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Understanding Anova (car) output

2 messages · Ista Zahn, John Fox

Ista Zahn
# 
Dear professor Fox and R helpers,
I have a quick question about the Anova function in the car package.
When using the default "type II" SS I get results that I don't
understand (see below).

library(car)
Data <- data.frame(y=rnorm(10), x1=factor(c(rep("a",4), rep("b",6))),
x2 = factor(c(rep("j", 2), rep("k", 3), rep("j", 2), rep("k", 3))))
Data$c1 <- recode(Data$x1, '"a"=-1;"b"=1', as.factor.result=FALSE)
Data$c2 <- recode(Data$x2, '"j"=-1;"k"=1', as.factor.result=FALSE)
Data$c3 <- Data$c1*Data$c2

options(contrasts=c("contr.Sum", "contr.poly"))

m1 <- lm(y ~ x1*x2, data=Data)
m2 <- lm(y ~ c1 + c2 + c3, data=Data)

Anova(m1)#Why is this
Anova(m2)#Different from this?

I would greatly appreciate it if someone can help me understand why
the results differ.

Thanks,
Ista
John Fox
# 
Dear Ista,
Zahn
How could Anova() possibly know in m2 that c1 and c2 are marginal to c3 (as
in can tell in m1 that x1 and x2 are marginal to x1:x2)? What you're
getting, therefore, is equivalent to "type-III" tests:
Anova Table (Type II tests)

Response: y
           Sum Sq Df F value Pr(>F)
x1         0.3313  1  0.1599 0.7031
x2         0.6800  1  0.3282 0.5875
x1:x2      0.1160  1  0.0560 0.8209
Residuals 12.4321  6
Anova Table (Type II tests)

Response: y
           Sum Sq Df F value Pr(>F)
c1         0.2715  1  0.1310 0.7298
c2         0.5891  1  0.2843 0.6131
c3         0.1160  1  0.0560 0.8209
Residuals 12.4321  6
Anova Table (Type III tests)

Response: y
             Sum Sq Df F value Pr(>F)
(Intercept)  0.5384  1  0.2598 0.6284
x1           0.2715  1  0.1310 0.7298
x2           0.5891  1  0.2843 0.6131
x1:x2        0.1160  1  0.0560 0.8209
Residuals   12.4321  6
Regards,
 John