Bioequivalence and Bioavailability Forum • PowerTOST question

jag009
★★★

NJ,
2013-08-21 18:06
(4683 d 01:13 ago)

Posting: # 11332
Views: 4,430

PowerTOST question [Software]

Hi all,

Can someone run powerTOST in R to confirm the ISCV from the following:

1. CVfromCI(lower=0.8341, upper=0.9536, n=55, design="2x3x3", alpha=0.05, robust=TRUE)

2. CVfromCI(lower=0.7909, upper=0.9258, n=55, design="2x3x3", alpha=0.05, robust=TRUE)

3. CVfromCI(lower=0.7985, upper=0.9240, n=55, design="2x3x3", alpha=0.05, robust=TRUE)

My Results from R:

1. 0.2456493 or 24.56%

2. 0.2905871 or 29.06%

3. 0.2685729 or 26.86%

The result for 2 and 3 is a bit strange. The CIs are very similar and yet the ISCVs differ by 0.032 or in percentage = 3.2%?

Here are the T/R ratios:

I am running the latest PowerTOST....

Thanks
John

d_labes
★★★

Berlin, Germany,
2013-08-22 11:21
(4682 d 07:58 ago)

@ jag009
Posting: # 11335
Views: 3,493

PowerTOST CVfromCI

Post reply

Dear John,

❝ Can someone run powerTOST in R to confirm the ISCV from the following:

Of course

.

Without repeating the numbers, my results are of course exactly the same as yours. Wouldn't expect something different since we are using the same software.

❝ The result for 2 and 3 is a bit strange.

Why? What is wrong with a 3.2% difference. IMHO not really big. Humans usually recognize 10% as a difference, f.i. 10 cent of a dollar, to desist from nickel nursers or from the Scotch (Beg your pardon Scotch ;-)

).

One caveat! From the documentation of CVfromCI():"It is assumed that the sequence groups in a crossover study or the treatment arms in a parallel-group study are balanced. The estimated CV is conservative (i.e. greater than actually observed) in case of unbalanced studies." Emphasis by me.

You may check this (your studies are unbalanced) if you recalculate the CI's from the point estimators, n and the obtained CV. Example 2 of your post:

n  <- 55

df <- n-3 # robust

bk <- 1.5 # design constant for 2x3x3

CV <- 0.2905871

point <- 0.8557

# CI in the log-transformed domain

CI <- log(point) + c(-1,1)*qt(0.95,df)*sqrt(bk*CV2mse(CV)/n)

CI <- exp(CI) # back transformation

gives nearly the same interval than the original:

0.7909039 0.9258046 # back calculated

0.7909    0.9258    # original

Seems the un-balancedness is not a problem here.

—
Regards,

Detlew