Bioequivalence and Bioavailability Forum

Hi Vaibhav,

» Completed N =52
» 1. Cmax
» T/R 104.13
» CI: 97.89-110.77
» CV: 18.97
» Power: 99.99
» 2. AUCo-t
» T/R 102.20
» CI: 100.51- 103.92
» CV: 5.08
» Power:100

You should base the sample size estimation on C_max because it was worse than AUC_0–t (T/R-ratio more deviating from 100%, larger CV).

» Now are planning to conduct another BE study with same Test formulation (used in EMA) for ANVISA.
» […] can we proceed with sample size i.e. 54 or should re-calculate sample size based on EMA study results?

With 52 subjects power was extremely high. It’s questionable whether the IEC/IRB will accept a study with such a sample size. The same holds for the ANVISA (you have to submit the protocol before initiating the study). Possibly > 90% power will not be accepted (like other agencies ANVISA recommends 80–90%).

» T/R- 105%
» Power- 90%
» CV- Approx. 19%

Essentially there are two extreme approaches. One is all too often used – but stupid – and the other conservative. There are others in between. For details see this article.
I hope you have and the package PowerTOST. If not, see this article how to download/install them.
Start with:

library(PowerTOST) # attach it
# results of the previous study (eligible subjects, CV, T/R-ratio)
m      <- 52
CV     <- 0.1897
theta0 <- 1.0413
# target (desired) power of the planned study
target <- 0.90
You don’t have to specify alpha = 0.05 and design = "2x2" because they are defaults of the functions.

1. The ‘Carved in Stone’ approach, where you assume that in the next study you will get exactly the same results (T/R-ratio, CV) like in the previous one. Strong assumptions. Risky, at least.
   • sampleN.TOST(CV = CV, theta0 = theta0, targetpower = target)
   Gives:
   • +++++++++++ Equivalence test - TOST +++++++++++
            Sample size estimation
-----------------------------------------------
Study design: 2x2 crossover
log-transformed data (multiplicative model)

alpha = 0.05, target power = 0.9
BE margins = 0.8 ... 1.25
True ratio = 1.0413,  CV = 0.1897

Sample size (total)
 n     power
20   0.900234
   With your slightly more conservative assumptions you would get n = 22.
   
   
2. A Bayesian approach, where you take the uncertainties of the T/R-ratio and the CV into account. Remember, the T/R-ratio and CV are estimates and not natural constants.
   • expsampleN.TOST(CV = CV, theta0 = theta0, targetpower = target, prior.type = "both",
                prior.parm = list(m = m, design = "2x2"), details = FALSE)
   Gives:
   • ++++++++++++ Equivalence test - TOST ++++++++++++
  Sample size est. with uncertain CV and theta0
-------------------------------------------------
Study design:  2x2 crossover
log-transformed data (multiplicative model)

alpha = 0.05, target power = 0.9
BE margins = 0.8 ... 1.25
Ratio = 1.0413 with 50 df
CV = 0.1897 with 50 df

Sample size (ntotal)
 n   exp. power
26   0.909949
   Even such a conservative approach requires only ½ the sample size of the previous study and you can expect ≈91% power.

One caveat: Although you plan to use the same test product as in the EMA study, the reference will be different. Make sure that it is not worse (measured content, dissolution similarity). Otherwise, you should assume a more conservative T/R-ratio.