similar to: bug(?) in chisq.test

Hello all, I would like to know what the difference is between chisq.test and fisher.test when using the Monte Carlo method with simulate.p.value=TRUE? Thank you -- View this message in context: http://r.789695.n4.nabble.com/Difference-in-Monte-Carlo-calculation-between-chisq-test-and-fisher-test-tp2322494p2322494.html Sent from the R help mailing list archive at Nabble.com.

Dear all, I have a question about the chisq.test command. As an option one can chose the computation of p-values by Monte-Carlo simulation (simulate.p.value=T). Is there any documentation available how this calculations are done and how this simulation based test behaves in small samples? Thanks Klaus Abberger University of Konstanz, Germany [[alternate HTML version deleted]]

>>>>> dmurdoch writes: >> Date: Wed, 16 Jul 2003 01:27:25 +0200 (MET DST) >> From: shitao@ucla.edu >>> x >> [,1] [,2] >> [1,] 149 151 >> [2,] 1 8 >>> c2x<-chisq.test(x, simulate.p.value=T, B=100000)$p.value >>> for(i in (1:20)){c2x<-c(c2x,chisq.test(x, >> simulate.p.value=T,B=100000)$p.value)}

Where does the value 2.2e-16 come from in p-values for chisq tests such as those reported below? > Anova(cm.mod2) Analysis of Deviance Table (Type II tests) Response: Freq LR Chisq Df Pr(>Chisq) B 11026.2 1 < 2.2e-16 *** W 7037.5 1 < 2.2e-16 *** Age 886.6 8 < 2.2e-16 *** B:W 3025.2 1 < 2.2e-16 *** B:Age 1130.4 8 < 2.2e-16 *** W:Age 332.9 8 < 2.2e-16 *** --- Signif.

Full_Name: Reginaldo Constantino Version: 2.8.0 OS: Ubuntu Hardy (32 bit, kernel 2.6.24) Submission from: (NULL) (189.61.88.2) For many tables, chisq.test with simulate.p.value=TRUE gives a p value that is obviously incorrect and inversely proportional to the number of replicates: > data(HairEyeColor) > x <- margin.table(HairEyeColor, c(1, 2)) >

Full_Name: Tao Shi Version: 1.7.0 OS: Windows XP Professional Submission from: (NULL) (149.142.163.65) > x [,1] [,2] [1,] 149 151 [2,] 1 8 > c2x<-chisq.test(x, simulate.p.value=T, B=100000)$p.value > for(i in (1:20)){c2x<-c(c2x,chisq.test(x, simulate.p.value=T,B=100000)$p.value)} > c2tx<-chisq.test(t(x), simulate.p.value=T, B=100000)$p.value > for(i in

Hello, I have a question concerning the R-function chisq.test. For example, I have some count data which can be categorized as follows class1: 15 observations class2: 0 observations class3: 3 observations class4: 4 observations I would like to test the hypothesis whether the population probabilities are all equal (=> Test for discrete uniform distribution) If you have a small sample size

R Help on 'chisq.test' states that "if 'simulate.p.value' is 'TRUE', the p-value is computed by Monte Carlo simulation with 'B' replicates. In the contingency table case this is done by random sampling from the set of all contingency tables with given marginals, and works only if the marginals are positive... In the

Full_Name: Kjetil Kjernsmo Version: 1.1.1.1 (patched the rpois :-)) OS: osf4.0e Submission from: (NULL) (129.240.28.227) Dear all, I have just gotten through a debugging session of my code, being confused for a few days. I sent what I thought was a straightforward matrix to chisq.test() and once in a while got the error message: > chisq.test(t2) Error in chisq.test(t2) : missing value where

Dear List, recently tried to reproduce the results of some custom model selection function after updating R, which unfortunately failed. However, I ultimately found the issue to be that testing with pchisq() in drop1() seems to have changed. In the below example, earlier versions (e.g. R 2.4.1) produce a missing P-value for the variable x, while newer versions (e.g. R 2.7.1) produce 0 (2.2e-16).

Hello everybody, I'm seeing some strange behavior on R 1.8.1 on Intel/Linux compiled with gcc 3.2.2. The p-value calculated from the chisq.test function is incorrect for some input values: > chisq.test(matrix(c(0, 1, 1, 12555), 2, 2), simulate.p.value=TRUE) Pearson's Chi-squared test with simulated p-value (based on 2000 replicates) data: matrix(c(0, 1, 1,

** I am using the following way to get p-values from fiser exact test. However, I do need to print for each pair the values "n00, n01, n10, n11". How can I print that as a table and not a matrix as below along with the p-value? Any help will be greatly appreciated fish <- function(y, x) {n00 = sum((1-x)*(1-y)); n01 = sum((1-x)*y); n10 = sum(x*(1-y)); n11 = sum(x*y); a =

Hi I am running chisq as below and getting a warning. Can anyone tell me the significance or the warning? > chisq.test(c(10 ,4 ,2 ,6 ,5 ,3 ,4 ,4 ,6 ,3 ,2 ,2 ,2 ,4 ,7 ,10 ,0 ,6 ,19 ,3 ,2 ,7 ,2 ,2 ,2 ,1 ,32 ,2 ,3 ,10 ,1 ,3 ,9 ,4 ,10 ,2 ,2 ,4 ,5 ,7 ,6 ,3 ,7 ,4 ,3 ,3 ,7 ,1 ,4 ,2 ,2 ,3 ,3 ,5 ,5 ,4 ), p =c(0.01704142 ,0.017988166 ,0.018224852 ,0.017751479 ,0.017988166 ,0.018224852 ,0.017278107

Dear list! I would like to calculate "chisq.test" on simple data set with 70 observations, but the output is ''Warning message:'' Warning message: In chisq.test(tabele) : Chi-squared approximation may be incorrect Here is an example:         tabele <- matrix(c(11, 3, 3, 18, 3, 6, 5, 21), ncol = 4, byrow = TRUE)         dimnames(tabela) <- list(        

Hi, This report deals with p-values coming from chisq.test using the simulate.p=TRUE option. The issue is numerical accuracy and was brought up in previous bug reports 3486 and 3896. The bug was considered fixed but apparently was only mostly fixed. Just the typical problem of two values that are mathematically equal not ending up numerically equivalent. Consider this series of three 2x2

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

Dear R users, I have now tried out several options of obtaining p-values for (quasi)poisson lmer models, including Markov-chain Monte Carlo sampling and single-term deletions with subsequent chi-square tests (although I am aware that the latter may be problematic). However, I encountered several problems that can be classified as (1) the quasipoisson lmer model does not give p-values when

chisq.test with simulate.p.value=TRUE uses the Patefield algorithm, this is not documented, and the original reference is not given, as it ought to be. The reference is: Patefield,W. M. (1981) An efficient method of generating r * c tables with given row and column totals (algorithm AS 159). Applied Statistics 30, 91-97. Kjetil Halvorsen

Dear useRs, First off, sorry about the long post. Figured it's better to give context to get good answers (I hope!). Some time ago I wrote an R function that will get all pairwise interactions of variables in a data frame. This worked fine at the time, but now a colleague would like me to do this with a much larger dataset. They don't know how many variables they are going to have in the

Is Pearson's Chi-Square test for contingency tables asymptotically unbiased for large tables (large degrees of freedom) regardless of the expected values in each cell? The rule of thumb is that Pearson's Chi-square should not be used when large numbers of cells have expected values < 5. However, I compared the results on 4x4 contingency tables for R's chisq.test using chi-square