Bioequivalence and Bioavailability Forum

Dear DLabes,

I could be wrong, but.........

❝ My interpretation of the code is: The MSE is taken from the lm() call, but

❝ point estimate is calculated by the ordinary means m_T and

❝ m_R.

❝ Any body out there: Am I right?

Better let the authors answers here, but if you are right then Bear has a potential problem in the case of imbalance. But you could also ask yourself: Do I want to test if the means differ or do I want to test if the LSMeans differ?
I once heard that some statisticians think that a diference in LSMeans should be tested with type III SS. After all LSMeans are means that are adjusted for all other effects, while type III SS for a factor X, corresponds to the variation caused by factor X given the presence of all other factors (thus 'single term deletions') which may be the as saying that a type III SS is the variation for factor X adjusted for the variation caused by everything else. So I may digress and ask: If type III SS provide tests for LSMeans, then which SS are appropriate for ordinary means? And why would the former be more relevant than the latter from a practical standpoint?

❝ If I am right: This is ok, as long as the design is balanced between

❝ sequences. but for unbalanced data there is a difference

Certainly.

❝ In SAS this is done usually via LSMeans or an ESTIMATE statement.

❝

❝ Astonishing enough there seems no counterpart to LSMeans in R. After

❝ reading tons of Web pages it seems this another mine field like Type III

❝ sums of squares.

It is quite easy to find websites with renowned statisticians who do not really believe that LSMeans are more relevant than ordinary normal old-fashioned boring means.
The people who write R do it for a reason. For example some of them do not like SAS. Therefore they don't like LSMeans and type III SS. Therefore these things are not really implemented in R. It's just like getting tired of one type of random number generator and then coming up with a new one which has better attributes (better is in this regard completely in the eyes of the beholder).
So can we extract LSMeans from an lm fit? No!
Can we in stead extract the difference in LSMeans (test minus R) from an lm fit? Yes! And this is exactly what we need for the BE evaluation, because log(LSMean(T)/LSMean(R))=log(LSMean(T)-log(LSMean(R)).
As far as I know the reason has to do with differences in the way model matrices are specified in R and SAS. I was at some point thinking about asking a q about this here because I noticed that some model matrices in R are not full rank. Anyone with insight, please comment!

❝ But nevertheless there is the task to estimate the treatment effect with

❝ 90% confidence interval (from the parameter of the fitted model), i.e.

❝ from a counterpart of the ESTIMATE statement.

Extract the difference in LSMeans, then you have your estim T/R ratio. Extract the MSE, then you get your sigma. Calculate the 90% CI, leave from work early and enjoy the good weather.


EM.