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?

• Is there any reference which can explain the concept in detail – user77003 May 13 '15 at 6:10

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.
• In fact, even using a $1-2\alpha$ confidence interval, the procedure is still slightly conservative. – rvl Nov 25 '14 at 13:44
• @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 $\alpha=0.01$. – Alexis Apr 7 '15 at 4:56