similar to: Power of test

Hello, Is there a built-in test in R for hypothesis testing with samples of known variance? For example, I've got a set of data, a mean to compare against, and a known variance, and I want to determine the p-value for which I can reject the null hypothesis (mu_1 = mu_0) and accept the alternative (mu_1 > mu_0). I've found that JMP and Minitab can both do this (in JMP, it's a

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

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

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

Hi, I'm trying to calculate the value of the variable, dp, below, in the argument to the integral of dnorm(x-dp) * pnorm(x)^(m-1). This corresponds to the estimate of the sensitivity of an observer in an m-alternative forced choice experiment, given the probability of a correct response, Pc, a Gaussian assumption for the noise and no bias. The function that I wrote below gives me an error:

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,

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]]

Hi R users. I am new to the community and have got myself into a little problem. I have a dataset of birth weights recorded by nurses at a delivery clinic in an developing country. The weights are entered in KiloGrams with one decimal. However there is substantial heaping at each 500g when looking at the sample in a histogram. Do anyone of you know a easy way to adjust for this and if it exists

I am converting some matlab code into R that use inverse of the complementary error function, erfcinv and did not find an equivalent in R, is there such a function in some contributed modules? Thanks.

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

Dear Contributors I have a problem with the collection of data from the results of a test. I need to perform a comparative test over groups of data , recall the value of the pvalue and create a table. My problem is in the way to replicate the analysis over and over again over subsets of data according to a condition. I have this database, called y: gg t1 t2 d 40 1 1

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

Hi Krzysztof, Sure, please see below. DAG.dump.() before and after, annotated with what I believe the DAG means. I've spent some time debugging the method but it's proving difficult to determine where the logic is misfiring. Disabling the entire combine causes a lot of failing x86-64 tests - I may have to learn an upstream vector ISA to make progress on this. Thank you >From your

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

Hi everyone, I have dataset which I make a sample of it couple of times and each time I get the mean and standard deviation of each row for each sample. I have a function for that, which takes the name of the file and number of times to sample and then returns the mean and standard deviation for each row in each sample. Sample=function(name, n){

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