7
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Is there any reference which can explain the concept in detail $\endgroup$ – user77003 May 13 '15 at 6:10
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ In fact, even using a $1-2\alpha$ confidence interval, the procedure is still slightly conservative. $\endgroup$ – rvl Nov 25 '14 at 13:44
  • $\begingroup$ but in the end, we can only be sure 90% which is less than with a normal signifance test, right? $\endgroup$ – 00schneider Dec 11 '14 at 17:24
  • 1
    $\begingroup$ @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 Apr 7 '15 at 4:56
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.