similar to: chisq.test and anova problems

Using anova of a glm with test = "Chisq", I get this: Analysis of Deviance Table Model: poisson, link: log Response: Days Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev P(>|Chi|) NULL 373 370.56 Block 3 71.05 370 299.51 2.543e-15 Variety 1 94.04 369

Hello, I am using R to analyze a large multilevel data set, using lmer() to model my data, and using anova() to compare the fit of various models. When I run two models, the output of each model is generated correctly as far as I can tell (e.g. summary(f1) and summary(f2) for the multilevel model output look perfectly reasonable), and in this case (see below) predictor.1 explains vastly more

Full_Name: Robert King Version: 1.5.0 OS: windows Submission from: (NULL) (134.148.4.19) Also occurs in 1.6.0 on linux anova.glm(fitted.object,test="Chisq") is giving strange answers in this situation > resptime sex task time 1 m s 210 2 m s 300 3 m s 420 4 f s 250 5 f s 310 6 f s 390 7 m c 310 8 m c 400 9 m c 600

Hi, I was wondering why anova() and drop1() give different tail probabilities for F tests. I guess overdispersion is calculated differently in the following example, but why? Thanks for any advice, Tom For example: > x<-c(2,3,4,5,6) > y<-c(0,1,0,0,1) > b1<-glm(y~x,binomial) > b2<-glm(y~1,binomial) > drop1(b1,test="F") Single term deletions Model: y ~

Hi, New to both R and SEM, so this may be a very simple question. I am trying to run a very simple path analysis using the sem package. There are 2 exogenous (FARSCH, LOCUS10) and 2 endogenous (T_ATTENT, RMTEST) observed variables in the model. The idea is that T_ATTENT mediates the effect of FARSCH and LOCUS10 on RMTEST. The RAM specification I used is FARSCH -> T_ATTENT, y1x1, NA

This is a follow-up to a query that was posted regarding some problems that emerge when running anova analyses for cox models, posted by Mathias Gondan: Matthias Gondan wrote: >* Dear List,*>**>* I have tried a stratified Cox Regression, it is working fine, except for*>* the "Anova"-Tests:*>**>* Here the commands (should work out of the box):*>**>*

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

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 following data and code; cov <- c (1.670028 ,-1.197685 ,-2.931445,-1.197685,1.765646,3.883839,-2.931445,3.883839,12.050816) cov.matrix <- matrix(cov, 3, 3, dimnames=list(c("y1","x1","x2"), c("y1","x1","x2"))) path.model <- specify.model() x1 -> y1, x1-y1 x2 <-> x1, x2-x1 x2 <->

In this years biostat teaching I will include Rcommander (it indeed simplifies syntax problems that makes students frequently miss the core statistical problems). But I could not find how to make a simple chisquare comparison between observed frequencies and expected frequencies (eg in genetics where you expect phenotypic frequencies corresponding to 3:1 in standard dominant/recessif

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

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.

You should send this to r-help@stat.math.ethz.ch. On 03/09/2012 09:21 AM, Andrea Sica wrote: > Hello everybody, I'm looking for someone who is able with MCA and > would like to gives some help. > > If what I'm doing is not wrong, according to the purpose I have, I > need to understand how to create a dependence matrix, where I can > analyze the > dependence between

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: foo ba baz Version: R2.2.0 OS: Mac OS X (10.4) Submission from: (NULL) (219.66.32.183) chisq.test(matrix(c(9,10,9,11),2,2)) Chi-square value must be 0, and, P value must be 0 R does over correction when | a d - b c | < n / 2 &#65292;chi-sq must be 0

Hi, I want to determine the confidence interval on the sum of two sigma's. Is there an easy way to do this in R? I guess I have to use some sort of chisquare convolution algorithm??? Thanx, Roy -- The information contained in this communication and any atta...{{dropped}}

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(        

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]]

Hi, this is more related to understanding some statistics while using R; I've see such output in a paper: out <- glm(response~Var1+Var2+Var3..,family=binomial,data=mydata) summary(out) stepAIC(out) anova(out, test='Chisq') I understand that stepAIC is used to select the model with the lowest AIC (the best model) but can someone explain what is the purpose of doing the anova:

Dear List, I have tried a stratified Cox Regression, it is working fine, except for the "Anova"-Tests: Here the commands (should work out of the box): library(survival) d = colon[colon$etype==2, ] m = coxph(Surv(time, status) ~ strata(sex) + rx, data=d) summary(m) # Printout ok anova(m, test='Chisq') This is the output of the anova command: > Analysis of Deviance Table