search for: chies

I have written an application in perl, to run multiple rsyncs, retrying those which fail. I have started getting this error on very large modules since i moved to the latest CVS version. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ ToolSyncProgress Succeeded Getting big/tools/vlsi_local/etc from willy at 1004767211:2001:11:3:6:0:11. ToolSyncProgress Getting big1/cadappl1/hpux

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 approximation may be incorrect." Does it mean the same as

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

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

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

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) |

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 context:

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

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 the results on 4x4 contingency tables for R's chisq.test using chi-square

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)

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(4,1,6,9,9,10,2,4,8,2,3,0,1,2,3,1,3,4,5,4,4,4,9,5,4,3,11,8,12,3,10,0,7) lambda<- mean(No_of_Frauds) # Chi-Squared

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

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(diag(S)))) I think

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

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

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

Dear R Helpers, First, I apologize for asking for help on the first of my topics. I have been looking at the posts and pages for apply, tapply etc, and I know that the solution to this must be ridiculously easy, but I just can't seem to get my brain around it. If I want to produce a set of tables for all the variables in my data, how can I do that without having to type them into the table

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

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 <-