similar to: Have problem to do loop to generate transformed chi-squared variates

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))

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

Hello R-helpers.. I want to ask about how I can sample data sets without having the infinite numbers coming out. For example, set.seed(1234) a<-rnorm(15,0,1) b<-rnorm(15,0,1) c<-rnorm(15,0,1) d<-rnorm(15,0,36) After come out with the sample, I need to do a transformation (by Hoaglin, 1985) for each data set. Actually I need to measure the skewness and kurtosis, that's why I

Dear R People, This question has nothing to do with R directly, but it is a simulation question. I need to generate a random variable distributed as gamma(\alpha,\beta), with the additional proviso that it must be a function of random variable(s) which do not depend on \alpha, \beta. In this case, I have \alpha = (T-1)/2, where T is a positive integer. So, it seems reasonable to first simulate

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

Hello R-helpers, I want to ask your opinion since I am not so sure how to do it. This is regarding one part of my paper project and my situation is: Stage I I have 2 groups and for each group I need to compute the following steps; i) Generate 3 random numbers from normal distribution and square them. ii) Repeat step 1 for 15 times and at the end I will get 15 random numbers. I already done

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:

Does anybody have any experience fitting data to the non-central chi-squared or chi-squared distribution? I am trying to fit some data to this distribution but there is error after error. audaces fortuna iuvat --------------------------------- [[alternative HTML version deleted]]

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

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]]

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]]

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

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

Hi all, I know Chi-squared test can be done with the frequency data by R function "chisq.test()", but I am not sure if it can be applied to the percentage data ? The example of my data is as follow: ############################################# KSL MHL MWS CLGC LYGC independent (%) 96.22 92.18 68.54 93.80 85.74

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,

Hello, Can you tell me what R function to use to do a two-sample chi-squared test? I want to see if two distributions are significantly different from each other, and I don't specify the theoretical distribution of either. For example, I have the following fake count data: x <- sample(1:10,50,replace=TRUE) y <- sample(1:10,100,replace=TRUE) I saw chisq.test in the stats package, but

I've got a dataset which looks like this in the beginning: cbr dust smoking expo 1 0 0.20 1 5 2 0 0.25 1 4 3 0 0.25 1 8 4 0 0.25 1 4 5 0 0.25 1 4 (till no. 1240, anyway, a huge set) I have to analyse cbr and smoking, I know it works with chisq.test() for the whole set, but I only need cbr and smoking, and I

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, 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))