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Back to the forum  2018-06-25 17:09 CEST (UTC+2h)

Leave-One-Out (IBD) [Design Issues]

posted by Helmut Homepage - Vienna, Austria, 2018-06-12 12:42  - Posting: # 18888
Views: 430

Hi Irene,

» After I tried to run the SAS code (Example 4.5), I got the CI, but there is only one Residual value. I guess I suppose to calculate the intra-subject CV from this value, right? So, My question is, Do we only get one Intra-subject CV from three-comparison of the treatment (T-R, S-T, S-R) while we got three CI (T-R, S-T, S-R)?

I’m not equipped with [image] … Hence, my understanding of the code is limited.
You are right, in the model you get only one (pooled) estimate of the variance. That’s not a good idea, since it will “work” only if intra-subject variances would be identical and the treatment differences at least very similar. Otherwise, the treatment estimates will be biased and the type I error is not controlled. See also this post.
At the 2nd Conference of the Global Bioequivalence Harmonisation Initiative (Rockville, September 2016) Pina D’Angelo gave a presentation “Testing for Bioequivalence in Higher‐Order Crossover Designs: Two‐at‐a‐Time Principle Versus Pooled ANOVA” showing exactly that. Here some of her slides:

Purpose

  • To determine which method of statistical analysis
    is more appropriate to conclude bioequivalence in
    higher‐order crossover studies:
    1. The two‐at‐a‐time principle using two separate
      incomplete block design ANOVAs
    2. A pooled approach using one ANOVA and a common
      error term for the two contrasts

Introduction


Statistical Concerns:

  1. Different means (point estimates) between formulations
  2. Different variances between formulations
  • If either situations exist, which method of analysis
    reduces bias the most: two‐at‐a‐time principle or
    pooled ANOVA?

Simulated Data: Summary

  • When all three treatments have similar means and there is
    homogeneity of variances, both methods give very similar
    results.
  • When treatment means differ but there is homogeneity of
    variances, both methods give very similar results. With higher
    variability, the power is slightly increased when using the pooled
    ANOVA method.
  • When treatment means are similar but variances are not
    homogeneous, the two‐at‐a‐time method gives higher power to
    detect BE for the treatment with lower variability
  • When treatment means differ and variances are not
    homogeneous, the two‐at‐a‐time method increases power to
    detect BE for the treatment with lower variability. Moreover,
    type I error is higher when using the pooled ANOVA method.

Closing Remarks

  • Using a two‐at‐a‐time principle for statistical analysis of a
    higher‐order pilot study will have more value for decision-
    making on which multiple tests lots will be selected for use
    in a pivotal study based on the pilot study results. The
    intra‐subject variability of a specific test‐to‐reference
    comparison can be determined using the two‐at‐a‐time
    principle, which may be an important factor in selecting a
    test product considering test products are generally
    formulated to show different characteristics for testing in a
    pilot study.
(my emphasis)

If you are interested in all pairwise comparisons, generate three data sets (values or T&R, S&R, T&S, i.e., exclude all values of the respective other treatment S, T, R) whilst keeping the codes for sequence and period. This will give you data sets which represent an IBD (incomplete block design). Run the usual model on them.
This approach is also recommended in the EMA’s BE-GL:

In studies with more than two treatment arms (e.g. a three period study including two references, one from EU and another from USA […], the analysis for each comparison should be conducted excluding the data from the treatments that are not relevant for the comparison in question.

(my emphasis)
Example 4.5 (Phoenix/WinNonlin 8, mixed effects, Satterthwaite’s degrees of freedom):
                           PE (%)    90% CI (%)       s²w    CVw (%)
pooled model T vs. R  AUC  116.15 (108.97, 123.81)  0.043954  21.20
             T vs. R  Cmax 129.65 (118.84, 141.45)  0.084569  29.71
             S vs. R  AUC  140.63 (131.93, 149.90)  0.043954  21.20
             S vs. R  Cmax 159.75 (146.42, 174.30)  0.084569  29.71
             T vs. S  AUC   82.60 ( 77.46,  88.07)  0.043954  21.20
             T vs. S  Cmax  81.16 ( 74.35,  88.56)  0.084569  29.71
IBD models   T vs. R  AUC  116.05 (108.92, 123.65)  0.042525  20.84
             T vs. R  Cmax 129.54 (119.26, 140.71)  0.074744  27.86
             S vs. R  AUC  141.09 (131.39, 151.51)  0.053706  23.49
             S vs. R  Cmax 160.48 (145.20, 177.36)  0.109492  34.02
             T vs. S  AUC   83.12 ( 78.37,  88.15)  0.035788  19.09
             T vs. S  Cmax  81.63 ( 75.33,  88.46)  0.069372  26.80

It’s clear that the variances are not identical. The ones of T/R are smaller than the ones of S/R. If we apply the pooled model the CIs of T/R will be wider than in the IBD model and the CIs of S/R narrower.

BTW, I don’t understand what the purpose of the carry variable in Byron’s code is.

Cheers,
Helmut Schütz
[image]

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