Bioequivalence and Bioavailability Forum

Dear Detlew!

❝ Above alpha ~0.5545 the confidence interval does not cover the point estimate. This is astonishing to me and fascinating, because in my belief the point estimate was always bracketed by a confidence interval.

❝ Is this correct? Or is there a hidden bug? The latter I can't imagine because the calculation is simple enough.

What’s going on here? I always followed blindly Chow & Liu (Chapter 7); calculated a two-sided interval and used only the upper limit. I tried not to fall in the trap of using your code (which is fine, IMHO), but recalculated an example (the famous Clayton & Leslie data set from Table 6.9.1). Straightforward analysis in log-scale, SS_intra 4.94398, MS_intra 0.308999, 16 df). Now the confusion starts already. For CV_intra I get 0.601715. They write

“The estimated CV is about 36.55%…”

I beg your pardon? Do they mean CV_inter? No, that would be 0.334785. Anyhow, maybe just another typo.*
I’m using their symbols.

alpha   <- 0.05
n1      <- 9; n2 <- 9
df      <- n1+n2-2
MSintra <- 0.308999
CVintra <- sqrt(exp(MSintra)-1)
SSintra <- MSintra*df
Chi2L   <- qchisq(alpha/2, df)
Chi2U   <- qchisq(1-alpha/2, df)
Le      <- SSintra/Chi2U
Ue      <- SSintra/Chi2L
cat("(",round(Le,2),",",round(Ue,2),")\n")

0.17 – 0.72 agrees with their Table 7.3.1. But this is the 95% CI of σ²_e and we want the CI of CV_intra. Therefore

CVL <- sqrt(exp(Le)-1)
CVU <- sqrt(exp(Ue)-1)
cat("alpha",alpha,":",round(CVL,4),"]", round(CVintra,4),"[",round(CVU,4),"\n")
alpha 0.05 : 0.4324 ] 0.6017 [ 1.0226

Fine.
Give it a try with alpha <- 0.9065

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• I guess (!) they calculated the CV from log-transformed data like in the untransformed (!) case with \(CV=\sqrt{MSE}/\bar{x}_R=\sqrt{0.308999}/1.520898=0.36549235.\) Crazy.