BE parallel design [General Sta­tis­tics]

posted by Jaime_R – Barcelona, 2007-04-04 15:14 (5408 d 15:18 ago) – Posting: # 623
Views: 15,442

Dear PKPDPKPD!

» Should AUC and Cmax be also calculated from the log transformed data?

See my first post. You should only transform the calculated PK parameters (not the concentrations).
  1. Calculate AUC from your analytical results by any method you like (trapezoidal rule preferred, but not limited to).
  2. Cmax is simply the highest measured concentration.
  3. For the comparison you log-transform AUC and Cmax.
Second an apology: yesterday I was too fast (referring to my memory and not a textbook - I'm also using software and not explicit formulas). Therefore a little correction (valid for unequal group sizes):
Let's misuse Helmut's data (download here) ;-)
Although his data are from a cross-over study, we will use only period 1.
  1. Change the header of the first column from 'Seq' to 'Trt'
  2. Delete the column 'Rand'
  3. Delete the column 'P2'
Now we have data from a parallel study where Trt 1 = reference and Trt 2 = test, Group 1 = Subjects 1-12, Group 2 = Subjects 13-24, and Response (your PK parameter) in column 'P1'.
  1. log-transform 'P1'
  2. calculate separately for each treatment:
    • arithmetic mean: (1: 3.56227, 2: 3.38383)
    • exp(arithmetic mean): (1: 35.24321, 2: 29.48349),
    • note: these are the geometric means of untransformed data!
    • standard deviations: (1: 0.35950, 2: 0.42377)
    • variances = SD²: (1: 0.12924, 2: 0.17958)
    • n1,2 (group sizes): (1: 12, 2: 12)
    • Q1,2 = variance × (n1,2-1): (1: 1.42165, 2: 1.97539)
  3. calculate R = sqrt[(n1+n2)/(n1×n2)×(Q1+Q2)/(n1+n2-2)]: 0.16042
  4. look up the critical value of the t-distribution for alpha=0.05 with n1+n2-2 degrees of freedom: t 1.71714
  5. calculate t × R: 0.27547
  6. calculate Delta (difference of means Trt 2 - Trt 1): -0.17844
  7. antilog Delta (= point estimate): 0.83657
  8. calculate lower/upper 90% confidence limit = Delta ± t × R: lo: -0.45391, hi: 0.09702
  9. antilog lo and hi: exp(lo): 0.63514, exp(hi): 1.10189
So the final results are (point estimate and 90% confidence interval):
83.657% (63.514% - 110.189%)

I checked the 'manual' calculation in WinNonlin and EquivTest:
WinNonlin: 83.6572% (63.5100% - 110.1958%)
EquivTest: 83.66% (63.51% - 110.18%)

Slight differences seen in results are not uncommon... :-D

Regards, Jaime

Complete thread:

Activity
 Admin contact
21,844 posts in 4,571 threads, 1,555 registered users;
online 2 (0 registered, 2 guests [including 2 identified bots]).
Forum time: Monday 05:33 CET (Europe/Vienna)

Do, or do not.
There is no ‘try’.    Yoda

The Bioequivalence and Bioavailability Forum is hosted by
BEBAC Ing. Helmut Schütz
HTML5