Bioequivalence and Bioavailability Forum

Dear d_labes

❝ What is PASS simulator? What does it simulate?

I meant NCSS PASS- special soft for sample size calculation.
Why simulation?
I have always thought that PASS calculates via simulation. Am I mistaken?

❝ Sample size (total)

❝  n     power

❝ 16   0.926021

PASS returns the same result!

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Edit: Full quote removed. Please delete everything from the text of the original poster which is not necessary in understanding your answer; see also this post! [Helmut]

Hi Balaga,

❝ ❝ Observed Pilot study ISCV% - 12.29%

❝ We […] considered the 13 subjects for final statistical analysis.

❝ ❝ […] is it possible to expect the 15% ISCV or not for pivotal study design?

Let’s consider the CV first. Calculate the upper confidence limit of the CV you observed in the pilot. Then you could calculate the probability of a CV which is ≥15%. R-code:

library(PowerTOST)
CV     <- 0.1229 # observed
n      <- 13     # sample size (crossover)
alpha  <- 0.05   # default
target <- 0.15   # to explore
# 1. Upper 1-alpha CL of the CV
# -----------------------------
ucl    <- CVCL(CV, df=n-2, alpha=alpha)[["upper CL"]]
# 2. Find the alpha which gives an upper CL = target CV
# -----------------------------------------------------
f      <- function(x) CVCL(CV, df=n-2, alpha=x)[["upper CL"]]-target
p      <- uniroot(f, interval=c(1e-8, 0.5), tol=1e-6)$root
cat(paste0("Sample size: ", n, ", observed CV: ", 100*CV, "%",
  "\nUpper confidence limit of CV: ", round(100*ucl, 2),
  "% (\u03b1 ", alpha, ")", "\nProbability to expect a CV of \u2265",
  round(100*target, 2), "%: ~", signif(p, 2), "\n"))

I got:

Sample size: 13, observed CV: 12.29%
Upper confidence limit of CV: 19.16% (α 0.05)
Probability to expect a CV of ≥15%: ~0.24

In other words, if you plan the pivotal study with a CV of 12.29%, there is a ~24% chance that the ob­served one will be at least 15% (thus compromising power). You can plug in any α you want in the function CVCL(). Let’s say you want to have a 90% chance (α 0.1):

cat(sprintf("%.2f%%%s", 100*CVCL(CV, df=n-2, alpha=0.1)[["upper CL"]], "\n"))
17.32%

Some people use an α of 0.2 (similar to the commonly used producer’s risk):

cat(sprintf("%.2f%%%s", 100*CVCL(CV, df=n-2, alpha=0.2)[["upper CL"]], "\n"))
15.45%

❝ Why we need to expect the Null (true) ratio = 0.95 only?

I never said that. The θ₀ is unforgiving. Power curves are relatively flat close to 1, but become in­creas­ing­ly steep further away than ±5%. Had you any chance to look at my presentations as Detlew sug­gest­ed? What does the 90% CI of a BE-study tell us? That the true ratio lies with a probability of 0.9 some­where within its CI. Let’s assume the best (you observed a GMR of 1 in the pilot):

round(100*CI.BE(alpha=0.05, pe=1, CV=0.1229, n=13, design="2x2"), 2)
 lower  upper
 91.74 109.01

The true ratio might by ~9% away from 1. If you plugin a point estimate of 0.95, the lower CL is ~13% away and with a PE of 0.90 already ~17%… Which ratio and CV you use in your sample size estimation is a mix­ture of statistics, guesswork, and gut-feeling. Sorry. See what Anders Fuglsang wrote* on the topic.

Yet another story is the measured content. The EMA allows a dose-correction only if you provide a sound justi­fi­ca­tion that it was impossible to find a batch of the reference which differs ≤5% from the test. Otherwise, any measured difference is ignored (i.e., D_T ≡ D_R has to be assumed). No analytical method is “perfect”. Let’s presume an amazing method (accuracy 100% and imprecision ±2.5%):

alpha <- 0.05
mean  <- c(97.5, 100, 102.5)
CV    <- rep(2.5, 3)
sd    <- 100*CV/mean
lower <- qnorm(p=alpha/2,   mean=mean, sd=sd)
upper <- qnorm(p=1-alpha/2, mean=mean, sd=sd)
cat("Measured contents (%)   :", sprintf("%6.2f ", mean),
"\nCV of the method (%)    :  ", unique(CV),
"\nSD of means             :", sprintf("%6.2f ", sd),
paste0("\n\u03b1 level                 :  ", alpha, " (", 100*(1-alpha), "% CI)"),
"\nLower CL of contents (%):", sprintf("%6.2f ", lower),
"\nUpper CL of contents (%):", sprintf("%6.2f ", upper), "\n")

I got:

Measured contents (%)   :  97.50  100.00  102.50
CV of the method (%)    :   2.5
SD of means             :   2.56    2.50    2.44
α level                 :   0.05 (95% CI)
Lower CL of contents (%):  92.47   95.10   97.72
Upper CL of contents (%): 102.53  104.90  107.28

Do you get the idea?
Rule of thumb for a ~95% CI: x±2(100CV%/x). Talk to your analyst(s) about what to expect…

I think it is much more interesting to live with uncertainty
than to live with answers that might be wrong.    Richard Feynman

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• Fuglsang A. Pilot and Repeat Trials as Development Tools Associated with Demonstration of Bioequivalence. AAPS J. 2015;17(3):678–83. doi 10.1208/s12248-015-9744-6