Bioequivalence and Bioavailability Forum

Hi ElMaestro,

❝ to round this off, can it be mathematically proven that upper type 1 error asymptotes exist

Yes. At least this is what my stats friends told me. Note that Bonferroni is the most conservative method in adjusting for multiplicity.*
IMHO, you could gain a little solving the quadratic equation \(\small{1-(1-x)^2=\alpha}\), which leads for α 0.05 to \(\small{(2-\sqrt{4-4\alpha})/2=0.02532057\ldots}\)
Others (f.i. Holm, Hochberg) are less conservative but may lead to a CI which doesn’t include the PE any more. Counterintuitive, but my stats friends just smile about it. See also EMA’s 2012 workshop on multiplicity. If you – really! – have nothing better to do, watch the six (‼) hours of video.

❝ for any level of common alpha smaller than 1?

Smaller than one?

❝ I bet that should be possible, too. I'd love to see how. I imagine that to deliver the proof one would have to one way or another get aorund to Owen's Q and non-central t. So I am very certain that would improve yours truly's understanding of statistics for BE.

Yep. People are working on it (for three years now)…

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• Since I’m not gifted with enough knowledge to proof anything I can only offer extreme simulations:
  
  library(Power2Stage)
power.2stage.GS(alpha=rep(0.025, 2), n=c(10, 100000),
  CV=0.05, theta0=1.25, fCrit="PE", fClower=0, nsims=1e7)

TSD with 2x2 crossover
 non-adaptive group sequential with
alpha (s1/s2) = 0.025 0.025
No futility criterion
BE acceptance range = 0.8 ... 1.25

CV= 0.05; n(s1, s2)= 10 1e+05

1e+07 sims at theta0 = 1.25 (p(BE) = TIE 'alpha').
p(BE)    = 0.0493985
p(BE) s1 = 0.0250372
Studies in stage 2 = 97.5%
  
  That’s pretty close to what we would expect from Bonferroni (0.049375).
  The Japanese stuff gives me 0.0975317 (expected 0.0975).

❝ Bonferroni in a Guidance! Wow! Didn't know that!

Gone with the Wind…