similar to: Fitting data to chi-squared or noncentral chi-squared distributions

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 am trying to fit the chi-squared distribution to a set of data using the fitdistr function found in the MASS4 library, the data set is called ONES3, I have loaded it using the command ONES3<-read.table("ONES3.pdf",header=TRUE,na="NA") I print out the dataset ONES3 to the screen to make sure it has loaded Then I try to fit this data using the command

I am just trying to teach myself how to use the mle function in R because it is much better than what is provided in MATLAB. I am following tutorial material from the internet, however, it gives the following errors, does anybody know what is happening to cause such errors, or does anybody know any better tutorial material on this particular subject. >

Dear all, I have a question regarding performing test if the data fits chi-squared distribution. For example, using ks.test() I found in the examples how to fit it to gamma or weibull x<-rnorm(100) ks.test(x, "pweibull", shape=2,scale=1) for the gamma, pgamma can be used But I cannot find the value of this second parameter for the chi-squared distribution. Maybe someone

Hola a todos. Un problema dos pasos. Quiero desde RStudio trabajar con dos versiones de R (pongamos R2.13 y R2.15) desde Linux, Ubuntu 10.04 (lucid). Ya tengo instalada la R.2.15.1. Por defecto trabajo con ella. Pasos que creo he de dar: 1º. Instalar dos versiones de R en la misma maquina. ¿Como se hace?. Me estoy metiendo con el PATH y tal... hay algún protocolo clarito por ahí? 2º.

¿conocéis esta web: http://www.rseek.org/? quizás soy el único que no la conocía... por si un caso ya me diréis si la veis útil... Un saludo. -- Antonio Maurandi López Sección Apoyo Estadístico. Servicio de Apoyo a la Investigación (SAI) Vicerrectorado de Investigación e Internacionalización. Universidad de Murcia Edif. CAID. Campus de Espinardo. 30100 Murcia @. amaurandi en um.es T. 868

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 =

Hi, Actually I am facing a similar problem. I would like to fit both an ordinary (symmetric) and a non-central t distribution to my (one-dimensional) data (quite some values.. > 1 mio.). For the symmetric one, fitdistr or funInfoFun (using fitdistr) from the qAnalyst package should do the job, and for the non-central one.. am I right to use gamlss(x ~ 1, family=GT()) ? Anyway, I am a little

Hi, I would like to know if there is a program written in R to get the CDF (cumulative distribution function) of a bivariate non-central chi-square distribution. Hope someone will reply. Thank you, Rossita M Yunus yunus@usq.edu.au This email (including any attached files) is confidentia...{{dropped:19}}

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, Attention chi-squared distribution, unlike F distribution, has only df1 as parameter, not df1 and df2. So correct into: outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1)) ^^^^^^^^^^^^^^^^^^^^ Regards, Vito you wrote: Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F

Hi, pchisq -> distribution function dchisq -> density function pval is the area under the curve, to calculte it you use distribution function which is the integral of density function. See: http://www.itl.nist.gov/div898/handbook/eda/section3/eda362.htm http://mathworld.wolfram.com/DistributionFunction.html f(x) density function F(x) distribution function =Pr(X<x)= integral(f(x))

I have troubles understanding how goodfit() function in the vcd package computes the Pearson coefficient. Can anybody provide more information on the computation? In particular, for HorseKicks data in vcd package, goodfit() yields > oo <- goodfit(HorseKicks,type="poisson",method="MinChisq") > summary(oo) Goodness-of-fit test for poisson distribution

Hello I could not find whether there is any R-function that implements the inverse of a noncentral Beta. Could someone out there tell me where I can find it? Or how to implement it? Many thanks Ed [[alternative HTML version deleted]]

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! I have some data that looks like this up down percentaje uew_21 20 14 58.82 uew_20_5 27 40 40.29 uew_20 8 13 38.09 uew_19_5 17 42 28.81 So I have 4 experimental conditions and I am counting number of animals in the up and down compartment and the calculating the percentage, I want to know which one of the conditions is different from each other. If the data wouldn't be percentage

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]]

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 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 R-helpers, I need to generate standard variates normal to 'create' chi-squared variates. To make you more understand, (1)   a<-rnorm(3,0,1) *after do (1), I need to squared and summed the three values. My problem is, how am I going to continue the programming if I had to repeat the process for 15 times, which in the end I will get 15 values from the whole programme.Hope you can