By looking at R thread, it seems that the approach is: (1) cut the data into bins (you can use hist() to do this); (2) calculate the expected numbers in each bin using the differences of the CDF (pnorm, pexp, etc.); (3) calculate sum((exp-obs)^2/exp); (4) find the tail probability of the chi-square distribution (pchisq).
I am a newbie in R. Your help will be greatly appreciated. Thx ej On 12/5/06, Don McKenzie <[EMAIL PROTECTED]> wrote: > Ethan Johnsons wrote: > > If we use this data as an example, does ks.test still valid? > > > > E.Coli Group Observed Expected > > A 57 77.9 > > B 330 547.1 > > C 2132 2126.7 > > D 4584 4283.3 > > E 4604 4478.5 > > F 2119 2431.1 > > G 659 684.1 > > H 251 107.2 > You can use the test with any numeric data I believe. Whether it is > valid is more a question > for a statistician than for R. :-) > > Don > > -- > ___________________________________ > > Don McKenzie, Research Ecologist > Pacific Wildland Fire Sciences Lab > USDA Forest Service > 400 N 34th St. #201 > Seattle, WA 98103, USA > (206) 732-7824 > [EMAIL PROTECTED] > > Affiliate Assistant Professor > College of Forest Resources > CSES Climate Impacts Group > University of Washington > [EMAIL PROTECTED] > __________________________________ > > ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
