similar to: historical significance of Pr(>Chisq) < 2.2e-16

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).

This is a message for whoever maintains "chisq.test": For an outcome more extreme than 2000 simulations, a Monte Carlo p-value of "< 2.2e-16" was printed. Ripley said the proper p-value for such cases should be 1/(B+1) = 1/2001. This can be easily fixed by adding "if(PVAL==0)PVAL <- 1/(B+1)" right after the following line in the code for chisq.test (in R

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,

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)) >

Hi list, i've got a question about the chisq.test function. in the use of the "given probabilities" method (p= ...), normally there should be typed in probabilities in the range of 0 to 1 with the absolute sum of 1.0 (r-help) But it is possible to use probabilities > than 1. or the sum <1.! without any warning message Ok, now the question, what does r calcutate in these

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

Hi All, I was wondering if someone could help me to solve this issue with lmer. In order to understand the best mixed effects model to fit my data, I compared the following options according to the procedures specified in many papers (i.e. Baayen <http://www.google.it/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CDsQFjAA

Dear Colleagues, I run this model: mod1 <- lmer(x~category+subcomp+category*subcomp+(1|id),data=impchiefsrm) obtain this summary result: Linear mixed-effects model fit by REML Formula: x ~ category + subcomp + category * subcomp + (1 | id) Data: impchiefsrm AIC BIC logLik MLdeviance REMLdeviance 4102 4670 -1954 3665 3908 Random effects: Groups Name Variance

Hi I am trying to do this: chisq.test(c(11, 13, 12, 18, 21, 43, 15, 12, 9, 10, 5, 28, 22, 11, 15, 11, 18, 28, 16, 8, 15, 19, 44, 18, 11, 23, 15, 23, 2, 5, 4, 14, 3, 22, 9, 0, 6, 19, 15, 32, 3, 16, 14, 10, 24, 16, 24, 31, 29, 28, 16, 26, 11, 11, 4, 17, 16, 13, 20, 26, 16, 19, 34, 19, 17, 14, 22, 25, 17, 12, 23, 14, 19, 30, 18, 10, 23, 21, 17, 16, 10, 14, 6, 17, 17, 10, 21, 25, 20, 4, 11, 4,

Hello, I'm trying to test my data for normality. I enter the data (95ish species counts) run >ks.test (data,pnorm) and get a p- value <2.2e-16 But this seems to be the p-value no matter what the data I enter. (I have multiple datasets and am testing them all for normality). [Actually, I just entered a vector of 1's and the p-value changed.] When I use the >Shapiro.test command,

Hi, In the chisq.test(), if the expected frequency for some categories is <5, there will be a warning message which says Warning message: Chi-squared approximation may be incorrect in: chisq.test(x, p = probs) I am wondering whether there are some methods to get rid of this mistake... Seems the ?chisq.test() doesn''t provide more options to solve this problem. Or, the only choice is

> On 28 Dec 2017, at 13:08 , Kurt Hornik <Kurt.Hornik at wu.ac.at> wrote: > >>>>>> Jan Motl writes: > >> The chisq.test on line 57 contains following code: >> STATISTIC <- sum(sort((x - E)^2/E, decreasing = TRUE)) > > The preceding 2 lines seem relevant: > > ## Sorting before summing may look strange, but seems to be >

In an effort to select the most appropriate number of clusters in a mixture analysis I am comparing the expected and actual membership of individuals in various clusters using the Fisher?s exact test. I aim for the model with the lowest possible p-value, but I frequently get p-values below 2.2e-16 and therefore does not get exact p-values with standard Fisher?s exact tests in R. Does anybody know

I am getting "Chi-squared approximation may be incorrect in: chisq.test(x)" with the data bleow. Frequency distribution of number of male offspring in families of size 5. Number of Male Offspring N 0 518 1 2245 2 4621 3 4753 4 2476 5

I'm using R1.7.0 runing with Win XP. Thanks, ...Tao ???????????????????????????????????????????????????????? >x [,1] [,2] [1,] 149 151 [2,] 1 8 >t(x) [,1] [,2] [1,] 149 1 [2,] 151 8 >chisq.test(x, simulate.p.value=T, B=100000) Pearson's Chi-squared test with simulated p-value (based on 1e+05 replicates) data: x X-squared = 5.2001, df =

Hello, I received the following warning when running chi-square; n Is there a way to catch the 'error' code of 'warning' after run chisq.test(x)? n What does this error mean? Thank you for your help. [[alternative HTML version deleted]]

for example, an expression such as chisq(df=1,ncp=0) ? thanks -- View this message in context: http://www.nabble.com/how-to-calculate-chisq-value-in-R-tp15338943p15338943.html Sent from the R help mailing list archive at Nabble.com.

I am very new to R. I have some data from a CVS stored in vdata with 4 columns labeled: X08, Y08, X09, Y09. I have created two new "columns" like so: Z08 <- (vdata$X08-vdata$Y08) Z09 <- (vdata$X09-vdata$Y09) I would like to use chisq.test for each "row" and output the p-value for each in a stored variable. I don't know how to do it. Can you help? so far I have

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(        

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