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 <dmck at u.washington.edu> 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 donaldmckenzie at fs.fed.us Affiliate Assistant Professor College of Forest Resources CSES Climate Impacts Group University of Washington dmck at u.washington.edu __________________________________