similar to: chi square table

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 <-

Hello all, Here's my query: Running R 1.6.2 on FreeBSD 5.0, and on WinXP, and I find that the following hangs the process: dchisq(alpha=0.01, df=1, ncp=295) but it does work for ncp < 294.92. Is this general? Best wishes to all, Andrew Andrew Robinson Ph: 208 885 7115 Department of Forest Resources Fa: 208 885 6226 University of Idaho E : andrewr at uidaho.edu PO

Hello, I want to use the quantile function so I read the doc but I don't understand with this > qchisq(seq(0.05,0.95,by=0.05),df=(length(don)-1)) [1] 62667.11 62795.62 62882.42 62951.47 63010.74 63064.00 63113.39 63160.27 63205.65 63250.33 63295.04 63340.48 63387.48 63437.03 63490.53 63550.14 63619.68 [18] 63707.24 63837.16 Can you help me please?

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

There is a problem with `rpois'. It does seem to take care about the order of the arguments. This is an example: > rpois(n=1,lambda=2) [1] 3 > rpois(lambda=2,n=1) [1] 2 0 It obviously uses the first argument as the number of samples to be drawn, which is wrong. I used Version 0.49 Beta (April 23, 1997). Fredrik

There is a problem with `rpois'. It does seem to take care about the order of the arguments. This is an example: > rpois(n=1,lambda=2) [1] 3 > rpois(lambda=2,n=1) [1] 2 0 It obviously uses the first argument as the number of samples to be drawn, which is wrong. I used Version 0.49 Beta (April 23, 1997). Fredrik

Hi all, I'm trying to superimpose a rchisq density line over a histogram with something like: hist(alnlength) lines(density(rchisq(length(alnlength), 4)),col="red") But the rchisq line won't appear anywhere, Anyone knows what I am missing here? Thanks in advance, Albert.

(1) ...need to speed up a monte-carlo sampling...any suggestions about how I can get R to use all 8 cores of a mac pro would be most useful and very appreciated... (2) spent the last few hours trying to get pnmath to compile under os- x 10.5.4... using gcc version 4.2.1 (Apple Inc. build 5553) as downloaded from CRAN, xcode 3.0... ...xcode 3.1 installed over top of above after

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

Hi, I think I have identified a inaccuracy in dchisq when the non-centrality parameter is non-zero and large. Here's a little test: sys.dchisq.test <- function(N = 100000,mean = 0) { z <- rnorm(N,mean = mean, sd = 1) x <- z^2 xmin <- min(x) xmax <- max(x) br <- seq(xmin,xmax,length = 101) dbr <- br[2]-br[1] hist(x,br) p <- dchisq(br,df = 1,ncp =

Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F distribution results with the table, the answers were perfect. But I need to see for chi-sqaured distribution. When I employed the similar formula outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) , I am getting unexpected results. I need to see the following values: p=0.750 ..... 1 1.323

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

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:

Please could you tell me how to make code to make chi-square distribution table? Please help

Hi there, How accurate is the aproximation R makes to the Chi-Square distribution? For example, if I run: > qchisq(1/1000000,6) [1] 0.03650857 how accurate is 0.0365 compared to the theoretical percentile? What kind of approximations have been made in the software's algorithm? It woudl be useful to know since I am working with tiny percentiles such as one one-millionth and one

Dear help list. I would like to plot the chi-square distribution for n=1 through 10, on one plot. How do I go about doing that? I know how to plot one chi-square (using curve() for example), but plotting 10 on one plot, that is too difficult for me. I greatly appreciate your help. -- fb isabellesup at fastmail.fm --

Hi, I have written out the log-likelihood function to fit some data I have (called ONES20) to the non-central chi-squared distribution. >library(stats4) >ll<-function(lambda,k){x<-ONES20; 25573*0.5*lambda-25573*log(2)-sum(-x/2)-log((x/lambda)^(0.25*k-0.5))-log(besselI(sqrt(lambda*x),0.5*k-1,expon.scaled=FALSE))} > est<-mle(minuslog=ll,start=list(lambda=0.05,k=0.006))

I have dowloaded the Source Code of R,and I want to know the process of chi-sqared test,but how can I found it? [[alternative HTML version deleted]]

Hi to all I found an formular of an ** ***p-Value Calculator for the Chi-Square test* *http://www.danielsoper.com/statcalc/calc11.aspx* *with the formula* *http://www.danielsoper.com/statkb/topic11.aspx* *what's the gamma function of this formula in r?* *df=5* *ch2=25.50878* *the following code does not give the result <0.001 for the values above * *p=

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