From Equivalence Limits to a Precision Requirement
An average bioequivalence conclusion depends on both the estimated test-to-reference ratio and its uncertainty. A confidence interval can be narrow yet fail to establish equivalence if its center is too close to, or beyond, a limit. Conversely, an estimate near the center of the acceptable range can tolerate a wider interval. Translating the limits into a precision requirement makes this relationship explicit.
The quantity of interest is commonly the geometric mean ratio (GMR) for a pharmacokinetic endpoint such as AUC or Cmax. In the standard log-scale analysis, the estimated treatment contrast is exponentiated to obtain the GMR. The limits and confidence interval must be compared on the same scale: either compare log-scale bounds with log-scale endpoints, or compare ratio-scale bounds with ratio-scale endpoints.
For illustration, this tutorial uses the familiar ratio limits \(L=0.80\) and \(U=1.25\). These are examples of limits, not a claim that the same limits apply to every product or regulatory context. On the log scale they are symmetric about zero:
Assume the treatment contrast has a valid, symmetric \(100(1-2\alpha)\%\) confidence interval of the form \(\hat{\theta}\pm c\,SE\), where \(SE\) is the standard error and \(c\) is the appropriate critical value. With a t-based interval, \(c=t_{1-\alpha,df}\). For the usual two one-sided tests at \(\alpha=0.05\), the corresponding interval is a 90% confidence interval, so \(c=t_{0.95,df}\). The t-based calculation assumes that the estimated contrast and standard error come from an appropriate analysis model and that its stated t inference is justified. If the interval is asymmetric, use its actual endpoints rather than this symmetric formula.
The Allowable Half-Width
For the confidence interval to fit within the equivalence limits, its lower endpoint must be at least \(\ell\), and its upper endpoint must be at most \(u\):
Rearranging each inequality gives an upper bound on the interval half-width. Both conditions must hold, so the binding one is the smaller of the two distances from the estimate to the limits.
This is the precision requirement for a symmetric interval centered at \(\hat{\theta}\). It is meaningful only if the estimate is itself inside the limits; otherwise at least one distance is negative and no nonnegative half-width can make the interval fit. When limits are symmetric on the log scale and the estimate is exactly at their midpoint, the allowable half-width is greatest. Moving the estimate toward either limit reduces the available precision margin.
Why the Confidence-Interval Criterion Matches the Tests
The equivalence hypotheses can be expressed as \(H_0:\theta\leq\ell\) or \(\theta\geq u\), against \(H_a:\ell<\theta<u\). The null is a union of two non-equivalence possibilities. The two one-sided tests each address one boundary: the lower-bound test asks whether there is evidence that \(\theta>\ell\), and the upper-bound test asks whether there is evidence that \(\theta<u\).
At level \(\alpha\), the lower one-sided t test rejects its boundary null when the one-sided lower confidence bound exceeds \(\ell\). That bound is \(\hat{\theta}-t_{1-\alpha,df}SE\). The upper one-sided test rejects its boundary null when the one-sided upper confidence bound is below \(u\), namely when \(\hat{\theta}+t_{1-\alpha,df}SE<u\). Requiring both rejections therefore requires both endpoints of the corresponding \(100(1-2\alpha)\%\) two-sided interval to fall within the limits. This establishes the interval criterion under the stated t-based inference; it is not a separate rule that can be applied regardless of how the interval was constructed.
The comparison can be made on the ratio scale as well. Exponentiation preserves order, so the log-scale interval \([\hat{\theta}-h,\hat{\theta}+h]\) becomes \([\hat{R}\exp(-h),\hat{R}\exp(h)]\), where \(h=c\,SE\). Its containment condition is equivalent to the log-scale condition. The ratio-scale interval is multiplicatively, not additively, symmetric around \(\hat{R}\).
Worked Example: Maximum Precision at a GMR of 1.00
Suppose a hypothetical average bioequivalence analysis uses limits of 0.80 and 1.25, has \(df=24\), and produces an estimated GMR of 1.00. Find the maximum allowable standard error for a 90% t confidence interval, then check a proposed \(SE=0.120\).
The log-scale estimate is \(\hat{\theta}=\ln(1.00)=0\). The distance to either log-scale limit is \(0.22314\). For \(df=24\), \(t_{0.95,24}\) is approximately 1.711. Therefore:
The proposed standard error, 0.120, is below 0.1304. Its log-scale half-width is \(1.711(0.120)=0.2053\), giving endpoints \(-0.2053\) and \(0.2053\). Exponentiating gives a 90% interval of approximately \([0.8144,1.228]\), which lies inside \([0.80,1.25]\). Under the stated assumptions, this interval meets the equivalence criterion.
Worked Example: A GMR Closer to the Lower Limit
In a second hypothetical analysis, the estimated GMR is 0.95, \(df=30\), and the same ratio limits apply. Find the maximum allowable \(SE\) for the 90% interval. Then assess an observed \(SE=0.095\).
The log estimate is \(\hat{\theta}=\ln(0.95)=-0.05129\). Its distances to the lower and upper log limits are:
The lower limit is closer, so it controls the precision requirement. With \(t_{0.95,30}\approx1.697\):
The observed \(SE=0.095\) is smaller than this maximum. The half-width is \(1.697(0.095)=0.1612\), so the log interval is approximately \([-0.2125,0.1099]\). Exponentiation gives approximately \([0.8085,1.116]\), within the limits. The maximum standard error is smaller here than in the first example because the estimated ratio is closer to the lower boundary, even though the equivalence limits are unchanged.
Worked Example: Checking an Interval and Its Remaining Margin
A hypothetical study reports a 90% confidence interval for the GMR of \([0.84,1.18]\), with ratio limits of 0.80 and 1.25. Check equivalence and quantify how much log-scale half-width could be added before the interval would touch a limit.
The reported ratio interval is inside the limits because \(0.84\geq0.80\) and \(1.18\leq1.25\). On the log scale, the endpoints are \(\ln(0.84)=-0.1744\) and \(\ln(1.18)=0.1655\). The interval center and half-width are:
The estimate’s available half-width is the smaller distance to a log limit:
The observed half-width is below the maximum by approximately \(0.2187-0.1699=0.0488\) log units. Thus the interval satisfies the criterion, with the lower limit being the limiting boundary. This calculation describes the margin for this interval; it does not change the confidence level or imply a probability that the true GMR lies inside it.
Using the Requirement Carefully
A precision requirement is a constraint on uncertainty for a specified estimate or planning value. Before data collection, a planning calculation may use an anticipated GMR to determine how small the standard error needs to be. After data collection, the actual estimate and its interval endpoints determine whether equivalence has been demonstrated. The earlier material on defining the power target for equivalence studies addresses a related but distinct planning question: power concerns the probability of meeting a decision rule under specified conditions, whereas the precision inequality here describes the interval width needed at a particular center.
Several checks help detect a misapplied requirement:
- Check that the estimate is inside the limits. A narrow interval centered outside a limit cannot establish equivalence.
- Use the correct critical value. The confidence level and degrees of freedom determine \(c\); changing either changes the allowable \(SE\).
- Keep scales consistent. Compare log endpoints with \(\ell,u\), or exponentiated endpoints with \(L,U\). Do not compare a log estimate directly with ratio limits.
- Use both boundaries. The nearer limit determines the maximum half-width; checking only one endpoint can incorrectly label an interval equivalent.
- Use actual endpoints for asymmetric intervals. The half-width formula relies on a symmetric interval around \(\hat{\theta}\).
The log-scale formulation is often convenient because the contrast is additive and a symmetric log interval exponentiates to a multiplicatively symmetric ratio interval. A direct ratio-scale endpoint check is equally valid when the interval has been constructed appropriately. A width-only rule, such as requiring an interval to be “narrow enough” without considering its center, is not an alternative equivalence criterion: the same width can pass or fail depending on where the interval lies. The calculations also rely on the limits and analysis model being specified in advance and appropriate to the endpoint; they do not establish that those choices are clinically or regulatorily justified.
In practice, the required standard error is a useful bridge from an equivalence margin to study precision, but it does not by itself determine the number of participants. That requires additional design and variability information. Nor does satisfying a planned precision target guarantee that the final interval will fit within the limits: the realized estimate may be farther from the anticipated center, or the realized uncertainty may differ from its planning value.
Check Your Understanding
Answer each question, then compare your work with the solutions.
- With log limits \(-0.20\) and \(0.20\), an estimate of \(0.05\), and critical value \(c=1.70\), what is the maximum allowable \(SE\)?
- For a 90% interval with ratio limits \([0.80,1.25]\), an estimated GMR of 0.78 can it meet the symmetric-interval equivalence criterion at any nonnegative standard error?
- A log-scale 90% interval is \([-0.10,0.18]\), with log limits \([-0.223,0.223]\). Does it meet the interval criterion? Which boundary is closer to its interval endpoint?
- Why does a 90% confidence interval correspond to two one-sided tests at \(\alpha=0.05\), under the standard t-based equivalence procedure?
Solutions. (1) The distances to the limits are \(0.05-(-0.20)=0.25\) and \(0.20-0.05=0.15\), so \(h_{\max}=0.15\). Thus \(SE_{\max}=0.15/1.70=0.0882\). (2) No. Since \(0.78<0.80\), the estimate is outside the lower limit; a symmetric interval centered there cannot be wholly inside the limits. (3) Yes: \(-0.10\geq-0.223\) and \(0.18\leq0.223\). The upper endpoint is \(0.223-0.18=0.043\) from the upper boundary, while the lower endpoint is \(-0.10-(-0.223)=0.123\) from the lower boundary, so the upper boundary is closer. (4) Each one-sided test at level 0.05 uses a one-sided 95% confidence bound. Requiring both bounds to lie inside the equivalence limits is equivalent to requiring the corresponding two-sided 90% interval, with endpoints formed using \(t_{0.95,df}\), to lie entirely inside them.