Do you think p values are bad? That's not my understanding. P values may not be reported by some software, because the algorithm developers didn't know how to efficiently compute a reasonably accurate p value. And if you do a thousand or a million statistical tests and report only the results with the smallest p value, that's ultimately fraudulent. (http://en.wikipedia.org/wiki/Multiple_comparisons).
However, the concept of a significance probability or p value is quite valuable (http://en.wikipedia.org/wiki/P-value), though like any tool or concept, it can be misused. Hope this helps. Spencer Graves Monica Pisica wrote: > Hi Spencer and David, > > Thanks for your answers .... first - yes it is deviance but just > before i just spoke and explain that it is the equivalent of r square > from the "normal" regression..... > > I hope i can do the comparison and show that the model is significant > and hopefully i am off the hook. Sincerely i try to avoid all this > business with p-values but certainly some are quite found of it. The > problem is that you get almost by default a p-value from an F test if > you use lm for example, so ..... quite few times i was asked to > provide a similar thing for quite different models. > > Thanks again, > > Monica > > > Date: Thu, 27 Mar 2008 16:38:12 -0700 > > From: [EMAIL PROTECTED] > > To: [EMAIL PROTECTED] > > CC: r-help@r-project.org > > Subject: Re: [R] dreaded p-val for d^2 of a glm / gam > > > > I assume you mean 'deviance', not 'squared deviance'; if the > > latter, then I have no idea. > > > > If the former, then a short and fairly quick answer to your > > question is that 2*log(likelihood ratio) for nested hypotheses is > > approximately chi-square with numbers of degrees of freedom = the > number > > of parameters in the larger model fixed to get the smaller model, under > > standard regularity conditions, the most important of which is that the > > maximum likelihood is not at a boundary. > > > > For specificity, consider the following modification of the first > > example in the 'glm' help page: > > > > counts <- c(18,17,15,20,10,20,25,13,12) > > outcome <- gl(3,1,9) > > treatment <- gl(3,3) > > glm.D93 <- glm(counts ~ outcome + treatment, family=poisson()) > > glm.D93t <- glm(counts ~ treatment, family=poisson()) > > anova(glm.D93t, glm.D93, test="Chisq") > > > > The p-value is not printed by default, because some people would > > rather NOT give an answer than give an answer that might not be very > > accurate in the cases where this chi-square approximation is not very > > good. To check that, you could do a Monte Carlo, refit the model with, > > say, 1000 random permutations of your response variable, collect > > anova(glm.D93t, glm.D93)[2, "Deviance"] in a vector, and then find out > > how extreme the deviance you actually got is relative to this > > permutation distribution. > > > > Hope this helps. > > Spencer Graves > > p.s. Regarding your 'dread', please see fortune("children") > > > > Monica Pisica wrote: > > > OK, > > > > > > I really dread to ask that .... much more that I know some > discussion about p-values and if they are relevant for regressions > were already on the list. I know to get p-val of regression > coefficients - this is not a problem. But unfortunately one editor of > a journal where i would like to publish some results insists in giving > p-values for the squared deviance i get out from different glm and gam > models. I came up with this solution, but sincerely i would like to > get yours'all opinion on the matter. > > > > > > p1.glm <- glm(count ~be+ch+crr+home, family = 'poisson') > > > > > > # count - is count of species (vegetation) > > > # be, ch, crr, home - different lidar metrics > > > > > > # calculating d^2 > > > d2.p1 <- round((p1.glm[[12]]-p1.glm[[10]])/p1.glm[[12]],4) > > > d2.p1 > > > 0.6705 > > > > > > # calculating f statistics with N = 148 and n=4; f = > (N-n-1)/(N-1)(1-d^2) > > > f <- (148-4-1)/(147*(1-0.6705)) > > > f > > > [1] 2.952319 > > > > > > #calculating p-value > > > pval.glm <- 1-pf(f, 147,143) > > > pval.glm > > > [1] 1.135693e-10 > > > > > > So, what do you think? Is this acceptable if i really have to give > a p-value for the deviance squared? If it is i think i will transform > everything in a fuction .... > > > > > > Thanks, > > > > > > Monica > > > _________________________________________________________________ > > > Windows Live Hotmail is giving away Zunes. > > > > > > M_Mobile_Zune_V3 > > > ______________________________________________ > > > R-help@r-project.org 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. > > > > > > ------------------------------------------------------------------------ > Watch “Cause Effect,” a show about real people making a real > difference. Learn more. > <http://im.live.com/Messenger/IM/MTV/?source=text_watchcause> ______________________________________________ R-help@r-project.org 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.