similar to: Accuracy (PR#13999)

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

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)

Hi all, I am having a trouble with this function I wrote ################################################### p26=function(x,alpha){ # dummy variable j=1 ci=matrix(ncol=2,nrow=3) while (j<4){ if (j==2) {x=x+c(-1,1)*0.5} ci[j,]= x+qnorm(1-alpha/2)^2/2+ c(-1,1)*qnorm(1-alpha/2)* sqrt(x+qnorm(1-alpha/2)^2/4) j=j+1 if (j==3) { # exact x=x-c(-1,1)*0.5

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 want to compute the quantiles of Chi^2 distributions with different degrees of freedom like x<-cbind(0.005, 0.010, 0.025, 0.05, 0.1, 0.5, 0.9, 0.95, 0.975, 0.99, 0.995) df<-rbind(1:100) m<-qchisq(x,df) and hoped to get back a length(df) times length(x) matrix with the quantiles. Since this does not work, I use x<-c(0.005, 0.010, 0.025, 0.05, 0.1, 0.5, 0.9, 0.95, 0.975,

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

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 <-

Greetings! I suspect that there is an error in the code for the function ci.pd() in the Epi package. This function is for computing confidence intervals for a difference of proportions between two independent groups of 0/1 responses, and implements the Newcombe ("Nc") method and the Agrasti-Caffo "AC" method. I think there is an error in the computation for the Newcombe

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, 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 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

Dear R community. I am using the identify() function to identify outliers in my dataset. This is the code I am using: #################################################################### # Function to allow identifying points in the QQ plot (by mouseclicking) qqInteractive <- function(..., IDENTIFY = TRUE) { qqplot(...) -> X abline(a=0,b=1) if(IDENTIFY) return(identify(X))

Folks, I'm fairly sure that I'm doing something stupid, but I'm getting a few really strange results from *some* of the distributions, but by no means all, when I link directly to the R shared library. I've tried this on both Windows with the precompiled Mingw binary of R-2.5.0 (compiling my code with MinGW-3.4.2), and by building R-2.5.0 on Mandriva Linux with gcc-3.4.4 and

Hello, I got a rather simple question: Can I find somewhere in R the significance values for a Kolmogorov distribution (I know the degrees of freedom and I have already the maximum deviation). ks.test is not really doing what I want. All I need is the values, like one can get the values for a chi-squared distribution by 'qchisq(0.05, 375)'. tnx, Kurt.

I compiled the 1.9.1 src.rpm with the standard gnu tools and it works. I tried compiling the 1.9.1 src.rpm with the Intel 8 C and FORTRAN compilers and it bombs out during the testing phase: comparing 'd-p-q-r-tests.Rout' to './d-p-q-r-tests.Rout.save' ...267c267 < df = 0.5[1] "Mean relative difference: 5.001647e-10" --- > df = 0.5[1] TRUE

Hello everyone I'm a beginner in Stats and R, I'm using R 2.10.1. I need to create a multivariate qq plot, there is 8 variable group with each has 55 number of input. An example of what I did so far, just to get my point out: > data=read.csv(file.choose(),header=T) > data country village group av_expen P2ary_ed no_fisher 1 Cook Islands Aitutaki D

(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

Dear colleague, I not sure this R code is correctly ? I would to show the number of Sample Size at Sample Size Axis that line draw from Power Axis (80%) from R code. How I show this and select the most appropriate of this power (.79955687 - 80983575). Thank for your help and answer. Best Regards, Nikom Thanomsieng, Email: nikom at kku.ac.th .... #Power analysis: Sample size for

Hi: Does Ellipse and dataellipse function in R produce the same ellipse? I wanted to see how the radius for the Ellipse function in R calculated. Also what is the var-covariance matrix, if any, assumed for the dataellipse function? Heres an example of the code where I am generating Multivariate normal data and creating ellipse using the 2 functions: library(car) library(mvtnorm) mu =

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