Bioequivalence and Bioavailability Forum

❝ […]: why does the sample size estimation with R with package PowerTOST differ between sampleN.TOST and sampleN.scABEL when CV=30%?

❝ So why do the sample size estimations differ? They have the same arguments (design="2x3x3", theta0=0.95 ...). CV is 30% so there should be no scaling (conventional BE limits, i.e., 80.00-125.00).

❝ Does it have to do anything with simulations? But even when I increase number of simulations, the differences aren't that big.

❝ Or if I reformulate the question, why dont the following powers match:

❝ a) power.TOST(CV=0.3, theta0=0.95,design="2x3x3", n=30)


❝ [1] 0.8204004

❝ b) power.scABEL(CV=0.3, theta0=0.95, design="2x3x3", n=30)


❝ [1] 0.85977

1. ABE (exact)
   sampleN.TOST(CV = 0.25, theta0 = 0.95, design = "2x3x3")
+++++++++++ Equivalence test - TOST +++++++++++
            Sample size estimation
-----------------------------------------------
Study design: 2x3x3 (partial replicate)
log-transformed data (multiplicative model)

alpha = 0.05, target power = 0.8
BE margins = 0.8 ... 1.25
True ratio = 0.95,  CV = 0.25

Sample size (total)
 n     power
21   0.814342
   
   
2. ABEL (simulations)
   Note that for the partial replicate design you should use subject simulations by the function sampleN.scABEL.sdsims() instead of simulating the associated statistics by the function sampleN.scABEL().
   sampleN.scABEL.sdsims(CV = 0.25, theta0 = 0.95, design = "2x3x3", details = FALSE)
+++++++++++ scaled (widened) ABEL +++++++++++
            Sample size estimation
   (simulation based on ANOVA evaluation)
---------------------------------------------
Study design: 2x3x3 (partial replicate)
log-transformed data (multiplicative model)
1e+05 studies for each step simulated.

alpha  = 0.05, target power = 0.8
CVw(T) = 0.25; CVw(R) = 0.25
True ratio = 0.95
ABE limits / PE constraint = 0.8 ... 1.25
Regulatory settings: EMA

Sample size
 n   power
21   0.8223

• The simulated variance \(\small{s_\text{wR}^2}\) depends on its associated \(\small{\chi^2}\)-distribution with \(\small{n-2}\) degrees of freedom. The \(\small{\chi^2}\)-distribution is skewed to the right. Hence, when simulating any \(\small{CV_\text{wR}=x}\), you will obtain slightly more values \(\small{>x}\) than \(\small{<x}\).
  Try this to show the asymmetric distribution:
  n  <- 1e7
df <- 27 - 2
set.seed(1234567)
x  <- rchisq(n = n, df = df)
y  <- hist(x, breaks = "FD", plot = FALSE)
plot(y, freq = FALSE, col = "lightblue", border = NA,
     xlim = c(0, max(y$mids)), cex.main = 1, las = 1,
     xlab = bquote("Number of random samples of "*chi^2 == .(n)),
     main = bquote(italic(df) == .(df)))
abline(v = y$mids[y$density == max(y$density)], col = "blue")
box()