similar to: pnorm how to decide lower-tail true or false

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:

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

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

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

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 to all I would to determinate whether bits is a binary code and I would to find out the which bit is set to 1 bits <-"00110110" I found to detect whether there are only numbers all.digits(bits) but is there any function to detect whether there are only 0 and 1 in the string And how could I get the f.e the third "bit" from the right hand side With regards Carmen

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 to all, I have some problems to get the times-scale to the x-axis the times are coming from an excel sheet f. e [1] "0:01:00" "0:02:00" "0:03:00" "0:04:00" "0:05:00" "0:06:00" "0:07:00" [8] "0:08:00" "0:09:00" "0:10:00" "0:11:00" "0:12:00" "0:13:00"

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,

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

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 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

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

How do u calculated p values for a z test.. so far i ve done this A = read.table("cw3_data.txt") xbar = mean(A) s = 1 n = 20 mu = 0 z.test = (xbar-mu)/(s/sqrt(n)) p.value = pnorm(abs(z.test)) error = qnorm(0.99)*s/sqrt(n) left = xbar - error right = xbar + error and have got values off of it...but the values for p dont match up with other sites that i have used to check it

Hi to all I did not found the right hints for functions with the dot-dot-dot argument. Is it possible to write own functions with the tree dots and if yes what's wrong with the following example? test <- function(x, ...) { print (x) if (exists("y"))print(y) if (exists("z"))print(z) } test(4,y=2) With regards Carmen

I am trying calculate a probability using numerical integration. The first program I ran spit out an answer in a very short time. The program is below: ## START PROGRAM trial <- function(input) { pmvnorm(lower = c(0,0), upper = c(2, 2), mean = input, sigma = matrix(c(.1, 0, 0, .1), nrow = 2, ncol = 2, byrow = FALSE)) } require(mvtnorm) require(adapt) bottomB <- -5*sqrt(.1) topB <-

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

sorry the integration was from -Inf to 1.96 The integrate function in R is not taking Inf (infinity). How do you use infinity in R. I was doing: integrate(dnorm,- Inf, 1.96) and I was getting Error: NA/NaN/Inf in foreign function call (arg 2). Obviously this should be equal to pnorm(1.96)= 0.9750021. How do you get around the infinity problem in R?

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

--=-YFjXKq8/D/t1qWmIzQ9D Content-Type: text/plain Content-Transfer-Encoding: quoted-printable I was going over the source in src/nmath/pnorm.c and noticed a little bug in pnorm_both (in R 1.7.0). The else-if on line 205 covers the entire real line. Seems you want an &&, not an ||. Doesn't make a big difference (you still get a 0 or 1 from extreme starting values) but your log