Hi Fabian,
You my find my discussion of "types" of SS helpful. My website has
been down for some time, but you can retrieve it from
http://psychology.okstate.edu/faculty/jgrice/psyc5314/SS_types.pdf
among other places.
Best,
Ista
On Thu, May 12, 2011 at 10:33 AM, Fabian <fabian_ro...@gmx.de> wrote:
> #subject: type III sum of squares - anova() Anova() AnovaM()
> #R-version: 2.12.2
>
> #Hello everyone,
>
> #I am currently evaluating experimental data of a two factor
> experiment. to illustrate de my problem I will use following #dummy
> dataset: Factor "T1" has 3 levels ("A","B","C") and factor "T2" has 2
> levels "E" and "F". The design is #completly balanced, each factor
> combinations has 4 replicates.
>
> #the dataset looks like this:
>
> T1<-(c(rep(c("A","B","C"),each=8)))
> T2<-(c(rep(rep(c("E","F"),each=4),3)))
> RESPONSE<-c(1,2,3,2,2,1,3,2,9,8,8,9,6,5,5,6,5,5,5,6,1,2,3,3)
> DF<-as.data.frame(cbind(T1,T2,RESPONSE))
> DF$RESPONSE<-as.numeric(DF$RESPONSE)
>
> > DF
> T1 T2 RESPONSE
> 1 A E 1
> 2 A E 2
> 3 A E 3
> 4 A E 2
> 5 A F 2
> 6 A F 1
> 7 A F 3
> 8 A F 2
> 9 B E 7
> 10 B E 6
> 11 B E 6
> 12 B E 7
> 13 B F 5
> 14 B F 4
> 15 B F 4
> 16 B F 5
> 17 C E 4
> 18 C E 4
> 19 C E 4
> 20 C E 5
> 21 C F 1
> 22 C F 2
> 23 C F 3
> 24 C F 3
>
> library(biology)
> replications(RESPONSE ~ T1*T2,data=DF)
> T1 T2 T1:T2
> 8 12 4
> is.balanced(RESPONSE ~ T1*T2,data=DF)
> [1] TRUE
>
>
> #Now I would like to know whether T1, T2 or T1*T2 have a significant
> effect on RESPONSE. As far as I know, the #theory says that I should use
> a type III sum of squares, but the theory also says that if the design
> is completely #balanced, there is no difference between type I,II or III
> sum of squares.
>
> #so I first fit a linear model:
>
> my.anov<-lm(RESPONSE~T1+T2+T1:T2)
>
> #then I do a normal Anova
>
> > anova(my.anov)
>
> Analysis of Variance Table
>
> Response: RESPONSE
> Df Sum Sq Mean Sq F value Pr(>F)
> T1 2 103.0 51.500 97.579 2.183e-10 ***
> T2 1 24.0 24.000 45.474 2.550e-06 ***
> T1:T2 2 12.0 6.000 11.368 0.000642 ***
> Residuals 18 9.5 0.528
>
> #When I do the same with the Anova() function from the "car" package I
> get the same result
>
> Anova(my.anov)
>
> Anova Table (Type II tests)
>
> Response: RESPONSE
> Sum Sq Df F value Pr(>F)
> T1 103.0 2 97.579 2.183e-10 ***
> T2 24.0 1 45.474 2.550e-06 ***
> T1:T2 12.0 2 11.368 0.000642 ***
> Residuals 9.5 18
>
> #(type two sees to be the default and type="I" produces an error (why?))
>
> #yet, when I specify type="III" it gives me something completely different:
>
> Anova(my.anov,type="III")
> Anova Table (Type III tests)
>
> Response: RESPONSE
> Sum Sq Df F value Pr(>F)
> (Intercept) 16.0 1 30.316 3.148e-05 ***
> T1 84.5 2 80.053 1.100e-09 ***
> T2 0.0 1 0.000 1.000000
> T1:T2 12.0 2 11.368 0.000642 ***
> Residuals 9.5 18
>
> #an the AnovaM() function from the "biology" package does the same for
> type I and II and produces the following #result:
>
> library(biology)
> AnovaM(my.anov,type="III")
> Df Sum Sq Mean Sq F value Pr(>F)
> T1 2 84.5 42.250 80.053 1.10e-09 ***
> T2 1 24.0 24.000 45.474 2.55e-06 ***
> T1:T2 2 12.0 6.000 11.368 0.000642 ***
> Residuals 18 9.5 0.528
>
> #Is type 3 the Type I should use and why do the results differ if the
> design is balanced? I am really confused, it would #be great if someone
> could help me out!
>
> #Thanks a lot for your help!
>
> #/Fabian
> #University of Gothenburg
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> ______________________________________________
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org
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