In testing for equivalence via the two one-sided test approach with confidence intervals, a (1–2α) × 100% confidence interval is calculated to check for equivalence. I assume this is because you calculate a CI for mean of group a and mean of group b.

But why is not it possible to calculate a 95%-CI of the difference between the groups? Can you explain why (1–2α) × 100% is used here all the time?

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    $\begingroup$ Is there any reference which can explain the concept in detail $\endgroup$
    – user77003
    Commented May 13, 2015 at 6:10

2 Answers 2


The $1-2\alpha$ is not because you calculate the CI for each group separately. It is because you calculate the "inequivalence" to the upper and to the lower end separately. The parameter $\theta$ lies in the equivalence interval $[\epsilon_L, \epsilon_U]$ iff $$\theta \geq \epsilon_L \wedge \theta \leq \epsilon_U.$$

Each part is tested separately by a one sided test at level $1-\alpha$. Only if both tests are significant, we can conclude equivalence. (This is the very intuitive intersection-union-principle.) Turning this into a single confidence interval, we must remove $\alpha$ from both the upper and the lower probability mass of the CI. So we end up with $1-2\alpha$. The TOST-CI is simply the intersection of the one-sided CIs.

By the way, it is still possible to do the TOST with a $1-\alpha$ CI, but it would be unnecessarily conservative.

  • $\begingroup$ In fact, even using a $1-2\alpha$ confidence interval, the procedure is still slightly conservative. $\endgroup$
    – Russ Lenth
    Commented Nov 25, 2014 at 13:44
  • $\begingroup$ but in the end, we can only be sure 90% which is less than with a normal signifance test, right? $\endgroup$ Commented Dec 11, 2014 at 17:24
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    $\begingroup$ @00schneider No: $\alpha$ is the researcher's choice of nominal Type I error. One could choose $\alpha=0.1$ as in your example, or one could another value such as $\alpha=0.01$. $\endgroup$
    – Alexis
    Commented Apr 7, 2015 at 4:56

The answer to this question is that 90% is possible because of a logical fact that makes this "bonus" in confidence possible. In the TOST procedure, two one-tailed tests are conducted at a 5% level. The type 1 error rate stills remains at 5% because if one test decision is a type 1 error, the other one cannot be a type 1 error anymore. For example if one test falsely states that the difference is larger than -3 (i.e. in fact it is smaller than -3), the other test which tests if it is smaller than 3 cannot produce a type 1 error because the value is in fact smaller than 3.


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