similar to: Inverse noncentral Beta

Dear List: The square of the noncentral t-statistic with noncentrality parameter \delta is a noncentral F with noncentrality parameter \lambda=\delta^2. So, t^2_{\nu,\delta} = F_{1,\nu,\lambda=\delta^2}. Consequently, it should follow that t^2_{1-\alpha/2,\nu,\delta} = f_{1-alpha,1,\vu,\lambda=\delta^2}. However, this is not what is happening with the following code. The central

I have a data with 236 observations. After plotting the histogram, I found that it looks like non-central t distribution. I would like to get MLE for mu and df. I found an example to find MLE for gamma distribution from "fitting distributions with R": library(stats4) ## loading package stats4 ll<-function(lambda,alfa) {n<-200 x<-x.gam

hello R-friends, I'm looking for a quantile function for the noncentral f-distribution in the area of equivalence hypotheses testing. Can somebody help me? Many thanks ----------------------------------------------------------------- Dipl. Inform. J. Hedderich Institut f?r Medizinische Informatik Phone : 0431 / 5973182 und Statistik im Klinikum an der CAU

Full_Name: Long Qu Version: 2.3.1 OS: Windows XP Submission from: (NULL) (64.113.93.235) The QQ-plot of two versions of simulating noncentral F-distributed random numbers has quite different scales: > qqplot(rf(1000,2,15,3),qf(runif(1000),2,15,3)) The rf() function reads: > rf function (n, df1, df2, ncp = 0) { if (ncp == 0) .Internal(rf(n, df1, df2)) else rchisq(n, df1,

Hi there: I'd thought these two versions of noncentral t-distribution are essentially the same: > qqplot(rt(1000,df=20,ncp=3),qt(runif(1000),df=20,ncp=3)) But, the scales of the x-axis and the y-axis are quite different according to the QQ-plot. Did I make any mistakes somewhere? Thanks, Long ---------------------------------

Hi everyone. I have a problem that I have been unable to determine either the best way to proceed and why the methods I'm trying to use sometimes fail. I'm using the pf() function in an optimization function to find a noncentrality parameter that leads to a specific value at a specified quantile. My goal is to have a general function that returns the noncentrality parameter that

Hello all. It seems that the pf() function when used with noncentral parameters can behave badly at times. I've included some examples below, but what is happening is that with some combinations of df and ncp parameters, regardless of how large the quantile gets, the same probability value is returned. Upon first glance noncentral values greater than 200 may seem large, but they are in

Full_Name: Mike Meredith Version: 2.3.0 OS: WinXP SP2 Submission from: (NULL) (210.195.228.29) Using dt() with a non-centrality parameter and near-zero values for 'x' results in erratic output. Try this: tst <- c(1e-12, 1e-13, 1e-14, 1e-15, 1e-16, 1e-17, 0) dt(tst,16,1) I get: 0.2381019 0.2385462 0.2296557 0.1851817 0.6288373 3.8163916 (!!) 0.2382217 The 0.238 values are okay,

Hi, I have written out the log-likelihood function to fit some data I have (called ONES20) to the non-central chi-squared distribution. >library(stats4) >ll<-function(lambda,k){x<-ONES20; 25573*0.5*lambda-25573*log(2)-sum(-x/2)-log((x/lambda)^(0.25*k-0.5))-log(besselI(sqrt(lambda*x),0.5*k-1,expon.scaled=FALSE))} > est<-mle(minuslog=ll,start=list(lambda=0.05,k=0.006))

Despite the need to focus on pbeta, I'm still willing to put in some effort. But I find it really helps to have 2-3 others involved, since the questions back and forth keep matters moving forward. Volunteers? Thanks to Martin for detailed comments. JN On 2020-03-26 10:34 a.m., Martin Maechler wrote: >>>>>> J C Nash >>>>>> on Thu, 26 Mar 2020

In the help file for the non-central t, the following appears: ncp: non-centrality parameter delta; currently `ncp <= 37.62'. I assume that this means the ncp cannot exceed 37.62. Is this still the case and is there any plans to increase this restriction? Thanks! Jeff Jeff Morris Design Support Clinical Chemistry R&D Ortho-Clinical Diagnostics email: jmorris6 at ocdus.jnj.com

The problem occurs when the sample odds ratio is Inf, such as in the following example. Given the fact that both upper bounds of the two 95% confidence intervals are Inf, I would have expected that the two lower bounds be equal, but they aren't. x <- matrix(c(9,4,0,2),2,2) x # [,1] [,2] #[1,] 9 0 #[2,] 4 2 rbind("two.sided.95CI"=fisher.test(x)$conf.int,

I sent a message a couple days ago about doing calculations for power of the ANOVA. Several people got back to me very quickly which I really appreciated. I'm working now on a similar problem, but instead of a balanced ANOVA, I have an unbalanced one. The first part of the question was: You assume that the within-population standard deviations all equal 9. You set the Type 1 error rate at á

Hi, Actually I am facing a similar problem. I would like to fit both an ordinary (symmetric) and a non-central t distribution to my (one-dimensional) data (quite some values.. > 1 mio.). For the symmetric one, fitdistr or funInfoFun (using fitdistr) from the qAnalyst package should do the job, and for the non-central one.. am I right to use gamlss(x ~ 1, family=GT()) ? Anyway, I am a little

Full_Name: Version: 1.9.0 OS: Windows XP Submission from: (NULL) (134.126.93.191) I believe the noncentrality parameter, lambda, in power.anova.test is incorrect. The noncentrality paramter is defined as lambda <- (groups - 1) * n * (between.var/within.var). The n should not be there. The function pf defines the noncentrality parameter as the sum of squares of the means. Therefore, the

Hi, I think I have identified a inaccuracy in dchisq when the non-centrality parameter is non-zero and large. Here's a little test: sys.dchisq.test <- function(N = 100000,mean = 0) { z <- rnorm(N,mean = mean, sd = 1) x <- z^2 xmin <- min(x) xmax <- max(x) br <- seq(xmin,xmax,length = 101) dbr <- br[2]-br[1] hist(x,br) p <- dchisq(br,df = 1,ncp =

Greetings all, I'm trying to figure out how to calculate the inverse CDF (i.e. a quantile) for a non-central F distribution. I could put together a quick numerical solver routine using the CDF, but I wonder if there's a function that I've missed that would be more efficient? Thank-you, Andrew Andrew Robinson Ph: 208 885 7115 Department of Forest Resources Fa: 208 885

Hello Everyone, I've been told that the memory allocation in R and S-Plus is different, with R being better at doing simulations for whatever reason(s). I was wondering if anyone could enlighten me on the differences in the memory usage/allocation (preferably in layman's terms!). The reason I ask this is because of a simulation I am trying to run. In it, I need to use the pt( )

For those who are interested, I have made available a R function for noncentral hypergeometric distribution at http://www.geocities.com/jg_liao/software/Hypergeometric/hypergeometric_in_R.txt The paper that describes the algorithm will appear in The American Statistician. The function does not run on S-plus as the R's scoping rule is used. Here is how the function can be used: > n1

help(qt) states that: "ncp non-centrality parameter delta; currently except for rt(), only for abs(ncp) <= 37.62" so I would expect that calling qt with non-centrality parameter exceeding 37.62 should fail, instead e.g. calling > mapply(function(x) qt(p = 0.9, df = 55, ncp = x),35:45) gives: [1] 40.21448 41.35293 42.49164 43.68862 44.82945 45.97048 47.11170 48.25310 [9]