Lack of ISR / only 5% ISR in study with 656 samples [Regulatives / Guidelines]

posted by ElMaestro  – Denmark, 2013-04-24 01:31 (4420 d 18:20 ago) – Posting: # 10480
Views: 14,592

Cool code Helmut.

Run your code, then do shapiro.test(data) and discuss if the high p-value proves that the data are Gaussian.
Next do this:
data   <- c(-1.05,  -6.38, -5.30,  -5.55,  -2.99,  -2.24, +10.75, +0.90,
            -6.66,  -5.01, -2.74, -10.98,  +4.43,  -3.53,  -8.85, -7.38,
            -8.35, +11.13, -4.07, -15.58,  +2.05, -14.86,  -3.34, +2.66,
            -6.80,  +5.60, -2.48,  +5.61, -11.71, +10.61,  -4.00, -1.73)
                            # Observed deviations
size   <- 656               # No. of samples in a study
ifelse(size < 1000,         # EMA: 10% if <1000; +5% for >1000
  ISR.no <- ceiling(size*0.1),
  ISR.no <- 100 + ceiling((size-1000)*0.05))
sims   <- 1e6               # No. of simulated studies
failed <- 0                 # ISR failed criterion (≤1/3 of repeats >20% dev.)
for (i in 1:sims) {
  xboot = sample(data, replace=T)
  y      <- length(xboot[abs(xboot)>20]) # No. of abs. deviations > 20%
  y      <- y/ISR.no        # Fraction of repeats with > 20% deviation
  if (y > 1/3) failed <- failed + 1
}
cat(sprintf("%i",sims),"simulated studies with", size, "samples:\n",
  "Method: Bootstrapping (Mersenne-Twister) of ISR data\n",
  "Failed criterion (deviations >20% in >1/3 of repeats) seen in",
  sprintf("%.2f%% %s", 100*failed, "of studies.\n"))


...and argue that even when you do not make any assumption of ISR distribution you are still very clearly very far from any sort of trouble.

Pass or fail!
ElMaestro

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