CV from CI [Power / Sample Size]
❝ […] how to get the CV(intra) when little or no (or not reliable) prior info available. I went over the "Biostatistics Sample Size Estimation for BE Studies" presentation […] and at one point in one of the "Hints" slide it says that "All you need is the 90% geometric confidence interval and the sample size."
I guess you are referring to this presentation (slide 24), right?
❝ Am I understanding correctly that the CLlow and CLhigh would be the 90% confidence limits for a prior CV intra…
Essentially yes.
<nitpick>
The 100(1–2α) CI is calculated from
- the geometric mean ratio (GMR),
- the number of subjects / sequence,
- the t-value (for given α and degrees of freedom specific for the design), and
- the mean square error (which is an estimate of the variance).
It’s not rocket science to perform the calculation “backwards” (get the unknown #4 from the known #1–#3 and the CI) and then \(CV_{intra} = \sqrt{e^{MSE} - 1}\). If the GMR is not known, use \(\sqrt{CL_{lower} \times CL_{upper}}\).
</nitpick>
❝ I'm asking that because of the "...geometric confidence interval...". Why are we talking here about that?
I wanted to make clear that we need for the calculations outlined in slides 25–28 the confidence interval derived from log-transformed data (which is required for most PK metrics like AUC, Cmax,…).
Note that the calculation in the presentation is outlined for 2×2 crossover designs. It can be a little bit tricky to perform the calculations “manually” if you are unsure how the degrees of freedom in other designs are defined. I recommend the package
PowerTOST
for R (open source & free of costs). 
Examples:
- 2×2 crossover, subjects in the two sequence (n1, n2) 11 and 10 (planned for 22 but one dropout), α 0.05 (90% CI), lower CL 0.91, upper CL 1.15: Calculate the CVintra and then the CI in order whether the input was correct (should give the original CI).
library(PowerTOST)
alpha <- 0.05
lower <- 0.91
upper <- 1.15
n <- c(11, 10)
design <- "2x2"
CV <- CVfromCI(alpha=alpha, lower=lower, upper=upper,
n=n, design=design)
CI <- CI.BE(alpha=alpha, pe=sqrt(lower*upper), CV=CV,
n=n, design=design)
cat("\n CV =", sprintf("%.2f%%", 100*CV),
"\n", sprintf("%g%%", 100*(1-2*alpha)), "CI =",
paste(CI, collapse = " to "), "\n")
Should give
CV = 22.17%
90% CI = 0.91 to 1.15
- As above but the study was a parallel design in 2×20 subjects. Use
n <- c(20, 20)
anddesign <- "parallel"
to get
CV = 22.22%
90% CI = 0.91 to 1.15
Note that this is the total (or pooled) CV and >2 groups are not supported by the function.
- If the study was a replicate design for reference-scaling evaluated for Average Bioequivalenc with Expanding Limits (ABEL: EMA, Health Canada, Russia, and many others): You need only the upper expanded limit (neither the CI, the design, or the sample size).
U <- 1.38
CV <- CVwRfromU(U=U)
cat("\n CVwR =", sprintf("%.2f%%", 100*CV), "\n")
Should give
CVwR = 44.35%
Note that CVwR is the within-subject CV of the reference product.
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Helmut Schütz
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Science Quotes
Complete thread:
- Finding the CI limits Louis52 2018-02-01 19:57 [Power / Sample Size]
- CV from CIHelmut 2018-02-02 00:56
- CV from CI Louis52 2018-02-03 17:12
- CV from CI Helmut 2018-02-03 21:48
- CV from CI Louis52 2018-02-04 14:38
- CV from CI Helmut 2018-02-05 11:39
- CV from CI Louis52 2018-02-04 14:38
- CV from CI Helmut 2018-02-03 21:48
- CV from CI Louis52 2018-02-03 17:12
- CV from CIHelmut 2018-02-02 00:56