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chisq.test vs manual calculation - why are different results produced?
2 messages · Louise Mair, David Winsemius
On Feb 20, 2012, at 5:57 AM, Louise Mair wrote:
Hello, I am trying to fit gamma, negative exponential and inverse power functions to a dataset, and then test whether the fit of each curve is good. To do this I have been advised to calculate predicted values for bins of data (I have grouped a continuous range of distances into 1km bins), and then apply a chi-squared test. Example:
data <- data.frame(distance=c(1,2,3,4,5,6,7), observed=c(43,13,10,6,2,1),
predicted=c(28, 18, 10, 5 ,3, 1, 1))
There's an error with that code.
chisq.test(data$observed, data$predicted)
Which gives:
Pearson's Chi-squared test
data: data$observed and data$predicted
X-squared = 35, df = 25, p-value = 0.0882
Warning message:
In chisq.test(data$observed, data$predicted) :
Chi-squared approximation may be incorrect
I understand this is due to having observed/predicted values of less
than
five, however I am interested to know firstly why R uses such a large
number of degrees of freedom (when by my understanding there should
only be
4 df), and secondly whether using the following manual calculation is
therefore inappropriate -
Read the help page Details section .... end of second paragraph. You probably wanted: chisq.test(cbind(data$observed, data$predicted))
X2 <- sum(((data$observed - data$predicted)^2)/data$predicted) 1-pchisq(X2,4)
[1] 0.04114223 If chi-squared is unsuitable, what other test can I use to determine whether my observed and predicted data come from the same distribution? The frequently recommended fisher's test doesn't seem to be any more appropriate as it requires values of greater than 5 for contingency tables larger than 2 x 2. Thanks for your help. Louise [[alternative HTML version deleted]]
Plain text is requested as the mail format. David Winsemius, MD West Hartford, CT