## LSMeans, still causing lots of people headaches [General Sta­tis­tics]

Hi ElMaestro,

I guess LSMeans still gives a lot of people headaches which explains why your thread has been avoided like a plague.

Well I am not an expert but lemme try and say what I know....my widow's might. I am responding to questions (1) and (3).

I think LSMeans is a more accurate estimate of the mean of Ln[T] and Ln[R] in crossover studies. This is because the other mean estimates do not take into cognizance of the fact that the var(Ln[T]) in the sequence TR is different from the var(Ln[T]) in the sequence RT. Same for Ln[R]. In effect, the two population (sequence TR and RT) are different.

I usually imagine a situation where a drug has a lower AUC in period II than period I (perhaps due to the action of antibodies that was developed during period I). In such case, it would not be advisable to find the means for Ln[T] and Ln[R] without accounting for the period effect. If not done, the mean Ln[T] might be underestimated when compared to mean Ln[R] if the study was not balanced (RT having more subjects than TR).

The Model Residual is the variance for the difference of LSMeans between Ln[T] and Ln[R]. Lemme do some little mathematical manipulations (I might be wrong though) to buttress my point:

Diff of LSM = {(Ln[T] in TR) + (Ln[T] in RT)}/2 - {(Ln[R] in TR) + (Ln[R] in RT)}/2
Diff of LSM = 1/2{(Ln[T] in TR) + (Ln[T] in RT) - (Ln[R] in TR) - (Ln[R] in RT)}
Diff of LSM = 1/2{(Ln[T] in TR) - (Ln[R] in TR) + (Ln[T] in RT) - (Ln[R] in RT)}
Diff of LSM = {(Ln[T] in TR) - (Ln[R] in TR)}/2 + {(Ln[T] in RT) - (Ln[R] in RT)}/2
Diff of LSM = Residuals in TR + Residuals in RT

Thus the Mean Sum of Squares for the Residuals is the variance for the difference of LSMeans. This is my arithmetolophystics and Al Jabra. I know there are lots of magic in it.

For your other questions, I guess the headaches remains until someone offers us some panadol or Tylenol.

Regards

Scopy Ing. Helmut Schütz 