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Confidence Intervals and the TOST Procedure · Tutorial 239 of 399

Distinguishing Joint TOST Decisions From Multiplicity Adjustments

You will learn when requiring every endpoint to pass TOST controls the joint error rate without an endpoint-count adjustment, and when multiplicity adjustment is needed instead.

Intermediate 12 min read

What You'll Learn

  • State the global null and alternative when equivalence is required for every endpoint.
  • Explain why valid endpoint-level TOSTs at level alpha control the joint decision without an independence assumption.
  • Combine endpoint TOST p-values into a joint intersection-union p-value.
  • Distinguish an all-endpoints-required claim from a claim supported by any one positive result.
  • Identify when endpoint selection, separate claims, or simultaneous coverage changes the multiplicity question.
  • Diagnose common errors in choosing or interpreting multiplicity adjustments.

Two Different Questions About Several Endpoints

A bioequivalence analysis may evaluate more than one exposure endpoint, such as AUC and Cmax. The multiplicity question depends first on the scientific decision rule. If the conclusion is permitted only when every specified endpoint is equivalent, the overall claim is a conjunction. If a conclusion can be made when any one endpoint gives a positive result, the claim is a disjunction. These decision rules have different null hypotheses and different error-control requirements.

For an endpoint \(k\), let \(\theta_k\) denote its contrast on the analysis scale, and let \(L_k\) and \(U_k\) be its prespecified lower and upper equivalence limits, with \(L_k<U_k\). For a ratio endpoint, the contrast may be on the log scale, with the limits expressed on that same scale. The endpoint is equivalent when \(L_k<\theta_k<U_k\). TOST tests the two one-sided boundary hypotheses for that endpoint; the endpoint passes only if both are rejected at the stated level.

Definition: An intersection-union test (IUT) tests a global alternative that requires all of several component alternatives to hold. The global test rejects only when every component test rejects. By contrast, a procedure for a claim supported by any one of several positive results must account for the opportunity to obtain a false positive from at least one test.

Why the All-Endpoints Decision Needs No Endpoint-Count Adjustment

Suppose there are \(m\) prespecified endpoints, and the intended conclusion is that all are equivalent. For endpoint \(k\), define its null as non-equivalence: at least one of its two boundaries is not crossed into the equivalence region. The global null is that at least one endpoint is non-equivalent; the global alternative is that every endpoint is equivalent.

$$ H_{0,\mathrm{global}}=\bigcup_{k=1}^{m}H_{0k}, \qquad H_{a,\mathrm{global}}=\bigcap_{k=1}^{m}H_{ak}. $$

The global decision rejects \(H_{0,\mathrm{global}}\) only when every endpoint-specific TOST rejects its endpoint null. Assume each endpoint test is valid at level \(\alpha\): whenever its endpoint null is true, its probability of rejection is no greater than \(\alpha\). No assumption that endpoints are independent is needed.

$$ \begin{aligned} H_{0,\mathrm{global}}\text{ true} &\implies \text{there is at least one true }H_{0j},\\ \{\text{reject }H_{0,\mathrm{global}}\} &=\{\text{reject every }H_{0k}\} \subseteq \{\text{reject }H_{0j}\},\\ P(\text{reject }H_{0,\mathrm{global}}) &\leq P(\text{reject }H_{0j})\leq\alpha. \end{aligned} $$

This event-inclusion argument is the central justification: a false declaration that all endpoints are equivalent requires falsely rejecting at least one true endpoint null. The global false-positive probability is therefore bounded by the level of that true component test. The bound holds for any dependence structure among endpoints, including strong correlation from measurements on the same participants.

The same logic applies within a single TOST. Equivalence requires rejecting both one-sided boundary nulls, so a false equivalence decision requires rejecting a true boundary null. TOST uses level \(\alpha\) for each one-sided test; it does not divide \(\alpha\) by two for those two required rejections. The multi-endpoint conjunction extends that logic: all required tests must pass.

If \(p_k\) is the endpoint-specific TOST p-value, defined as the larger of its two one-sided p-values, then the joint IUT p-value is the largest endpoint p-value:

$$ p_{\mathrm{joint}}=\max_{1\leq k\leq m}p_k. $$

The joint decision is to declare all endpoints equivalent when \(p_{\mathrm{joint}}\leq\alpha\), equivalently when every \(p_k\leq\alpha\). This does not mean each endpoint is individually “more significant” after a correction. It means the global alternative itself requires every component to pass.

Worked Decisions With Multiple Endpoints

Worked Example: Both Exposure Endpoints Must Pass

Hypothetical data. Suppose AUC and Cmax are analyzed as log test-to-reference ratios, with prespecified ratio limits 0.80 and 1.25. On the log scale, these are \(L=\log(0.80)=-0.2231\) and \(U=\log(1.25)=0.2231\). Assume each estimated contrast is approximately normal with its stated standard error, the standard errors are valid, the margins were set independently of these results, and the two one-sided tests use the standard normal reference distribution. The endpoints may be correlated; that does not affect the IUT argument.

For an estimate \(\hat{\theta}\) and standard error \(SE\), the lower-boundary statistic is \(z_L=(\hat{\theta}-L)/SE\), with one-sided p-value \(1-\Phi(z_L)\). The upper-boundary statistic is \(z_U=(\hat{\theta}-U)/SE\), with one-sided p-value \(\Phi(z_U)\). The endpoint TOST p-value is the larger of these two p-values.

For AUC, suppose \(\hat{\theta}=0.0300\) and \(SE=0.0700\). Then \(z_L=(0.0300-(-0.2231))/0.0700=3.616\), giving a lower-test p-value of about 0.00015. Also, \(z_U=(0.0300-0.2231)/0.0700=-2.759\), giving an upper-test p-value of about 0.0029. Thus \(p_{\mathrm{AUC}}=\max(0.00015,0.0029)=0.0029\), and AUC passes at \(\alpha=0.05\).

For Cmax, suppose \(\hat{\theta}=-0.0400\) and \(SE=0.0800\). Then \(z_L=(-0.0400-(-0.2231))/0.0800=2.289\), giving a lower-test p-value of about 0.0110. Also, \(z_U=(-0.0400-0.2231)/0.0800=-3.289\), giving an upper-test p-value of about 0.0005. Thus \(p_{\mathrm{Cmax}}=0.0110\), which also passes at \(\alpha=0.05\).

The joint p-value is \(\max(0.0029,0.0110)=0.0110\). Since it is below 0.05, the all-endpoints-required rule supports the conclusion that both endpoints meet their prespecified equivalence limits. No division of 0.05 by two is needed for the two endpoints in this conjunctive decision.

Worked Example: One Endpoint Fails, So the Joint Claim Fails

Hypothetical data. Keep the same log-scale margins and normal-reference assumptions. Suppose the AUC estimate is 0.0200 with \(SE=0.0600\), while the Cmax estimate is 0.1500 with \(SE=0.0800\). The scientific decision still requires equivalence for both endpoints.

For AUC, \(z_L=(0.0200+0.2231)/0.0600=4.052\), with lower-test p-value about 0.00003. The upper statistic is \(z_U=(0.0200-0.2231)/0.0600=-3.385\), with upper-test p-value about 0.0004. Therefore \(p_{\mathrm{AUC}}\approx0.0004\), and AUC passes.

For Cmax, \(z_L=(0.1500+0.2231)/0.0800=4.664\), with a negligible lower-test p-value. But \(z_U=(0.1500-0.2231)/0.0800=-0.914\), so the upper-test p-value is about 0.180. Therefore \(p_{\mathrm{Cmax}}\approx0.180\), and Cmax does not pass.

The joint p-value is \(\max(0.0004,0.180)=0.180\). The all-endpoints conclusion is not supported at 0.05. This does not establish that Cmax is non-equivalent; it establishes that the required evidence of equivalence was not obtained for that endpoint. The fact that AUC passes cannot compensate for the failing Cmax result.

When Multiplicity Adjustment Is the Right Question

Now change the decision rule: a positive overall claim is allowed if any one among several tests is positive. The global null is then that none of the candidate claims is true, and the alternative is that at least one is true. If each test is run at level \(\alpha\) without adjustment, the chance of one or more false rejections under the global null can exceed \(\alpha\). This is the setting for familywise error control, such as Bonferroni or Holm procedures.

For two independent tests, each with false-positive probability 0.05 under its null, the probability of at least one false positive is \(1-(1-0.05)^2=0.0975\). Dependence changes that particular calculation, but testing each at 0.05 does not in general guarantee a familywise error rate of 0.05. Bonferroni instead tests each at \(\alpha/m\); the probability of any false rejection is at most the sum of the individual probabilities, and hence at most \(\alpha\). Holm’s step-down procedure can also control familywise error and is often less conservative.

Worked Example: Any One Positive Result Is Sufficient

Hypothetical data. Suppose a separate exploratory decision permits a claim if either of two prespecified endpoint tests is positive. The valid endpoint p-values are 0.030 and 0.200. This is an any-one-positive rule, not the all-endpoints-equivalent rule.

Without adjustment, the first test is significant at 0.05, so an unadjusted procedure would make a positive claim. With Bonferroni control for two tests, the threshold is \(\alpha/m=0.05/2=0.025\). The smaller p-value, 0.030, exceeds 0.025, so neither test is significant under this procedure.

Holm gives the same decision here: order the p-values as 0.030 and 0.200, then compare the smaller with \(0.05/2=0.025\). Since 0.030 is larger, stop and reject neither. The adjustment is relevant because either test alone could trigger the overall claim. It would not be appropriate simply to apply this any-one-positive logic to a decision that requires all endpoints to pass.

What the Decision Rule Does—and Does Not—Guarantee

For a conjunctive equivalence claim, passing all endpoint TOSTs at level \(\alpha\) controls the type I error of the global all-equivalent decision at no more than \(\alpha\). It does not guarantee that the collection of endpoint confidence intervals has joint coverage of \(1-\alpha\). Each endpoint’s usual TOST-compatible interval can have its own specified coverage, while the probability that all such intervals simultaneously cover their parameters is a different quantity.

Likewise, the IUT result does not automatically justify every possible statement about individual endpoints. If the analysis will make separate endpoint-specific claims, highlight whichever endpoint passes, or choose endpoints after inspecting the data, the claim structure may have changed. The protocol and analysis plan should make clear which endpoints are required, which margins apply, and whether the intended conclusion is conjunctive or disjunctive.

For a valid IUT, the component TOSTs must themselves be valid at the chosen level, with appropriate models, standard errors, and prespecified margins. The error-control argument does not repair a misspecified analysis, data-dependent margins, selective endpoint omission, or invalid p-values. Nor does “no multiplicity adjustment” mean that endpoint choice is unimportant: it means the all-must-pass decision has a different mathematical error structure from an any-one-positive decision.

Common Mistakes and Diagnostics

  • Applying Bonferroni automatically because there are several endpoints. First write the global alternative. If it requires every endpoint to be equivalent, the intersection-union argument controls the global false-positive rate without an endpoint-count adjustment, assuming valid component tests.
  • Using an unadjusted 0.05 threshold when any endpoint can trigger the claim. This can inflate the chance of at least one false positive. Use an appropriate familywise procedure or another prespecified multiplicity strategy.
  • Assuming correlation makes the IUT argument fail. It does not. The global rejection event is a subset of the rejection event for each true component null, so independence is unnecessary. Dependence can matter for some multiplicity procedures and power calculations, but not for this basic size bound.
  • Calling a failed joint test proof that an endpoint is non-equivalent. Failure means the all-required evidence was not achieved. It may be caused by an endpoint that is truly outside the limits, or by uncertainty too large to establish equivalence.
  • Confusing endpoint-wise intervals with simultaneous coverage. Check whether the requested statement is an all-endpoints TOST decision or a joint confidence statement about all parameters. They are not interchangeable.

A useful diagnostic is to write, before calculating, whether one failure prevents the scientific conclusion or whether one success is enough. Then verify that the reported decision follows that rule: the joint IUT p-value is the largest endpoint TOST p-value, whereas any-one-positive claims require attention to multiplicity across the opportunities for rejection.

Check Your Understanding

For each question, identify the decision rule and explain the error-control implication. Solutions follow each question.

  1. A study declares equivalence only if all three prespecified exposure endpoints pass valid TOSTs at 0.05. Is a Bonferroni correction across endpoints required for this global decision? Solution: No. This is a conjunction. Under the global null, at least one endpoint null is true, and rejecting the global null requires rejecting that true component. The global type I error is therefore at most 0.05.
  2. Three endpoint TOST p-values are 0.008, 0.042, and 0.031. What is the joint IUT p-value, and does the all-endpoints decision pass at 0.05? Solution: \(p_{\mathrm{joint}}=\max(0.008,0.042,0.031)=0.042\). All three are at most 0.05, so the joint decision passes.
  3. A team may claim a positive result if any one of four endpoint tests is significant. What is the Bonferroni threshold at familywise \(\alpha=0.05\)? Solution: \(0.05/4=0.0125\). Each test is compared with 0.0125 under this procedure.
  4. In an all-endpoints TOST analysis, one endpoint has p-value 0.12 and the other two have p-values below 0.01. What can be concluded? Solution: The joint p-value is 0.12, so the all-endpoints equivalence claim does not pass at 0.05. This is insufficient evidence for the required joint claim, not proof that the endpoint with p-value 0.12 is non-equivalent.