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

posted by Rocco_M – Mexico, 2019-09-18 00:02 (2039 d 08:14 ago) – Posting: # 20608
Views: 8,061

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:

UA Flag
Activity
 Admin contact
23,424 posts in 4,927 threads, 1,710 registered users;
29 visitors (0 registered, 29 guests [including 6 identified bots]).
Forum time: 08:16 CEST (Europe/Vienna)

Do not worry about your difficulties in mathematics.
I can assure you mine are still greater.    Albert Einstein

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