similar to: pnorm conditional (PR#2883)

Hi, I am facing a problem which i think i need to explain it to you with some background. I need to use the Project R pnorm function in Visual Basic 6.0. I have already installed R and this is how i perform and get back the result: > pnorm(2, 15) [1] 6.117164e-39 which is what i need. I have already installed R, i generated the Rmath.DLL file out so i can import it into my VB6 and use it.

Trenkler, Dietrich said: > > I found the following strange behavior using qnorm() and pnorm(): > > > x<-8.21;x-qnorm(pnorm(x)) > [1] 0.0004638484 > > x<-8.28;x-qnorm(pnorm(x)) > [1] 0.07046385 > > x<-8.29;x-qnorm(pnorm(x)) > [1] 0.08046385 > > x<-8.30;x-qnorm(pnorm(x)) > [1] -Inf > qnorm(1-.Machine$double.eps) [1] 8.12589

sorry for the late reply. i dunno if anyone ever wrote you back. although perhaps not written for clarity, i think what appears in the code is correct. the routine only guarantees that one of cum or ccum has a valid return value depending on i_tail (the other might as well be garbage, but that isn't what's returned). j On Wed, 9 Jul 2003 Henrik.Seidel@schering.de wrote: >

As to the function"pnorm",the default degree of freedom(df) is infinite. I wanna know how to set the df as I want. Help on pnorm doesn't have df setting.The only choice are:"mean, sd, lower.tail, log.p",but no df. For instance: sample size=6 df=6-1=5 t value=9.143 I wanna to the corresponding p value by using function "pnorm". How can I do it? Thanks a lot

Dear R users Is there any way to increase the decimal accuracy for the normal probability distribution? When one needs an accurate p-value for instance this is provided by pnorm(10,lower.tail=F) [1] 7.619853e-24 However, what happens when instead of a P[X<x], a more accurate P[X>=x] is the objective. Thank you in advance for your responses. Dimitris [[alternative HTML version

Hello, Looking over the swap_tail macro in pnorm.c, the comment inside the code indicates it is to swap the values of '*cum' and '*ccum'. According to the code below that is taken from the source file, the swap functionality is dependent on what values are provided for 'x' and 'lower'. #define swap_tail \ if (x > 0.) {/* swap ccum <--> cum */

Full_Name: James Michael Rath Version: all (I think) OS: doesn't matter Submission from: (NULL) (129.116.226.162) The code for pnorm in R was adapted from a Fortran library published in the ACM TOMS journal. The published version had a typographical error, though, which was pointed out in a second article published three years after the original. The error was that a macro/variable named

Full_Name: Julian Faraway Version: R-Release (June 15) OS: Linux (redhat) Submission from: (NULL) (141.211.66.172) R fails to compile on the current released version. Some constants are undefined in pnorm.c. It appears that adding #include "nmath.h" to pnorm.c solves this problem. -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-devel mailing list

I found the following strange behavior using qnorm() and pnorm(): > x<-8.21;x-qnorm(pnorm(x)) [1] 0.0004638484 > x<-8.22;x-qnorm(pnorm(x)) [1] 0.01046385 > x<-8.23;x-qnorm(pnorm(x)) [1] 0.02046385 > x<-8.24;x-qnorm(pnorm(x)) [1] 0.03046385 > x<-8.25;x-qnorm(pnorm(x)) [1] 0.04046385 > x<-8.26;x-qnorm(pnorm(x)) [1] 0.05046385 > x<-8.27;x-qnorm(pnorm(x))

I stumbled across this and I am wondering if this is unexpected behavior or if I am missing something. > pnorm(-1.0e+307, log.p=TRUE) [1] -Inf > pnorm(-1.0e+308, log.p=TRUE) [1] NaN Warning message: In pnorm(q, mean, sd, lower.tail, log.p) : NaNs produced > pnorm(-1.0e+309, log.p=TRUE) [1] -Inf I don't know C and am not that skilled with R, so it would be hard for me to look into

Hi to all, maybe the last question was not clear enough. I did not found any hints how to decide whether it should use lower.tail or not. As it is an extra R-feature ( written in http://finzi.psych.upenn.edu/R/Rhelp02a/archive/66250.html ) I do not find anything about it in any statistical books of me. Regards Carmen

Hi, I'm trying to evaluate a character vector within pnorm. I have a vector with values and names x = c(2,3) names(x) = c("mean", "sd") so that i tried the following temp = paste(names(x), x, sep = "=") #gives #> temp #[1] "mean=2" "sd=3" #Problem is that both values 2 and 3 are taken as values for the mean argument in pnorm pnorm(0,

Hi, I've read in a csv file (test.csv) which gives me the following table: Hin1 Hin2 Hin3 Hin4 Hin5 Hin6 HAI1 9534.83 4001.74 157.16 3736.93 484.60 59.25 HAI2 13272.48 1519.88 36.35 33.64 46.68 82.11 HAI3 12587.71 5686.94 656.62 572.29 351.60 136.91 HAI4 15240.81 10031.57 426.73 275.29 561.30 302.38 HAI5 15878.32 10517.14 18.93 22.00 16.91

Dear R list, I calculated a two-sided p values according to 2*(1-pnorm(8.104474)), which gives 4.440892e-16. However, it appears to be 5.30E-16 by a colleague and 5.2974E-16 from SAS. I tried to get around with mvtnorm package but it turns out to be using pnorm for univariate case. I should have missed some earlier discussions, but for the moment is there any short answer for a higher

Dear R people The function below should be decreasing, convex, and tend to zero when x tends to infinity. curve((1-pnorm(x))/dnorm(x),from=0, to=9) >From the plot we see that for x between 8.0 and 8.3 the function is fluctuating. As far as I understand, this is due to the function pnorm() not being sufficiently accurate in the tails. I am using pnorm() in a way that has probably not been

Hi, I would have thought that these two constructions would produce the same result but they do not. Resp <- rbinom(10, 1, 0.5) Stim <- rep(0:1, 5) mm <- model.matrix(~ Stim) Xb <- mm %*% c(0, 1) ifelse(Resp, log(pnorm(Xb)), log(1 - pnorm(Xb))) pnorm(as.vector(Xb), lower.tail = Resp, log.p = TRUE) > ifelse(Resp, log(pnorm(Xb)), log(1 - pnorm(Xb))) [1] -0.6931472 -1.8410216