On 8/2/06, John Fox <[EMAIL PROTECTED]> wrote: > The argument y is the response variable and group is a factor defining > groups (as ?levene.test says). If you have more than one factor, then you > can use interaction() to create from them a factor with levels given by the > product set of the levels of the individual factors. Here's an example > > > library(car) > > data(Moore) > > attach(Moore) > > levene.test(conformity, interaction(fcategory, partner.status)) > Levene's Test for Homogeneity of Variance > Df F value Pr(>F) > group 5 1.4694 0.2219 > 39 > > levels(interaction(fcategory, partner.status)) > [1] "high.high" "low.high" "medium.high" "high.low" "low.low" > [6] "medium.low" > > levels(fcategory) > [1] "high" "low" "medium" > > levels(partner.status) > [1] "high" "low" > > I'll add a couple of examples to the help page.
Thanks, John. Now, I understand how to use levene.test. There is only a question remaining: is the null hypothesis corresponding to homogeneity of variances, i.e., should one conclude that Levene's Test for Homogeneity of Variance Df F value Pr(>F) group 95 3.5919 < 2.2e-16 *** 864 tell us that the hypothesis that the variances are equal is (highly) significant? Paul ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.