Ibrahim Komeil
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Egypt,
2019-10-06 13:56
(1635 d 09:28 ago)

Posting: # 20666
Views: 3,145
 

 Sample size and randomization plan [Power / Sample Size]

Dear ALL

Thank you for this helpful forum , i would like to ask about how can i calculate a sample size for full replicated study design for sodium valproate bioequivalence study ?

My second question is how can i generate a randomization plan with william's design using SAS software ?

Thank you for your kind help

Yours,
Ibrahim A.Komeil


Edit: Category changed; see also this post #1[Helmut]
Helmut
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Vienna, Austria,
2019-10-06 20:00
(1635 d 03:23 ago)

@ Ibrahim Komeil
Posting: # 20669
Views: 2,629
 

 Sample size and randomization plan

Salam Ibrahim,

❝ Thank you for this helpful forum


Welcome!

❝ […] sample size for full replicated study design for sodium valproate bioequivalence study ?


Why do you want to do that? Sodium valproate has an extremely low variability. In the 12 studies I performed the intra-subject CVs were 6.2–14.2% (AUC) and 5.3–16.1% (Cmax). In the European applications I’m aware of, so­di­um valproate was not classified as a NTID and consequently studies performed as 2×2×2 crossovers. Then ~16 subjects are sufficient to show BE for an assumed T/R-ratio of 0.95 and 80% power. A full replicate does not make sense.

However, the FDA classified it as a NTID and recommends reference-scaling. That’s another cup of tea and requires a 4-period full replicate study indeed. Another obstacle is the additional comparison of \(\sigma_{wT}/\sigma_{wR}\). My data are the CVw, which are pooled from the – unknown – CVwT and CVwR.
If CVwR < CVw the BE limits are downscaled and the sample size has to be increased. You have also to increase the sample size if CVwT > CVwR. Difficult to predict…
Exploring my data: I assumed a nice case (swT = 0.5×swR), equal variances, and a worse case (swT = 2×swR). As usual a conservative T/R-ratio of 0.975 since the FDA requires for NTIDS more strict batch release spec’s.

library(PowerTOST)
var.ratio <- c(0.5, 1, 2)
CV.AUC    <- c(0.062, 0.142)
CV.Cmax   <- c(0.053, 0.161)
CVw.AUC   <- CVw.Cmax <- data.frame(CVwT = rep(NA, 2*length(var.ratio)),
                                    CVwR = NA)
i         <- 0
for (j in seq_along(CV.AUC)) {
  for (k in seq_along(var.ratio)) { # split variances
    i <- i + 1
    CVw.AUC[i, 1:2]  <- CVp2CV(CV.AUC[j],  ratio = var.ratio[k])
    CVw.Cmax[i, 1:2] <- CVp2CV(CV.Cmax[j], ratio = var.ratio[k])
  }
}
res <- data.frame(metric = c(rep("AUC", length(CV.AUC)*length(var.ratio)),
                             rep("Cmax", length(CV.AUC)*length(var.ratio))),
                  CVw = c(rep(CV.AUC, each = length(var.ratio)),
                          rep(CV.Cmax, each = length(var.ratio))),
                  CVwT = signif(c(CVw.AUC$CVwT, CVw.Cmax$CVwT), 4),
                  CVwR = signif(c(CVw.AUC$CVwR, CVw.Cmax$CVwR), 4),
                  n = NA, power = NA)
for (j in 1:nrow(res)) {
  res[j, 5:6] <- sampleN.NTIDFDA(CV = c(res$CVwT[j], res$CVwR[j]),
                                 print=FALSE, details=F)[8:9]
}
print(res, row.names = FALSE)

metric   CVw    CVwT    CVwR  n   power
   AUC 0.062 0.05061 0.07161 18 0.83691
   AUC 0.062 0.06200 0.06200 24 0.81550
   AUC 0.062 0.07161 0.05061 44 0.81187
   AUC 0.142 0.11570 0.16420 12 0.82560
   AUC 0.142 0.14200 0.14200 16 0.83656
   AUC 0.142 0.16420 0.11570 28 0.83189
  Cmax 0.053 0.04326 0.06121 20 0.81502
  Cmax 0.053 0.05300 0.05300 30 0.81917
  Cmax 0.053 0.06121 0.04326 58 0.80154
  Cmax 0.161 0.13120 0.18630 12 0.83440
  Cmax 0.161 0.16100 0.16100 16 0.84655
  Cmax 0.161 0.18630 0.13120 26 0.80828

Given the wide range of sample sizes (12–58) it is difficult to predict what is suitable in your case. Even if you are optimistic and assume CVwT = CVwR, you end up with 16–30. Risky business without a pilot study.

❝ My second question is how can i generate a randomization plan with william's design using SAS software ?


No idea about SAS. I recommend Detlew’s package randomizeBE for [image].

Dif-tor heh smusma 🖖🏼 Довге життя Україна! [image]
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