sampleN.TOST and CI.BE [Power / Sample Size]

posted by Rocco_M – Mexico, 2019-09-17 22:02 (581 d 04:58 ago) – Posting: # 20608
Views: 3,709

(edited by Rocco_M on 2019-09-17 23:43)

Okay gracias. That was a brain fart on my end. I now see that it is equivalent to 1/n1 + 1/n2.

One last question. If I use sampleN.tost and input
sampleN.TOST(CV=.3, theta0=1.0, theta1= 0.8, theta2=1.25, logscale=TRUE, alpha=0.05, targetpower=0.9, design="parallel")

I get sample size minimum be 78 total.

But then if I run CI.BE(pe=1.0, CV=.3, design="parallel", n=24)
I get CI = approx [0.81, 1.23].

This confuse me. Shouldn’t I need to enter n at least 78 in order to get CI within [.8, 1.25]?

Maybe I confuse concepts. In other words, what are implications if study *meet* bioequivalence but is underpowered?

Complete thread:

Activity
 Admin contact
21,419 posts in 4,475 threads, 1,510 registered users;
online 11 (0 registered, 11 guests [including 2 identified bots]).
Forum time: Wednesday 03:00 CEST (Europe/Vienna)

In the Middles Ages the lingua franca of science was Latin.
Nowadays the language of science is bad English.    Anonymous

The Bioequivalence and Bioavailability Forum is hosted by
BEBAC Ing. Helmut Schütz
HTML5