Howdy,
I am interested in testing the power of a pre-existing survey structure which has produced considerable data. The ultimate survey aim is determine if poison baiting reduces the number of wild dogs. However, at the moment I am only interested in being able to determine the power of the survey technique to detect the presence of dogs on sandplots prior to baiting. I will consider the comparison between pre and post baiting next. Sandplots are replicated X times in a survey area and are sampled across Y days. As there is no baiting I am not considering days as time-series but I am considering plots to be spatially correlated. I have therefore chosen the following model (using lme4) with day and plot as random effects.
model1<-lmer(dog~1+(1|day)+(1|plot), "data")
summary(model1)
Linear mixed model fit by REML
Formula: dog ~ 1 + (1 | day) + (1 | plot)
Data: june09
AIC   BIC logLik deviance REMLdev
270.6 282.9 -131.3    258.6   262.6

Random effects:
Groups   Name        Variance Std.Dev.
plot     (Intercept) 0.077326 0.27808
day      (Intercept) 0.000000 0.00000
Residual             0.238994 0.48887

Number of obs: 159, groups: plot, 53; day, 3
Fixed effects:
Estimate Std. Error t value
(Intercept)  0.27673    0.05443   5.085

The data is strongly zero inflated with a mean of 0.2 dogs over the 53 plots on day 1 for example and an overall mean of 1.26 dogs per plot over the 3 days (max=2). I interpret this result, assuming it is implemented correctly, to mean practically zero variation between days, almost no variation between plots with the vast majority being due to random error. Presumably due to the high level of non-detection across the majority of plots.

If correct so far, testing the detection power with different numbers of days and plots would be, as we say in Australia, 'grouse'. Hence PAMM.
I am assuming I use EAMM but do no not understand it's implementation, especially in respect to the above results. 
Following the example in Martin: "ours=EAMM(numsim=10,group=10,repl=4,fixed=c(0,1,1),VI=seq(0.05,0.3,0.05),VS=seq(0.05,0.2,0.05) )"
I am presumming my 'group'=53 (i.e.plots), 'replicate=3 (days), fixed=(mean (0.27673???), variance(0.05443????), estimate of effect to simulate ???), 
and not sure about what the 3 factors in "VI=seq(0.05,0.3,0.05)" refer to.
Help appreciated.

Dr Graeme Armstrong
adjunct Research Fellow
Griffith School of Environment
Brisbane Australia


 		 	   		  
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