Bioequivalence and Bioavailability Forum

Dear Vitamin K, (what a nick name )

❝ First of all, I set in the proc mixed to get residuals in a data set. Then I calculate MSE from the residuals.

❝ Then I applied the formula: CV=100*sqrt(exp(MSE)-1), where I get a result, but I do not know it is the right method to get the real CV?

I'm not Helmut but I hope I can answer your question to some extent.
Sorry, but I do not understand what you do with Proc Mixed from your description .

Here is the SAS code I would employ for a classical 2x2 cross over (use search in this forum to find a lot of discussions!):

Proc MIXED data=yours;
class subject treatment period sequence;

model logPK=treatment period sequence;
random subject(sequence);
Estimate 'T-R' treatment -1 1/Cl alpha=0.1;

Run;

logPK is the log-transformed parameter under analysis (AUC, Cmax or so).
The estimate statement assumes that treatment is coded with T and R.

Le's use Helmuts data with 24 subjects found here.

Among the lengthy output you will find covariance parameter estimates
                 The Mixed Procedure
[...]
           Covariance Parameter Estimates

            Cov Parm              Estimate

            Subject(Sequence)      0.1002
            Residual               0.04102

[...]
Subject(Sequence) is the inter-subject variability s²_inter and residual the within-subject (or intra-subject) variability s²_within.
Convert them to CV with the formula CV=100*sqrt(exp(s²_{inter or within})-1) with the corresponding variance. But cave! This only applies if your PK parameter under analysis is log-transformed.

BTW: Why do you use Proc MIXED here?