# Bioequivalence and Bioavailability Forum 01:13 CET

## N(μ, σ²) [Software]

Hi Yong,

» » » I prefer the following formula ：
» » » intersubject CV = sqrt(Var(Sequence*Subject)) / RefLSM
» » Why?
» Because this is also used for ln_transformed data.

No it isn’t! Back to the basics. The model of a 2×2×2 crossover assumes [sic] IID. For log-transformed data:
log(μTμR) = μT − μR, which is estimated by the difference of the LSMs of log-transformed data xT − xR. Now it gets interesting (i.e., the assumption!): In the balanced case for simplicity (where n1 = n2 and n = n1 + n2), xT − xR follows a normal distribution N(log(μTμR), 2σ2n)).
Since σ2 is unknown, it is estimated by the MSE from ANOVA. Then we can estimate CV = √MSE − 1. Do you see xR in this derivation? I don’t. Remember that the normal distribution is described by two parameters, μ and σ2, which are independent. If you are interested in the variance component, please leave the mean(s) completely out of it (as it it correctly done in PHX/WNL for log-transformed data).

» Like one formula for intrasubject CV for both transformed and none transformed data, I think this formula is generally accepted both for transformed and none transformed data.

I think that for untransformed data dividing MSE by LSMR goes back to Kem Phillips, who wrote*

σ being expressed as a percentage of a reference mean, that is, as a coefficient of variation; values of the difference in means are expressed as percentages of the same reference mean.

That’s unfortunate and IMHO, not correct at all (Wolfgang Pauli would say: That is not only not right; it is not even wrong!). Again: The mean and variance are independent. So, the more I think about it: Even my idea of using the weighted global mean does not make sense. Paraphrasing Ste­phen Senn: Proving that apples are oranges by comparing the weight.

• Phillips KM. Power of the Two One-Sides Tests Procedure in Bioequivalence. J Pharmacokinet Biopharm. 1990; 18(2): 137–44. doi:10.1007/BF01063556.

Cheers,
Helmut Schütz

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