car::linearHypothesis Sum of Sqaures Error?

Dear John,
-----Original Message-----
From: John Jay Wiley Jr. [mailto:jwileyjr at syr.edu]
Sent: Tuesday, October 09, 2012 9:17 AM
To: John Fox
Cc: r-help at r-project.org
Subject: RE: [R] car::linearHypothesis Sum of Sqaures Error?

John,

Thank you for the reply.

The data are balanced; I double-checked. I believe the contrasts are
orthogonal. Sum of squares in summary(aov) with the contrasts split out
add to the main effect. I am still unsure of where the error is for the
sum of squares calculation.

I have written some code with a parallel model structure that may help
(see below).

Can linearHypothesis return type II tests? It seems to only return type
III with no option to set 'type'.

Cheers,
John

y<-runif(36,0,100)
block<-factor(rep(c("A","B","C"),each=12))
a<-factor(rep(c("A","B"),times=3,each=6))
b<-factor(rep(c("A","B"),times=6,each=3))
c<-factor(rep(c("A","B","C"),times=12,each=1))
covar<-0.6*y+rnorm(36,10,25)
data<-data.frame(y,block,a,b,c,covar)

c_contrasts<-matrix(c(-1,2,-1,1,0,-1),3,2)
dimnames(c_contrasts)<-list(levels(data$c),c("B vs. A&C","A vs. C"))
contrasts(data$c)<-c_contrasts

model<-lm(y~block+a*b*c+covar,data=data)
summary.aov(model,split=list(c=list("B vs. A&C"=1,"A vs. C"=2)))
#Sum of squares add here, but factorial ANCOVA non-orthogonal in type I
SS

Anova(model,type=2)
Anova(model,type=3)
linearHypothesis(model,c("cB vs. A&C","cA vs. C"))
There are two problems here: (1) the covariate isn't balanced; (2) you
didn't pay attention to the contrasts for factors other than c. The default
is contr.treatment, which produces terms whose contrasts are not orthogonal
in the row basis of the design.

So, the following two ANOVAs (not ANCOVAs) give the same result, but the
third doesn't. For balanced data, types I (sequential), II, and III tests
should all be the same.

----------- snip --------------

model.2 <- lm(y~block+a*b*c, data=data)
anova(model.2)
Anova(model.2)
Anova(model.2, type=3) # incorrect

----------- snip --------------

Here's one way to get correct type III tests:

----------- snip --------------

options(contrasts=c("contr.sum", "contr.poly"))
model.3 <- update(model.2)
anova(model.3)
Anova(model.3)
Anova(model.3, type=3) # now all the same

----------- snip --------------
	#Anova and linear hypothesis produce equal sum of squares for c
main effect in type III
linearHypothesis(model,"cB vs. A&C")
linearHypothesis(model,"cA vs. C")
	#Sum of squares of the individual contrasts do not add to the main
effect of c
This is a manifestation of the same problem:

----------- snip --------------

linearHypothesis(model.3, c("cB vs. A&C","cA vs. C")) # SS equal to the sum
of the next two

linearHypothesis(model.3, "cB vs. A&C")
linearHypothesis(model.3, "cA vs. C")

----------- snip --------------

It doesn't make sense to ask whether linearHypothesis() does "type II" or
"type III" tests -- it just tests directly specified linear hypotheses
concerning the parameters of the model as parametrized. Anova() can
formulate sets of type II and III linear hypotheses (but for the latter, you
do have to pay attention to the parametrization).

Best,
 John

John J. Wiley, Jr.
PhD Candidate
State University of New York
College of Environmental Science and Forestry
Department of Environmental and Forest Biology
460 Illick Hall
Syracuse, NY 13210
315.470.4825 (office)
740.590.6121 (cell)

________________________________________
From: John Fox [jfox at mcmaster.ca]
Sent: Tuesday, October 09, 2012 7:15 AM
To: John Jay Wiley Jr.
Cc: r-help at r-project.org
Subject: Re: [R] car::linearHypothesis Sum of Sqaures Error?

Dear John

On Tue, 9 Oct 2012 02:07:07 +0000
 "John Jay Wiley Jr." <jwileyjr at syr.edu> wrote:
I am working with a RCB 2x2x3 ANCOVA, and I have noticed a difference
in the calculation of sum of squares in a Type III calculation.

For type III tests, you should use contrasts that are orthogonal in the
row basis of the design. Perhaps you've done that (by setting the
contrasts for the factors directly), but I suspect not. Why not just use
type II tests? They're hard to screw up.

As well, I assume that the variables that enter additively are the
covariates. If not, and a covariate is involved in the interaction, the
type III tests aren't sensible unless the 0 point of the covariate is
where you want to test a "main effect" or lower-order interaction.

Anova output is a follows:

Anova(aov(MSOIL~Forest+Burn*Thin*Moisture+ROCK,data=env3l),type=3)
Anova Table (Type III tests)

Response: MSOIL
                    Sum Sq Df F value    Pr(>F)
(Intercept)        22.3682  1 53.2141 3.499e-07 ***
Forest              1.0954  2  1.3029   0.29282
Burn                2.6926  1  6.4058   0.01943 *
Thin                0.0494  1  0.1176   0.73503
Moisture            1.2597  2  1.4984   0.24644
ROCK                2.1908  1  5.2119   0.03296 *
Burn:Thin           0.2002  1  0.4764   0.49763
Burn:Moisture       1.0612  2  1.2623   0.30360
Thin:Moisture       1.6590  2  1.9734   0.16392
Burn:Thin:Moisture  1.1175  2  1.3292   0.28605
Residuals           8.8272 21

However, I would like to calculate some a priori contrasts within the
Moisture factor as follows:

Transect_moisture_contrasts<-matrix(c(-1,2,-1,1,0,-1),3,2)
dimnames(Transect_moisture_contrasts)<-list(levels(env$Moisture),c("I
vs. X&M","X vs. M"))
contrasts(env$Moisture)<-Transect_moisture_contrasts
contrasts(env3l$Moisture)
  I vs. X&M X vs. M
X        -1       1
I         2       0
M        -1      -1

soilmodel<-lm(MSOIL~Forest+Burn*Thin*Moisture+ROCK,data=env3l)
linearHypothesis(soilmodel,"MoistureI vs. X&M")
Linear hypothesis test

Hypothesis:
MoistureI vs. X&M = 0

Model 1: restricted model
Model 2: MSOIL ~ Forest + Burn * Thin * Moisture + ROCK

  Res.Df    RSS Df Sum of Sq      F Pr(>F)
1     22 9.4106
2     21 8.8272  1   0.58333 1.3877  0.252
linearHypothesis(soilmodel,"MoistureX vs. M")
Linear hypothesis test

Hypothesis:
MoistureX vs. M = 0

Model 1: restricted model
Model 2: MSOIL ~ Forest + Burn * Thin * Moisture + ROCK

  Res.Df    RSS Df Sum of Sq      F Pr(>F)
1     22 9.6359
2     21 8.8272  1   0.80871 1.9239   0.18

The sum of squares for these two contrasts do not add up to the sum of
squares of the main effect Moisture
.80871+.58333
[1] 1.39204
1.39204-1.2596
[1] 0.13244

Checking them together produces the correct sum of squares for the
main effect
linearHypothesis(soilmodel,c("MoistureI vs. X&M","MoistureX vs. M"))
Linear hypothesis test

Hypothesis:
MoistureI vs. X&M = 0
MoistureX vs. M = 0

Model 1: restricted model
Model 2: MSOIL ~ Forest + Burn * Thin * Moisture + ROCK

  Res.Df     RSS Df Sum of Sq      F Pr(>F)
1     23 10.0869
2     21  8.8272  2    1.2596 1.4984 0.2464

So my question is:
Should the sum of squares for the two contrasts add to the main effect
here?

Only if the data are balanced.

I hope this helps,
 John

------------------------------------------------
John Fox
Sen. William McMaster Prof. of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox/

If they should, maybe we can figure out why mine do not.

Thanks in advance for any assistance.

Cheers,
John

John J. Wiley, Jr.
PhD Candidate
State University of New York
College of Environmental Science and Forestry
Department of Environmental and Forest Biology
460 Illick Hall
Syracuse, NY 13210
315.470.4825 (office)
740.590.6121 (cell)

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car::linearHypothesis Sum of Sqaures Error?

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