## 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, C

_{max},…).

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 (n
_{1}, n_{2}) 11 and 10 (planned for 22 but one dropout), α 0.05 (90% CI), lower CL 0.91, upper CL 1.15: Calculate the CV_{intra}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)`

and`design <- "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 CV_{wR}is the within-subject CV of the reference product.

*Dif-tor heh smusma*🖖🏼 Довге життя Україна!

_{}

Helmut Schütz

The quality of responses received is directly proportional to the quality of the question asked. 🚮

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