similar to: Fitting a Weibull Distribution

All, Can someone help me create a skewed distribution, mean = 30, with probability of selecting a random number from the distribution greater than or equal 60 equal to 10%? I need the probability density function to equal zero at zero, and have a maximum height at or near 30. Is this possible? And if it is possible, how can I adjust the distribution so that the probability of selecting a

Hi all, I wondered whether anyone has some advice on a stats-related 'sanity check', as I ran a nonparametric multivariate test (mulrank function as decribed by R. Wilcox, 2005) on both systems, but got different results (please see below for the system-specific outputs)! The functions I used are attached as well. Any advice would be much appreciated! Thanks in advance for getting back to

Dear all, I get the following error message. And I cannot quite work out what is wrong. I think the optim gets infinite values. Certainly my data do not have any infinite values. How can I solve this? fitdistr(A1, "weibull") Error in optim(start, mylogfn, x = x, hessian = TRUE, ...) : non-finite value supplied by optim I am using R version 1.9.1 on RedHat Linux, Kernel 2.6.8.

Dear R users, Does anyone know how to write function for Kumaraswamy distribution in R? Since I cannot write dkumar, pkumar, etc. in R. Please help. Thanks a lot, Debbie _________________________________________________________________ [[elided Hotmail spam]] [[alternative HTML version deleted]]

Hi all, I suspect the answer to this query will be the tongue-in-cheek "use a quantum computer", but I thought my understanding might be sufficiently limited that I'm missing a simpler option. I'm looking for a way to cycle through all possible combinations of 2 groups of data. For 10 data points, that's 2^10 combinations, for 20 data points it's 2^20, etc. combn() from

Hello, is there a quick way of estimating Weibull parameters for some data points that are assumed to be Weibull-distributed? I guess I'm just too lazy to set up a Maximum-Likelihood estimation... ...but maybe there is a simpler way? Thanks for any hint (and yes, I've read help(Weibull) ;) Kaspar Pflugshaupt -- Kaspar Pflugshaupt Geobotanical Institute ETH Zurich, Switzerland

Hello, I would like to fit a mixture of two Weibull distributions to my data, estimate the model parameters, and compare the fit of the model to that of a single Weibull distribution. I have used the mix() function in the 'mixdist' package to fit the mixed distribution, and have got the parameter estimates, however, I have not been able to get the log-likelihood for the fit of this model

Dear All, I have two questions regarding distribution fitting. I have several datasets, all left-truncated at x=1, that I am attempting to fit distributions to (lognormal, weibull and exponential). I had been using fitdistr in the MASS package as follows: fitdistr<-(x,"weibull") However, this does not take into consideration the truncation at x=1. I read another posting in this

Dear R-users I have some datasets, all left-censoring, and I would like to fit distributions to (weibull,exponential, etc..). I read one solution using the function survreg in the survival package. i.e survreg(Surv(...)~1, dist="weibull") but it returns only the scale parameter. Does anyone know how to successfully fit the exponential, weibull etc... distributions to left-censoring

Hi folks, I tried to retain the Weibull distribution using the `nls' function and proceeding along the lines of the example provided in the `SSweibull' help (at least I thought so): t <- (1:200)/100 v <- pweibull(t, shape=3, scale=1) df <- data.frame(Time=t, Value=v) Asym <- 1.0; Drop <- 1.0; lrc <- 0; pwr <- 1 df.estimate <- nls(Value ~ SSweibull(Time,

Hi, I'm currently doing some analyses on time of failure of a product. I found on the internet some article about the Arrhenius equation, and I can calculate the results with R. Equation: k=A*exp(-Ea/R*T) I can fit a model with lm for this purpose, so far no problem. But for the confidence interval, how can I use the Weibull distribution? Or should I use the weibull distribution at all? Or

Hi, I'm dealing with wind data and I'd like to model their distribution in order to simulate data to fill-in missing values. Wind direction are typically following a vonmises distribution and wind speeds follow a weibull distribution. I'd like to build a joint distribution of directions and speeds as a VonMises-Weibull bivariate distribution. First is this a stupid question? I'm

Hi all, I'm trying to generate a Weibull distribution including four covariates in the model. Here is the code I used: T = rweibull(200, shape=1.3, scale=0.004*exp(-(-2.5*b1+2.5*b2+0.9*x1-1.3*x2)/1.3)) C = rweibull(n, shape=1.5, scale=0.008) #censoring time time = pmin(T,C) #observed time is min of censored and true event = time==T # set to 1 if event is observed

HI Every one Could some one provide me definitions of following bivariate distributions gamma, exponencial, Weibull, half-normal , Rayleigh, Erlang,chi-square thanks A.S. Qureshi

The ez package was developed to aid those that are new to statistical programming. Over the course of several years of helping colleagues and students learn R, I observed that folks are often initially turned off R because they have difficulty obtaining SPSS-like results quickly (SPSS is the dominant environment in my field, psychology). ez attempts to fill this gap, providing quick and easy

The ez package was developed to aid those that are new to statistical programming. Over the course of several years of helping colleagues and students learn R, I observed that folks are often initially turned off R because they have difficulty obtaining SPSS-like results quickly (SPSS is the dominant environment in my field, psychology). ez attempts to fill this gap, providing quick and easy

Dear Sir/madam, I'm getting a problem with a R-code which calculate Fisher Information Matrix for Hybrid Censored Weibull Distribution. My problem is that: when I take weibull(scale=1,shape=2) { i.e shape>1} I got my desired result but when I take weibull(scale=1,shape=0.5) { i.e shape<1} it gives error : Error in integrate(int2, lower = 0, upper = t) : the integral is probably

Hi guys, I have a data-set that fits well into a Weibull model y = a-b*exp(-c*x^d). I want to estimate the parameters of the coefficients a, b, c and d, given x and y. Can you guys help me? Just as an example, I fit the data y <- c(1,2,3,4,10,20) and x <- c(1,7,14,25,29,30) According to this model above, and using the software CurveExpert, and I got the estimates of a (2.95),b (2.90),c

Hi, I have a problem fitting the following Weibull Model to a set of data. The model is this one: a-b*exp(-c*x^d) If I fitted the model with CurveExpert I can find a very nice set of coefficients which create a curve very close to my data, but when I use the nls.lm function in R I can't obtain the same result. My data are these: X Y 15 13 50 13 75 9 90 4 With the commercial

Dear All; I tried to use fitdistr() in the MASS library to fit a mixture distribution of the 3-parameter Weibull, but the optimization failed. Looking at the source code, it seems to indicate the error occurs at if (res$convergence > 0) stop("optimization failed"). The procedures I tested are as following: >w3den <- function(x, a,b,c)