LJJ
2024-Sep-17 02:58 UTC
[Rd] Degree of freedom issue on chi-square test of goodness of fit
I am running chi-square test of goodness of fit, the observation data was well groupped to observation freq. I assume the data obey the normal distribution N(? ?^2), while ? and ?^2 are unknown. So I use their estimation. Consequently, I calculated out the probabilities and input in vector p>x<-c(6,13,14,27,25,19,10,6) #observation freq>p<-c(0.0505, 0.0874, 0.1533, 0.2088, 0.2088, 0.1533, 0.0874, 0.0505)#probabilities from normal distribution test_result<-chisq.test(x=x,p=p) test_result Chi-squared test for given probabilities data: x X-squared = 1.8468, df = 7, p-value = 0.9678 from the out put, the degree of freedon is 7 (df=7), because the data was in 8 groups so df=8-1=7 That is perfect if the parameters (? and ?^2) in normal distribution is precise. However, in this case, they are unknow and I used ther estimations. If the observation group is n, number of unknow parameters in distribution is r, then in chi-square test, the degree of freedom should be n-r-1. But in chisq.test() output, it is always n-1. In this case, the degree of freedom should be 8-2-1=5 (there are two unknow parameters). I have to run function pchisq() for rest calculation manually. I propose to enhance function chisq.test() in degree of freedom. In practice, there are unknow parameters in distribution function. [[alternative HTML version deleted]]