Testing for strength of fit using R
Steve Murray wrote:
Dear all, I am trying to validate a model by comparing simulated output values against observed values. I have produced a simple X-y scatter plot with a 1:1 line, so that the closer the points fall to this line, the better the 'fit' between the modelled data and the observation data. I am now attempting to quantify the strength of this fit by using a statistical test in R. I am no statistics guru, but from my limited understanding, I suspect that I need to use the Chi Squared test (I am more than happy to be corrected on this though!). However, this results in the following:
chisq.test(data$Simulation,data$Observation)
Pearson's Chi-squared test
data: data$Simulation and data$Observation
X-squared = 567, df = 550, p-value = 0.2989
Warning message:
In chisq.test(data$Simulation, data$Observation) :
Chi-squared approximation may be incorrect
The ?chisq.test document suggests that the objects should be of vector or matrix format, so I tried the following, but still receive a warning message (and different results):
chisq.test(as.matrix(data[,4:5]))
Pearson's Chi-squared test
data: as.matrix(data[, 4:5])
X-squared = 130.8284, df = 26, p-value = 6.095e-16
Warning message:
In chisq.test(as.matrix(data[, 4:5])) :
Chi-squared approximation may be incorrect
What am I doing wrong and how can I successfully measure how well the simulated values fit the observed values?
If it's of any help, here are how my data are structured - note that I am only using columns 4 and 5 (Observation and Simulation).
str(data)
'data.frame': 27 obs. of 5 variables: $ Location : Factor w/ 27 levels "Australia","Brazil",..: 8 2 13 19 22 14 16 23 6 7 ... $ Vegetation : Factor w/ 21 levels "Beech","Broadleaf evergreen laurel",..: 17 21 2 16 15 16 9 16 3 4 ... $ Vegetation.Class: Factor w/ 4 levels "Boreal and Temperate Evergreen",..: 3 3 4 1 1 1 4 1 4 1 ... $ Observation : num 24 8.9 14.7 26.7 42.4 31.7 30.8 7.5 14 22 ... $ Simulation : num 33.9 7.8 9.74 7.6 11.8 10.7 12 28.1 1.7 1.7 ...
The chisquare test is not the right thing here. You may have been fooled by the "goodness-of-fit" phrase associated with the test. I would do a cor.test(). But if the above is the real data, then there probably isn't much to test; you have very little agreement for the first 10 pairs. -Peter Ehlers
I hope someone is able to point me in the right direction. Many thanks, Steve
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