[Study As­sess­ment]

posted by Helmut Homepage – Vienna, Austria, 2019-07-27 23:53  – Posting: # 20456
Views: 529

Hi Datacollector,

looks familiar. :-D

Since we are in the same boat and out of curiosity: p-values of the sequence effect were given by the CRO and also by the BfArM with 0.0077 (AUC) and 0.0252 (Cmax).
Recalculated in Phoenix WinNonlin 8.1 and R 3.6.1 (function lm() of stats):

PK metric  p (period)  p (sequence)
  AUC        0.0321      2.42·10–5
  Cmax       0.0358      2.36·10–5

What‽


Extending what I wrote at the end of this post (not for you – as an obvious initiate – but the archive). Let’s assume that the two groups randomized to sequences TR and RT differ (by chance) in their body weights. Might happen cause we don’t stratify in a crossover for anything. Both T and R have a relative BA of 1. Hence, T/R should be 100%. I assumed that the response with a body weight of 70 would be exactly 1. Due to different volumes of distribution the response will be higher in the group with low BW and vice versa. Two cases (the responses are the means of groups):
  1. No period effects:
    group  BW  sequence  period  treatment  per. effect  response
      1    75     TR       1         T          1.00      0.933
      1    75     TR       2         R          1.00      0.933
      2    50     RT       1         R          1.00      1.400
      2    50     RT       2         T          1.00      1.400
    crossover
      T   = √0.933 × 1.400 = 1.143
      R   = √0.933 × 1.400 = 1.143
      T/R = 100% (unbiased)
    period 1
      T   = 0.933
      R   = 1.400
      T/R =  67% (bias –33%)
    period 2
      T   = 1.400
      R   = 0.933
      T/R = 150% (bias +50%)


  2. Extremely unequal period effects:
    group  BW  sequence  period  treatment  per. effect  response
      1    75     TR       1         T          1.25      1.167
      1    75     TR       2         R          1.25      1.167
      2    50     RT       1         R          0.50      0.700
      2    50     RT       2         T          0.50      0.700
    crossover
      T   = √1.167 × 0.700 = 0.904
      R   = √1.167 × 0.700 = 0.904
      T/R = 100% (unbiased)
    period 1
      T   = 1.167
      R   = 0.700
      T/R = 167% (bias +67%)
    period 2
      T   = 0.700
      R   = 1.167
      T/R =  60% (bias –40%)
Apart from Freeman’s theoretical stuff this shows clearly that is futile to analyze periods of a crossover separately. Given, such different body weights are unlikely. However, try it with a smaller difference. Estimates will always be biased.
One might be tempted to be prepared for the worst (i.e., bizarre deficiency letters) and aim at a stratified randomization keeping body weights as close as possible. But where will it end? Having a single slow metabolizer in one group and none in the other could already be the killer. Doesn’t make any sense.

Cheers,
Helmut Schütz
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