'CVcap' [Power / Sample Size]

posted by Helmut Homepage – Vienna, Austria, 2024-02-06 14:51 (73 d 20:31 ago) – Posting: # 23859
Views: 754

Hi Mahmoud,

❝ what do mean about 'CVcap' = 0.2142

Quoting my previous post:

If the CVwR >21.42% the additional criterion “must pass 80–125%” becomes increasingly important and power drops.

It’s not explicitly stated in the FDA’s guidances (therefore, we give it in quotes) but can be easily derived with a little algebra. In a nutshell: Limits are scaled based on \(\small{CV_\text{wR}}\) with the FDA’s regulatory constant \(\small{\theta_0}\) in the first place. But for any \(\small{CV_\text{wR}>21.42\%}\) that would result in implied limits \(\small{\left\{L,U\right\}}\), which are wider than 80.00 – 125.00%. That’s not we want for an NTID.$$\eqalign{
\Delta&=1/0.9\approx 1.11111\\
\left\{L,U\right\}&=100\exp(\mp\theta_0\cdot s_\text{wR})
}$$Practically the limits are scaled indeed, but if the study would have passed, additionally inclusion within the conventional 80.00 – 125.00% limits is assessed as well.
That’s numerically the same as if scaling would be ‘capped’ at \(\small{CV_\text{wR}=21.42\%}\).

If algebra is not your thing, try this:

fun <- function(x) {
  Delta   <- 1.11111 # aproximate; only the FDA knows why
  sigma.0 <- 0.10
  theta.0 <- log(Delta) / sigma.0
  swR     <- sqrt(log(x^2 + 1))
  # the implied upper limit which is ≈1.25
  U       <- exp(theta.0 * swR)
  return(U - 1.25)
CVcap  <- uniroot(fun, interval = c(0.01, 0.3),
                  extendInt = "upX")$root # numerically find the CV where U ≈1.25
cat(sprintf("\'CVcap\' = %.4g\n", CVcap))

'CVcap' = 0.2142

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