similar to: using Chi-square test with a certain number of degrees of freedom ?

Dear list, I had some confusion regarding what function too use in order too relate results from spec.pgram() too a chi-square distribution. The documentation indicates that the PSD estimate can be approximated by a chi-square distribution with 2 degrees of freedom, but I am having trouble figuring out how to do it in R, and figuring out what specifically that statement in the documentation

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

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

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

hello, I work since a few time on R and i wanted to know how to obtain the Wald chi square value when you make a binary logistic regression. In fact, i have the z value and the signification but is there a script to see what is the value of Wald chi square. You can see my model below, Best regards, S??verine Erhel [Previously saved workspace restored] > m3 = glm(reponse2 ~ form +

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

Dear R experts, I was wondering if there are any R functions that give the tail area of a sum of chisquare distributions of the type: a_1 X_1 + a_2 X_2 where a_1 and a_2 are constants and X_1 and X_2 are independent chi-square variables with different degrees of freedom. Thanks, Klaus -- "Feel free" - 5 GB Mailbox, 50 FreeSMS/Monat ...

Dear list I have a vector of values that allegedly have a chi-squared distribution. I want to create a plot that shows the values I have obtained, and the chi-squared distribution curve for the specified number of degrees of freedom to show what should have been obtained. At the moment I am plotting the values I have obtained as a histogram and somehow want to put on to this plot the

System: R 2.3.1 on Windows XP machine. I am building a logistic regression model for a sample of 100 cases in dataframe "d", in which there are 3 binary covariates: x1, x2 and x3. ---------------- > summary(d) y x1 x2 x3 0:54 0:50 0:64 0:78 1:46 1:50 1:36 1:22 > fit <- glm(y ~ x1 + x2 + x3, data=d, family=binomial(link=logit)) >

I have been trying to use lavaan (version 0.4-7) for a simple path model, but the program seems to be computing far less degrees of freedom for my model then it should have. I have 7 variables, which should give (7)(8)/2 = 28 covariances, and hence 28 DF. The model seems to only think I have 13 DF. The code to reproduce the problem is below. Have I done something wrong, or is this something I

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:

I'm doing some chi-square tests, and I recall some arbitrary rule that says each band must have at least 5 events in order for the test to be meaningful. Is there some way to do the banding automagically in R ? For instance, in the following survdiff, I'm trying to see if ADL affects survival. But when ADL=3,5 and 6, the number observed is too little. Anyway for me to tell R how to group

Hello all. It seems that the pf() function when used with noncentral parameters can behave badly at times. I've included some examples below, but what is happening is that with some combinations of df and ncp parameters, regardless of how large the quantile gets, the same probability value is returned. Upon first glance noncentral values greater than 200 may seem large, but they are in

Dear R users, we proudly announce the latest release of our R package plsdof: Degrees of Freedom and Statistical Inference for Partial Least Squares. Features include: * Degrees of Freedom estimates for Partial Least Squares (PLS) Regression * Model selection for PLS based on various information criteria and on cross-validation * approximate confidence intervals and significance tests for PLS

Dear R users, we proudly announce the latest release of our R package plsdof: Degrees of Freedom and Statistical Inference for Partial Least Squares. Features include: * Degrees of Freedom estimates for Partial Least Squares (PLS) Regression * Model selection for PLS based on various information criteria and on cross-validation * approximate confidence intervals and significance tests for PLS

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 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

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

Dear R-help,   Chi Square Test for Goodness of Fit     Problem Faced :   I have got a discrete data as given below (R script)   No_of_Frauds <-c 1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,2,1,2,2,2,1,1,2,1,1,1,1,4,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,5,1,2,1,1,1,1,1,1,1,3,2,1,1,1,2,1,1,2,1,1,1,1,1,2,1,3,1,2,1,2,14,2,1,1,38,3,3,2,44,1,4,1,4,1,2,2,1,3)   I am trying to fit

Dear R-help, ? ? Chi Square Test for Goodness of Fit ? I have got a discrete data as given below (R script) ? No_of_Frauds<-c(1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,2,1,2,2,2,1,1,2,1,1,1,1,4,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,5,1,2,1,1,1,1,1,1,1,3,2,1,1,1,2,1,1,2,1,1,1,1,1,2,1,3,1,2,1,2,14,2,1,1,38,3,3,2,44,1,4,1,4,1,2,2,1,3) ? I am trying to fit Poisson distribution to