similar to: In chisq.test(x) : Chi-squared approximation may be incorrect

Hi All, I'm new with R; this is a basic question. I was given a matrix I of (nrow, ncol), I would like to create another matrix A with some data in the matrix I, say [1,4] (row 1, column 4) to [271,19000] (row 271, column 19000). How do I do this? Please help. Thank you very much. --mc [[alternative HTML version deleted]]

Dear R-Helpers In R 2.5.1, the command library(Rgraphviz) fails on my Windows (XP SP2) system with error popup "The procedure entry point Rf_allocString could not be located in the dynamic link library R.dll". Thanks in advance for any suggestion in solving the error. My D. Coyne Imagination is more important than knowledge... (Albert Einstein) mcoyne@boninc.com

I don't know if this is the correct forum to ask the following question; however, when I search the aroma.affymetrix discussion group, it suggested that I should posted the question to r-help. Here it goes. I followed the instructions on aroma.affymetrix trying to install the packages; following are the steps: > install.packages(c("R.oo", "R.utils"),

Hello, Newbie question and hope you can help . I have two vector V1 and V2, where length(V2) = length of (V1) * 2; length(V1) ~ 16,000. For each member in V1, I need to compare 2 element of V2 for equality i.e. for (I in 1:length (V1)) { if ( v2[i] == v1[i] & v2[i+1]==v1[i] ){ statement_1 statement_2 . } } This for-loop is too slow

R-newbie question I have 2 matrices (a) P1 has only one column of 32K rows (b) PC has 2 column {P, C} of 3200 rows Every values in P1 matches with a value in PC[,p] (column p). I would like to use Which to search for all value in P1 that matchex PC[,p] and get the PC[,c]. However because P1 and PC does not have the same number of rows, I got lots of 'NA'. Thanks for your

Hi, I place a t.test in a loop and would like to continue to process the loop even when t.test encounter error. How do I do that? For example, in one iteration, the data is completely constant and t.test gives error, the entire program terminates. I would like to write the information out to a file, and the loop should continue. Thanks My D. Coyne [[alternative HTML version

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 would like to run a series of chisq.test() and store results (x-square, df, p-value) in a matrix for further analysis, but don't know how to do such. Currently I run the command line for each set of data at a time and it is time consuming. Thank you for your help. --My Coyne [[alternative HTML version deleted]]

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: >

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

Dear UseRs, I'm running a chi-squared test where the expected matrix is the same as the observed, after rounding. R reports a X-squared of zero with a p value of one. I can justify this because any other result will deviate at least as much from the expected because what we observe is the expected, after rounding. But the formula for X-squared, sum (O-E)^2/E gives a positive value. What

hi all - is there a package or library that contains a function for partitioning the chi-square statistic of an I X J contingency table into its respective independent parts? i looked around for this, but i didn't find anything. perhaps there's another name for this sort of analysis? i know it as "g-squared". thanks, chris. [[alternative HTML version deleted]]

Full_Name: Sahotra Sarkar Version: 2.2.0 OS: Windows XP Submission from: (NULL) (128.83.34.44) This is not a bug: I'm just wondering why chisq.test does not allow the specification of the degree of freedom (df).

Hello, I'm trying to calculate a chi-squared test to see if my data are different from the theoretical distribution or not: chisq.test(rbind(c(79, 52, 69, 71, 82, 87, 95, 74, 55, 78, 49, 60),c(80,80,80, 80, 80, 80, 80, 80, 80, 80, 80, 80))) Pearson's Chi-squared test data: rbind(c(79, 52, 69, 71, 82, 87, 95, 74, 55, 78, 49, 60), c(80, 80, 80, 80, 80, 80, 80, 80, 80, 80, 80,

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)

I've got a dataset which looks like this in the beginning: cbr dust smoking expo 1 0 0.20 1 5 2 0 0.25 1 4 3 0 0.25 1 8 4 0 0.25 1 4 5 0 0.25 1 4 (till no. 1240, anyway, a huge set) I have to analyse cbr and smoking, I know it works with chisq.test() for the whole set, but I only need cbr and smoking, and I

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 >

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)) >