8 Multivariate Analysis of Variance (MANOVA)

Suppose in the discrimination problem encountered earlier we wished to test the hypothesis equality of means. Specifically we wish to test \(H_0:\underline {\mu }_1 = \underline {\mu }_2 = \cdots = \underline {\mu }_n\) the way of the univariate case of testing \(H_0: \mu _1 = \mu _2 = \cdots = \mu _n\).

How do we proceed?

Using the univariate analog we set up a multivariate model \begin {equation} \tag {1} \underline {X}_{ij}= \underline {\mu } + \tau _i + \underline {e}_{ij}, j=1,2,\ldots ,n_i\qquad i=1,2,\ldots ,g \end {equation} where \(e_{ij}\) are independent \(N_p\big (\underline {0},\Sigma \big )\) (multivariate normal) variables (vectors) we have the parameter vector \(\underline {\mu }\) is an overall mean and \(\tau _i\) represents the \(i^{\text {th}}\) treatment effect with \(\sum \limits ^g_{i=1}n_i\tau _i= \underline {0}\).

The model in (1) may also be represented as \begin {equation} \tag {2} \underline {X}_{ij} =\underline {\mu }_i + \underline {e}_{ij}, e_{ij}\thicksim ^{iid}N_p\big (\underline {0},\Sigma \big ) \end {equation} where \(\mu _i = \underline {\mu } + \underline {\tau }_i\)

In model (1) we note the following

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each component of the observation vector \( \underline {X}_{ij}= \begin {pmatrix} X_{ij1} & X_{ij2} & \cdots& X_{ijp}\\ \end {pmatrix}'\,\) satisfies the univariate model \[X_{ij} = \mu + \tau _i + e_{ij} \quad e_{ij} \thicksim ^{iid} N(0,\sigma ^2)\]

\(-\)

the errors of the components of \(\underline {X}_{ij}\) are correlated but \(\Sigma \) covariance is the same for all populations.

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