[R] (no subject)

[Ista Zahn]

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