Re: Hotelling T^2 Test and MANOVA with 2 Groups Equivalent?

Hi

The title says it all really: am I correct in
thinking that these two
procedures are essentially equivalent?

Thanks in advance




For MANOVA, there are various statistics based on functions of the eigenvalues of SE^(-1). However, in the case of p = 2 groups, there is only one nonzero eigenvalue and it's Hotelling T^2. Thus, they are all equivalent.

Jack
.