similar to: Chi-squared test when observed near expected

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 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]]

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

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,

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).

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

Can anyone tell me how to calculate a mid-p value for a chi-squared test in R? Many thanks, Andrew Wilson

How i can perform a Pearson's Chi-squared Test in this data set: | Outcome -----------------+-----------+----------------------------------+ Treatment | Sex | None |Some | Marked | Total -----------------+------------+--------+--------+-------------+ Active | Female | 6 | 5 | 16 | 27

Hi all, I know Chi-squared test can be done with the frequency data by R function "chisq.test()", but I am not sure if it can be applied to the percentage data ? The example of my data is as follow: ############################################# KSL MHL MWS CLGC LYGC independent (%) 96.22 92.18 68.54 93.80 85.74

Hello, Can you tell me what R function to use to do a two-sample chi-squared test? I want to see if two distributions are significantly different from each other, and I don't specify the theoretical distribution of either. For example, I have the following fake count data: x <- sample(1:10,50,replace=TRUE) y <- sample(1:10,100,replace=TRUE) I saw chisq.test in the stats package, but

I was asked by my boss to do an analysis on a large data set, and I am trying to convince him to let me use R rather than SPSS. I think Sweave could make my life much much easier. To get me a little closer to this goal, I ran my analysis through R and SPSS and compared the resulting values. In all but one case, they were the same. Given the matrix [,1] [,2] [1,] 110 358 [2,] 71 312 [3,]

I want to calculate chi square test of goodness of fit to test, Sample coming from Poisson distribution. please copy this script in R & run the script The R script is as follows ########################## start ######################################### No_of_Frouds<- c(4,1,6,9,9,10,2,4,8,2,3,0,1,2,3,1,3,4,5,4,4,4,9,5,4,3,11,8,12,3,10,0,7) N <- length(No_of_Frouds) # Estimation of

R 2.12 windows 7 I am summary data that I would like to make into a 2x2 table representing counts positive vs. negative counts: 28/289 20/276 My table should look something like the following: group1 group2 Positive 28 20 Negative 289 276 How can a (1) create the 2x2 table (2) run a chi square test on the table? I have tried the following code, but I

a warning message appears when i use the chisq.test ,but it doesnt appear everytime, why? "Warning message: Chi-squared approximation may be incorrect in: chisq.test(matrix(c(20.1, 18.5, 2.6, 32.9, 23.5, 5.4), nc = 2)) " why does the warning message appear, please? Thank you very much here is the data which I have tried appear and not appear warning message have warning message

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 R users, I am a master student in Mathematics and I am writing my thesis in statistics. I need to use R and unfortunately I do not have any experience with a computer program. Could you please help me about chi squared goodness of fit test with R? In R-help website I saw a message about how to do that but I do not know how to cut the data into bins and calculate the expected numbers in each

An organization has asked me to comment on the validity of their recent all-employee survey. Survey responses, by geographic region, compared with the total number of employees in each region, were as follows: > ByRegion All.Employees Survey.Respondents Region_1 735 142 Region_2 500 83 Region_3 897 78

TOPIC My question regards the philosophy behind how R implements corrections to chi-square statistical tests. At least in recent versions (I'm using 2.13.1 (2011-07-08) on OSX 10.6.8.), the chisq.test function applies the Yates continuity correction for 2 by 2 contingency tables. But when used as a goodness of fit test (GoF, aka likelihood ratio test), chisq.test does not appear to implement

Full_Name: Jerry W. Lewis Version: 2.6.1 OS: Windows XP Professional Submission from: (NULL) (24.147.191.250) pchisq(0,0,ncp=lambda) returns 0 instead of exp(-lambda/2) pchisq(x,0,ncp=lambda) returns NaN instead of exp(-lambda/2)*(1 + SUM_{r=0}^infty ((lambda/2)^r / r!) pchisq(x, df + 2r)) qchisq(.7,0,ncp=1) returns 1.712252 instead of 0.701297103 qchisq(exp(-1/2),0,ncp=1) returns 1.238938