similar to: how to use chisq.test with 2 row matrix having 0 in it?

Hi Tao: The P-values for 2x2 table are generated based on a random (discrete uniform distribution) sampling of all possible 2x2 tables, conditioning on the observed margin totals. If one of the cells is extremely small, as in your case, you get a big difference in P-values. Suppose, you changed the cell with value 1 to, say, 5 or 6, then the two P-values are nearly the same. However, I

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,

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 =

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

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 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 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)) >

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

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

Hi All I am trying to translate a proc mixed into a lme() syntax. It seems that I was able to do it for part of the model, but a few things are still different. It is a 2-level bivariate model (some call it a pseudo-3-level model). PROC MIXED DATA=psdata.bivar COVTEST METHOD = ml; CLASS cluster_ID individual_id variable_id ; MODEL y = Dp Dq / SOLUTION NOINT; RANDOM Dp Dq / SUBJECT = cluster_ID

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, I have a data set with 999 observations, for each of them I have data on four variables: site, colony, gender (quite a few NA values), and cohort. This is how the data set looks like: > str(dispersal) 'data.frame': 999 obs. of 4 variables: $ site : Factor w/ 2 levels "1","2": 1 1 1 1 1 1 1 1 2 2 ... $ gender: Factor w/ 2 levels "0","1":

Hi All, I woul like to ask you a couple of questions on chisq.test. First, I have 40 flies, 14 males and 26 females and I want to test for an a priori hypothesis that the sex ratio is 1:1 sex<-c(14,26) pr<-c(1,1)/2 chisq.test(se, p=pr, correct=TRUE) Chi-squared test for given probabilities data: sex X-squared = 3.6, df = 1, p-value = 0.05778 If my calculations are correct, this is

Hi, I want to do a chi square test and I have two tab delimited text files with Expected and Observed values to compare. Each file contains only the values and are 48 rows by 116 columns. I have managed to do something with them, but I don't think it is right as I got a p value of 1. In this case I used the read.table() function to read the values from the files. But I don't know if

Hi, I have a genotype data for both case and controls and would like to calculate the HW p-value. However, since the number of one genotype is 0, I got wired result. Would someone help me to figure it out? Or confirm it's right? Thanks a lot. ============ > library( "genetics" ) NOTE: THIS PACKAGE IS NOW OBSOLETE. The R-Genetics project has developed an set of enhanced

Dear all, I am trying to validate a model by comparing simulated output values against observed values. I have produced a simple X-y scatter plot with a 1:1 line, so that the closer the points fall to this line, the better the 'fit' between the modelled data and the observation data. I am now attempting to quantify the strength of this fit by using a statistical test in R. I am no

Dear R help I have been trying to conduct a chi square test on two groups of variables to test whether there is any relationship between the two sets of variables chisq.test(oxygen, train) Pearson's Chi-squared test data: oxygen X-squared = 26.6576, df = 128, p-value = 1 > chisq.test(oxygen) Pearson's Chi-squared test data: oxygen X-squared = 26.6576, df = 128,

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

The Chisquare test is based upon a normal approx of the (essentially) binomial distribution for the data in question. Small EXPECTED (not observed) values (<5) suggest a asymetric distribution and potential errors in inferential conclusions. The alternative is the exact test, which calculates the exact probabilities of the observed distribution, or a more extreme one, given the constraining