From: Douglas Bates <bates_at_stat.wisc.edu>

Date: Thu 07 Sep 2006 - 15:32:06 GMT

R-help@stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Fri Sep 08 02:46:34 2006

Date: Thu 07 Sep 2006 - 15:32:06 GMT

On 07 Sep 2006 17:20:29 +0200, Peter Dalgaard <p.dalgaard@biostat.ku.dk> wrote:

> Martin Maechler <maechler@stat.math.ethz.ch> writes:

*>
**> > >>>>> "DB" == Douglas Bates <bates@stat.wisc.edu>
**> > >>>>> on Thu, 7 Sep 2006 07:59:58 -0500 writes:
**> >
**> > DB> Thanks for your summary, Hank.
**> > DB> On 9/7/06, Martin Henry H. Stevens <hstevens@muohio.edu> wrote:
**> > >> Dear lmer-ers,
**> > >> My thanks for all of you who are sharing your trials and tribulations
**> > >> publicly.
**> >
**> > >> I was hoping to elicit some feedback on my thoughts on denominator
**> > >> degrees of freedom for F ratios in mixed models. These thoughts and
**> > >> practices result from my reading of previous postings by Doug Bates
**> > >> and others.
**> >
**> > >> - I start by assuming that the appropriate denominator degrees lies
**> > >> between n - p and and n - q, where n=number of observations, p=number
**> > >> of fixed effects (rank of model matrix X), and q=rank of Z:X.
**> >
**> > DB> I agree with this but the opinion is by no means universal. Initially
**> > DB> I misread the statement because I usually write the number of columns
**> > DB> of Z as q.
**> >
**> > DB> It is not easy to assess rank of Z:X numerically. In many cases one
**> > DB> can reason what it should be from the form of the model but a general
**> > DB> procedure to assess the rank of a matrix, especially a sparse matrix,
**> > DB> is difficult.
**> >
**> > DB> An alternative which can be easily calculated is n - t where t is the
**> > DB> trace of the 'hat matrix'. The function 'hatTrace' applied to a
**> > DB> fitted lmer model evaluates this trace (conditional on the estimates
**> > DB> of the relative variances of the random effects).
**> >
**> > >> - I then conclude that good estimates of P values on the F ratios lie
**> > >> between 1 - pf(F.ratio, numDF, n-p) and 1 - pf(F.ratio, numDF, n-q).
**> > >> -- I further surmise that the latter of these (1 - pf(F.ratio, numDF,
**> > >> n-q)) is the more conservative estimate.
**> >
**> > This assumes that the true distribution (under H0) of that "F ratio"
**> > *is* F_{n1,n2} for some (possibly non-integer) n1 and n2.
**> > But AFAIU, this is only approximately true at best, and AFAIU,
**> > the quality of this approximation has only been investigated
**> > empirically for some situations.
**> > Hence, even your conservative estimate of the P value could be
**> > wrong (I mean "wrong on the wrong side" instead of just
**> > "conservatively wrong"). Consequently, such a P-value is only
**> > ``approximately conservative'' ...
**> > I agree howevert that in some situations, it might be a very
**> > useful "descriptive statistic" about the fitted model.
**>
**> I'm very wary of ANY attempt at guesswork in these matters.
**>
**> I may be understanding the post wrongly, but consider this case: Y_ij
**> = mu + z_i + eps_ij, i = 1..3, j=1..100
**>
**> I get rank(X)=1, rank(X:Z)=3, n=300
**>
**> It is well known that the test for mu=0 in this case is obtained by
**> reducing data to group means, xbar_i, and then do a one-sample t test,
**> the square of which is F(1, 2), but it seems to be suggested that
**> F(1, 297) is a conservative test???!
*

It's a different test, isn't it? Your test is based upon the between group sum of squares with 2 df. I am proposing to use the within group sum of squares or its generalization.

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