Equivalence testing - tost method - why CI of 90%? 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?
 A: 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.
A: 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.
