Hi:
Look at the structure of the experiment.
The six blocks represent different replications of the experiment.
No treatment is assigned at the block level.
Within a particular block, there are three plots, to which each
variety is randomly assigned to one of them. Ideally, separate
randomizations
of treatments to plots take place in each block.
Each plot is divisible into four subplots, to which the nitrogen levels are
randomly assigned. Again, separate randomizations in each plot is ideal.
For a particular block, then, there are 12 subplots altogether, each
producing a
single measurement.
Over all replicates, we have
Six blocks
18 plots
72 subplots
which determine how the degrees of freedom (and corresponding SS) are
allocated
at each size of experimental unit (or stratum):
Block level:
Replicate (block) 5
Error(block) 0
Plot level:
Variety 2
Variety x block 10 (whole plot error)
Subplot level:
Nitrogen 3
Nitrogen x Variety 6
N x V x B 45 (split plot error)
Observe how the degrees of freedom add up at each size of EU.
The expected mean squares of the random effects terms are used to get their
variance component estimates. Since the data are balanced, the ANOVA (or
method of moments) approach can be used to match the observed and expected
mean squares, from which the variance component estimates are computed.
In lme() and lmer(), the variance or standard deviation component estimates
are reported rather than the mean squares, but in a balanced model
one can reconstruct the observed mean squares for each random term
by plugging in the estimated variance components into the expected
mean square formulas. [In lmer(), you need to add both variance estimates
of block to get it right.] You will see that they agree.
lme():
Random effects:
Formula: ~1 | block
(Intercept)
StdDev: 14.64496
Formula: ~1 | variety %in% block
(Intercept) Residual
StdDev: 10.29863 13.30727
lmer():
Random effects:
Groups Name Variance Std.Dev.
variety:block (Intercept) 106.06 10.299
block (Intercept) 107.24 10.356
block (Intercept) 107.24 10.356
Residual 177.08 13.307
The error variance component is estimated by the split plot error mean
square.
Since there are four observations per plot, the expected whole plot mean
square is
sigma_e^2 + 4 * sigma_w^2, which is estimated by the observed whole plot
MSE.
Finally, there are 12 observations per block, so its expected mean square is
sigma_e^2 + 4 * sigma_w^2 + 12 * sigma_b^2, which is estimated by the block
mean
square. From the ANOVA approach, one should be able to estimate the variance
components reported in lme() and lmer() by equating observed and expected
mean
squares, solving the resulting linear system from the bottom up. Conversely,
given the variance component estimates, you should be able to reconstruct
the mean squares of the three random effects terms, at which point you
should
deduce that all three are doing the right thing in their own way.
This happy circumstance obtains because in balanced, normal theory models,
the methods of least squares and (restricted) maximum likelihood coincide.
In more general unbalanced data situations, the results from LS and (RE)ML
do
not necessarily agree, and in fact they may even disagree about what are the
degrees of freedom for certain terms in a given model.
HTH,
Dennis
On Fri, Nov 5, 2010 at 3:50 AM, ian m s white <i.m.s.wh...@ed.ac.uk> wrote:
> Like James Booth, I find the SSQ and MSQ in lmer output confusing. The
> F-ratio (1.485) for Variety is the same for aov, lme and lmer, but
> lmer's mean square for variety is 1.485 times the subplot residual mean
> square. In the conventional anova table for a split-plot expt, the
> variety mean square is 1.485 times *main-plot* residual mean square.
> --
> ian m s white <i.m.s.wh...@ed.ac.uk>
>
>
> The University of Edinburgh is a charitable body, registered in
> Scotland, with registration number SC005336.
>
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