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...al transfer time but is still active. It also ends by killing its calling process. The error looks different, though, but it's on a local nfs-to-nfs, instead of rsyncd-to-nfs. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ ToolSyncProgress Replicating big1/cadappl1/hpux from chi-tools1 to chi-tools2 at 1004913115:2001:11:4:22:31:55. io timeout after 3600 seconds - exiting rsync error: timeout in data send/receive (code 30) at io.c(75) ToolSyncProgress chinetsvr failed Replicating big1/cadappl1/hpux from /wan/chi-tools1/chi-tools1 to /wan/chi-tools2/chi-tools2 at 1004925...

Sir, I have a problem here while applying chisquare test to the following Data ( below the subject of this mail) ...when I wanted to test the significance using three different free statistical packages, here R, EpiInfo and OpenEpi. *Only OpenEpi accepts the test based on Cochran's Recommendations. * R says " chi squared approximatio...

Dear All, I have a problem understanding the difference between the outcome of a fisher exact test and a chi-square test (with simulated p.value). For some sample data (see below), fisher reports p=.02337. The normal chi-square test complains about "approximation may be incorrect", because there is a column with cells with very small values. I therefore tried the chi-square with simulated p-val...

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,...

Hi. I have made 100 experiments of an M/M/1 queue, and for each one I have calculated both, mean and variance of the queue size. Now, a professor has told me that variance is usually chi-squared distributed. Is there a way in R that I can find the parameter that best fits a chi-square to the variance data? I know there's fitdistr()m but this function doesn't handle chi-square. I believe the mean estimator for the chi-square is df (degrees of freedom). The theoretical varian...

I got some R code that I don't understand. Question as comment in code //where is t comming from, what is phi inverse rAC <- function(name, n, d, theta){ #generic function for Archimedean copula simulation illegalpar <- switch(name, clayton = (theta < 0), gumbel = (theta < 1), frank = (theta < 0), BB9 = ((theta[1] < 1) | (theta[2] < 0)), GIG = ((theta[2] < 0) | (theta[3] < 0) | ((theta[1]>0) & (theta[3]==0)) | ((theta[1]<0) & (theta[2]==0)...

if i want to plot the chi-square distribution with a different degree of freedom how can i plot it in the graph?Sometimes i plot the histogram and cut it in a lot of piece.It's distribution like a chi-square.So i want to plot the chi-square with a different degree of freedom to compare it . -- View this message in cont...

Hi! This is going to be a real newbie question, but I can't figure it out. I'm trying to plot densities of various functions of chi-square. A simple chi-square plot I can do with dchisq(). But e.g. chi.sq/degrees of freedom I only know how to do using density(rchisq()/df). For example: plot(1, type="n", xlab="", ylab="", xlim=c(0,2), ylim=c(0,7)) for (i in c(10,50,100,200,500)){ lines(density(rc...

Is Pearson's Chi-Square test for contingency tables asymptotically unbiased for large tables (large degrees of freedom) regardless of the expected values in each cell? The rule of thumb is that Pearson's Chi-square should not be used when large numbers of cells have expected values < 5. However, I compared...

Hi, I have a function that generates a set of data but I am having problems determining the parameters using the nls fitting procedure. #### "MH"<-function(field,diameter,mu=10e-7,sig=0.1,Ms=100,chi=0){ #variables mu, sig, chi, Ms #input: field and diameter #all in CGS rho <- 5 kb <- 1.38e-16 t <- 300 length.d<-length(diameter) length.H<-length(field) M<-double(length.H) for (i in 1:length.H){ S1<-0 S2<-0 H <- field[i] for (j in 1:length.d){ d<-diameter[j]...

Hi, I have calculated chi-square goodness of fit test,Sample coming from Poisson distribution. please copy this script in R & run the script The R script is as follows ########################## start ######################################### No_of_Frauds<- 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,...

...336 52 Region_12 1823 312 Region_13 80 9 Region_14 774 121 Region_15 561 24 Region_16 834 134 How well does the survey represent the employee population? Chi-square test says, not very well: > chisq.test(ByRegion) Pearson's Chi-squared test data: ByRegion X-squared = 163.6869, df = 15, p-value < 2.2e-16 By striking three under-represented regions (3,6, and 15), we get a more reasonable, although still not convincing, result: &gt...

Dear all, I met a problem while doing SEM by sem package. I got a negative chi-square of null model. Because the theoretical value of chi-square cannot be negative, I checked the source code of sem.R in sem package and I found the Chi-square of null model was computed by the following expression: result$chisqNull <- (N - 1) * (sum(diag(S %*% diag(1/diag(S)))) + log(prod(d...

I am not sure whether this is really a bug, but seems to be as my program works pretty well in S+, for small iterations. I am creating a random sample from a chi^2 distribution with 23.5 df. Thus I have to sample from a chi^2 with 23 or 24 df with probability 0.5. The other alternative is to create a weighted version of the two distribution, which is where my problem lies. I use the following commands while(iter<100){ old.seed<-.Random.seed #samp...

Hi, How do we get the value of a chi square as we usually look up on the table on our text book? i.e. Chi-square(0.01, df=8), the text book table gives 20.090 > dchisq(0.01, df=8) [1] 1.036471e-08 > pchisq(0.01, df=8) [1] 2.593772e-11 > qchisq(0.01, df=8) [1] 1.646497 > nono of them give me 20.090 Thanks, cruz

I am running lrm() with a single factor. I then run anova() on the fitted model to obtain a p-value associated with having that factor in the model. I am noticing that the "Model L.R." in the lrm results is almost the same as the "Chi-Square" in the anova results, but not quite; the latter value is always slightly smaller. anova() calculates the p-value based on "Chi-Square", but I have independent evidence that "Model L.R." is the actual -2*log(LR), so should I be using that? Why are the values differ...

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

...(t? s? r?)apply to use one command instead of the following set of table commands: data(infert, package = "datasets") attach(infert) table.education<-table(education) table.age<-table(age) table.parity<-table(parity) etc. To make matters worse, what I subsequently need is the chi-square for each and all of the pairs of variables. Such as: chi.education.age<-chisq.test(table(education,age)) chi.education.parity<-chisq.test(table(education,parity)) chi.age.parity<-chisq.test(table(age,parity)) etc. Your guidance would be much appreciated. --John J. Sparks, Ph.D.

Timothy, I believe you are mistaken. Fisher's exact test give the correct answer even in the face of small expected values for the cell counts. Pearson's Chi-square approximates Fisher's exact test and can give the wrong answer when expected cell counts are low. Chi-square was developed because it is computationally "simple". Fisher's exact test, particularly with tables larger than 2 by 2 can be computationally complex. The value of t...

Dear R helpers: Thanks for the previous reply. I am using Friedman racing test. According the the book "Pratical Nonprametric Statistic" by WJ Conover, after computing the statistics, he suggested to use chi-squared or F distribution to accept or reject null hypothesis. After looking into the source code, I found that R uses chi-sqaured distribution as below: PVAL <- pchisq(STATISTIC, PARAMETER, lower = FALSE) but still I cant figure out why they are using this pschisq insted of dchisq. Sorry I am...