ABEL vs ABE, randomization [🇷 for BE/BA]

posted by Helmut Homepage – Vienna, Austria, 2022-06-04 15:11 (33 d 03:59 ago) – Posting: # 23042
Views: 236

Hi Bebac user,

» For a drug that has CV = 25%, we need to design a full replicated study

Why? See also this article.
CV 25% of what? Unusual that Cmax and AUC have the same one. If your authority accepts reference-scaling for Cmax, generally AUC drives the sample size because you have to use conventional ABE for it.

» To estimate the sample size, which one is the right to use in R
» SampleN.tost OR sampleN.scabel??

None would work. [image] is case-sensitive. You are interested in sampleN.TOST() and sampleN.scABEL().

If really both metrics have 25% (please check), chances are pretty low that ABEL can be applied (only if CVwR > 30%):

library(PowerTOST)
sampleN.TOST(CV = 0.25, design = "2x2x4")

+++++++++++ Equivalence test - TOST +++++++++++
            Sample size estimation
-----------------------------------------------
Study design: 2x2x4 (4 period full 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
14   0.813985


sampleN.scABEL(CV = 0.25, theta0 = 0.95, design = "2x2x4")

+++++++++++ scaled (widened) ABEL +++++++++++
            Sample size estimation
   (simulation based on ANOVA evaluation)
---------------------------------------------
Study design: 2x2x4 (4 period full 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
EMA regulatory settings
- CVswitch            = 0.3
- cap on scABEL if CVw(R) > 0.5
- regulatory constant = 0.76
- pe constraint applied

Sample size search
 n     power
12   0.7659
14   0.8266


Therefore, you would need the same sample size, though you would achieve a slightly higher power with ABEL.

» I am using a RandomizBE to make randomization but I want to divide them into period 1, period 2 and so on. How?

Extract the treatment from the sequence and create additional columns:

library(randomizeBE)
n       <- 14
seqs    <- c("TRTR", "RTRT")   # any crossover or replicate design
pers    <- nchar(seqs)[1]      # number of periods
periods <- paste0("p", 1:pers) # column names
random  <- RL4(nsubj = n, seqs = seqs, randctrl = TRUE)
random$rl[ , periods] <- ""    # empty period columns
for (i in 1:pers) {            # extract treatment / period
  random$rl[i + 3] <- substring(random$rl$sequence, i, i)
}
print(random)

Randomization table          created: 2022-06-04 15:32:05
(seed: 1892562 blocksize: 4 4 4 2 )

 subject seqno sequence p1 p2 p3 p4
       1     2     RTRT  R  T  R  T
       2     1     TRTR  T  R  T  R
       3     2     RTRT  R  T  R  T
       4     1     TRTR  T  R  T  R
       5     2     RTRT  R  T  R  T
       6     1     TRTR  T  R  T  R
       7     2     RTRT  R  T  R  T
       8     1     TRTR  T  R  T  R
       9     2     RTRT  R  T  R  T
      10     2     RTRT  R  T  R  T
      11     1     TRTR  T  R  T  R
      12     1     TRTR  T  R  T  R
      13     1     TRTR  T  R  T  R
      14     2     RTRT  R  T  R  T


Dif-tor heh smusma 🖖 [image]
Helmut Schütz
[image]

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