similar to: chi-squared with zero df (PR#10551)

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).

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)

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

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

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

Full_Name: Viktor Witkovsky Version: 2.9.2 OS: Windows XP Submission from: (NULL) (78.98.89.227) Hello, I have found strange behavior of the function qchisq (the non-central qchisq is based on inversion of pchisq, which is further based on pgamma). The function gives wrong results without any warning. For example: qchisq(1e-12,1,8.94^2,lower.tail=FALSE) gives 255.1840972465858 (notice that

Hi, I want to determine the confidence interval on the sum of two sigma's. Is there an easy way to do this in R? I guess I have to use some sort of chisquare convolution algorithm??? Thanx, Roy -- The information contained in this communication and any atta...{{dropped}}

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

for example, an expression such as chisq(df=1,ncp=0) ? thanks -- View this message in context: http://www.nabble.com/how-to-calculate-chisq-value-in-R-tp15338943p15338943.html Sent from the R help mailing list archive at Nabble.com.

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]]

Hi all, I would like to estimate power and necessary sample size for a GLM with a response variable that has a poisson distribution. Do you have any suggestions for how I can do this in R? Thank you for your help. Sincerely, Craig -- Craig A. Faulhaber Department of Forest, Range, and Wildlife Sciences Utah State University 5230 Old Main Hill Logan, UT 84322 (435)797-3892

https://bugzilla.samba.org/show_bug.cgi?id=10551 Summary: Daemon infinite loop when no matched user in secrets Product: rsync Version: 3.1.1 Platform: x64 OS/Version: Linux Status: NEW Severity: major Priority: P5 Component: core AssignedTo: wayned at samba.org ReportedBy: ryan at

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

The Intel compiled version also fails the below test: > ###------------ Very big and very small > umach <- unlist(.Machine)[paste("double.x", c("min","max"), sep='')] > xmin <- umach[1] > xmax <- umach[2] > tx <- unique(outer(-1:1,c(.1,1e-3,1e-7)))# 7 values (out of 9) > tx <- unique(sort(c(outer(umach,1+tx))))# 11 values

Craig, Thanks for your follow-up note on using the asypow package. My problem was not only constructing the "constraints" vector but, for my particular situation (Poisson regression, two groups, sample sizes of (1081,3180), I get very different results using asypow package compared to my other (home grown) approaches. library(asypow) pois.mean<-c(0.0065,0.0003) info.pois <-

Hi all, I have noticed incorrect parsing of very small hexadecimal numbers like "0x1.00000000d0000p-987". Such a hexadecimal representation can can be produced by sprintf() using the %a flag. The return value is incorrectly reported as 0 when coercing these numbers to double using as.double()/as.numeric(), as illustrated in the three examples below:

R-1.7.0 built on NetBSD 1.6, but the validation test suite failed: Machinetype: Intel Pentium III (600 MHz); NetBSD 1.6 (GENERIC) Remote gcc version: gcc (GCC) 3.2.2 Remote g++ version: g++ (GCC) 3.2.2 Configure environment: CC=gcc CXX=g++ LDFLAGS=-Wl,-rpath,/usr/local/lib make[5]: Entering directory `/local/build/R-1.7.0/src/library' >>> Building/Updating

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,

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

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?