similar to: normal distribution and floating point traps (?): unexpected behavior

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

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]]

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))

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()

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 am looking for high precision values for the normal distribution in the tail,(1e-10 and 1 - 1e-10) as the R package that I am using sets any number which is out of this range to these values and then calls the qnorm and qt function. What I have noticed is that the qnorm implementation in R is not symmetric when looking at the tails. This is quite surprising to me, as it is well known

hello I'm learning mgcv and would like to obtain numerical output corresponding to plot.gam. I can do so when seWithMean=FALSE (the default) but only approximately when seWithMean=TRUE. Can anyone show how to obtain the exact values? Alternatively, can you clarify the explanation in the manual "Note that, if seWithMean=TRUE, the confidence bands include the uncertainty about the

how R implement qnorm() I wonder anyone knows the mathematical process that R calculated the quantile? The reason I asked is soly by curiosity. I know the probability of a normal distribution is calculated through integrate the Gaussian function, which can be implemented easily (see code), while the calculation of quantile (or Zα) in R is a bit confusing as it requires inverse error function (X

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

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

Dear R helper, I hope that u can help me to sort out my problem because I sent an E-mail last night to R-list but I have not receive any help and at the same time I think this problem is not so hard. I have used the following functions before > K<-10 > prime<-c(2,3,5,7,11,13,17) > UN<-seq(1:K)%*%t(sqrt(prime)) > U1<-UN-as.integer(UN) > U<-matrix(qnorm(U1),K,7)

A recent message on ansari.test() prompted me to play with the examples. This doesn't work for me in R version 2.4.1 R> ansari.test(rnorm(100), rnorm(100, 0, 2), conf.int = TRUE) Error in uniroot(ab, srange, tol = 1e-04, zq = qnorm(alpha/2, lower = FALSE)) : object "ab" not found It looks like there's a small typo in ccia() inside ansari.test.default() in which

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

>>>>> William Dunlap via R-devel >>>>> on Sun, 23 Jun 2019 10:34:47 -0700 writes: >>>>> William Dunlap via R-devel >>>>> on Sun, 23 Jun 2019 10:34:47 -0700 writes: > include/Rmath.h declares a set of 'logspace' functions for use at the C > level. I don't think there are core R functions that call

I'm looking for a function in R similar to t.test() which was generously pointed out to me yesterday, but which can be used for normally distributed data. To recap yesterday: > x <- scan() 1: 62 52 68 23 34 45 27 42 83 56 40 12: Read 11 items > alpha<- .05 > t.test(x) One Sample t-test data: x t = 8.8696, df = 10, p-value = 4.717e-06 alternative hypothesis: true

Hi. I'm trying to create an Agresti-Coull confidence interval without using the binom package. Despite many trials, I keep getting the same problem- see below. > y=334 > n=1160 > alpha=.05 > b=(y+.5*qnorm(1-alpha/2)**2)/(n+qnorm(1-alpha/2)**2) > b [1] 0.288631 > ac=b+qnorm(1-alpha/2)*sqrt(b(1-b)/(n+qnorm(1-alpha/2)**2)) Error: could not find function "b" What am I

Dear all, Suppose I plot two normal distributions (A and B) side by side and add vertical line which hipotheticaly represent alpha value; e.g.: x <- seq(-3.5,5, length=1000) y <- dnorm(x) # Plot distribution A plot(y~x, type='l',axes=F,xlab="",ylab="",lwd=2) # Plot distribution B y2 <- dnorm(x-1.5) lines(y2~x,lwd=2) # Plot vertical line for alpha value

Sorry to all that are angry about the form of my previous mail. I didn't realise what would happen :((. Here it is in (hopefully) plain text (if my mailer doesn't spoil it again): ############## Dear developers, I have a problem with some discrepancy between R 1.2.1 for Windows and R 1.2.2 (and less) for Linux. While trying to correct the wilcox.test (see my previous bug report) I

It's probably toadally elementary (and, like, duhhhhh) but I can't figure out why the following doesn't work: curve(function(x){qnorm(x,4,25)},from=0,to=1) I get the error: Error in xy.coords(x, y, xlabel, ylabel, log) : 'x' and 'y' lengths differ But if I do foo <- function(x){qnorm(x,4,25)} curve(foo,from=0,to=1) it goes like a train. Also

I think qqline does not do exactly what it is advertised to do ("`qqline' adds a line to a normal quantile-quantile plot which passes through the first and third quartiles."). Consider the graph: tmp <- qnorm(ppoints(10)) qqnorm(tmp) qqline(tmp) The line (which I expected go through all the points), has a slightly shallower slope than does the points plotted by qqnorm. I think