Bioequivalence and Bioavailability Forum

❝ I have a question regarding sample size estimation for parallel designs with more than 2 treatment arms (with more than 2 groups). Package PowerTOST in R covers sample size estimation for parallel design for just 2 parallel groups (known.designs()) and not for more.

❝ Lets say we are conducting a (pilot) parallel study with 3 test formulations and 1 reference. […] I would say that we just double the necessary sample size estimated (in R with sampleN.TOST) for 2 parallel groups (to get 4 independent groups for 3 test formulations and 1 reference). For example for CV=0.2 (parallel design and theta0=0.95) for 2 parallel groups we get sample size of 36 subjects (total sample size for 2 groups), so if we double it we get 72 subjects (for 4 groups?). Is this assumption and sample size estimation a correct one?

Tests                   : 3
Groups                  : 4
alpha                   : 0.05
Sample size per group   : 18
Total sample size       : 72
Achieved power          : 0.80994

❝ […] I have read this article regarding multiple treatments comparison in parallel design and I dont exactly understand why should we just control Type I Error and change alpha when estimating sample size with multiple treatments?

❝ So in our case (when using Bon­ferroni adjustment) we would get the total sample size (for 4 groups) of 50 subjects (parallel design, CV=0.2, theta0=0.95, alpha=0.0167)? Is this correct?

Tests                   : 3
Groups                  : 4
alpha                   : 0.05
Multiplicity adjustment : Bonferroni
Adjusted alpha          : 0.0166667 (alpha / 3)
Sample size per group   : 25
Total sample size       : 100
Achieved power          : 0.80304

❝ What is more, …

❝ … 50 is not even dividable by 4.

❝ I am confused and I probably didnt get this explanation in the article right. Nevertheless I wouldnt mind this being the right way as the needed sample size is smaller

❝ Regarding alpha adjustment in the first case (for doubling sample size from 36 to 72) I would say that it is not necessary as it is a pilot study.

❝ On the other hand, if this was a pivotal study and we would want to show equivalence for just 1 test formulation (among 3) then we would have to adjust alpha (to 0.0167) and then I would say we would double the estimated sample size for 2 groups the same way as described above (so 50x2=100 subjects)?

1. If all tests should pass (AND-condition), you don’t have to adjust \(\alpha\).
   
2. If any of the \(k\) tests should pass (OR-condition), you could use Bonferroni’s \(\alpha / k\), which is the most conservative method.

library(PowerTOST)
alpha0 <- 0.05  # nominal (overall) level
CV     <- 0.2   # assumed total CV
theta0 <- 0.95  # assumed T/R-ratio(s)
target <- 0.8   # target (desired) power
k      <- 3     # number of tests
adj    <- FALSE # FALSE in a pilot study and in a pivotal study if all tests should pass
                # TRUE in a pivotal study if any of the tests should pass
if (!adj) {     # unadjusted
  alpha <- alpha0
} else {        # Bonferroni
  alpha <- alpha0 / k
}
# Total sample size and power for two groups
tmp    <- sampleN.TOST(alpha = alpha, CV = CV, theta0 = theta0, targetpower = target,
                       design = "parallel", print = FALSE)
N      <- tmp[["Sample size"]]
pwr    <- tmp[["Achieved power"]]
# Total sample size for k (tests) + 1 (reference) groups
n      <-  N * (k + 1) / 2
txt    <- paste0("\nTests                   : ", k,
                 "\nGroups                  : ", k + 1)
if (adj) {
  txt <- paste0(txt, "\nalpha                   : ", alpha0,
                     "\nMultiplicity adjustment : Bonferroni",
                     "\nAdjusted alpha          : ", signif(alpha, 6),
                     " (alpha / ", k, ")")
} else {
  txt <- paste0(txt, "\nalpha                   : ", alpha0)
}
txt    <- paste0(txt, "\nSample size per group   : ", N / 2,
                      "\nTotal sample size       : ", n,
                      "\nAchieved power          : ", signif(pwr, 5), "\n")
cat(txt)