Could you please explain a bit more about how p values are calculated for the AMOVA test?
I have a dataset with 2 groups of 5 replicates each. This is 10!/2!(5!)^2 or 126 ways to make 2 groups of 5 where order does not matter. So, I am trying to understand how I can get a p-value that is <0.001? It seems like the lowest possible p value should be 1/126 or 0.008. When the row/column names of the distance matrix are shuffled, does order matter?
Thanks very much for your help!
Anderson MJ (2001). A new method for non-parametric multivariate analysis of variance. Austral Ecol 26: 32-46.
The smallest p-value you should be able to get should be 1/126 ~ 0.008. What are you getting? It could be that if all your Frand values are greater than your Fobs that it woudl come back as 0.000 or <0.001.
I am getting:
VD-VN Among Within Total
SS 0.329998 0.876101 1.2061
df 1 8 9
MS 0.329998 0.109513
This is 2 groups of 5 replicates (10 samples total).