[R] How do anova() and Anova(type="III") handle incomplete designs?
John Fox
jfox at mcmaster.ca
Mon Jun 18 20:25:07 CEST 2012
Dear Justin,
On Mon, 18 Jun 2012 11:24:33 -0400
Justin Montemarano <jmontema at kent.edu> wrote:
> Thanks for your response, John. That was helpful.
>
> I was using Type III from Anova() as a comparison to some results I had
> obtained JMP, which I've lost access to and have moved on to R, and I was
> confused by the error. Given that I do have a continuous covariate, the
> analyses are not likely comparable, considering your response.
If you look more carefully, you'll see that the reference here was to type-II tests. I believe that the definition used by Anova() is more sensible.
>
> I am still confused about interpretation of interactions within an anova()
> with an incomplete design, as mine is. Is the interaction term still
> informative?
I don't think that these matters are easily discussed on an email list. Briefly, I'd argue that the type-II tests (as defined by Anova) still have a straightforward interpretation since the test for a (say, two-way) interaction represent a contrast to a model that's additive with respect to the predictors involved in the interaction.
Best,
John
>
> -
> Justin Montemarano
> Graduate Student
> Kent State University - Biological Sciences
>
> http://www.montegraphia.com
>
>
> On Sat, Jun 16, 2012 at 9:20 PM, John Fox <jfox at mcmaster.ca> wrote:
>
> > Dear Justin,
> >
> > anova() and Anova() are entirely different functions; the former is part
> > of the standard R distribution and the second part of the car package. By
> > default, Anova() produces an error for type-III tests conducted on
> > rank-deficient models because the hypotheses tested aren't generally
> > sensible.
> >
> > From ?Anova:
> >
> > "singular.ok
> > defaults to TRUE for type-II tests, and FALSE for type-III tests (where
> > the tests for models with aliased coefficients will not be
> > straightforwardly interpretable); if FALSE, a model with aliased
> > coefficients produces an error."
> >
> > and
> >
> > "The designations "type-II" and "type-III" are borrowed from SAS, but the
> > definitions used here do not correspond precisely to those employed by SAS.
> > Type-II tests are calculated according to the principle of marginality,
> > testing each term after all others, except ignoring the term's higher-order
> > relatives; so-called type-III tests violate marginality, testing each term
> > in the model after all of the others. This definition of Type-II tests
> > corresponds to the tests produced by SAS for analysis-of-variance models,
> > where all of the predictors are factors, but not more generally (i.e., when
> > there are quantitative predictors). Be very careful in formulating the
> > model for type-III tests, or the hypotheses tested will not make sense."
> >
> > 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/
> >
> >
> > On Fri, 15 Jun 2012 15:01:27 -0400
> > Justin Montemarano <jmontema at kent.edu> wrote:
> > > Hello all:
> > >
> > > I am confused about the output from a lm() model with an incomplete
> > > design/missing level.
> > >
> > > I have two categorical predictors and a continuous covariate (day) that
> > > I am using to model larval mass (l.mass):
> > >
> > > leaf.species has three levels - map, syc, and oak
> > >
> > > cond.time has two levels - 30 and 150.
> > >
> > > There are no response values for Map-150, so that entire, two-way, level
> > > is missing.
> > >
> > > When running anova() on the model with Type I SS, the full factorial
> > > design does not return errors; however, using package:car Anova() and
> > > Type III SS, I receive an singularity error unless I used the argument
> > > 'singular.ok = T' (it is defaulted to F).
> > >
> > > So, why don't I receive an error with anova() when I do with Anova(type
> > > = "III")? How do anova() and Anova() handle incomplete designs, and how
> > > can interactions of variables with missing levels be interpreted?
> > >
> > > I realize these are fairly broad questions, but any insight would be
> > > helpful. Thanks, all.
> > >
> > > Below is code to illustrate my question(s):
> > >
> > > > lmMass <- lm(log(l.mass) ~ day*leaf.species + cond.time, data =
> > > growth.data) #lm() without cond.time interactions
> > > > lmMassInt <- lm(log(l.mass) ~ day*leaf.species*cond.time, data =
> > > growth.data) #lm() with cond.time interactions
> > > > anova(lmMass); anova(lmMassInt) #ANOVA summary of both models
> > > with Type I SS
> > > Analysis of Variance Table
> > >
> > > Response: log(l.mass)
> > > Df Sum Sq Mean Sq F value Pr(>F)
> > > day 1 51.373 51.373 75.7451 2.073e-15
> > > leaf.species 2 0.340 0.170 0.2506 0.7786
> > > cond.time 1 0.161 0.161 0.2369 0.6271
> > > day:leaf.species 2 1.296 0.648 0.9551 0.3867
> > > Residuals 179 121.404 0.678
> > > Analysis of Variance Table
> > >
> > > Response: log(l.mass)
> > > Df Sum Sq Mean Sq F value Pr(>F)
> > > day 1 51.373 51.373 76.5651 1.693e-15
> > > leaf.species 2 0.340 0.170 0.2533 0.77654
> > > cond.time 1 0.161 0.161 0.2394 0.62523
> > > day:leaf.species 2 1.296 0.648 0.9655 0.38281
> > > day:cond.time 1 0.080 0.080 0.1198 0.72965
> > > leaf.species:cond.time 1 1.318 1.318 1.9642 0.16282
> > > day:leaf.species:cond.time 1 1.915 1.915 2.8539 0.09293
> > > Residuals 176 118.091 0.671
> > > > Anova(lmMass, type = 'III'); Anova(lmMassInt, type = 'III')
> > > #ANOVA summary of both models with Type III SS
> > > Anova Table (Type III tests)
> > >
> > > Response: log(l.mass)
> > > Sum Sq Df F value Pr(>F)
> > > (Intercept) 39.789 1 58.6653 1.13e-12
> > > day 3.278 1 4.8336 0.02919
> > > leaf.species 0.934 2 0.6888 0.50352
> > > cond.time 0.168 1 0.2472 0.61968
> > > day:leaf.species 1.296 2 0.9551 0.38672
> > > Residuals 121.404 179
> > > Error in Anova.III.lm(mod, error, singular.ok = singular.ok, ...) :
> > > there are aliased coefficients in the model
> > > > Anova(lmMassInt, type = 'III', singular.ok = T) #Given the error
> > > in Anova() above, set singular.ok = T
> > > Anova Table (Type III tests)
> > >
> > > Response: log(l.mass)
> > > Sum Sq Df F value Pr(>F)
> > > (Intercept) 39.789 1 59.3004 9.402e-13
> > > day 3.278 1 4.8860 0.02837
> > > leaf.species 1.356 2 1.0103 0.36623
> > > cond.time 0.124 1 0.1843 0.66822
> > > day:leaf.species 2.783 2 2.0738 0.12877
> > > day:cond.time 0.805 1 1.1994 0.27493
> > > leaf.species:cond.time 0.568 1 0.8462 0.35888
> > > day:leaf.species:cond.time 1.915 1 2.8539 0.09293
> > > Residuals 118.091 176
> > > >
> > >
> > >
> > >
> > > -
> > > Justin Montemarano
> > > Graduate Student
> > > Kent State University - Biological Sciences
> > >
> > > http://www.montegraphia.com
> > > <http://www.montegraphia.com/>
> > > --
> > > Justin Montemarano
> > > Graduate Student
> > > Kent State University - Biological Sciences
> > >
> > > http://www.montegraphia.com
> > >
> > >
> > > [[alternative HTML version deleted]]
> > >
> > > ______________________________________________
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