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Confidence Intervals and the TOST Procedure · Tutorial 214 of 399

Making the TOST Decision From Both Tests

Use both one-sided test results—not either one alone—to decide whether the estimated effect meets the prespecified equivalence criteria.

Intermediate 10 min read

What You'll Learn

  • State the two component null hypotheses and the equivalence alternative.
  • Apply the lower- and upper-bound rejection criteria using test statistics or p-values.
  • Interpret outcomes when both tests, exactly one test, or neither test rejects.
  • Explain why equivalence requires rejection of both component null hypotheses.
  • Check the model and standard error assumptions needed for the decision.
  • Identify common decision errors and diagnose results near an equivalence boundary.

Why the Decision Requires Both Tests

An equivalence conclusion is a joint claim: the treatment effect is greater than a lower margin and less than an upper margin. Evidence against only one boundary does not establish the other part of that claim. For example, an estimate can be convincingly above the lower margin while remaining too uncertain to rule out effects at or beyond the upper margin.

Let \(\theta\) denote the population treatment effect on the analysis scale, and let \(L\) and \(U\) be the prespecified lower and upper equivalence margins, with \(L<U\). The estimate is \(\hat{\theta}\), its standard error is \(SE\), and the reference distribution has \(df\) degrees of freedom. The two one-sided tests (TOST) assess the lower and upper boundaries separately at significance level \(\alpha\).

Definition: Equivalence is established by TOST only when both component null hypotheses are rejected at the prespecified level \(\alpha\). The lower-bound test assesses \(H_{0L}:\theta\le L\) against \(H_{aL}:\theta>L\). The upper-bound test assesses \(H_{0U}:\theta\ge U\) against \(H_{aU}:\theta<U\). The overall equivalence alternative is \(L<\theta<U\).

The overall null is the union of the component nulls: at least one of \(\theta\le L\) or \(\theta\ge U\) is true. To conclude equivalence, the data must provide evidence against both possibilities. Rejecting just one component leaves the other part of the overall null unrefuted. This is why the decision is not based on an average of the two p-values, nor on whether one test is especially convincing.

Apply Each Rejection Criterion

For the standard one-sided \(t\) procedure, the statistics and directional p-values are

$$ t_L=\frac{\hat{\theta}-L}{SE}, \qquad p_L=P(T_{df}\ge t_L), \qquad t_U=\frac{\hat{\theta}-U}{SE}, \qquad p_U=P(T_{df}\le t_U). $$

Here \(T_{df}\) has a Student \(t\) distribution with \(df\) degrees of freedom. Large positive values of \(t_L\) support \(\theta>L\), while large negative values of \(t_U\) support \(\theta<U\). The p-values therefore use opposite tails. At level \(\alpha\), reject the lower component null when \(p_L\le\alpha\), and reject the upper component null when \(p_U\le\alpha\). Equivalently, if \(t_{1-\alpha,df}\) is the \(1-\alpha\) quantile of the \(t\) distribution, the criteria are \(t_L\ge t_{1-\alpha,df}\) and \(t_U\le -t_{1-\alpha,df}\).

These criteria presume that the treatment-effect estimate and standard error come from an appropriate model, the stated degrees of freedom are valid for the chosen procedure, and the test-statistic reference distribution is justified under each boundary null. In a crossover analysis, for instance, the model and its standard error must reflect the design and the analysis of the treatment effect. The margins and \(\alpha\) must also be specified before interpreting the two results; selecting them after seeing the estimates would change the meaning of the tests.

Worked Example: Both Tests Reject

Suppose a hypothetical analysis estimates a treatment effect of \(\hat{\theta}=0.0\) analysis-scale units, with \(SE=4.0\), \(df=20\), and equivalence margins \(L=-10\) and \(U=10\) units. Let \(\alpha=0.05\). Assume the model-based \(t\) reference distribution and standard error are appropriate for both one-sided tests.

For the lower boundary,

$$ t_L=\frac{0.0-(-10)}{4.0}=2.50, \qquad p_L=P(T_{20}\ge2.50)\approx0.01062. $$

Because \(0.01062<0.05\), reject \(H_{0L}\). For the upper boundary,

$$ t_U=\frac{0.0-10}{4.0}=-2.50, \qquad p_U=P(T_{20}\le-2.50)\approx0.01062. $$

Because \(0.01062<0.05\), reject \(H_{0U}\) as well. Both required rejections have occurred, so equivalence is established at the 5% level for these margins and this analysis. The decision does not say that the true effect is known exactly or that it has a 95% probability of lying between the margins. It says that both specified one-sided null hypotheses were rejected using the stated procedure.

When Exactly One Test Rejects

The mixed outcomes are important because they are not equivalence conclusions. One test may reject clearly while the other does not. In that situation, the appropriate report identifies which boundary was ruled out and which was not; it does not turn the one rejection into an overall success.

Worked Example: Lower Test Rejects, Upper Test Does Not

Consider a hypothetical estimate \(\hat{\theta}=9\) units, \(SE=5\) units, \(df=20\), margins \(L=-10\) and \(U=10\) units, and \(\alpha=0.05\). Assume the stated standard error and \(t\) reference distribution are suitable. For the lower boundary,

$$ t_L=\frac{9-(-10)}{5}=3.80. $$

For \(df=20\), the one-sided 5% critical value is approximately \(1.725\). Thus \(3.80>1.725\), so the lower test rejects \(H_{0L}\) (its p-value is less than 0.001). For the upper boundary,

$$ t_U=\frac{9-10}{5}=-0.20, \qquad p_U=P(T_{20}\le-0.20)\approx0.4218. $$

Since \(0.4218>0.05\), the upper test does not reject \(H_{0U}\). Therefore, equivalence is not established. The analysis supports the conclusion that the effect exceeds the lower margin, but it has not ruled out effects at or above the upper margin. This is not the same as proving that the effect is outside the equivalence range; the upper test simply did not provide sufficient evidence to exclude its null.

Worked Example: Upper Test Rejects, Lower Test Does Not

Now consider a second hypothetical analysis with \(\hat{\theta}=-9\) units, \(SE=5\) units, \(df=20\), the same margins \(-10\) and \(10\) units, and \(\alpha=0.05\). Under the same model assumptions, the lower-bound calculation is

$$ t_L=\frac{-9-(-10)}{5}=0.20, \qquad p_L=P(T_{20}\ge0.20)\approx0.4218. $$

The lower test does not reject because \(0.4218>0.05\). For the upper boundary,

$$ t_U=\frac{-9-10}{5}=-3.80. $$

Since \(-3.80<-1.725\), reject \(H_{0U}\); its p-value is less than 0.001. Only the upper test rejects, so equivalence is again not established. The data support an effect below the upper margin but have not ruled out effects at or below the lower margin. The mirror-image outcomes in these two examples emphasize that both sides of the claim must be supported, regardless of which side is uncertain.

Decision Logic and a Short Justification

The two-test decision can be justified directly from the hypotheses. Equivalence requires \(\theta>L\) and \(\theta<U\) simultaneously. The null says at least one of those inequalities fails. Thus, evidence against the overall null requires evidence against each of its two component possibilities.

$$ H_0:\{\theta\le L\}\ \text{or}\ \{\theta\ge U\}, \qquad H_a:\{\theta>L\}\ \text{and}\ \{\theta<U\}. $$

If either component test fails to reject, its corresponding boundary remains compatible with the test outcome at level \(\alpha\). It follows that the conjunction of the two rejection criteria is required. If both component tests are conducted at level \(\alpha\), the probability of falsely declaring equivalence when the overall null is true is controlled at no more than \(\alpha\): whenever the overall null is true, at least one component null is true, and a false equivalence declaration requires falsely rejecting that true component null. No adjustment that treats the two tests as independent is needed for this intersection-union decision.

1
Check the setup.
Confirm the margins, \(\alpha\), effect estimate, standard error, degrees of freedom, and model assumptions.
2
Evaluate the lower test.
Reject \(H_{0L}\) only if \(p_L\le\alpha\), or equivalently if \(t_L\ge t_{1-\alpha,df}\).
3
Evaluate the upper test.
Reject \(H_{0U}\) only if \(p_U\le\alpha\), or equivalently if \(t_U\le -t_{1-\alpha,df}\).
4
Make the joint decision.
Declare equivalence only if both tests reject. Otherwise, state which test did not reject and avoid a positive equivalence conclusion.

Diagnostics, Alternatives, and Common Pitfalls

Before finalizing a decision, verify that each p-value uses the correct tail. The lower test is a right-tail test; the upper test is a left-tail test. A sign error or using the same tail for both tests can reverse a decision. Check the calculations against the critical-value form as well: the lower statistic must be sufficiently positive, and the upper statistic sufficiently negative. If the p-value and critical-value decisions disagree, inspect the signs, degrees of freedom, standard error, and rounding.

Results near \(\alpha\) deserve particular care. A p-value close to the threshold can change its displayed rounded value, so retain adequate precision for the decision and report enough digits to make the comparison transparent. A test statistic near its critical value is also a cue to verify that the standard error and model specification are appropriate. Do not change the margins, significance level, or analysis model merely because one result is inconvenient; any justified model change needs to be evaluated and reported on its own terms.

The principal alternative summary is a confidence interval constructed at the level corresponding to TOST. The earlier material on the confidence-interval rule establishes the equivalent interval criterion. Here the practical distinction is one of presentation: the two-test approach displays each boundary-specific result, while the interval approach displays the range of effects compatible with the data. Neither presentation permits a conclusion when one boundary criterion fails. A conventional test of no difference is not a substitute for TOST: failing to detect a difference does not, by itself, establish that effects outside the equivalence margins have been ruled out.

A common mistake is to reason, “The estimated effect lies between the margins, and one test rejects, so the treatments are equivalent.” The estimate’s location alone does not account for uncertainty, and one rejection addresses only one boundary. The correction is to evaluate both directional tests and require both p-values to be at most \(\alpha\). If exactly one rejects, report that mixed result and conclude that equivalence was not established.

Key takeaway: TOST is a two-part decision. Equivalence is established only when the lower-bound test rejects \(H_{0L}\) and the upper-bound test rejects \(H_{0U}\) at the prespecified \(\alpha\). One rejection, however strong, is not enough.

Check Your Understanding

For each question, identify the component-test outcome and the correct overall conclusion.

  1. At \(\alpha=0.05\), the lower-test p-value is 0.03 and the upper-test p-value is 0.04. What is the TOST decision?
  2. The lower test rejects at \(\alpha=0.05\), but the upper-test p-value is 0.20. Can equivalence be concluded? What can be said about the upper boundary?
  3. For \(df=20\), a lower statistic is \(t_L=1.60\) and an upper statistic is \(t_U=-1.90\). Using a one-sided 5% critical value of approximately 1.725, make the decision.
  4. Why is “the estimate is between the margins” not sufficient to establish equivalence?

Solutions. (1) Both p-values are below 0.05, so both component nulls are rejected and equivalence is established at the stated level, subject to the model assumptions. (2) Equivalence is not established because the upper test did not reject. The data have not ruled out effects at or above the upper margin; this is not proof that such effects are true. (3) The lower test does not reject because \(1.60<1.725\); the upper test rejects because \(-1.90<-1.725\). Exactly one test rejects, so equivalence is not established. (4) The point estimate does not show how uncertain the estimate is. TOST requires evidence against both boundary null hypotheses, and an estimate inside the margins can still be too imprecise for one or both tests to reject.