Re: [R] Nested ANOVA with a random nested factor (how to use the lme function?)

2005-07-18 Thread Simon Blomberg
At 01:59 PM 18/07/2005, Addison, Prue wrote:
Hi,

I am having trouble using the lme function to perform a nested ANOVA
with a random nested factor.

My design is as follows:

Location (n=6) (Random)

Site nested within each Location (n=12) (2 Sites nested within each
Location) (Random)

Dependent variable: sp (species abundance)

By using the aov function I can generate a nested ANOVA, however this
assumes that my nested factor is fixed.
  summary(aov(sp~Location/Site, data=mavric))

Df Sum Sq Mean Sq F value Pr(F)

Location   4 112366   28092  1.2742 0.2962

Location:Transect  5 121690   24338  1.1039 0.3736

Residuals 40 881875   22047

Yes, this is equivalent to aov(sp ~ Location + Location:Site,...)

I have tried the following lme function to specify that Site is random:

  lme1 - lme(sp~Location, random=~1|Site, data=mavric)

Here, Location is fixed, and Site is a grouping factor. There are fixed and 
random components to the intercept.


  lme2 - lme(sp~Location, random=~1|Location/Site, data=mavric)

Here you have Location as a fixed effect, the intercept is random and the 
grouping is Site %in % Location.

  anova(lme1)

 numDF denDF  F-value p-value

(Intercept) 140 3.418077  0.0719

Location4 5 1.152505  0.4294

How can you have 6 sites, but only 4 df for Location (should be 5?)


This gives me the correct F-value for Location from
MSLocation/MSLocation:Transect, but the p-value doesn't seem to be
correct (by my calculations in Microsoft Excel it should be 0.345)

I don't know your data, or your calculations. Microsoft Excel does not fill 
me with confidence.

  anova(lme2)

Warning in pf(q, df1, df2, lower.tail, log.p) :

  NaNs produced

Warning: NAs introduced by coercion

 numDF denDF  F-value p-value

(Intercept) 140 0.459966  0.5015

Location4 0 0.155091 NaN

? I don't know what this output means

Division by zero, somewhere? :-)

  anova(lme1,lme2)

  Model df  AIC  BIClogLik   Test  L.Ratio p-value

lme1 1  7 603.7534 616.4000 -294.8767

lme2 2  8 605.7534 620.2067 -294.8767 1 vs 2 1.815674e-05  0.9966


? I also don't know what this output means.

you are testing the difference between the models, using a likelihood ratio 
test. The difference is not significant, so the conventional wisdom is to 
choose the simpler model (lme1). Note that the AIC and BIC are lower 
(better) for lme1, too.


Can anyone tell me if there is a way to use the lme() function in order
to obtain the same output as the aov() function (above), but so it
correctly calculates the MS, F and p values for my main Location factor?

The following code shows agreement:

dat - data.frame(Location=factor(rep(1:6, each=2)), Site=factor(rep(1:2, 
12)), sp=rnorm(12))

fit - aov(sp ~ Location + Error(Site), data=dat)
fit2 - lme(sp ~ Location, random=~1|Site, data=dat)

summary(fit)
anova(fit2)

HTH,

Simon.

Thanks,



Prue

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Simon Blomberg, B.Sc.(Hons.), Ph.D, M.App.Stat.
Centre for Resource and Environmental Studies
The Australian National University
Canberra ACT 0200
Australia
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[R] Nested ANOVA with a random nested factor (how to use the lme function?)

2005-07-17 Thread Addison, Prue
Hi,

 

I am having trouble using the lme function to perform a nested ANOVA
with a random nested factor.

 

My design is as follows:

 

Location (n=6) (Random)

Site nested within each Location (n=12) (2 Sites nested within each
Location) (Random)

Dependent variable: sp (species abundance) 

 

 

By using the aov function I can generate a nested ANOVA, however this
assumes that my nested factor is fixed.

 

 summary(aov(sp~Location/Site, data=mavric))

  

Df Sum Sq Mean Sq F value Pr(F)

Location   4 112366   28092  1.2742 0.2962

Location:Transect  5 121690   24338  1.1039 0.3736

Residuals 40 881875   22047   

 

 

I have tried the following lme function to specify that Site is random:

 

 lme1 - lme(sp~Location, random=~1|Site, data=mavric)

 lme2 - lme(sp~Location, random=~1|Location/Site, data=mavric)

 

 anova(lme1)

numDF denDF  F-value p-value

(Intercept) 140 3.418077  0.0719

Location4 5 1.152505  0.4294

 

This gives me the correct F-value for Location from
MSLocation/MSLocation:Transect, but the p-value doesn't seem to be
correct (by my calculations in Microsoft Excel it should be 0.345)

 

 anova(lme2)

Warning in pf(q, df1, df2, lower.tail, log.p) : 

 NaNs produced

Warning: NAs introduced by coercion

numDF denDF  F-value p-value

(Intercept) 140 0.459966  0.5015

Location4 0 0.155091 NaN

 

? I don't know what this output means

 

 anova(lme1,lme2)

 Model df  AIC  BIClogLik   Test  L.Ratio p-value

lme1 1  7 603.7534 616.4000 -294.8767

lme2 2  8 605.7534 620.2067 -294.8767 1 vs 2 1.815674e-05  0.9966  

 

? I also don't know what this output means.

 

Can anyone tell me if there is a way to use the lme() function in order
to obtain the same output as the aov() function (above), but so it
correctly calculates the MS, F and p values for my main Location factor?

 

Thanks,

 

Prue 
  
This e-mail is solely for the named addressee and may be confidential. 
You should only read, disclose, transmit, copy, distribute, act in reliance 
on or commercialise the contents if you are authorised to do so.  If you 
are not the intended recipient of this e-mail, please notify 
[EMAIL PROTECTED] by e-mail immediately, or notify the sender 
and then destroy any copy of this message.  Views expressed in this e-mail 
are those of the individual sender, except where specifically stated to be 
those of an officer of Museum Victoria.  Museum Victoria does not represent, 
warrant or guarantee that the integrity of this communication has been 
maintained nor that it is free from errors, virus or interference. 
 
Museum Victoria  
+61 3 8341  
11 Nicholson St 
Carlton 
Victoria 
www.museum.vic.gov.au 
  

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