similar to: precision problems in testing with Intel compilers

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

Hi, I am building R off the source file "The latest release (2004-06-21): R-1.9.1.tgz <http://cran.stat.ucla.edu/src/base/R-1.9.1.tgz> " on RH Linux 7.3. After the build the "make check" failed with an error which I judge to be minor, but would like to hear from the experts in the field. Is it harmeless? Can it indicate some problem with my computer setup (not R)?

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).

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

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 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

Hello developers and users: My system fails (the computer freezes) when I use the ncp parameter, with the lower.tail=FALSE option in the qchisq function. qchisq(0.025,31,ncp=1,lower.tail=FALSE) Thank you very much for your help. Kenneth Cabrera Universidad Nacional de Colombia ICNE Sede Medellin krcabrer at perseus.unalmed.edu.co PS I am using: $platform "i386-pc-mingw32"

Kenneth Cabrera <krcabrer@epm.net.co> writes: > Hello developers and users: > > My system fails (the computer freezes) when I use the ncp parameter, > with the lower.tail=FALSE option in the qchisq function. > > qchisq(0.025,31,ncp=1,lower.tail=FALSE) Yup, that's a bug. We have in pnchisq.c 48 for (ux = 1.0; pnchisq(ux, n, lambda, lower_tail, log_p) <

>>>>> "PD" == p dalgaard <p.dalgaard@biostat.ku.dk> writes: PD> Kenneth Cabrera <krcabrer@epm.net.co> writes: >> Hello developers and users: >> >> My system fails (the computer freezes) when I use the ncp parameter, >> with the lower.tail=FALSE option in the qchisq function. >> >>

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: 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}}

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

R-2.10.1.tar.gz was built a short while ago. This is a maintenance release and fixes a number of mostly minor issues. See the full list of changes below. You can get it from http://cran.r-project.org/src/base/R-2/R-2.10.1.tar.gz or wait for it to be mirrored at a CRAN site nearer to you. Binaries for various platforms will appear in due course. For the R Core Team Peter

R-2.10.1.tar.gz was built a short while ago. This is a maintenance release and fixes a number of mostly minor issues. See the full list of changes below. You can get it from http://cran.r-project.org/src/base/R-2/R-2.10.1.tar.gz or wait for it to be mirrored at a CRAN site nearer to you. Binaries for various platforms will appear in due course. For the R Core Team Peter

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.

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)

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]]

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 <-