P Ehlers wrote: > > Marc Schwartz wrote: > >> On Thu, 2005-11-24 at 21:55 +0000, Ted Harding wrote: >> >>> On 24-Nov-05 P Ehlers wrote: >>> >>>> Bianca Vieru- Dimulescu wrote: >>>> >>>>> Hello, >>>>> I'm trying to calculate a chi-squared test to see if my data are >>>>> different from the theoretical distribution or not: >>>>> >>>>> chisq.test(rbind(c(79,52,69,71,82,87,95,74,55,78,49,60), >>>> >>> >>> c(80,80,80,80,80,80,80,80,80,80,80,80))) >>> >>>>> Pearson's Chi-squared test >>>>> >>>>> data: rbind(c(79, 52, 69, 71, 82, 87, 95, 74, 55, 78, 49, 60), >>>>> c(80, 80, 80, 80, 80, 80, 80, 80, 80, 80, 80, 80)) >>>>> X-squared = 17.6, df = 11, p-value = 0.09142 >>>>> >>>>> Is this correct? If I'm doing the same thing using Excel I obtained >>>>> a different value of p.. (1.65778E-14) >>>>> >>>>> Thanks a lot, >>>>> Bianca >>>> >>>> >>>> It would be unusual to have 12 observed frequencies all equal to 80. >>>> So I'm guessing that you have a 12-category variable and want to >>>> test its fit to a discrete uniform distribution. I assume that your >>>> frequencies are >>>> >>>> x <- c(79, 52, 69, 71, 82, 87, 95, 74, 55, 78, 49, 60) >>>> >>>> Then just use >>>> >>>> chisq.test(x) >>>> >>>> (see the help page). >>>> >>>> (If those 80's are expected cell frequencies, they should sum to >>>> sum(x) = 851.) >>>> >>>> I don't know what Excel does. >>>> >>>> Peter >>>> >>>> Peter Ehlers >>>> University of Calgary >>> >>> >>> I'm rather with Peter on this question! I've tried to infer what >>> you're really trying to do. >>> >>> My a-priori plausible hypothesis was that you have >>> >>> k<-12 >>> >>> independent observations which have equal expected values >>> >>> m<-rep(80,k) >>> >>> and are observed as >>> >>> x<-c(79,52,69,71,82,87,95,74,55,78,49,60) >>> >>> On this basis, a chi-squared test Sum((O-E)^2/E) gives >>> >>> C2<-sum(((x-m)^2)/m) >>> >>> so C2 = 41.1375, and on this hypothesis the chi-squared would >>> have k=12 degrees of freedom. Then: >>> >>> 1-pchisq(C2,k) >>> ## [1] 4.647553e-05 >>> >>> which is nowhere near the 1.65778E-14 you report from Excel. >>> Also, the result from Peter's chisq.test(x) is p = 0.0006468, >>> even further away. >> >> >> >> It's late on Turkey Day here, but shouldn't that be: >> >> >>> 1 - pchisq(C2, k - 1) # 11 df >> >> >> [1] 2.282202e-05 >> >> which is what I get using OO.org's Calc 2.0 with the CHITEST function >> using the two vectors as the observed (x) and expected (m) values. I >> also get this result from Gnumeric 1.4.3 using the same CHITEST >> function. >> > [snip] > > Marc, it's a bit sad to see that OO.org copies Excel's behaviour > to a _fault_. As Peter D. points out, we would expect the expected > frequencies and the observed frequencies to sum to the same value. > Excel (and Calc) blithely ignores that. R, OTH, gives an error > message when the probabilities don't sum to 1. > > Turkey soup for a few days now?
Thanks a lot for your answers! I have a fault in my Excel sheet:(, sorry. I corrected it and indeed I obtained 2.282202e-05 As I want to make a comparaison between independent observations which have equal expected values, I will do as Marc suggested and give up at the idea of using excel:) Bianca ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html