CI for Transformed data Unbalanced study [General Sta­tis­tics]

posted by Helmut Homepage – Vienna, Austria, 2007-03-07 12:04 (5434 d 08:40 ago) – Posting: # 565
Views: 27,196

Dear USFDA_EMEA!

» » Let's call Xt the LSM of the test, and Xr the LSM of the reference (log-scale).
»
» Can you help me with the formula for LSM (I wish to do a manual calculation, for verification purposes) and a brief explaination about why this method is used for unbalanced study?

The calculation is not different from a balanced study.

I’ll give you an example for the reference:
Calculate the (arithmetic) mean of log-transformed values in sequence 1 (if seq 1 = RT from period 1) as
Xr1 = sum ( Xr1 … n1 ) / n1

Calculate the (arithmetic) mean of log-transformed values in sequence 2 (if seq 2 = TR from period 2) as
Xr2 = sum ( Xr1 … n2 ) / n2

LSM for the reference is the arithmetic mean of Xr1 and Xr2
Xr = ( Xr1 + Xr2 ) / 2 )


The difference Xt – Xr is also called MLE (the Maximum Likelihood Estimator) of the true difference Mu(t) – Mu(r), which is slightly biased according to the degree of ‘unbalancedness’.
A better estimate of the true difference is MVUE (the Minimum Variance Unbiased Estimator), however, its calculation is a little bit tricky (involves the Γ-distribution).
For details see Chow & Liu’s book.

Dif-tor heh smusma 🖖
Helmut Schütz
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