similar to: chisq

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.

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).

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.

Hi all, I have done a backward stepwise selection on a full binomial GLM where the response variable is gender. At the end of the selection I have found one model with only one explanatory variable (cohort, factor variable with 10 levels). I want to test the significance of the variable "cohort" that, I believe, is the same as the significance of this selected model: >

Hi, I have a text mining project and currently I am working on feature generation/selection part. My plan is selecting a set of words or word combinations which have better discriminant capability than other words in telling the group id's (2 classes in this case) for a dataset which has 2,000,000 documents. One approach is using "contrast-set association rule mining" while the

Hi there - I am using R1.5.1 on WinNT and the latest MASS (Venables and Ripley) library. Running the following code: >minimod<-glm(miniSF~gtbt*f.batch+log(mxjd),data=gtbt,family="binomial") >summary(minimod,cor=F) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 0.91561 0.32655 2.804 0.005049 ** gtbtgt 0.47171

Hi everyone, Suppose I have a count the occurrences of positive results, and the total number of occurrences: pos <- 14 total <- 15 testing that the proportion of positive occurrences is greater than 0.5 gives a p-value and confidence interval: prop.test( pos, total, p=0.5, alternative='greater') 1-sample proportions test with continuity correction data: 14 out of

Dear all, I was wondering if anyone could help solve a problem of a missing interaction effect!! I carried out a 2 x 2 factorial experiment to see if eggs from 2 different locations (Origin = 1 or 2) had different hatching success under 2 different incubation schedules (Treat = 1 or 2). Six eggs were taken from 10 females (random = Female) at each location and split between the treatments,

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(        

Hello all, I am trying to understand the different results I am getting from the following 3 commands: chisq.test(c(62,50), p = c(0.512,1-0.512), correct = F) # p-value = 0.3788 binom.test(x=62,n=112, p= 0.512) # p-value = 0.3961 2*(1-pbinom(62,112, .512)) # p-value = 0.329 Well, the binom.test was supposed to be "exact" and give the same results as the pbinom, while the chisq.test

Dear all, I noticed the following while playing around with fitting log-linear models to contingency tables using R 1.8.1, but the problem also exists under R 1.9.0. A reproducible example uses the following contingency table: > library(MASS) > data(quine) > tmp <- with(quine, expand.grid(Eth=levels(Eth), Sex=levels(Sex), + Lrn=levels(Lrn), Age=levels(Age)))

Hello, I am trying to see whether there has been a significant difference in whether people experienced damages from wildlife in two different years. I therefore have two columns: year 1: yes no no no yes yes no year 2: no yes no yes I wanted to do a chisq.test, but if I enter it this way: chisq.test(year1, year2) I get the error saying the columns are two different lengths. So then I tried

Hello, I want to use the quantile function so I read the doc but I don't understand with this > qchisq(seq(0.05,0.95,by=0.05),df=(length(don)-1)) [1] 62667.11 62795.62 62882.42 62951.47 63010.74 63064.00 63113.39 63160.27 63205.65 63250.33 63295.04 63340.48 63387.48 63437.03 63490.53 63550.14 63619.68 [18] 63707.24 63837.16 Can you help me please?

Hi, I have used lme4 and I have found a significant result when using anova to compare model 1 and model 2 (where I took out an interaction). The result looks like this: model.3: DIFFERENCE ~ (1 | MALE.ID) model.2: DIFFERENCE ~ MALE.SPECIES + (1 | MALE.ID) Df AIC BIC logLik Chisq Chi Df Pr(>Chisq) model.3 3 1379.7 1387.1 -686.86 model.2 4 1374.1 1384.0 -683.05 7.6235 1

Hello, I am using the following model model1=lmer(PairFrequency~MatingPair+(1|DrugPair)+(1|DrugPair:MatingPair), data=MateChoice, REML=F) 1. After reading around through the R help, I have learned that the above code is the right way to analyze a mixed model with the MatingPair as the fixed effect, DrugPair as the random effect and the interaction between these two as the random effect as well.

I should hope that there is trouble, since "type III" is an undefined concept for a Cox model. Since SAS Inc fostered the cult of type III they have recently added it as an option for phreg, but I am not able to find any hints in the phreg documentation of what exactly they are doing when you invoke it. If you can unearth this information, then I will be happy to tell you whether

Dear List, How do I extract the approximate Wald test for the frailty (in the following example 17.89 value)? What about the P-values, other Chisq, DF, se(coef) and se2? How can they be extracted? ######################################################> kfitm1 Call: coxph(formula = Surv(time, status) ~ age + sex + disease + frailty(id, dist = "gauss"), data = kidney)

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,

Hi all, I am having issues comparing models with lmer. As an example, when I run the code below the model summaries (AIC, BIC, loglik) differ between the summary() and anova() commands. Can anyone clear up what's wrong? Thank you! Darren Ward library(lme4) data(sleepstudy) fm1<-lmer(Reaction ~ Days + (1|Subject), sleepstudy) summary(fm1) fm2<-lmer(Reaction ~ Days +

Dear R gurus I am running a GLMM that looks at whether chimpanzees spend time in shade more than sun (response variable 'y': used cbind() on counts in the sun and shade) based on the time of day (Time) and the availability of shade (Tertile). I've included some random factors too which are the chimpanzee in question (Individual) and where they are in a given area (Zone). There are