Bioequivalence and Bioavailability Forum

Dear Detlew and all!

❝ Of course you can formulate the Potvin's decision scheme B without them.

❝ But then it will happen that the sample size estimation step will give you a number which is smaller or at best equal to that you have already used for step 1.

❝

❝ Now what then?

❝ Natural would be: STOP study. Bioequivalence not proven!

❝ Exactly this is what the intermediate power calculation is for in scheme B.

Not for you, but a numerical example for non-initiates. Let’s consider Example 1 from Potvin’s paper (n₁ 12 failed in Stage 1 and n_total 14):

• Method B:
  Stage 1:
  94.12% CI: 104.27–134.17%, CV_intra 14.56%
  Power at T/R 95%: 76.26% (exact; in the paper 75.6% based on shifted t)
  sampleN.TOST(alpha=0.0294, CV=0.1456, theta0=0.95) > n_total 14 (n₂ 2)
  Pooled data:
  94.12% CI: 102.83–129.71%, CV_intra 14.65%
  
• No intermediate power:
  Outcome exactly as above.

Now I manipulated part of the data:
                        original
Subject Treatment Response manipulated
   1        T        3.3   →   2.9
   1        R        2.9       2.9
   2        T        5.3   →   6.1
   2        R        6.1       6.1
   3        T        6.2   →   6.3
   3        R        7.1       7.1

• Method B:
  Stage 1:
  94.12% CI: 105.21–133.57%, CV_intra 13.77%
  Power at T/R 95%: 80.99% (80.5% based on shifted t)
  Since power in Stage 1 ≥80% (fail)
  
• No intermediate power:
  sampleN.TOST(alpha=0.0294, CV=0.1377, theta0=0.95) > n_total 12 (n₂ 0) Yess!

❝ I've heard some rumour from regulators that you should not stop but recruit 2 more subjects for stage 2 in that case. But this lacks any scientific justification IMHO.

Yeah, but why not 2.718282 or 3.141593?

BTW, one regulator called Montague’s paper a “correction” of Potvin’s.