Stata The Stata listserver
[Date Prev][Date Next][Thread Prev][Thread Next][Date index][Thread index]

Re: st: -manova- question

From   [email protected]
To   [email protected]
Subject   Re: st: -manova- question
Date   Sat, 26 Mar 2005 09:06:02 -0600

Jack Buckley <[email protected]> asks:

> I have a quick question about how -manova- calculates the Roy's
> largest root statistic. I am not a big MANOVA user, but I need to
> teach it in a multivariate course, so I have been working through
> some problems by hand from Rencher's excellent text, _Meth ods of
> Multivariate Analysis_. I found in the Statalist archive that if
> I run -manovatest- after estimation, the eigenvalues of the
> matrix [E^-1]H are saved in r(eigvals). Rencher states that Roy's
> largest root, theta, is computed simply by:
> theta = largest eigenvalue of [E^-1]H / (1+largest eigenvalue)
> when I work through problems and check the answers numerically in
> Stata, I find that everything is as expected except that Stata
> appears to report theta = largest eigenvalue of [E^-1]H
> Is this an alternative variant of Roy's statistic (i.e. is
> Rencher only describing one method in use) or is Stata computing
> a different quantity for some other reason, or (option three) am
> I just missing something obvious?

I too enjoy Rencher's books.  I took my multivariate statistics
class from Rencher before the books existed -- his class notes
were the basis for his books.

Page 373 of "[R] manova" says (substitute greek symbols for
lambda and theta below):

    "... Some authors report lambda_1, and others (including
    Rencher) report theta = lambda_1/(1 + lambda_1) for Roy's
    largest root.  Stata reports lambda_1."

I hope you enjoy teaching the course.

Ken Higbee    [email protected]
StataCorp     1-800-STATAPC

*   For searches and help try:

© Copyright 1996–2024 StataCorp LLC   |   Terms of use   |   Privacy   |   Contact us   |   What's new   |   Site index