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_roger at 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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--
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org