Dear Ralf, Unfortunately, Anova.mlm(), and indeed Anova() more generally, won't handle a model with only a constant. As you point out, this isn't reasonable for repeated-measures ANOVA, where it should be possible to have only within-subjects factors. When I have a chance, I'll see what I can do to fix the problem -- my guess is that it shouldn't be too hard.
Thanks for pointing out this limitation in Anova.mlm() John On Tue, 16 Oct 2007 17:20:07 +0200 Ralf Goertz <[EMAIL PROTECTED]> wrote: > Hi, > > sorry if this is explained somewhere but I didn't find anything. > > How can I use "Anova" from the car package to test a modell without > between subject's factors? Suppose I have the following data > > mat.1 mat.2 mat.3 di ex > 1 85 85 88 1 1 > 2 90 92 93 1 1 > 3 97 97 94 1 1 > 4 80 82 83 1 1 > 5 91 92 91 1 1 > 6 83 83 84 2 1 > 7 87 88 90 2 1 > 8 92 94 95 2 1 > 9 97 99 96 2 1 > 10 100 97 100 2 1 > 11 86 86 84 1 2 > 12 93 103 104 1 2 > 13 90 92 93 1 2 > 14 95 96 100 1 2 > 15 89 96 95 1 2 > 16 84 86 89 2 2 > 17 103 109 90 2 2 > 18 92 96 101 2 2 > 19 97 98 100 2 2 > 20 102 104 103 2 2 > 21 93 98 110 1 3 > 22 98 104 112 1 3 > 23 98 105 99 1 3 > 24 87 132 120 1 3 > 25 94 110 116 1 3 > 26 95 126 143 2 3 > 27 100 126 140 2 3 > 28 103 124 140 2 3 > 29 94 135 130 2 3 > 30 99 111 150 2 3 > > Using > > > > Anova(lm(mat~di*ex,data=data),idata=data.frame(zeit=ordered(1:3)),idesign=~zeit) > > Type II Repeated Measures MANOVA Tests: Pillai test statistic > Df test stat approx F num Df den Df Pr(>F) > di 1 0.377 14.524 1 24 0.0008483 *** > ex 2 0.800 47.915 2 24 4.166e-09 *** > di:ex 2 0.281 4.695 2 24 0.0190230 * > zeit 1 0.782 41.209 2 23 2.491e-08 *** > di:zeit 1 0.252 3.865 2 23 0.0357258 * > ex:zeit 2 0.836 8.611 4 48 2.538e-05 *** > di:ex:zeit 2 0.518 4.189 4 48 0.0054586 ** > --- > Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 > > works as expected. But every once in a while I have a model without > between subject's factors. So I thought of > > > > Anova(lm(mat~1,data=data),idata=data.frame(zeit=factor(1:3)),idesign=~zeit) > Fehler in L %*% B : nicht passende Argumente > > (Error in L %*% B : non matching arguments) > > On the other hand using anova.mlm I get > > > > anova.mlm(lm(mat~1,data),idata=data.frame(zeit=factor(1:3)),X=~1,test="Spherical") > Analysis of Variance Table > > > Contrasts orthogonal to > ~1 > > Greenhouse-Geisser epsilon: 0.7464 > Huynh-Feldt epsilon: 0.7777 > > Df F num Df den Df Pr(>F) G-G Pr H-F Pr > (Intercept) 1 11.767 2 58 5.1375e-05 3.1183e-04 2.4939e-04 > Residuals 29 > > > How can achieve this with Anova? > > > Thanks in advance, > > Ralf > > ______________________________________________ > R-help@r-project.org 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. -------------------------------- John Fox, Professor Department of Sociology McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox/ ______________________________________________ R-help@r-project.org 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.
