How do anova() and Anova(type="III") handle incomplete designs?

Dear Justin,

On Mon, 18 Jun 2012 11:24:33 -0400
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

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How do anova() and Anova(type="III") handle incomplete designs?

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