Why Equivalence Needs Two Boundaries
A treatment comparison can be statistically inconclusive without showing that the treatments are sufficiently similar. For example, failing to detect a difference does not establish that any difference is small enough to be unimportant. Equivalence testing asks a more targeted question: is the treatment contrast confined to a range judged acceptable in advance?
That range has two boundaries. A contrast below the lower boundary is too small in one direction; a contrast above the upper boundary is too large in the other. The test must therefore address both possibilities. This tutorial develops the hypotheses for those two boundary tests and explains what their outcomes mean.
For a bioavailability comparison, \(\theta\) may be a treatment contrast estimated from a log-scale crossover model. In that case, the parameter and margins are expressed on that log scale. The same hypothesis structure applies to another parameter, provided its scale, direction, and margins are clearly defined. Choosing scientifically and regulatorily justified margins is a separate design decision; the margins must not be selected after looking at the estimated treatment effect.
Formulating the Two One-Sided Tests
To support equivalence, the data must rule out contrasts at or below \(L\), and also rule out contrasts at or above \(U\). Each test treats one of these non-equivalent regions as its null hypothesis.
The inequality directions matter. For the lower-bound test, a value far below \(L\) is consistent with its null, while evidence sufficiently above \(L\) supports its alternative. For the upper-bound test, a value far above \(U\) is consistent with its null, while evidence sufficiently below \(U\) supports its alternative. Reversing either alternative would make that test answer the wrong question.
In this terminology, \(H_{0L}\) and \(H_{0U}\) are the two component null hypotheses. The possibility that \(\theta\) lies outside the equivalence region is captured by at least one of them: \(\theta\leq L\) or \(\theta\geq U\). The corresponding component alternatives each point toward the interior, but each addresses only one boundary. That is why a favorable result for just one test cannot establish equivalence.
Model and Test Decisions
The hypotheses specify the scientific claim; a statistical model supplies uncertainty for estimating \(\theta\). For illustration, suppose an analysis provides an estimate \(\hat{\theta}\), a standard error \(SE\), and degrees of freedom \(df\). Under the usual linear-model conditions, the standardized estimate relative to a boundary follows a \(t\) distribution with \(df\) degrees of freedom when the parameter is at that boundary. These conditions include an appropriate model for the treatment contrast, independent subject-level errors, and a residual variance estimate that gives a valid standard error. For a standard crossover analysis, the treatment estimate and uncertainty must come from the specified crossover model, with its assumptions checked using the relevant diagnostics.
With \(t_{1-\alpha,df}\) denoting the \(1-\alpha\) quantile of the \(t\) distribution, the lower-bound test rejects when the estimate is sufficiently above \(L\). The upper-bound test rejects when it is sufficiently below \(U\):
These are one-sided tests, each evaluated at the chosen one-sided level \(\alpha\). The first statistic measures how far the estimate is above the lower margin in standard-error units. The second measures how far it is below the upper margin. A large positive value supports the corresponding alternative. Equivalently, each test can be expressed using a one-sided \(p\)-value, with rejection when that \(p\)-value is below \(\alpha\).
Why Both Rejections Are Required
The parameter is in the equivalence region only if it satisfies both inequalities defining that region. A short logical check makes the two-test requirement clear. If the lower-bound null remains plausible, the data have not ruled out \(\theta\leq L\); if the upper-bound null remains plausible, the data have not ruled out \(\theta\geq U\). Either unresolved possibility is incompatible with a supported equivalence conclusion.
Thus the two alternatives must both be supported: the lower test must reject \(H_{0L}\), and the upper test must reject \(H_{0U}\). If either test fails to reject, the analysis has not established equivalence. This does not prove that the parameter is outside the margins; it means the evidence is insufficient to rule out at least one non-equivalent region.
Worked Example: Both Tests Support Equivalence
Suppose a hypothetical crossover study estimates a treatment contrast on the log-AUC scale. The prespecified margins are \(L=-0.10\) and \(U=0.10\) log units. The fitted model gives \(\hat{\theta}=0.020\), \(SE=0.025\) log units, and \(df=23\). Use \(\alpha=0.05\).
The hypotheses are \(H_{0L}:\theta\leq-0.10\) versus \(H_{aL}:\theta>-0.10\), and \(H_{0U}:\theta\geq0.10\) versus \(H_{aU}:\theta<0.10\). For \(df=23\), the one-sided 0.05 critical value is approximately \(1.714\).
For the lower-bound test:
Since \(4.80>1.714\), reject \(H_{0L}\). For the upper-bound test:
Since \(3.20>1.714\), reject \(H_{0U}\) as well. Both component null hypotheses are rejected at the 0.05 one-sided level, so the results support equivalence within the stated margins. This conclusion concerns the log-AUC contrast and these specific margins; it is not a claim that the estimated contrast is exactly zero.
Worked Examples at the Boundaries
Worked Example: A Central Estimate With Adequate Precision
Consider a hypothetical treatment contrast measured in log units, with margins \(L=-0.15\) and \(U=0.15\). The estimate is \(\hat{\theta}=-0.040\), \(SE=0.050\), and \(df=18\). At \(\alpha=0.05\), the one-sided critical value is approximately \(1.734\).
For the lower test, the statistic is \(( -0.040-(-0.15))/0.050=0.110/0.050=2.20\). Because \(2.20>1.734\), reject \(H_{0L}:\theta\leq-0.15\).
For the upper test, the statistic is \((0.15-(-0.040))/0.050=0.190/0.050=3.80\). Because \(3.80>1.734\), reject \(H_{0U}:\theta\geq0.15\). Both tests reject, so the estimate and its precision provide evidence of equivalence within the prespecified log-unit margins.
The estimate is not exactly zero, nor does that matter: equivalence concerns whether the parameter is within the accepted region, not whether it equals zero. Precision matters because the tests compare the estimate with each boundary relative to its standard error.
Worked Example: One Boundary Still Cannot Be Ruled Out
Now suppose a hypothetical log-scale contrast has margins \(L=-0.10\), \(U=0.10\), estimate \(\hat{\theta}=0.060\), \(SE=0.040\), and \(df=15\). The one-sided 0.05 critical value is approximately \(1.753\).
The lower-bound statistic is \((0.060-(-0.10))/0.040=0.160/0.040=4.00\). Since \(4.00>1.753\), reject \(H_{0L}\). The upper-bound statistic is \((0.10-0.060)/0.040=0.040/0.040=1.00\). Since \(1.00<1.753\), do not reject \(H_{0U}\).
The estimate itself lies between the margins, but the upper-bound test does not rule out a parameter at or above \(U\) with the required level of evidence. Therefore, equivalence is not established. The correct conclusion is not that the treatments are known to be non-equivalent; rather, this analysis has not demonstrated that the upper margin is satisfied.
Assumptions, Diagnostics, and Alternatives
The two-test structure does not repair a poorly defined estimand or an invalid standard error. Before interpreting either test, check that the treatment contrast, analysis population, scale, and margins match the scientific question. The margins should be justified and fixed independently of the observed treatment estimate. The chosen model must also provide a defensible estimate and uncertainty for that contrast.
In crossover work, diagnostics for residual patterns, influential subjects, unequal variance, and the suitability of the selected model can affect the credibility of \(SE\) and \(df\). A test statistic may be computed even when model assumptions are questionable, but its nominal significance level may then be unreliable. In particular, an estimate close to a boundary is not automatically evidence of equivalence: uncertainty must be small enough for the corresponding one-sided test to reject.
Equivalence testing answers a different question from a conventional test of no treatment difference. A conventional test commonly uses \(H_0:\theta=0\); failing to reject that null does not show that \(\theta\) lies between \(L\) and \(U\). A noninferiority test usually addresses one prespecified direction and one margin. It does not, by itself, rule out effects beyond a second boundary. The two one-sided tests are appropriate when the claim requires both limits to be respected.
Common Mistakes
- Using “no significant difference” as evidence of equivalence. A test against zero can fail because the estimate is imprecise. Equivalence requires evidence against both boundary null hypotheses.
- Rejecting only one component null. Passing the lower test does not establish that the upper margin is met, and vice versa. Both must be rejected.
- Reversing an inequality. The lower alternative is \(\theta>L\); the upper alternative is \(\theta<U\). These directions point inward from their respective non-equivalent regions.
- Treating failure to reject as proof of non-equivalence. Failure to reject means the data have not ruled out that boundary null. It does not establish that the parameter lies outside the equivalence region.
- Choosing margins after seeing the estimate. Data-dependent margins undermine the intended interpretation. Margins should be justified and prespecified.
Check Your Understanding
For each question, identify the relevant hypothesis or decision. Brief solutions follow the questions.
- Write the lower- and upper-bound hypotheses when the equivalence region is \(-0.08<\theta<0.08\).
- At \(\alpha=0.05\), the lower test rejects and the upper test does not reject. What conclusion is justified?
- Explain why failing to reject \(H_0:\theta=0\) does not establish equivalence.
- With \(L=-0.12\), \(U=0.12\), \(\hat{\theta}=0.010\), and \(SE=0.030\), calculate both standardized boundary statistics. Is equivalence supported if the critical value is \(1.70\)?
Solutions. (1) \(H_{0L}:\theta\leq-0.08\) versus \(H_{aL}:\theta>-0.08\); \(H_{0U}:\theta\geq0.08\) versus \(H_{aU}:\theta<0.08\). (2) Equivalence is not established because both null hypotheses must be rejected. (3) A nonsignificant test against zero may reflect uncertainty and does not rule out parameter values outside either equivalence margin. (4) The lower statistic is \((0.010-(-0.12))/0.030=0.130/0.030=4.33\); the upper statistic is \((0.12-0.010)/0.030=0.110/0.030=3.67\). Both exceed \(1.70\), so both boundary null hypotheses are rejected and equivalence is supported under the stated model and margins.