Dear Listserv community, I realize that this is more a statistical theory question, rather an R application question, but I hope those familiar the theory underlying manyglm and manylm could help me. In evaluating the overall response of a community to a treatment, I'm aware that one class of approaches involves doing univariate analyses for each species (e.g., ANOVA, t-test, chi square, logistic modeling, etc) and then "summing" the results across all species and evaluating statistical significance with a randomization procedure. My question is has anyone considered using a chi square test instead of randomization to obtain the significance value? The p value for any test statistic (F, t) can be converted to a chi square value with 1 df. Because chi square values are additive (assuming independence), it makes sense to me that you could simply add up the chi square values for all species and evaluate the significance of the resulting sum assuming a df equal to the number of tests (species). Presumably, one could use different tests for different species, depending on whichever is most appropriate (e.g., anova for common species that differ in abundance between treatments or chi square or a logistic model for species that differ in terms of frequency of occurrence between treatments). If one were confident that the univariate assumptions held for each species' test, other than the assumption of independence of responses among species, I'm wondering what if anything is wrong with such an approach for obtaining a significance value. Perhaps something similar is being done when the log likelihood value is calculated? If so, what are the similarities or differences? Thanks, and I apologize if this question is too basic or has already been answered. Steve J. Stephen Brewer Professor Department of Biology PO Box 1848 University of Mississippi University, Mississippi 38677-1848 Brewer web page - http://home.olemiss.edu/~jbrewer/ FAX - 662-915-5144 Phone - 662-202-5877
evaluating multiple responses using chi square
1 message · Steve Brewer