Bioequivalence and Bioavailability Forum

Dear Daniel

nice to read from you again!

❝ I would like to know if it is possible to calculate the CV total considering the point estimate (PE) and the confidence limits of a parallel BE study.

Yes. Package PowerTOST for R is very helpful.

❝ For this study Cmax - PE = 1.08 and CI = 0.82-1.18. Number of volunteers = 122.

Hhm, this is strange. The CI should be symmetrical around the PE in the log-domain, or in other words the geometric mean of the CI: \(PE=\sqrt{CL_{lower}\times CL_{upper}}\) which is 0.98 and not 1.08…
Your CI is symmetrical around 100%. Can you please check in the report whether these are the – obsolete for decades! – Westlake’s confidence intervals?* If yes, the following is useless. Try to find the classical (shortest) CI in the report.

library(PowerTOST)
100*CVfromCI(lower=0.82, upper=1.18, n=122, design="parallel")
[1] 66.653 (CV_total 66.7%)

❝ Another information is the root mean square error, Cmax = 0.592

Are you sure that this value comes from log-transformed data? If yes, CV_total would be \(100\sqrt{e^{0.592^2}-1}\sim 64.8\%\) or more comfortably:
100*se2CV(0.592)
[1] 64.78628

❝ I would like to confirm if a number of volunteers of 126 would be adequate for a parallel BE study considering these information.

Let’s see (power 80%, acceptance range 0.80–1.25, CV_total 66.7%); first sample size column with ‘fixed CV’ (n1)¹, second one taking uncertainty of the CV from the first study into account (n2)². Columns n3 and n4 are for CV 64.8%:
expected PE   n1   n2   n3   n4
    0.90     658  666  628  636
    0.95     318  324  304  308
    0.98     264  268  252  256
    1.00     256  260  244  248
    1.05     312  316  296  302
    1.08     428  434  408  414
    1.10     560  566  532  540

Seems that the first study passed by luck. No fun planning a new one.

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Example code for PE 0.95:

1. library(PowerTOST)
sampleN.TOST(targetpower=0.8, theta0=0.95, CV=0.667, design="parallel")
   
2. library(PowerTOST)
expsampleN.TOST(targetpower=0.8, theta0=0.95, CV=0.667, dfCV=122-2, design="parallel")

• The classical CI is T/R ± t_0.05·variability where α_lo = α_hi = 0.05 and total α = 0.10. Westlake suggested to select α_lo and α_hi in such a way that the resulting CI is symmetrical around 100%. If PE ≠ 1 that would require α_lo ≠ α_hi. By an iterative process t-values are found where α_lo + α_hi = 0.10 still holds.^a This procedure was soon criticized since no information about the location is available if only the CI is reported.^b With a little trial an error it should be possible to get the classical CI from Westlake’s if both the PE and the sample size are known.

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1. Westlake WJ. Symetrical Confidence Intervals for Bioequivalence Trials. Biometrics. 1976;32(4):741–4. doi:10.2307/2529259.
   
2. Mantel N. Do We Want Confidence Intervals Symetrical About the Null Value? Biometrics. 1977;33(4):759–60.