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Background

I have data from a field study in which there are four treatment levels and six replicates in each of two blocks. (4x6x2=48 observations)

The blocks are about 1 mile apart, and within the blocks, there is a grid of 42, 2m x 4m plots and a 1m wide walkway; my study only used 24 plots in each block.

I would like to evaluate evaluate spatial covariance.

Here is an example analysis using the data from a single block, without accounting for spatial covariance. In the dataset, plot is the plot id, x is the x location and y the y location of each plot with plot 1 centered on 0, 0. level is the treatment level and response is the response variable.

layout <- structure(list(plot = c(1L, 3L, 5L, 7L, 8L, 11L, 12L, 15L, 16L, 
17L, 18L, 22L, 23L, 26L, 28L, 30L, 31L, 32L, 35L, 36L, 37L, 39L, 
40L, 42L), level = c(0L, 10L, 1L, 4L, 10L, 0L, 4L, 10L, 0L, 4L, 
0L, 1L, 0L, 10L, 1L, 10L, 4L, 4L, 1L, 1L, 1L, 0L, 10L, 4L), response = c(5.93, 
5.16, 5.42, 5.11, 5.46, 5.44, 5.78, 5.44, 5.15, 5.16, 5.17, 5.82, 
5.75, 4.48, 5.25, 5.49, 4.74, 4.09, 5.93, 5.91, 5.15, 4.5, 4.82, 
5.84), x = c(0, 0, 0, 3, 3, 3, 3, 6, 6, 6, 6, 9, 9, 12, 12, 12, 
15, 15, 15, 15, 18, 18, 18, 18), y = c(0, 10, 20, 0, 5, 20, 25, 
10, 15, 20, 25, 15, 20, 0, 15, 25, 0, 5, 20, 25, 0, 10, 20, 
25)), .Names = c("plot", "level", "response", "x", "y"), row.names = c(NA, 
-24L), class = "data.frame")

model <- lm(response ~ level, data = layout)      
summary(model)

Questions

  1. How can I calculate a covariance matrix and include it in my regression?
  2. The blocks are very different, and there are strong treatment * block interactions. Is it appropriate to analyze them separately?
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  • 1
    $\begingroup$ Plots 37 and 39 are both at x=18, y=10. Typo? $\endgroup$ – Aaron - Reinstate Monica May 16 '11 at 15:54
  • $\begingroup$ @Aaron thank you for pointing that out. y= [0,10]. Fixed. $\endgroup$ – David LeBauer May 16 '11 at 15:59
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1) You can model spatial correlation with the nlme library; there are several possible models you might choose. See pages 260-266 of Pinheiro/Bates.

A good first step is to make a variogram to see how the correlation depends on distance.

library(nlme)
m0 <- gls(response ~ level, data = layout)  
plot(Variogram(m0, form=~x+y))

Here the sample semivariogram increases with distance indicating that the observations are indeed spatially correlated.

One option for the correlation structure is a spherical structure; that could be modeled in the following way.

m1 <- update(m0, corr=corSpher(c(15, 0.25), form=~x+y, nugget=TRUE))

This model does seem to fit better than the model with no correlation structure, though it's entirely possible it too could be improved on with one of the other possible correlation structures.

> anova(m0, m1)
   Model df     AIC      BIC    logLik   Test  L.Ratio p-value
m0     1  3 46.5297 49.80283 -20.26485                        
m1     2  5 43.3244 48.77961 -16.66220 1 vs 2 7.205301  0.0273

2) You could also try including x and y directly in the model; this could be appropriate if the pattern of correlation depends on more than just distance. In your case (looking at sesqu's pictures) it seems that for this block anyway, you may have a diagonal pattern.

Here I'm updating the original model instead of m0 because I'm only changing the fixed effects, so the models should both be fit using maximum likelihood.

> model2 <- update(model, .~.+x*y)
> anova(model, model2)
Analysis of Variance Table

Model 1: response ~ level
Model 2: response ~ level + x + y + x:y
  Res.Df    RSS Df Sum of Sq      F   Pr(>F)   
1     22 5.3809                                
2     19 2.7268  3    2.6541 6.1646 0.004168 **

To compare all three models, you'd need to fit them all with gls and the maximum likelihood method instead of the default method of REML.

> m0b <- update(m0, method="ML")
> m1b <- update(m1, method="ML")
> m2b <- update(m0b, .~x*y)
> anova(m0b, m1b, m2b, test=FALSE)
    Model df      AIC      BIC     logLik
m0b     1  3 38.22422 41.75838 -16.112112
m1b     2  5 35.88922 41.77949 -12.944610
m2b     3  5 29.09821 34.98847  -9.549103

Remember that especially with your knowledge of the study, you might be able to come up with a model that is better than any of these. That is, model m2b shouldn't necessarily be considered to be the best yet.

Note: These calculations were performed after changing the x-value of plot 37 to 0.

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  • $\begingroup$ thanks for your helpful answer; it is not clear why in part 2 you updated model instead of m0, eg. m2 <- update(m0, .~.+x*y) so that all three models can be compared using anova(m0,m1,m2); after doing this, m2 is a big looser (AIC = 64) it seems that your part $\endgroup$ – David LeBauer May 16 '11 at 16:51
  • $\begingroup$ p.s. the last line should be 'after changing y-value of plot 37 to 5'; the actual value is 0, but the results are equivalent. $\endgroup$ – David LeBauer May 16 '11 at 16:52
  • $\begingroup$ If you compare m0,m1, and m2 as you suggest you get the warning: Fitted objects with different fixed effects. REML comparisons are not meaningful. To compare fixed effects you have to use regular maximum likelihood instead of REML. See edit. $\endgroup$ – Aaron - Reinstate Monica May 16 '11 at 19:49
  • $\begingroup$ thank you for all of your help. I am not sure why, but I am getting errors when I try to fit other correlation structures, e.g. using corExp as in the Pinheiro and Bates example. I have opened a question on SO about this error, but your input would be appreciated. $\endgroup$ – David LeBauer May 16 '11 at 22:39
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1) What is your spatial explaining variable? Looks like the x*y plane would be a poor model for the spatial effect.

plot of treatments and responses

i=c(1,3,5,7,8,11,14,15,16,17,18,22,23,25,28,30,31,32,35,36,39,39,41,42)
l=rep(NA,42)[i];l[i]=level
r=rep(NA,42)[i];r[i]=response
image(t(matrix(-l,6)));title("treatment")
image(t(matrix(-r,6)));title("response")

2) Seeing as how the blocks are 1 mile apart and you expect to see effects for mere 30 meters, I would say it's entirely appropriate to analyze them separately.

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  • $\begingroup$ The visualization is helpful, but if you compare the lower right to the upper right of the figures, it appears to me that location has a stronger effect than level. (p.s I think l[i]=response should be r[i]=...) $\endgroup$ – David LeBauer May 15 '11 at 18:48
  • $\begingroup$ Yes. The location effect is remarkable, and so you'd really want a good model for that before you start estimating treatment effects. Unfortunately, there is a lot of missing data so it's difficult to say what it should be - the best I can come up with would be a modelling location effect as an average of the neighbours response + random component, and then trying the treatment on that. That's very suspect, so any additional domain knowledge would be valuable. typo fixed. $\endgroup$ – sesqu May 15 '11 at 20:14
  • $\begingroup$ @sesqu... there is no missing data, data from all 24 randomly located plots is there. $\endgroup$ – David LeBauer May 16 '11 at 14:19
  • $\begingroup$ There is missing data in the sense that not every x,y pair has data. $\endgroup$ – Aaron - Reinstate Monica May 16 '11 at 16:25

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