Bioequivalence and Bioavailability Forum

❝ But one thing worries: the article concludes that all tests are rather weak if sample size is below 30.

❝ What to think about BE-studies with samples 18, 24?

• The basic equation of PK:
    AUC = FD/CL,
  which can be linearized by taking the logarithms to
    ln(AUC) = ln(F) + ln(D) – ln(CL).
  Now we can apply an ANOVA, which requires additive effects.
  OK, we are interested in comparing the Fs of product (assuming equal doses and clearances). So,
    BA = AUC_T/AUC_R or ln(BA) = ln(AUC_T) – ln(AUC_R). Nice.
  
• Multiplicative errors in bioanalytics due to serial dilution of stock / spiking solutions.

• Strictly speaking the distribution of PK metrics is irrelevant (as long as they are identical). In the BE-context only the within-subject residuals of the model should be normal. It might well be that residuals of two extremely skewed (not normal, but identical) distributions still give nomal residuals.
  
• If these residuals are not normal, one could opt for a nonparametric method. Well, cough…

x1 <- rweibull(40, 1, 1) # Weibull distribution
x2 <- rweibull(40, 1, 1)
d  <- x1-x2
op <- par(no.readonly=T)
par(mfrow=c(3, 1))
hist(x1, freq=F, xlim=c(0, 8), xlab="", las=1, main="x1")
legend("topright", bty="n",
  legend=paste("p =", signif(shapiro.test(x1)$p.value, 3)))
hist(x2, freq=F, xlim=c(0, 8), xlab="", las=1, main="x2")
legend("topright", bty="n",
  legend=paste("p =", signif(shapiro.test(x2)$p.value, 3)))
hist(d, freq=F, xlim=c(-6, 6), xlab="", las=1, main="x1 \u2012 x2")
legend("topright", bty="n",
  legend=paste("p =", signif(shapiro.test(d)$p.value, 3)))
par(op)