## PowerTOST: CVfromCI [Study As­sess­ment]

¡Hola Rosy!

» Dear Helmut,

Not interested in other members’ opinions?

» If i found an article of a BABE study in which 3 formulation where evaluated against to reference (A) in a 4 way crossover study, …

So far, so good.

» … but in a bioequivalence table appears for example A vs B treatment and its respective Geometric Mean Ratio (IC 90%).

Being A the reference, it should be B vs A, right?

» If i wanna calculate CV% intrasubject ussing POWER.TOST which desing i have to pick up (4x4 or parallel)? Of course 4x4 is the real design but the comparative A vs B is technically a parallel study.

Parallel‽
You have to find out whether the study was evaluated with a “pooled ANOVA” or according to the “Two‐at‐a‐Time Principle” (see this post).
Example: 4×4 crossover, n 24, 90% CI 85.00–106.18%.

############################################# n    <- 24        # total sample size l    <- 0.8500    # lower 90% CL u    <- 1.0618    # upper 90% CL ############################################# library(PowerTOST) des  <- "4x4"     # design and 1st evaluation eval <- "2x2"     # 2nd evaluation pe   <- sqrt(l*u) # calculate the PE CV.1 <- CVfromCI(lower=l, upper=u, design=des, n=n)  # 1 CV.2 <- CVfromCI(lower=l, upper=u, design=eval, n=n) # 2 cat(paste0("\n", des, " design, n = ", n,            sprintf("%s %.2f%%%s", "\n90% CI =", 100*l, "\u2013"),            sprintf("%.2f%%", 100*u),            sprintf(" %s %.2f%%)", "(PE =", 100*pe),            sprintf("\n  Pooled ANOVA           : CVintra = %.2f%%",                    100*CV.1),            sprintf("\n  Two-at-a-Time Principle: CVintra = %.2f%%",                    100*CV.2)), "\n")

You will get:

4x4 design, n = 24 90% CI = 85.00%–106.18% (PE = 95.00%)   Pooled ANOVA           : CVintra = 23.41%   Two-at-a-Time Principle: CVintra = 22.73%

How to discover which method was used?
Work backwards, i.e., see with which CV you can reproduce the reported results for each comparison. (Nonsense: See Detlews post below)

res.1 <- CI.BE(pe=pe, CV=CV.1, n=n, design=des) res.2 <- CI.BE(pe=pe, CV=CV.2, n=n, design=eval) cat(paste0("\nBack-calculated 90% CI by",     "\n  Pooled ANOVA           : ",     sprintf("%.2f%%%s", 100*res.1[["lower"]], "\u2013"),     sprintf("%.2f%%", 100*res.1[["upper"]]),     "\n  Two-at-a-Time Principle: ",     sprintf("%.2f%%%s", 100*res.2[["lower"]], "\u2013"),     sprintf("%.2f%%", 100*res.2[["upper"]]), "\n")) Back-calculated 90% CI by   Pooled ANOVA           : 85.00%–106.18%   Two-at-a-Time Principle: 85.00%–106.18%

Dif-tor heh smusma 🖖
Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮
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