CI inclusion operationally identical to TOST [Regulatives / Guidelines]

posted by d_labes  – Berlin, Germany, 2020-12-07 11:11 (48 d 01:27 ago) – Posting: # 22120
Views: 663

Dear Helmut,

» ...
» Given, Donald used the phrase “operationally identical” on p.661 (right column, 2nd paragraph).
»
» However, for me (!) those are two different “operations”. Results of an example:
» 90% CI: lower CL = 0.8448
»         upper CL = 1.1003
»         CI within 0.8000 and 1.2500: passed BE
» TOST  : p(<0.8000) = 0.01239
»         p(>1.2500) = 0.001565
»         p(<0.8000) <0.05 and p(>1.2500) <0.05: passed BE

Of course the two calculations are different, no doubt about it.

I have understood “operationally identical” always as the fact that TOST and CI inclusion give the same answer with regard to the BE decision.

IMHO this is the meaning of the paragraph on page 661 in Donalds famous paper containing “operationally identical”:
"The two one-sided tests procedure turns out to be operationally identical to the procedure of declaring equivalence only if the ordinary 1 - 2α (not 1-α) confidence interval for µT-µR is completely contained in the equivalence interval [θ1, θ2]".
Emphasis by me.

Regards,

Detlew

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