similar to: Peculiarity in non-central qchisq for ncp > 294.92 ...

Greetings all, I'm trying to figure out how to calculate the inverse CDF (i.e. a quantile) for a non-central F distribution. I could put together a quick numerical solver routine using the CDF, but I wonder if there's a function that I've missed that would be more efficient? Thank-you, Andrew Andrew Robinson Ph: 208 885 7115 Department of Forest Resources Fa: 208 885

Full_Name: David Clayton Version: 1.8.1 OS: Linux Submission from: (NULL) (131.111.126.242) qchisq behaves very strangely when ncp is passed as zero (forcing internal qnchisq to be called) when first argument is small. Eg > qchisq(1-1e-6, 1, ncp=0, lower.tail=TRUE) qchisq(1-1e-6, 1, ncp=0, lower.tail=TRUE) [1] 1024 while, if ncp is unspecified, > qchisq(1-1e-6, 1) qchisq(1-1e-6, 1)

Full_Name: Drouilhet R?my Version: 1.8.1 OS: Linux Submission from: (NULL) (195.221.43.136) qchisq(1,10) works well but qchisq(1,10,ncp=0) does not work whereas ncp=0 is the default value of the function qchisq(1,10). (of course, 10 will be replaced by any integer value). Let us notice that this bug occurs only when applying probability one. (qchisq(seq(0,.9,.1),10,ncp=0) works very well).

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 =

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]]

Dear all, I want to see the source codes for "dchisq(x, df, ncp=0, log = FALSE)", but cannot find it. I input "dchisq" in the R interface, and then enter, the following message return: > dchisq /*****************************************************/ function (x, df, ncp = 0, log = FALSE) { if (missing(ncp)) .Internal(dchisq(x, df, log)) else

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 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

< Does anyone know why they're called random deviates, as opposed to random numbers?> Others will probably give you some technical reason about random numbers can be considered as random deviates from a mean (I think at least the 1875 Galton paper at http://www.mugu.com/galton/ uses similar terminology (I'm not claiming this is the earliest use - just the easiest to access at the

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

Hello, has anyone tried to get R running on a Sharp Zaurus 5500? Andrew Andrew Robinson Phone: 208-885-7115 Department of Forest Resources Fax: 208-885-6226 University of Idaho E: andrewr at uidaho.edu Po Box 441133 WWW: http://www.uidaho.edu/~andrewr Moscow, ID 83843 and: http://www.biometrics.uidaho.edu/ No statement above

I don't get the same result, do you have a package loaded that would change the default behavior (such as Hmisc)? > list(grp.1) [[1]] [1] 1 2 Levels: 1 2 > list(grp.1[mask]) [[1]] [1] 1 Levels: 1 2 > library(Hmisc) <<snip>> > list(grp.1[mask]) [[1]] [1] 1 Levels: 1 --Matt > version _ platform i386-pc-mingw32 arch i386 os

Dear R community, In the excellent nlme package the default diagnostic plot graphs the innermost residuals against innermost fitted values. I recently fit a mixed-effects model in which there was a very clear positive linear trend in this plot. I inferred that this trend occurred because my fixed effect was a two-level factor, and my random effect was a 12-level factor. The negative residuals

If I place par(new=T) before I create a plot in a script that is sent to a pdf device, the pdf is unopenable and reports itself as having no pages. The postscript device seems to ignore the par instruction. I guess one of these is a bug, but I don't know which one! --please do not edit the information below-- Version: platform = i386-pc-linux-gnu arch = i386 os = linux-gnu system =

Hi Folks, I note that, while the "chisq" functions dchisq(x, df, ncp=0, log = FALSE) pchisq(q, df, ncp=0, lower.tail = TRUE, log.p = FALSE) qchisq(p, df, ncp=0, lower.tail = TRUE, log.p = FALSE) rchisq(n, df, ncp=0) all have a slot for the non-centrality parameter "ncp", of the functions for the t and F distributions: dt(x, df, log = FALSE)

Dear R-community, I'm using Sweave to produce reports. The reports require the "axis" command. When I run axis the program returns NULL as well as creating the axis. > plot(1:4, rnorm(4), axes=FALSE) > axis(1, 1:4, LETTERS[1:4]) NULL > So, my Sweave tex files have \begin{Schunk} \begin{Soutput} NULL \end{Soutput} \end{Schunk} in front of each graphic that requires

Hi all, I'm using R 1.5.1 on Windows 2000. The following snippet of code doesn't seem to do anything - no error is reported, and there is no name change. names(myFrame[,c(1:3)]) <- c("name1", "name2", "name3") This code however works nicely: names(myFrame)[c(1:3)] <- c("name1", "name2", "name3") Can anyone suggest why

Hi dear R-community, I wonder if anyone has written code that will post-process the results of a tapply to create a dataframe that includes the outcome, the factor by which the function has been applied, and, if the factor is also an interaction, then also the levels of contributing factors? >From this outcome <- tapply(data, A.B, function) I'd like to arrange something like this:

Dear R community, I'm trying to understand this behavior of TukeyHSD. My goal is to obtain defensible, labelled multiple comparisons of an interaction term. Firstly, if I plot the TukeyHSD from the model that calculates its own interactions, then the y-axis labels appear to be reflected on their median when compared to the text output of the TukeyHSD statement. The labels are integers.

Greetings R-community, I'm running simulations within R that I wrote in C. The simulations require fitting that occasionally fails to finish. I was wondering if there is any kind of tool for process control in R, such that after e.g. 15 minutes I could kill the process, record the state for post-hoc analysis, and move to the next simulation? I'm running FreeBSD and could almost surely