similar to: R equivalent of erfcinv in matlab

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:

How do I generate a value in R from a poisson distribution with mean 20? Thanks! -- View this message in context: http://www.nabble.com/Generating-a-value-tf4922234.html#a14086120 Sent from the R help mailing list archive at Nabble.com.

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 would like to generate pseudo-random numbers between two numbers using R, up to a given distribution, for instance, rnorm. That is something like rnorm(HowMany,Min,Max,mean,sd) over rnorm(HowMany,mean,sd). I am wondering if dnorm(runif(HowMany, Min, Max), mean, sd) is good. Any idea? Thanks. -james

Dear all, I have the following problem with the uniroot function. I want to find roots for the fucntion "Fp2" which is defined as below. Fz <- function(z){0.8*pnorm(z)+p1*pnorm(z-u1)+(0.2-p1)*pnorm(z-u2)} Fp <- function(t){(1-Fz(abs(qnorm(1-(t/2)))))+(Fz(-abs(qnorm(1-(t/2)))))} Fp2 <- function(t) {Fp(t)-0.8*t/alpha} th <- uniroot(Fp2, lower =0, upper =1,

Hi, This problems has bothered me for the lase couple of hours. > 1e-100==0 [1] FALSE > (1-1e-100)==1 [1] TRUE How can I tell R that 1-1e-100 does not equal to 1, actually, I found out that > (1-1e-16)==1 [1] FALSE > (1-1e-17)==1 [1] TRUE The reason I care about this is that I was try to use qnorm() in my code, for example, > qnorm(1e-100) [1] -21.27345 and if I want to

Hi, does the shapiro wilk test in R-1.3.0 work correctly? Maybe it does, but can anybody tell me why the following sample doesn't give "W = 1" and "p-value = 1": R> x<-1:9/10;x [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 R> shapiro.test(qnorm(x)) Shapiro-Wilk normality test data: qnorm(x) W = 0.9925, p-value = 0.9986 I can't imagine a sample being

You may want to look into using the log option to qnorm e.g., in round figures: > log(1e-300) [1] -690.7755 > qnorm(-691, log=TRUE) [1] -37.05315 > exp(37^2/2) [1] 1.881797e+297 > exp(-37^2/2) [1] 5.314068e-298 Notice that floating point representation cuts out at 1e+/-308 or so. If you want to go outside that range, you may need explicit manipulation of the log values. qnorm()

Hello If i want to resampl from the tails of normal distribution , are these commans equivelant??   upper tail:qnorm(runif(n,pnorm(b),1))  if b is an upper tail boundary   or   upper tail:qnorm((1-p)+p(runif(n))  if p is the probability of each interval (the observatins are divided to intervals)   Regards [[alternative HTML version deleted]]

What would be the R formulae for a two-sided test? I have a formula for a one-sided test: powertest <- function(a,m0,m1,n,s){ t1 = -qnorm(1-a) num = abs(m0-m1) * sqrt(n) t2 = num/s pow = pnorm(t1 + t2) } Would you pls let me know if you know of? Thank you, ej

Dear list, has anyone written a package/function in R for computing a point- biserial resp. biserial correlation? Thanks in advance Bernd

I am using pnorm() with the log.p=T argument to get approximations to ln \Phi(x) and qnorm with the log.p=T argument to get estimates of \Phi^{-1}(exp(x)). What approximations are used in these two functions (I noticed in the source pnorm.c it doesn't look like Abramowitz and Stegen) and where can I find the citation? Thanks, Richard Morey

I agree with many the sentiments about the wisdom of computing very small p-values (although the example below may win some kind of a prize: I've seen people talking about p-values of the order of 10^(-2000), but never 10^(-(10^8)) !). That said, there are a several tricks for getting more reasonable sums of very small probabilities. The first is to scale the p-values by dividing the

Hello all, I have a question about basic statistics. Given a PDF value of 0.328161, how can I find out the value of -0.625 in R? It is like reversing the dnorm function but I do not know how to do it in R. > pdf.xb <- dnorm(-0.625) > pdf.xb [1] 0.328161 > qnorm(pdf.xb) [1] -0.444997 > pnorm(pdf.xb) [1] 0.628605 Many thanks, Edwin -- View this message in context:

Dear all, I need to do some calculation where the code used are below. I get error message when I choose k to be large, say greater than 25. The error message is "Error in integrate(temp, lower = 0, upper = 1, k, x, rho, m) : the integral is probably divergent". Can anyone give some help on resolving this. Thanks. Hannah m <- 100 alpha <- 0.05 rho <- 0.1 F0

Hello, Well, try it: p <- .Machine$double.eps^seq(0.5, 1, by = 0.05) z <- qnorm(p/2) pnorm(z) # [1] 7.450581e-09 1.228888e-09 2.026908e-10 3.343152e-11 5.514145e-12 # [6] 9.094947e-13 1.500107e-13 2.474254e-14 4.080996e-15 6.731134e-16 #[11] 1.110223e-16 p/2 # [1] 7.450581e-09 1.228888e-09 2.026908e-10 3.343152e-11 5.514145e-12 # [6] 9.094947e-13 1.500107e-13 2.474254e-14 4.080996e-15

I am wondering whether there is a bug in rnorm. When generating rnorm(1000000) and counting the cases > 4 and the cases < (-4) I get rather unexpectedly low counts for the latter. The problem goes away when using qnorm(runif(1000000)). Fritz Scholz, PhD Applied Statistics Group Boeing Phantom Works fritz.scholz at pss.boeing.com 425-865-3623 Tu/We 206-542-6545 (most likely)

Hi, Am not sure if my code itself is correct. Here's what am trying to do: Minimize integration of a function of gaussian distributed variable 'x' over the interval qnorm(0.999) to Inf by changing value of parameter 'mu'. mu is the shift in mean of 'x'. Code: # x follows gaussian distribution # fx2 to be minimized by changing values of mu # integration to be done over

Hi, I am working with some extremely small p-values and I want to capture the corresponding quantiles. I see the help file it says: 'qnorm' is based on Wichura's algorithm AS 241 which provides precise results up to about 16 digits. What happen after the 16th digits? If I am running R in a server 64-bit, can that improve the chances that beyond 16th digits to still have

I'm familiar with the quantile() command, but what if I have a specific number that I want to know its location in a vector? I know that in known distributions, (for example the normal distribution), there is pnorm and qnorm, but how can I do it with unknown vector? thanks in advance _________________________________________________________________ Walla! Mail - [1]Get