mean contrast is \(v=(0.28^2+0.25^2)/400=0.1409/400=0.00035225\). Substituting into each term of the variance formula gives:
Therefore the approximate power is \(\Phi[-1.645-(-0.01375/0.008796)]=\Phi(-0.0818)\approx 0.467\). The hypothetical design has about a 46.7% probability of passing under these parameter values. Meeting the population criterion does not guarantee high power; here the margin is not sufficiently far below zero relative to its sampling uncertainty.
Worked Example: A Mean Difference Outside the Criterion
Keep \(n=400\), \(\theta=0.50\), \(\sigma_T=0.28\), and \(\sigma_R=0.25\), but suppose the true mean difference is \(\delta=0.15\). The population margin is \(h=0.15^2+0.28^2-1.50(0.25^2)=0.0225+0.0784-0.09375=0.00715\), which is above zero. Under these values, the criterion is not met.
The mean-contrast variance remains \(v=0.00035225\). The mean-square contribution to \(V_h\) is \(4(0.15^2)(0.00035225)+2(0.00035225^2)=0.0000319507\). The sample-variance contributions are unchanged from the first example. Thus \(V_h=0.0000319507+0.00003081+0.00004406\approx0.00010682\), and \(SE_h\approx0.01034\).
The approximate probability of passing is \(\Phi[-1.645-(0.00715/0.01034)]=\Phi(-2.337)\approx0.0097\). This small probability is consistent with a true margin outside the PBE criterion. A nonzero chance of passing does not mean the criterion is met; it reflects sampling variability and the chosen one-sided test.
Worked Example: Greater Treatment Variability Can Reduce Power
Return to \(n=400\), \(\theta=0.50\), and \(\delta=0.04\), with \(\sigma_R=0.25\), but increase the hypothetical treatment standard deviation to \(\sigma_T=0.30\). The population margin is \(h=0.04^2+0.30^2-1.50(0.25^2)=0.0016+0.0900-0.09375=-0.00215\). It remains below zero, but is much closer to the boundary than in the first example.
Now \(v=(0.30^2+0.25^2)/400=0.1525/400=0.00038125\). The variance terms are:
The resulting power is \(\Phi[-1.645-(-0.00215/0.009348)]=\Phi(-1.415)\approx0.0785\). In this example, greater treatment variability both moves the population margin closer to zero and increases uncertainty. The calculation shows why variability can affect PBE power through the criterion and through the precision of its estimate.
How Sample Size and Components Affect Power
For fixed population parameters, increasing \(n\) reduces \(v\) and increases the degrees of freedom for estimating the variances. Those changes generally reduce \(SE_h\). If \(h<0\), a smaller standard error usually moves the power upward because the true margin is more clearly separated from the boundary. For example, in the first worked example, changing to \(n=900\) per group gives \(v=0.00015656\), \(V_h\approx0.00003428\), and \(SE_h\approx0.005855\). The approximate power is \(\Phi[-1.645-(-0.01375/0.005855)]=\Phi(0.703)\approx0.759\), compared with about 0.467 at \(n=400\). This is a planning comparison under the same model and population values, not a guarantee of a particular observed outcome.
The effect of changing a component should be checked in both places it enters: the true margin \(h\) and its variance \(V_h\). A larger absolute mean difference increases the squared-mean term in \(h\), and it increases \(4\delta^2v\) in the estimator variance. A larger treatment variance can move \(h\) toward or beyond zero while also increasing uncertainty. The reference variance affects the margin and the variance of its estimated contribution; it is also multiplied by the prespecified PBE constant. As a result, there is no general rule that changing one variance component affects power only through precision.
Diagnostics, Alternatives, and Limitations
Before using the approximation, check the assumptions and the arithmetic. Confirm that the parameter values, variance convention, and units are all on the same scale; verify that the stated values actually make \(h<0\) if the example is described as satisfying the criterion; and calculate each positive variance contribution separately before summing. A negative computed variance, or a power above one or below zero, indicates an error. Also check that increasing \(n\) produces a plausible reduction in the estimator variance. If a result seems counterintuitive, inspect both \(h\) and \(V_h\), not just the standard error.
The normal approximation can be poor with small samples, highly unequal variances, or parameter values for which the squared mean contrast makes the estimator strongly skewed. It also assumes independent groups and does not model within-subject correlation, sequence or period effects, dropout, or the details of a replicate crossover analysis. Those features change the joint sampling distribution of the mean and variance estimators, so the independent-groups formula should not be transferred unchanged to such a design.
An alternative is to calculate the pass probability from the full sampling distribution implied by the design. Under the independent normal model, the squared normal contrast and the independent scaled chi-square variance estimates can be combined numerically. A design-specific simulation can likewise generate data under specified parameters, apply the planned analysis and pass rule to each simulated study, and estimate power by the fraction that pass. These approaches can better reflect skewness or estimated standard errors, but require the correct sampling model and analysis rule. The normal approximation is useful for transparent planning and sensitivity checks; it is not a substitute for validating a complex design.
Check Your Understanding
Use the definitions and approximate power formula above. Unless a question says otherwise, use a one-sided level of 0.05, so \(z_{1-\alpha}=1.645\). Solutions are included for checking.
- For \(\delta=0\), \(\sigma_T=0.25\), \(\sigma_R=0.25\), and \(\theta=0.50\), is the population PBE criterion met? Calculate \(h\).
- Why does \(V_h\) contain \(2v^2\) as well as a term involving \(\delta^2v\)?
- In the stated model, what happens to the reference sample-variance contribution to \(V_h\) if \(\theta\) increases while all other quantities remain fixed?
- At the boundary \(h=0\), what is the approximate power of the one-sided test under the normal approximation?
Solutions. (1) \(h=0^2+0.25^2-1.50(0.25^2)=-0.03125\), so the criterion is met. (2) For \(D\sim N(\delta,v)\), the exact variance of \(D^2\) is \(4\delta^2v+2v^2\); the \(2v^2\) term remains even when \(\delta=0\). (3) The contribution is \(2(1+\theta)^2\sigma_R^4/(n-1)\), so it increases as \(\theta\) increases for nonnegative \(\theta\) and positive \(\sigma_R^2\). (4) Substituting \(h=0\) gives \(\Phi(-1.645)\approx0.050\), the nominal one-sided boundary probability under this approximation.