Dear r-helpers,
I am looking at a designed experiment in which one predictor variable
has 5 levels (0, ..., 4) and the other has 6 levels (1.1, ..., 1.6),
with 33 observations per cell. This design was given to 13 subjects.
0 1 2 3 4
1.1 32 33 0 0 0
1.2 33 33 33 0 0
1.3
Dear Prof. Logan:
Are you familiar with 'lme' in the 'nlme' package? Superlative
documentation for the 'nlme' package is Pinheiro and Bates (2000) Mixed
effects models in S and S-PLUS (Springer, listed as 'available' in the
catalog of your university library). The standard R distrib
Does anyone know of a way of dealing with unbalanced mixed effects
(fixed and random factors) for fully factorial designs.
An example of such data is given below;
The response variable is SQRTRECRUITS
SEASON is a random factor
DENSITY is a fixed factor
Thus DENSITY:SEASON is a fixed factor.
The
"Naiara S. Pinto" <[EMAIL PROTECTED]> writes:
> Dear all,
>
> I need to do a Manova but I have an unbalanced design. I have
> morphological measurements similar to the iris dataset, but I don't have
> the same number of measurements for all species. Does anyone know a
> procedure to do Manova wit
What have you tried? I just did help.searhc("manova"), which led me
to manova and summary.manova, which contained a balanced example, which
I unbalanced as follows:
tear <- c(6.5, 6.2, 5.8, 6.5, 6.5, 6.9, 7.2, 6.9, 6.1, 6.3,
6.7, 6.6, 7.2, 7.1, 6.8, 7.1, 7.0, 7.2, 7.5,
Dear all,
I need to do a Manova but I have an unbalanced design. I have
morphological measurements similar to the iris dataset, but I don't have
the same number of measurements for all species. Does anyone know a
procedure to do Manova with this kind of input in R?
Thank you very much,
Naiara.
alf Of Damián Cirelli
> Sent: Wednesday, December 01, 2004 2:03 PM
> To: [EMAIL PROTECTED]
> Subject: Re: [R] unbalanced design
>
>
> Thanks Peter,
>
> I still wonder why it thinks it's unbalanced...
>
>
__
[EMAIL PROT
Thanks Peter,
I still wonder why it thinks it's unbalanced...
The se's of the contrasts are different than the se's of the means,
which is the point of se=T in model.tables (type "means") I would have
thought. No big deal though, the following code makes a nice table with
the se's of the mean
Damián
I asked a similar question a few months ago (3 August 2004):
> temp.aov <- aov(S~rep+trt1*trt2*trt3, data=dummy.data)
> model.tables(temp.aov, type='mean', se=T)
>
> Returns the means, but states "Design is unbalanced - use se.contrasts
> for se's" which is a little surprising since the d
Hi all,
I'm new to R and have the following problem:
I have a 2 factor design (a has 2 levels, b has 3 levels). I have an
object kidney.aov which is an aov(y ~ a*b), and when I ask for
model.tables(kidney.avo, se=T) I get the following message along with
the table of effects:
Design is unbalanc
Arne,
For this sort of things, my suggestion is that you try to seek assistance
from local statistician(s), and I'd be very surprised if there's none for
you.
Data arising from unbalanced designs need no special tools for model
fitting. The garden variety linear models work just fine. The probl
Hello,
I'm wondering what's the best way to analyse an unbalanced design with a low number of
replicates. I'm not a statistician, and I'm looking for some direction for this
problem.
I've a 2 factor design:
Factor batch with 3 levels, and factor dose within each batch with 5 levels. Dose
leve
G'day,
The following code gives a two-way factorial anova, which is completely
balanced. However when model.tables is called, the response is:
"Design is unbalanced - use se.contrasts for se's", and SEs are not available.
Why doesn't R think my design is balanced and how do I convince it that i
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