similar to: plotting functions of chi square

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:

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

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.

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, 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

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 --

Hello, I'm taking samples from certain distributions and drawing a density distribution over the histogram of the samples It works fine for the chi-square and for the normal, but not for the cauchy. Any idea what I'm doing wrong? Thanks x <- rchisq(10000, df = 4) hist(x, freq = FALSE, breaks=100) curve(dchisq(x, df = 4), col = 2, add = TRUE) x <- rnorm(10000) hist(x, freq =

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=

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

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 <-

Dear all Just for fun, I have just downloaded the paper mentioned below and checked it with R-1.6.1. Everything is ok with exception of Table 2b, where I get always 1 instead of 0.5: > pbinom(1e15,2e15,0.5) [1] 1 Which value should be correct? Best regards Christian Stratowa ============================================== Christian Stratowa, PhD Boehringer Ingelheim Austria Dept NCE Lead

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 =

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

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?

Hi there! I'm not sure I can create a minimal example of my problem, so I'm linking to a minimal .RData file that has only two objects: obs and exp, each is a 6x9 matrix. http://dl.dropbox.com/u/10364753/test.RData link to dropbox file (I hope this is acceptable mailing list etiquette!) Here's what happens: > obs[1, 1] [1] 118 > exp[1, 1] [1] 118 > obs[1, 1]-exp[1, 1] [1]

There must be a very basic thing I am not getting... I'm working with some entropy functions and the convention is to use log(0)=0. So I wrote a function: llog<-function(x){ if (x ==0) 0 else log(x) } which seems to work fine for individual numbers e.g. >llog(0/2) [1] 0 but if I try whole vectors or tables: p<-c(4,3,1,0) q<-c(2,2,2,2) llog(p/q) I get this: [1] 0.6931472

Dear R useRs, i wanted to ask if the folded normal destribution (Y = abs(X) with X normal distributed) with density and random number generator is implemented in R or in any R-related package so far? Maybe i can use the non-central chi-square distribution and rchisq(n, df=1, ncp>0) here? Thanks and best regards Andreas

I am fairly new to log-linear modelling, so as opposed to trying to fit modells, I am still trying to figure out how it actually works - hence I am looking at the interpretation of parameters. Now it seems most people skip this part and go directly to measuring model fit, so I am finding very few references to actual parameters, and am of course clear on the fact that their choice is irelevant for

OK, this is really weird! here's an example code: t1<-c(1,2,3,4) t2<-c(4,2,4,2) plot(t1~t2, xlab="exp1", ylab="exp2") dev.copy2eps(file="test.eps") that all seems fine... until you look at the eps file created, where for some weird reason, if you scroll down to the end, the code reads: /Font1 findfont 12 s 0 setgray 214.02 18.72 (e) 0 ta -0.360 (xp1) tb

I'm trying to import a table into R the file is about 700MB. Here's my first try: > DD<-read.table("01uklicsam-20070301.dat",header=TRUE) Error: cannot allocate vector of size 15.6 Mb In addition: Warning messages: 1: In scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : Reached total allocation of 1535Mb: see help(memory.size) 2: In scan(file, what,