R help - Mar 2004 - two lme questions

1) I have the following data situation:

96 plots
12 varieties
2 time points
2 technical treatments

the experiment is arranged as follows:

a single plot has two varieties tested on it. if variety A on plot #1 has 
treatment T1 applied to it, then variety B on plot #1 has treatment T2 
applied to it.  across the whole experiment variety A is exposed to 
treatment T1 the same number of times as treatment T2.

with respect to time points, plots come in 3 kinds. (1) varietyA,
timepoint#1 vs. variety B, timepoint#1 (2) varietyA timepoint #2 vs. 
varietyB timepoint #2, and (3) varietyA timepoint #1 vs. variety A 
timepoint#2

plots and varieties are random samples from a population of plots and 
varieties, so they are random effects.  The technical treatment and 
timepoints are fixed effects.

i am particularly interested in the variance components for variety and 
timepoint within variety, in the estimate for the fixed stage effect and 
in the predictions (BLUP) for variety and stage within variety.

My First Question is about specifying the random part of the lme() 
statement

the fixed part is Measurement~Treatment+Time
the random part, i think, should include random=~1|variety/time or 
equivalently(?) list(variety=~1,Time=~1)

but how do i also specify that plot should be a random effect?  it's not 
nested within variety, nor is variety nested within it.



2) I have fixed effects as above where each only has two kinds (2 
Treatments, 2 Times).  When I use lme and look at the estimates for the 
fixed effects I get output that looks like:
>summary(asdf.lme)
...

Fixed effects: Measurement ~ Treatment+Time
		Value	 Std.Error ...
(Intercept)	xxx	xxx
TreatmentHot	xxx	xxx
TimeEarly	xxx	xxx
  Correlation
...


where my two treatments are Hot and Cold and my two times are Early and 
Late and the xxx are actual numbers.

Why isnt there any line for the Cold treatment and the late time?  is it 
because these necessarily are the opposite of the Hot and Early ones so 
putting them in would be redundant (i.e. Hot+Cold=0, Early+Late=0)?


Thanks much,
Scott Rifkin
scott.rifkin at yale.edu

This is an update to my previous post today after finding some previous 
posts about crossed random effects.  Any comments would be much 
appreciated.


I have two random factors (Plot and Variety), one fixed factor (Time).  I 
got rid of one of the fixed factors from my previous post for simplicity.  
Each Variety is tested at each of two Times, so I would like to include 
Time %in% Variety in my random factors.

I tried the following based on a suggestion from Douglas Bates 
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/26424.html
> Data = groupedData(Measurement ~Time|Variety, data = 
read.table("mvone.txt",header = TRUE))
> Data$vt = factor(paste(Data$Variety,Data$Time,sep="/"))
> Data$const=factor(rep(1,nrow(Data)))
>
Data.lme=lme(Measurement~Time,data=Data,random=list(const=pdBlocked(list(pdIdent(~Plot-1),pdIdent(~Variety-1),pdIdent(~vt-1)))))
I'd like some help interpreting the results.(the ... mean that I deleted 
repetitive stuff)
>summary(Data.lme)
...
 Block 1: Plot18Ea18La, Plot18Ea49Ea, Plot18Ea59Ea, Plot18Ea67Ea, 
Plot18La18Ea, Plot18La21La, Plot18La46La, Plot18La71La, Plot21Ea21La, 
...
Plot71La67La, Plot71La71Ea
 Formula: ~Plot - 1 | const
 Structure: Multiple of an Identity
        Plot18Ea18La Plot18Ea49Ea Plot18Ea59Ea Plot18Ea67Ea Plot18La18Ea
StdDev:     0.973655     0.973655     0.973655     0.973655     0.973655
        Plot18La21La Plot18La46La Plot18La71La Plot21Ea21La Plot21Ea59Ea
StdDev:     0.973655     0.973655     0.973655     0.973655     0.973655
...

I take it that these StdDevs are the sqrt of the variance component for 
Plot.  They are the diagonal of the pdBlocked matrix corresponding to 
pdIdent(~Plot-1)

same thing for 

 Block 2: VarietyL23, VarietyL54, VarietyL59, VarietyL71, VarietyL49, 
...
 Formula: ~Variety - 1 | const
 Structure: Multiple of an Identity
        VarietyL23 VarietyL54 VarietyL59 VarietyL71 VarietyL49 VarietyL21
StdDev: 0.04296328 0.04296328 0.04296328 0.04296328 0.04296328 0.04296328
...
and

 Block 3: vtL18/Early, vtL18/Late, vtL21/Early, vtL21/Late, vtL23/Early, 
...
vtL71/Late
 Formula: ~vt - 1 | const
 Structure: Multiple of an Identity
        vtL18/Early  vtL18/Late vtL21/Early  vtL21/Late vtL23/Early  vtL23/Late
StdDev: 0.006307627 0.006307627 0.006307627 0.006307627 0.006307627 0.006307627
...

and 

         Residual
StdDev: 0.1299986

is just the sigma(e)

Fixed effects: Measurement ~ Time 
                Value  Std.Error  DF  t-value p-value
(Intercept)  9.762629 0.10228970 190 95.44097  <.0001
TimeLate    -0.222656 0.03712913 190 -5.99680  <.0001

I'm still wondering why I only get TimeLate...is it because TimeEarly is 
just 0.222656?

when I do:
> intervals(Data.lme)
Approximate 95% confidence intervals

 Fixed effects:
                 lower       est.      upper
(Intercept)  9.5608597  9.7626290  9.9643984
TimeLate    -0.2958942 -0.2226559 -0.1494177
attr(,"label")
[1] "Fixed effects:"

 Random Effects:
  Level: const 
                       lower        est.     upper
sd(Plot - 1)    8.436375e-01 0.973655068 1.1237103
sd(Variety - 1) 1.577451e-02 0.042963279 0.1170143
sd(vt - 1)      4.186565e-06 0.006307627 9.5032935

 Within-group standard error:
    lower      est.     upper 
0.1119908 0.1299986 0.1509019 


These are my confidence intervals on the variance components, I assume.  
I'm slightly confused on this because in one of the posts I found:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/21857.html

it seems to be much more complicated to find the variance componenets.

Thanks much,
Scott Rifkin
scott.rifkin at yale.edu