Bioequivalence and Bioavailability Forum

Dear all,

As far as I know, when number of subjects in sequence RT equals to that of TR, LSM of R or T equals to Geometric Mean of all PKmetrics of R or T.

I want to calculate Least Square Means(LSM) for unequal sequence 2*2 Xover BE for further calculation of Diff(T-R) of LSM, and for further estimation of 90% CI of Diff(T-R) of LSM, and for further estimation of 90%CI of GMR(geometric means ratio).

Let me shown my calculation in details:

My data is extracted from Hauschke D's
Bioequivalence Studies in Drug Development: Methods and Applications
Page 70, (date of Subject 16 and 18 ws deleted to generate unequal sequences)

sub sequence   auc  period formulation ln(auc)
 5     1     249.94    1       1       5.521221
 2     1     339.03    1       1       5.826088
 8     1     184.32    1       1       5.216673
 4     1     242.64    1       1       5.491579
15     1     239.84    1       1       5.479972
13     1     207.4     1       1       5.334649
12     1     209.3     1       1       5.343769
 7     2     235.89    2       1       5.463366
17     2     169.7     2       1       5.134032
10     2     245.48    2       1       5.503215
11     2     134.89    2       1       4.904459
 1     2     288.79    2       1       5.6657
14     2     223.39    2       1       5.408919
 3     2     343.37    2       1       5.838809
 9     2     215.14    2       1       5.371289
 6     2     225.77    2       1       5.419517
 7     2     133.13    1       2       4.891326
11     2     163.93    1       2       5.09944
 6     2     217.97    1       2       5.384357
 1     2     228.04    1       2       5.429521
 3     2     288.21    1       2       5.66369
10     2     248.98    1       2       5.517373
 9     2     213.78    1       2       5.364947
14     2     245.92    1       2       5.505006
17     2     188.05    1       2       5.236708
12     1     231.98    2       2       5.446651
 2     1     329.76    2       2       5.798365
 8     1     249.64    2       2       5.52002
 5     1     201.56    2       2       5.306087
 4     1     258.19    2       2       5.553696
15     1     241.25    2       2       5.485834
13     1     234.19    2       2       5.456133
-----------------------------------------------
in which Treatment 1 was cconsidered as Reference.

First, I used the original data (no transformation) to do BE analysis with WinNonlin 5.2, some of the the results are:
Treatment   LSM      SE    CI level T_critical LowerCI  UpperCI
1         235.1521 12.7463   90       1.727    213.1332 257.1711
2         231.8667 12.7463   90       1.727    209.8477 253.8856

  Diff  Diff_SE
-3.2855 10.3873

My questions are:
1. How to calculate LSM and SE for each Treatment? I want to manually calculate it, can anyone give me some exact equation?
I have read WNL's user guide, it didn't tell me the exact equation. Only reads "(computed by LinMix)," on page 332 of WNL 5.1's User Guide
2. Why SE of the two treatments are identical?

Second, I setted "Ln(x)" transformation in WinNonlin's BE Wizard, some of the resuls are:
Treatment   LSM      SE    CI level T_critical LowerCI  UpperCI
1           5.4356  0.0563   90       1.721     5.3387   5.5326
2           5.4266  0.0563   90       1.721     5.3296   5.5235

LSMDiff    0.0091
Diff_SE    0.0513

I have the same questions as above mentioned.

I have tried to analysis this unequal sequences data in SAS and Stata, both were failed.

In SAS, I used the following code:
proc glm data=dose_equivalence;
class subject sequence period formulation;
model logAUC=sequence subject(sequence) period formulation;
random subject(sequence) / test;
lsmeans formulation/pdiff cl alpha=0.1;
run;
quit;

It returned:
ERROR: One or more variables are missing or freq or weight is zero on every observation.
182  quit;

In Stata, I used the following code:
pkequiv auc formulation period sequence sub
It returned:
must specify an equivalence comparison /*
r(198);

Is SAS or Stata not able to deal with unequal sequences, or I didn't use the right code?

Thank you very much for your help.

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