Bioequivalence and Bioavailability Forum

Dear Helmut,

good point, next question .

The simple, short answer:
The design constant for Baalam's design is 8 compared to 2 for the simple 2x2x2 crossover (in terms of N(total)). Thus the sample size for Baalam's design is approximately 4 times that of the 2x2x2 crossover, disregarding the different df for both designs.
Why? Because it is written¹⁾ .
To cite our Jewish Israeli guide with Viennese appeal: "Warum?! Schaun Sie: Steht geschrieben!"

The not so simple answer, longish:
Your expectation of doubling the sample size for the 2x2x2 crossover is totally right if you prefer to omit the data of the sequence groups TT and RR from the evaluation of T vs. R. Then the analysis of Baalam's design is the same as for the 2x2x2 crossover. And the sample size for the 2 omitted sequence groups is simply filled with the sample size per sequence group. This is written in Jone/Kenward³⁾ Chapter 3.3.

But using all data in Chow/Liu²⁾ Chapter 9.2.1 it is written that the sequence-by-period means (m_ik, i=sequence, k=period) have the expectations (omitting sequence effects) in an model with carry-over (cT, cR), log-transformed data:

        P1     P2        P2-P1=di
1 TT  µT+p1  µT+p2+cT           p2-p1-cT
2 RR  µR+p1  µR+p2+cR           p2-p1-cR
3 TR  µT+p1  µR+p2+cT   (µT-µR)+p2-p1-cT
4 RT  µR+p1  µT+p2+cR  -(µT-µR)+p2-p1-cR

Thus an estimate of µT-µR is
   est=(1/2)*((d3-d1)-(d4-d2))=(1/2)*((m32-m31)-(m21-m11)-(m42-m41)+(m22-m21))
This estimate has variance (assuming a common error variance)
   var(est)=(1/2)*(1/n1+1/n2+1/n3+1/n4)*s2e
In terms of all ni=n=N/4 (N= total sample size) this gives
   var(est)=(1/2)*(4/n)*s2e
           =(8/N)*s2e
Voilà! Here we have the design const. = 8.
But it's based on carry-over. And PowerTOST claims to deal with models without!
IMHO the same estimate and its variance applies if we omit the carry-over terms in the table above. But I'm not quite sure. The d1 and d2 are then used only for estimating the period effect and adjusting for it in d3,d4.
What puzzles me with this approach is the assumption of a common error variance. And this in a design which allows the assessment of intra-subject variances for T and R.

The pope of sample size has an other approach to the problem⁴⁾:
He combines the two possible estimates for µT-µR, namely the crossover estimate from the sequences TR/RT (variance proportional to error variance) and the repeated parallel group estimate from the sequences TT/RR (variance proportional to total variance).
He derived for the sample size within each sequence group
   nBaalams=n2x2(2*k+1)/(k+1)
where k=s²_b/s²_e, the ratio of the between-subject to error variance.
If k -> inf then n_Baalams=2*n_2x2. For 4 sequences this adds up to 4 times the total sample size. Thus design const. =8 here plays the role of a worst case in estimating the sample size.
If k=0 then n_Baalams=n_2x2, i.e. total sample size of Baalam's is doubled.
But I must confess that I didn't really understood St. Julious.

So what does this all tell us? Duno.
At least it seems that Baalam's design is very peculiar and special .

Hope I have managed all the super-, subscript indices.

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¹⁾K.-W. Chen, S.-C. Chow and G. Liu
"A Note on Sample Size Determination for Bioequivalence Studies with Higher-order Crossover Designs"
J. Pharmacokinetics and Biopharmaceutics, Vol. 25, No. 6, p753-765 (1997)

²⁾Shein-Chung Chow, Jen-Pei Liu
"Design and Analysis of Bioavailability and Bioequivalence Studies"
Third edition
Chapman & Hall/CRC Biostatistics, 2009

³⁾Byron Jones and Michael G. Kenward
"Design and Analysis of Cross-Over Trials"
Second edition
Chapman & Hall/CRC, 2003

⁴⁾Julious SA.
Tutorial in biostatistics
"Sample sizes for clinical trials with normal data."
Stat Med. 23(12):1921-1986, 2004