#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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