I am interested in testing for equivalence. I know about the TOST procedure, but many people who I work with do not, so I wanted to apply a method that they are more familiar with it. Specifically, I was about to use the two-sided $1-\alpha$ confidence interval of an estimate $\hat{\theta}$ to check whether this estimate lies within a predefined equivalence interval $(\theta_L,\theta_U)$.
But I am hesitating after reading the following statement at the website of the software Minitab:
[...] the confidence interval for equivalence also considers the additional information of the lower and upper limits of the equivalence interval. Because the confidence interval incorporates this additional information, a (1 – alpha) x 100% confidence interval for equivalence is in most cases tighter than a standard (1 – alpha) x 100% confidence interval that is calculated for a t-test.
In another post on Stack Overflow, Horst Grünbusch made an interesting statement:
The TOST-CI is simply the intersection of the one-sided CIs.
Assuming this statement is true, then my understanding is that for any test statistic $\theta$ with symmetrical distribution, the intersection of the one-sided $1-2\alpha$ confidence intervals should be identical with the one-sided $1-\alpha$ confidence interval. I would like to ask:
- Is the cited statement by Horst Grünbusch true?
- For which test statistics is the two-sided confidence interval not identical to the intersection of the one-sided confidence intervals (i.e. the TOST-CI)?
- Can one give an intuition, why the TOST-CI would "in most cases be tighter", as indicated by Minitab?
