Bioequivalence and Bioavailability Forum

Hi Imph

❝ I would like to know if this method can be applied to each pharmacokinetic measurement, for example to determine the duration of the sampling (5 x t1/2) or to calculate the predose limit of 5% Cmax when the Cmax is provided in the literature under mean+/-SD.

I don't know about "every" PK parameter, but FDA use this approach to plan partial AUC as well.

For example, for methylphenidate HCl ER tablet, among other parameters, FDA requests partial AUC_0--3h and AUC_0--4h for fasting and fed studies, respectively. The reason was that the Tmax was 2h under fasting and 3h under fed. the standard deviation (SD) of Tmax is about 0.5h, so 2 x SD is 1h, according to FDA:

> "For Tmax, two standard deviations = 1.0; Generally, approximately 95% of observations fall within two standard deviations of the mean; Thus, since the Tmax from the immediate-release portion of this formulation is about 2 hours under fasting conditions and 3 hours under fed conditions, pAUCs calculated to 0-3 hours in a fasting BE study and 0-4 hours in a fed BE study should capture the responses of 95% of the subjects. This should provide assurance that a test and reference product will be therapeutically equivalent over the early part of the daily dosing interval, corresponding to onset of response." (see ref linked here)

This is the same as what Helmut did for t_1/2 in his PPT cited in previous post.

❝ In the reference you gave, a normal distribution is assumed, but pharmacokinetic measurements do not generally follow a normal distribution but a log-normal one. Does this method still apply?

Well, Tmax is discrete so it is not normal distribution either. FDA still use it for the approximation for pAUC.

I'd use this 5 x (mean(PK) + 2xSD(PK)) instead of the empirical "7 x mean(PK)".

Hi Imph,

❝ I would like to know if this method can be applied to each pharmacokinetic measurement, for example to determine the duration of the sampling (5 x t1/2) or to calculate the predose limit of 5% Cmax when the Cmax is provided in the literature under mean+/-SD.

In principle, yes. I prefer to start with a single dose profile (even if read from a published figure) and perform nonparametric superposition.

❝ Why plan for 95% coverage (z0.95 = 1.96), is it related to alpha =5%.

Convention. You can use a more conservative value.

❝ In the reference you gave, a normal distribution is assumed, but pharmacokinetic measurements do not generally follow a normal distribution but a log-normal one. Does this method still apply?

You are right. But if the arithmetic mean ± SD is all you have… If you have results for the geometric mean, better, of course.

Hi Imph,

if you have individual half lives, you can calculate the geometric mean / SD and use any confidence interval you like. For planning the washout use the upper confidence limit.
-script (results in blue):

g.fun <- function(x, alpha = 0.05, digits = 4, print.only = FALSE) {
  # calculate geometric mean, SD, and CI of geometric mean
  if (sum(is.na(x)) > 0) { # Remove NA(s)
    message("NA(s) removed from the vector.")
    x <- x[-is.na(x)]
  }
  if (sum(x <= 0) > 0) {   # Only positive numbers are allowed
    message("Geometric mean applicable to",
            "\npositive numbers only; others removed.")
    x <- x[-which(x <= 0)]
  }
  mean.log <- mean(log(x))
  SD.log   <- sd(log(x))
  gMean    <- exp(mean.log)
  gSD      <- exp(SD.log)
  # confidence interval based on the t-distribution with n–1 degrees of freedom
  CI       <- exp(mean.log + c(-1, +1) *
                  qt(alpha / 2, length(x) - 1, lower.tail = FALSE) * SD.log)
  loc.dis  <- setNames(c(gMean, gSD, CI),
                       c("Geom. mean", "Geom. SD",
                         paste0("lower ", 100 * (1 - alpha), "%"),
                         paste0("upper ", 100 * (1 - alpha), "%")))
  if (print.only) {
    print(signif(loc.dis, digits))
  } else {
    return(loc.dis)
  }
}

# give your data in a t12 vector, e.g.,
# t12 <- c(7.15, 5.44, 5.33, ...) and proceed with descriptive statistics

# lognormal-distributed example data
set.seed(123456)
n   <- 24   # sample size
mue <- 6    # geometric mean of half life
CV  <- 0.25 # coefficient of variation
t12 <- rlnorm(n = n,
              meanlog = log(mue) - 0.5 * log(CV^2 + 1),
              sdlog = sqrt(log(CV^2 + 1)))

exp(summary(log(t12)))        # descriptive statistics
 Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
4.425   5.412   6.383   6.530   7.764  10.780

g.fun(t12, print.only = TRUE) # default: 95% CI
Geom. mean   Geom. SD  lower 95%  upper 95%
     6.530      1.275      3.951     10.790

alpha      <- 0.01            # conservative: 99% CI
g.fun(t12, alpha = alpha, print.only = TRUE)
Geom. mean   Geom. SD  lower 99%  upper 99%
     6.530      1.275      3.302     12.910

half.lives <- c(5, 7, 10)
col.name   <- paste0("upper ", 100 * (1 - alpha), "%")
res        <- data.frame(half.lives = half.lives,
                         washout = half.lives * g.fun(t12, alpha = alpha)[[col.name]])
res$days   <- ceiling(res$washout / 24) # round up washout (h) to days
print(round(res, 1), row.names = FALSE)
 half.lives washout days
          5    64.6    3
          7    90.4    4
         10   129.1    6

Hi Imph,

❝ […] Using the t1/2 +/-SD given in the SmPC, what sample size can be used? is it correct to use the sample size I plan to use in my study?

Yes.