[Multiple-Sample Testing of Variance Matrix]

14.6.3 Pillai's Test Statistic Let 21, . . ., , denote the s eigenvalues of the matrix E-1H. The Pillai test statistic is then given by S V(S) = tr[(E + H-)H] = > (14.22) i=1 1 + 1; The test procedure is to reject H if V(s) > V.". In R, we carry out the Pillai's method as below. > summary (root . manova, test = "Pillai") Df Pillai approx F num Df den Df Pr (> F) rs 5 1. 30 4. 07 20 168 2e-07 * * * Residuals 42 Signif. codes: 0 ' * **' 0.001 ' * *' 0.01 ' *' 0.05 '.' 0.1 ' ' 1 The Pillai's test statistic confirms the findings of the Wilks's and Roy's test that the mean vector for the six strata are significantly different. Finally, we look at the fourth test for testing the hypothesis H, that the strata mean vectors are equal to the Lawley-Hotelling test statistic.14.6.4 The Lawley-Hotelling Test Statistic The Lawley-Hotelling statistic is defined by: U(S) = tr (E 'H). (14.23) The test procedure is to reject the hypothesis H for large values of US). This is illustrated in R.> summary(root.manova, test = "Hate-Hing") Df Hotelling-Lawley approx F num Df den Df Pr(>F) rs 5 2.92 5.48 20 150 2.63-10 *** Residuals 42 Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Thus, it is concluded that all the four statistical tests lead to the same conclusion, that the mean vector for the six strata are different. From a theoretical point of View, there is no reason to prefer one of the four methods over the others. A general advice is to use all the four methods. Consider one more example before closing the MANOVA section. Problem 14.9 Repeat the above exercise for the Pillai and Lawley-Hotelling tests respec- tively given in Equations 14.22 and 14.23