similar to: chisq.test, basic question

I don't quite understand the difference between the two methods for performing a chi-squared test on contingency tables: summary(table()) and chisq.test() They may different results. E.g.: aa <- gl(2, 10) bb <- as.factor(c(1,2,2,2,1,2,1,2,2,2,1,2,2,2,1,1,1,2,1,1)) aa <- c(aa, aa) bb <- c(bb, bb) table(aa, bb) summary(table(aa, bb)) chisq.test(aa, bb) Could somebody give me

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

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

Hello, I am trying to fit gamma, negative exponential and inverse power functions to a dataset, and then test whether the fit of each curve is good. To do this I have been advised to calculate predicted values for bins of data (I have grouped a continuous range of distances into 1km bins), and then apply a chi-squared test. Example: > data <- data.frame(distance=c(1,2,3,4,5,6,7),

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 =

Dear Members, I have two estimated transition matrices and I want to compare them. In fact I want to check the hypothesis if they come from the same process. I tried to look for some test but all I found was independence test of contingency tables. The following code shows that the usual chi-squared test statistic does not follow chisq distribution. MCRepl <- 5000 khi12 <- rep(0,MCRepl)

Hello! What kind of calculation perform 'chisq.test', if its argument is a matrix? Thanks for any suggestion! ------------------------------------------------------------ Sincerely yours. Dr. Alessandro Semeria Tel. +39 544 536811 Models and Simulation Lab of Fax. +39 544 538663 The Environment Research Center - Montecatini

I have the following contingency table dat <- matrix(c(1,506,13714,878702),nr=2) And I want to test if their is an association between events A:{a,not(a)} and B:{b,not(b)} | b | not(b) | --------+-----+--------+ a | 1 | 13714 | --------+-----+--------+ not(a) | 506 | 878702 | --------+-----+--------+ I am worried that prop.test and chisq.test are not valid given the

Dear R-Users, How can I use chisq.test() as a goodness of fit test? Reading man-page I?ve some doubts that kind of test is available with this statement. Am I wrong? X2=sum((O-E)^2)/E) O=empirical frequencies E=expected freq. calculated with the model (such as normal distribution) See: http://www.itl.nist.gov/div898/handbook/eda/section3/eda35f.htm for X2 used as a goodness of fit test. Any

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, Found no measure of association or correlation for 2x2 contingency tables in fullrefman.pdf or google. Can someone point to a package that implements such calculations? Thanx. -- Alexandre Santos Aguiar - consultoria para pesquisa em sa??de - R Botucatu, 591 cj 81 tel 11-9320-2046 fax 11-5549-8760 www.spsconsultoria.com

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,

Dear List, If any of observed and/or expected data has less than 5 frequencies, then chisq.test (Pearson's Chi-squared Test for Count Data from package:stats) gives warning messages. For example, x<-c(10, 14, 10, 11, 11, 7, 8, 4, 1, 4, 4, 2, 1, 1, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1) y<-c(9.13112391745095, 13.1626482033341, 12.6623267638188, 11.0130706413029, 9.16415925139016,

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

Full_Name: nobody Version: 2.2.0 OS: any Submission from: (NULL) (219.66.34.183) 2 x 2 table, such as > x [,1] [,2] [1,] 10 12 [2,] 11 13 > chisq.test(x) Pearson's Chi-squared test with Yates' continuity correction data: x X-squared = 0.0732, df = 1, p-value = 0.7868 but, X-squared = 0.0732 is over corrected. when abs(a*d-b*c) <= sum(a,b,c,d), chisq.value

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(        

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 would like to do a goodness-of-fit test on my data to see if they follow a mixture of 2 poisson distributions. I have small numbers for observed values. Most of them <5. The chisq.test gives warning message: Chi-squared approximation may be incorrect in: chisq.test(x , p = prob). However, the option sim=TURE would suppress the warning message. Does that mean with the option

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]]