Bioequivalence and Bioavailability Forum

Dear all,

I stumbled across this statement in the MR guideline:

5.1.2. Variability
The inter-individual variability of the pharmacokinetic parameters of interest should be determined in the single dose or multiple dose studies […] and should be compared between the modified and immediate release formulation. The variability for the modified release formulation should preferably not exceed that for the immediate release formulation unless it is adequately justified in terms of potential clinical consequences.

(my emphases)
IMHO, that calls for a one-sided test (non-superiority). Between-subject variability is nasty. Even worse, if the drug is subjected to polymorphic metabolism it can be much larger than within-subject variability.
BTW, it’s funny to ask for it because one doesn’t obtain the between-subject variability from the EMA’s “all effects fixed” model.
Another ambiguity: It’s possible the extract CV_b and CV_w from a crossover study but not separate for the treatments. Only if we assume no period effects we can get the total (pooled) CV of T and R. Is that meant and only the all too common sloppy terminology is used by the EMA?

Let’s explore an example of a HVD(P) where reference-scaling is acceptable for C_max and C_τ. The within-subject CVs are 30, 50, and 60% for AUC, C_max, and C_τ, respectively. The between-subject variabilities are twice the within-subject variabilities. The assumed T/R-ratios are 0.95 for AUC and for 0.90 for the concentrations. The non-superiority margin is 1.25. 4 period full replicate design, ≥80% power. scripts at the end.
I got following sample sizes:

design metric variability  CV theta0 margin          method   n
 2x2x4    AUC      within 0.3 0.9500     NA             ABE  20
 2x2x4    AUC     between 0.6 1.0526   1.25 Non-superiority  84
 2x2x4   Cmax      within 0.5 0.9000     NA            ABEL  28
 2x2x4   Cmax     between 1.0 1.1111   1.25 Non-superiority 394
 2x2x4   Ctau      within 0.6 0.9000     NA            ABEL  32
 2x2x4   Ctau     between 1.2 1.1111   1.25 Non-superiority 506

I beg your pardon. If one takes this seriously, it’s a show stopper.

OK, seems that I’m on the wrong track. Here I’m testing non-superiority of the PK metrics. I’m not comparing their variabilities. How should we do that? The conventional F-test wouldn’t do cause it is for independent data. What about Pitman-Morgan? An example of a last year’s hybrid for Health Canada (T = MR 20 mg o.a.d [n 40], R = IR 10 mg b.i.d [n 39]; C_max and C_τ data, R-script at the end):

         Paired Pitman-Morgan test

data:  Cmax.T and Cmax.R
t = -0.10458, df = 37, p-value = 0.5414
alternative hypothesis: true ratio of variances is greater than 1
95 percent confidence interval:
 0.6826989       Inf
sample estimates:
variance of x variance of y
   0.09814356    0.10036611

 metric treatment   variance       CVp
   Cmax         T 0.09814356 0.3211248
   Cmax         R 0.10036611 0.3249240

        Paired Pitman-Morgan test

data:  Ctau.T and Ctau.R
t = 0.96519, df = 37, p-value = 0.1704
alternative hypothesis: true ratio of variances is greater than 1
95 percent confidence interval:
 0.852144      Inf
sample estimates:
variance of x variance of y
    0.3531756     0.2845929

 metric treatment  variance       CVp
   Cmax         T 0.3531756 0.6508311
   Cmax         R 0.2845929 0.5737777

In my hybrid applications I was never asked for a test so far (I just reported the CV).
What are your experiences?

NB, comparing pooled variances assumes no period effects. What about replicate designs? Work with the subjects’ geometric means / treatment? For the EMA’s imbalanced and incomplete data set I:

        Paired Pitman-Morgan test

data:  logDATA.T.means and logDATA.R.means
t = -1.1499, df = 75, p-value = 0.8731
alternative hypothesis: true ratio of variances is greater than 1
95 percent confidence interval:
 0.7506468       Inf
sample estimates:
variance of x variance of y
    0.7383189     0.8301172

 metric treatment    CVp     CVw
logDATA         T 1.0452 0.35157
logDATA         R 1.1374 0.46964
logDATA    pooled 1.0912 0.41362

As expected pooled CVs > within-subject CVs. 
Little bit cheating cause for subjects with only one observation I took it as it is. Not for this example but in the partial replicate it is possible to have one or two observations of R and none of T. Maybe this stuff helps.*

If you know a reference for calculating power of the Pitman-Morgan test, please let me know. Chow/Liu propose simulations in Chapter 7.5.

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• Derrick, B, Ruck A, Toher D, White P. Tests for Equality of Variances between Two Samples which Contain Both Paired Observations and Independent Observations. J App Quant Meth. 2018;13(2):36–47. Open access.

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################
# Sample sizes #
################
library(PowerTOST)
metric   <- c("AUC", "Cmax", "Ctau")
CVw      <- c(0.3, 0.5, 0.6)    # within-subject (AUC, Cmax, Ctau)
CVb      <- c(0.6, 1.0, 1.2)    # between-subject (AUC, Cmax, Ctau)
theta0e  <- c(0.95, 0.90, 0.90) # assumed T/R-ratio for equivalence
theta0s  <- 1/theta0e           # assumed T/R-ratio for non-superiority
margin   <- 1.25                # Non-superiority margin
design   <- "2x2x4"             # TRTR|RTRT
res      <- data.frame(design = design, metric = rep(metric, each = 2),
                       variability = rep(c("within", "between"), 3),
                       CV = c(CVw[1], CVb[1], CVw[2], CVb[2], CVw[3], CVb[3]),
                       theta0 = c(theta0e[1], theta0s[1],
                                  theta0e[2], theta0s[2],
                                  theta0e[3], theta0s[3]),
                       margin = rep(c(NA, margin), 3),
                       method = c("ABE", "Non-superiority",
                                  rep(c("ABEL", "Non-superiority"), 2)), n = NA,
                       stringsAsFactors = FALSE)
for (j in 1:nrow(res)) {
  if (j %%2 == 0) { # Non-superiority
    res$n[j] <- sampleN.noninf(CV = res$CV[j], design = design,
                               theta0 = res$theta0[j],
                               margin = res$margin[j], details = FALSE,
                               print = FALSE)[["Sample size"]]
  } else {
    if (j == 1) {    # ABE
      res$n[j] <- sampleN.TOST(CV = res$CV[j], design = design,
                               theta0 = res$theta0[j], details = FALSE,
                               print = FALSE)[["Sample size"]]
    } else {         # ABEL
      res$n[j] <- sampleN.scABEL(CV = res$CV[j], design = design,
                                 theta0 = res$theta0[j], details = FALSE,
                                 print = FALSE)[["Sample size"]]
    }
  }
}
res$theta0 <- signif(res$theta0, 5)
print(res, row.names = FALSE)

##############################
# Hybrid application example #
##############################
library(PowerTOST)
library(PairedData)
# requires normal distributed data; hence, log transform
df <- data.frame(subject = 1:40,
                 Cmax.T = log(c( 4.588, 4.056, 4.068,  5.222, 4.890, 8.051, 7.453,
                                 7.236, 6.057, 4.009,  5.658, 6.374, 6.062, 5.362,
                                11.468, 7.409, 6.548,  6.983, 7.708, 3.913, 6.971,
                                 4.867, 5.751, 3.766, 10.015, 4.134, 3.809, 4.956,
                                 3.380, 6.506, 7.700,  3.709, 4.148, 3.363, 6.491,
                                 4.869, 5.172, 4.532,  2.999, 3.494)),
                 Cmax.R = log(c( 5.787, 4.947, 4.113, 5.599, 5.857,    NA, 6.329,
                                 7.099, 4.114, 4.824, 4.330, 7.070, 5.950, 4.270,
                                13.264, 9.765, 6.709, 5.769, 6.277, 4.676, 6.662,
                                 5.295, 5.517, 4.425, 8.692, 3.794, 4.226, 5.009,
                                 2.816, 7.168, 4.386, 3.612, 5.539, 4.407, 4.615,
                                 8.683, 4.612, 3.537, 3.413, 3.457)),
                 Ctau.T = log(c(0.39, 0.13, 0.15, 0.56, 0.60, 0.24, 0.21, 0.14,
                                0.17, 0.31, 0.51, 0.19, 0.46, 0.37, 0.78, 0.37,
                                0.12, 0.21, 0.31, 0.51, 0.24, 0.22, 0.54, 0.34,
                                1.36, 0.47, 0.49, 0.26, 0.19, 0.13, 0.20, 0.14,
                                0.22, 0.42, 0.37, 0.11, 0.45, 0.21, 0.19, 0.12)),
                 Ctau.R = log(c(0.33, 0.16, 0.19, 0.12, 0.46,   NA, 0.19, 0.17,
                                0.19, 0.30, 0.60, 0.17, 0.45, 0.29, 0.81, 0.53,
                                0.15, 0.26, 0.37, 0.46, 0.20, 0.17, 0.59, 0.43,
                                0.93, 0.29, 0.37, 0.28, 0.19, 0.15, 0.26, 0.16,
                                0.33, 0.25, 0.22, 0.46, 0.38, 0.22, 0.14, 0.10)))
# create objects of class paired
paired.Cmax <- with(df, paired(Cmax.T, Cmax.R))
paired.Ctau <- with(df, paired(Ctau.T, Ctau.R))
# Pitman-Morgan tests
# only complete data are used, i.e., subject’s 6 missing R is dropped
PM.Cmax     <- Var.test(paired.Cmax, alternative = "greater")
PM.Ctau     <- Var.test(paired.Ctau, alternative = "greater")
# pooled variances and CVs
Cmax        <- data.frame(metric = "Cmax", treatment = c("T", "R"),
                          variance = c(PM.Cmax$estimate[["variance of x"]],
                                       PM.Cmax$estimate[["variance of y"]]),
                          CVp = c(mse2CV(PM.Cmax$estimate[["variance of x"]]),
                                  mse2CV(PM.Cmax$estimate[["variance of y"]])))
Ctau        <- data.frame(metric = "Cmax", treatment = c("T", "R"),
                          variance = c(PM.Ctau$estimate[["variance of x"]],
                                       PM.Ctau$estimate[["variance of y"]]),
                          CVp = c(mse2CV(PM.Ctau$estimate[["variance of x"]]),
                                  mse2CV(PM.Ctau$estimate[["variance of y"]])))
PM.Cmax; print(Cmax, row.names = FALSE); PM.Ctau; print(Ctau, row.names = FALSE)

##################
# EMA Data set I #
##################
library(replicateBE)
library(PowerTOST)
library(PairedData)
var.pool <- function(var, n) {
  if (!length(var) == length(n)) stop
  return(sum(var*(n-1))/(sum(n)-2))
}
EMA1 <- rds01[, c(1, 4, 6)]
names(EMA1)[3] <- "logDATA"
T.means <- aggregate(.~subject,
                     data = EMA1[EMA1$treatment == "T", ], mean)[, c(1, 3)]
R.means <- aggregate(.~subject,
                     data = EMA1[EMA1$treatment == "R", ], mean)[, c(1, 3)]
n       <- c(nrow(T.means), nrow(R.means))
names(T.means)[2] <- "logDATA.T.means"
names(R.means)[2] <- "logDATA.R.means"
EMA1.means        <- merge(T.means, R.means, by = "subject")
paired.logDATA    <- with(EMA1.means,
                          paired(logDATA.T.means, logDATA.R.means))
PM.logDATA        <- Var.test(paired.logDATA, alternative = "greater")
x <- as.numeric(method.A(data = rds01, print = FALSE,
                         details = TRUE)[c(11:12)])
options(digits = 7)
var.p <- c(PM.logDATA$estimate[["variance of x"]],
           PM.logDATA$estimate[["variance of y"]]) # pooled (total) T and R
var.w <- c(CV2mse(x[1]/100), CV2mse(x[2]/100))     # within T and R
res   <- data.frame(metric = "logDATA", treatment = c("T", "R", "pooled"),
                    CVp = c(mse2CV(var.p), mse2CV(var.pool(var.p, n))),
                    CVw = c(mse2CV(var.w), mse2CV(var.pool(var.w, n))))
res[, 3:4] <- signif(res[, 3:4], 5)
PM.logDATA; print(res, row.names = FALSE)