similar to: Display a very low p-value

Hello, Rlisters I have to compute p-values that are on the tail of the distribution, P-values < 10^-20. However, my current implementations enable one to estimate P-values up to 10^-12, or so. A typical example is found below, where t is my critical value. ########### example - code adapted from Rassoc ####################### rho01 = 0.5 rho105 = 0.5 rho005 = 0.5 t = 8 z = 2

Hi A very simple question on number formats. After some calculations, some variables I have come out looking like this: -1.892972e+00 Now apart from the fact that the "e+00" is completely redundant, I would rather have the number represented without the e bit. I want to export data like this to a text file and would rather have: -1.892972 than -1.892972e+00 I have tried

Hi to all, I found in the R-help archive how to calculate the p-value for a gee result: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/74150.html but there are two questions (I am afraid they are basic questions ...) 1. why is the result multiplicated with 2 2. how could I decide between lower.tail =TRUE and FALSE: example:

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))

Hi all, I have the following two function f1 and f2. f1 <- function(lambda,z,p1){ lambda*(p1*exp(-3*z-9/2)+(0.2-p1)*exp(4*z-8))-(1-lambda)*0.8} f2 <- function(p1,cl, cu){ 0.8*(pnorm(cl)+(1-pnorm(cu)))/(0.8*(pnorm(cl)+(1-pnorm(cu)))+p1*(pnorm(cl+3)+(1-pnorm(cu+3)))+(0.2-p1)*(pnorm(cl-4)+(1-pnorm(cu-4))))}-0.05 First fix p1 to be 0.15. (i) choose a lambda value, say lamda=0.6, (ii)

In checking over the solutions to some homework that I had assigned I observed the fact that in R (version 2.4.0) pnorm(-1.46) gives 0.07214504. The tables in the text book that I am using for the course give the probability as 0.0722. Fascinated, I scanned through 5 or 6 other text books (amongst the dozens of freebies from publishers that lurk on my shelf) and found that some agree with R

Hi all. Belove is the example from the cluster-help page wtih the output. I simply cannot figure out how to relate the estimate and robust Std. Err to the p-value. I am aware this a marginal model applying the sandwich estimator using (here I guess) an emperical (unstructered/exchangeable?) ICC. Shouldent it be, at least to some extend, comparable to the robust z-test, for rx :

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 all. I may be wrong, (and often am), but in trying to determine how to calculate the erf function, the documentation for 'pnorm' states: ## if you want the so-called 'error function' erf <- function(x) 2 * pnorm(x * sqrt(2)) - 1 ## and the so-called 'complementary error function' erfc <- function(x) 2 * pnorm(x * sqrt(2), lower=FALSE) Should, instead, it read:

Dear all, Can anyone take a look at my program below? There are two functions: f1 (lambda,z,p1) and f2(p1,cl, cu). I fixed p1=0.15 for both functions. For any fixed value of lambda (between 0.01 and 0.99), I solve f1(p1=0.15, lambda=lambda, z)=0 for the corresponding cl and cu values. Then I plug the calculated cl and cu back into the function f2. Eventually, I want to find the lambda value

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

Dear R user I follow the steps defined in Modern applied statistics page(453) to use optim. However, when I run the following code the parameters seems way off and the third parameter(p3) stayed as the initial value. below is the code: ## data da=c(418,401,416,360,411,425,537,379,484,388,486,380,394,363,405,383,392,363,398,526) ### initial values pars=c(392.25, 507.25, 0.80)

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

Hello all, I was wondering if there is an R function to do the following: [*] log(pnorm(x)-pnorm(y)), where x>y. I don't want all the area under the natural log of the normal pdf less than x, I only want the area between y and x. I am aware of the ability to specify log.p=TRUE, which gives me the log of the probability that X<=x. This does not help me, because the following code:

Hello r-group I have a question to the ks.test. I would expect different values for less and greater between data1 and data2. Does anybody could explain my point of misunderstanding the function? data1<-c(8,12,43,70) data2<- c(70,43,12,8) ks.test(data1,"pnorm") ks.test(data1,"pnorm",alternative ="less") #expected < 0.001

Dear R-people, I have to compute C - -(pnorm(B)*dnorm(B)*B + dnorm(B)^2)/pnorm(B)^2 This expression seems to be converging to -1 if B approaches to -Inf (although I am unable to prove it). R has no problems until B equals around -28 or less, where both numerator and denominator go to 0 and you get NaN. A simple workaround I did was C <- ifelse(B > -25, -(pnorm(B)*dnorm(B)*B

I believe that this may be more appropriate here in r-devel than in r-help. The normal hazard function, or reciprocal Mill's Ratio, may be obtained in R as dnorm(z)/(1 - pnorm(z)) or, better, as dnorm(z)/pnorm(-z) for small values of z. The latter formula breaks dowm numerically for me (running R 2.4.1 under Windows XP 5.1 SP 2) for values of z near 37.4 or greater. Looking at the pnorm

Hi all, I want to solve the following equation for x with rho <- 0.5 pnorm(-x)*pnorm((rho*dnorm(x)/pnorm(x)-x)/sqrt(1-rho^2))==0.05 Is there a function in R to do this? Thank you very much! Hannah [[alternative HTML version deleted]]

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'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