Post hoc power? [Power / Sample Size]

posted by Helmut Homepage – Vienna, Austria, 2016-12-04 15:16 (3490 d 17:59 ago) – Posting: # 16832
Views: 7,471

Hi BE-proff,

❝ Let's say I have the followind results after typical BE-study:

❝ n=30, CV=0.20, ratio=0.97, UL=1.02 and LL=0.92


The study passed by a nice margin. Why are you interested in the irrelevant post hoc power? BTW, I cannot reproduce your 90% CI:
library(PowerTOST)
round(CI.BE(alpha=0.05, pe=0.97, CV=0.2, n=30), 4)
 lower  upper
0.8892 1.0582

Only with a strange α:
round(CI.BE(alpha=0.147, pe=0.97, CV=0.2, n=30), 2)
lower upper
 0.92  1.02


1.

❝ if I want to calculate power using default settings of PowerTOST I see:

power.TOST(CV=0.20, n=30)

[1] 0.948603


Sure. Type help(power.TOST) in the console. You will find:

theta0

Defaults to 0.95.

theta1

Lower bioequivalence limit. Defaults to 0.8.

theta2

Upper bioequivalence limit. If not given theta2 will be calculated as 1/theta1.


So you are essentially asking which power can be expected with a CV of 0.20, assuming theta0 0.95, and the conventional acceptance range of 0.8–1.25. Works as designed.

2.

❝ If to add ratio I have:

power.TOST(CV=0.20, n=30, theta0=0.97)

[1] 0.9781158


That’s the post hoc (aka a posteriori) power of the study.

3.

❝ If to use also limits of CI result the result is:

power.TOST(CV=0.20, n=30, theta0=0.97, theta1=0.92, theta2=1.02)

[1] 2.334844e-05 :confused:


Now you asking for the power to show BE where the acceptance range practically equals the CI in the study. Of course this chance is very low.

❝ Do I use formulas correcly and which option is better? :-)

  1. Maybe. But it gives you only the power for another study with the same CV and sample size and theta0 0.95. Maybe more interesting: pa.ABE(CV=0.2, theta0=0.95, targetpower=0.9)
  2. Meaningless.
  3. Wrong.

Congratulations for your post #100! :ok:

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