Hello,
I am new to SAS (I am usually using R), and I was trying to redo some of my analyses on SAS using Fisher LSD multiple comparisons.
For my analysis, I don't need only the letters grouping the treatments, but also the LSD value.
In my dataset, I have most of the time a balanced dataset (same number of replicate n / treatment), but as I am working in plant science, sometime, one die, resulting in something unbalanced.
When everything is balanced, SAS and R both give me the same results (expected), and provide me with one LSD value. Knowing the LSD formula, if n is the same for both mean, in each comparison, all the LSD value calculated would be the same, so ONE value in the output.
But when it is unbalanced, R is not providing any LSD value while SAS does (which is why I am coming to SAS). I read online that if n was different between your pairs, the LSD value calculated will be different as compared to a pair with the same n. So you end up having multiple LSD values....But SAS is providing only ONE.
After searching and testing on my data, I am thinking that SAS is making an average of all the LSD values, weighted with how many time one LSD value appears versus another one (example: 10 pair comparisons with 8 using same LSD values lsd_1, and the 2 others with another one lsd_2 resulting in LSD = ( 8 * lsd_1 + 2 * lsd_2) / 10 ).
I would like to know if my hypothesis is good, and SAS is really doing that?
and because I never find any document on fisher LSD describing that method of estimating the LSD values for an unbalanced dataset, is it correct or not to use this "Average weighted LSD value" of SAS and how to justify it (in one way or another).
Thank you
Best
AiBee
