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Confidence Intervals and the TOST Procedure · Tutorial 209 of 399

The Intersection-Union Logic of TOST

Learn how the union of the two nonequivalence regions determines TOST’s decision rule and its type I error control.

Intermediate 10 min read

What You'll Learn

  • Express the equivalence alternative and its null as parameter sets.
  • Relate the overall equivalence null to the two one-sided component nulls.
  • Explain why both one-sided tests must reject before concluding equivalence.
  • Show why testing each component at level alpha controls the overall type I error without splitting alpha.
  • Distinguish failure to establish equivalence from evidence of nonequivalence.

Why TOST Requires Two Successful Tests

An equivalence analysis asks whether an unknown treatment contrast is sufficiently close to zero, not merely whether the data fail to show a difference. That distinction determines the logic of the test. A contrast can be too low, too high, or within the acceptable range; to support equivalence, the data must rule out both ways of being outside that range.

Let \(\theta\) denote the population parameter of interest, and let \(L\) and \(U\) be prespecified lower and upper equivalence bounds, with \(L<U\). For example, \(\theta\) could be a population mean difference or a mean log treatment contrast. The parameter is equivalent when it lies strictly between the bounds. TOST, the two one-sided tests procedure, tests whether the data support that condition by testing each boundary separately.

Definition: The equivalence alternative is \(H_a: L<\theta<U\). Its null is the complement of that interval: \(H_0: \theta\leq L\) or \(\theta\geq U\). Define the component nulls \(H_{01}:\theta\leq L\) and \(H_{02}:\theta\geq U\). Then \(H_0=H_{01}\cup H_{02}\), while \(H_a=\{\theta>L\}\cap\{\theta<U\}\).

The word “or” in the null is inclusive: a value at or below \(L\), a value at or above \(U\), or both would be in the null. Since \(L<U\), no parameter value can be in both component nulls at once, but every nonequivalent value is in at least one of them. Conversely, a parameter is in the equivalence alternative only if it is above \(L\) and below \(U\).

From Parameter Sets to a Decision Rule

The lower-bound test asks whether there is evidence that \(\theta>L\), against \(H_{01}:\theta\leq L\). The upper-bound test asks whether there is evidence that \(\theta<U\), against \(H_{02}:\theta\geq U\). Each test is one-sided and uses its corresponding boundary as the null boundary. At a chosen significance level \(\alpha\), the overall TOST decision is to claim equivalence only if both component nulls are rejected at level \(\alpha\).

This conjunction follows directly from the alternative. Rejecting \(H_{01}\) supports the lower condition, \(\theta>L\); rejecting \(H_{02}\) supports the upper condition, \(\theta<U\). Neither condition alone is enough. If the lower test rejects but the upper test does not, the results have not ruled out a parameter at or above \(U\). If the upper test rejects but the lower test does not, they have not ruled out a parameter at or below \(L\).

Key result: TOST rejects the overall null \(H_0=H_{01}\cup H_{02}\) if and only if it rejects both component nulls. Failure to reject either component null means that equivalence has not been established by TOST.

This is an instance of an intersection-union test. The alternative is an intersection of two requirements, so evidence must satisfy both. Equivalently, the null is a union of two possibilities, so rejecting the overall null requires ruling out every possibility in that union. The name describes the set logic; it does not require the two test statistics to be independent.

Why the Overall Type I Error Is Controlled

A type I error for the overall equivalence test occurs when the procedure claims equivalence even though \(\theta\) is outside the equivalence interval. Suppose each component test is valid at level \(\alpha\): under every parameter value in its own null, its probability of rejection is no greater than \(\alpha\). To claim equivalence, the procedure must reject both component nulls. If the overall null is true, at least one component null is true. The event of rejecting both is contained in the event of rejecting that true component null.

$$ \begin{aligned} H_0 \text{ true} &\implies H_{01}\text{ true or }H_{02}\text{ true},\\ \{\text{reject }H_{01}\text{ and reject }H_{02}\} &\subseteq \{\text{reject a true component null}\},\\ P_\theta(\text{claim equivalence}) &\leq \alpha. \end{aligned} $$

More explicitly, if \(\theta\leq L\), then \(H_{01}\) is true, and \(P_\theta(\text{reject both})\leq P_\theta(\text{reject }H_{01})\leq\alpha\). If \(\theta\geq U\), then \(H_{02}\) is true, and \(P_\theta(\text{reject both})\leq P_\theta(\text{reject }H_{02})\leq\alpha\). These cases cover the entire overall null. Thus the overall type I error is controlled at \(\alpha\), regardless of the dependence between the two tests.

The argument relies on valid component tests. Their test statistics, standard errors, degrees of freedom, and reference distributions must be appropriate for the model and data; an incorrect component test does not become valid merely because it is paired with another test. The margins must also be specified before interpreting the test, and the parameter and margins must use the same scale and units. For example, if \(\theta\) is a mean log contrast, then both bounds must be stated on that log scale.

Worked TOST Decisions

The following examples use hypothetical log-contrast estimates and illustrative margins, not data or margins from a real study. To keep the arithmetic transparent, suppose each estimate \( \hat{\theta} \) follows a normal model with known standard error \(SE\). The component test p-values are then calculated from the standard normal distribution. In applications, the valid model-based tests may use other appropriate reference distributions.

Worked Example: Both Boundary Nulls Are Rejected

Suppose the illustrative bounds are \(L=-0.10\) and \(U=0.10\) log units, the estimate is \(\hat{\theta}=0.02\) log units, and \(SE=0.04\) log units. Let \(\alpha=0.05\). Under the stated normal model, calculate the one-sided statistics and p-values for both tests.

For the lower-bound test \(H_{01}:\theta\leq -0.10\) against \(\theta>-0.10\), the standardized statistic at the boundary is \(z_L=(\hat{\theta}-L)/SE=(0.02-(-0.10))/0.04=0.12/0.04=3.00\). The one-sided p-value is \(1-\Phi(3.00)\approx 0.00135\), where \(\Phi\) is the standard normal cumulative distribution function.

For the upper-bound test \(H_{02}:\theta\geq 0.10\) against \(\theta<0.10\), the statistic is \(z_U=(\hat{\theta}-U)/SE=(0.02-0.10)/0.04=-0.08/0.04=-2.00\). The lower-tail p-value is \(\Phi(-2.00)\approx 0.0228\).

Both p-values are below 0.05, so both component nulls are rejected. Under this procedure, the result supports equivalence within the stated illustrative bounds. The conclusion is not that the true contrast is exactly zero; it is that the two one-sided tests reject the two specified forms of nonequivalence.

Worked Example: One Successful Test Is Not Enough

Keep \(L=-0.10\), \(U=0.10\), \(SE=0.04\), and \(\alpha=0.05\), but suppose the hypothetical estimate is \(\hat{\theta}=0.08\) log units. Check whether each condition for equivalence is supported.

For the lower-bound test, \(z_L=(0.08-(-0.10))/0.04=0.18/0.04=4.50\). Its upper-tail p-value is \(1-\Phi(4.50)\approx 0.0000034\), so the lower component null is rejected.

For the upper-bound test, \(z_U=(0.08-0.10)/0.04=-0.02/0.04=-0.50\). Its lower-tail p-value is \(\Phi(-0.50)\approx 0.3085\), which is not below 0.05. The upper component null is not rejected.

Since only one test rejects, TOST does not establish equivalence. This is not proof that the contrast is outside the acceptable range. It means that, with these data and this decision rule, the evidence did not rule out the upper nonequivalence region.

Worked Example: Each Test Uses the Full Significance Level

Suppose \(L=-0.10\), \(U=0.10\), \(\hat{\theta}=0.00\) log units, \(SE=0.05\) log units, and \(\alpha=0.05\), again under the illustrative known-standard-error normal model. Determine whether either p-value must be below \(0.025\).

The lower statistic is \(z_L=(0.00-(-0.10))/0.05=2.00\), giving a one-sided p-value \(1-\Phi(2.00)\approx 0.0228\). The upper statistic is \(z_U=(0.00-0.10)/0.05=-2.00\), giving a one-sided p-value \(\Phi(-2.00)\approx 0.0228\).

Both tests reject at 0.05, so TOST supports equivalence under these assumptions. Neither p-value is below 0.025, yet that does not prevent the TOST decision: the component tests are each compared with \(\alpha=0.05\), and both must pass. The error-control argument above—not a split of \(\alpha\) between the tests—is what justifies this rule.

Common Mistakes, Diagnostics, and Alternatives

A frequent error is to say that equivalence is supported when either one-sided test rejects. That reverses the logic: rejecting only one component leaves the other nonequivalence region unruled out. The correction is to check that both component nulls are rejected at the prespecified level. A second error is to divide \(\alpha\) by two automatically. TOST is not a procedure that rejects the overall null when either of two tests rejects; it requires both, and its type I error is controlled by the union-null argument.

When reviewing an analysis, first check that the contrast and both margins are on the same scale, that \(L<U\), and that the one-sided alternatives point in the correct directions. Then check that each component test is calibrated for the model used and that the decision requires both p-values to meet the stated threshold. A swapped tail, reversed inequality, or an “either test” decision can change the conclusion. If one test fails, report that equivalence was not demonstrated; do not convert that result into a claim of nonequivalence or equality.

These hypotheses answer a different question from a conventional test of equality, \(H_0:\theta=0\). A failure to reject equality does not show that \(\theta\) lies inside the equivalence bounds; it may reflect imprecise data. Conversely, TOST is designed to rule out values outside prespecified bounds. Another possible approach is to allocate a smaller error rate to each component test, but this is not needed for the intersection-union decision rule and can make it more conservative. Confidence-interval expressions of the TOST decision are a related way to present results, but they do not change the underlying union-null logic.

The method has limits. Its conclusion is relative to the chosen parameter, margins, model, and significance level; it does not establish equivalence on every outcome or scale. It also does not repair bias, poor measurements, inappropriate exclusions, or invalid standard errors. In crossover analyses, for example, the component tests rely on the treatment contrast and its uncertainty being estimated appropriately under the selected model. The diagnostic and model-assessment work remains essential to interpreting the two tests.

Key takeaway: Nonequivalence means \(\theta\leq L\) or \(\theta\geq U\), so the overall null is a union. Equivalence requires both \(\theta>L\) and \(\theta<U\), so TOST must reject both component nulls. If each component test has size at most \(\alpha\), this conjunction controls the overall type I error at \(\alpha\).

Check Your Understanding

Answer each question using the intersection-union logic. The solutions follow the questions.

  1. Write the overall null and alternative when the equivalence bounds are \(L\) and \(U\).
  2. At \(\alpha=0.05\), one component p-value is 0.01 and the other is 0.12. What is the TOST conclusion?
  3. Why does testing both component nulls at level \(\alpha\) not require an automatic Bonferroni split to \(\alpha/2\)?
  4. If the true parameter is \(\theta=L\), which component null is true, and why does the overall type I error remain controlled?
  5. Does failure to reject one component null establish that the true parameter is outside the equivalence interval?

Solutions. (1) \(H_0:\theta\leq L\) or \(\theta\geq U\), and \(H_a:L<\theta<U\). (2) Equivalence is not established, because both component tests must reject at 0.05. (3) The overall rejection event requires both rejections; whenever the overall null is true, at least one component null is true, and the event of rejecting both is contained in the event of rejecting that true component. Its probability is therefore no greater than \(\alpha\). (4) \(H_{01}:\theta\leq L\) is true. Claiming equivalence requires rejecting \(H_{01}\), so the probability of a false equivalence claim is no greater than the level-\(\alpha\) rejection probability of that true test. (5) No. It means equivalence was not demonstrated; the test alone does not prove the parameter is outside the interval.