Bioequivalence and Bioavailability Forum

Hi earlybird!

Sorry, I'm too short in time to do the algebra for you - but two hints:

• D. Labes' post about a partial replicate.
  
• Chapter 9.4.2 of Chow and Liu (2009).

Dear Earlybird,

you need to know the sample size additional to the numbers given by you.

Then you can use the formula
  [L,U]=log(point)+t(1-alpha,3*N-4)*s*sqrt(1/N)
where L=log(lower) and U=log(upper) (see Chapter 9.4.2 of Chow and Liu (2009)) to obtain s by a little algebra, easy for you .
N is the total sample size and it is assumed that each sequence has the same number of subjects assigned.
If not, the square root reads sqrt(1/16*(1/n1 + 1/n2 +1/n3 +1/n4)) where the n_i are the numbers of subjects assigned to the four sequences.

The degrees of freedom for the critical t-value are without carry over in the model. If your numbers are from a model with carry-over reduce them by 1.

From the obtained s you get the coefficient of variation according to the well known formula
  CV=sqrt(exp(s2)-1)*100   (in %)

Hope this helps.

Dear Earlybird, dear All,

for those who have access to R and have installed the package PowerTOST here an add-on function:
require(PowerTOST)
CVfromCI <- function(point,lower,upper,n,design="2x2",alpha=0.05)
{
  if (missing(lower) | missing(upper)) stop("Lower and upper CL must be given!")
  if (missing(n)) stop("Sample size must be given!")
 
  if (missing(point)) point <- sqrt(lower*upper)
 
  # get design characteristics
  d.no <- .design.no(design)
  if (is.na(d.no)) stop("Unknown design!")
  desi <- .design.props(d.no)
  dfe  <- parse(text=desi$df[1],srcfile=NULL) #degrees of freedom as expression

  tval <- qt(1-alpha,eval(dfe))
  s1 <- (log(point)-log(lower))/sqrt(desi$bk/n)/tval
  s2 <- (log(upper)-log(point))/sqrt(desi$bk/n)/tval
  sw <- 0.5*(s1+s2)
  # both estimates very different?
  if (abs(s1-s2)/sw>0.1) warning("sw1, sw2 very different. Check input.")
  return(se2CV(sw))
}

With this function you can obtain the CV from each design known within PowerTOST.
This function will be included in the next release of PowerTOST if the author of that package will accept the code . Any enhancement or comment welcome.

Taking the numbers from the first post in this thread and assuming 24 subjects one gets:
CVfromCI(point=1.03,lower=0.93,upper=1.15,n=24,design="2x4x4")
[1] 0.3196356
i.e. CV=32%.