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

posted by Rocco_M – Mexico, 2019-09-17 22:02 (376 d 04:26 ago) – Posting: # 20608
Views: 3,150

(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
21,079 posts in 4,396 threads, 1,468 registered users;
online 4 (0 registered, 4 guests [including 2 identified bots]).
Forum time: Monday 02:28 CEST (Europe/Vienna)

A central lesson of science is that to understand complex issues
(or even simple ones), we must try to free our minds of dogma and
to guarantee the freedom to publish, to contradict, and to experiment.
Arguments from authority are unacceptable.    Carl Sagan

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