similar to: random number generation

Hello R-Users, The following questions are not R-technical, but more of general statistical nature. 1. NORMALITY I built a normal linear regression model and now I want to check for the residual normality assumption. If I check the distribution graphically and look at the descriptive characteristics (skewness and kurtosis are below 1), I would confirm that the residuals are normally

Dear All, Can someone please tell me if there is a provision in R to fit a random coefficient multinomial logistic regression. Thanks in advance for your help and suggestions. Regards Murthy.N.M

I am using the describe function in prettyR. I would like to add the 25% 75% quartiles to the summary table how do I do this I have tried describe(x.f, num.desc=c("mean", "median", "sd", "min", "max", "skewness", "quantile(x.f, na.rm=T, probs=seq(0.25, 0.75))", "valid.n")) help -- Let's not spend our time

I have an external data (.txt) for annual peak flood. The first column is the year, second column is the observation date, and the last is the observed discharge. My task is to calculate the mean, skewness and kurtosis of the said data. I was advised to use read.table() to read the entire data. Please help me on how to perform the required computation. I am obviously a new user of this statistical

Greetings R users, My goal is to generate quartile groups of each variable in my data set. I would like each experiment to have its designated group added as a subsequent column. I can accomplish this individually with the following code: brks <- with(data_variables, cut2(var2, g=4)) #I don't want the actual numbers, I need a numbered group data$test1=factor(brks,

Hi I need some help getting results from multiple linear models into a dataframe. Let me explain the problem. I have a dataframe with ejection fraction results measured over a number of quartiles and grouped by base_study. My dataframe (800 different base_studies) looks like > afvtprelvefs basestudy quartile ef ef_std entropy CBP0908020 1 21.6 0.53 3.27

Dear R-Users and Developpers, Currently R does not include functions to compute the skewness and kurtosis. I programmed it myself in the following way, but probably *real* programmers/statisticians can do that better: mykurtosis <- function(x) { m4 <- mean((x-mean(x))^4) kurt <- m4/(sd(x)^4)-3 kurt } myskewness <- function(x) { m3 <- mean((x-mean(x))^3) skew <-

I have data that looks like lake,loglength,logweight 1,2.369215857,1.929418926 1,2.426511261,2.230448921 1,2.434568904,2.298853076 1,2.437750563,2.298853076 1,2.442479769,2.230448921 1,2.445604203,2.356025857 ... 102,2.722633923,3.310268367 102,2.781755375,3.502153893 102,2.836324116,3.683407299 102,2.802773725,3.583312152 102,2.790285164,3.546419267 102,2.806179974,3.599118565

Hi,everyone i need to calculate quartile values of a variable grouped by the other variable . same as in aggregate function(only median,mean or functions is possible-i think so) Could you please help me to achieve the same for other quartile values(5,10,25,75,90) as for median using aggregate. Thanks in advance. data : zip price 60000 567000 60001 478654 60004 485647 60001

Hi Dear, Can someone please inform me on genreating random variables of Loglogistic, and PERT beta distributions? Thanks for your time and help, in advance. Regards, Murthy.

Why am I getting a wrong result for quartiles? here is my code: > cbiomass = c(910, 1058, 929, 1103, 1056, 1022, 1255, 1121, 1111, 1192, > 1074, 1415) > summary(cbiomass) > IQR(cbiomass) The result R gives me is: For the summary > Min. 1st Qu. Median Mean 3rd Qu. Max. 910 1048 1088 1104 1139 1415 For IQR > 91.25 ********* The true Q1 is 1039

I've just started using R and am still a neophyte, but I found the following curious result. I'm using the current version of R (2.5.1 (2007-06-27) ). Why are the results for the third quartile different in the output from the summary and fivenum commands? For the following data set 457 514 530 530 538 560 687 745 745 778 786 790 792

Hi any, Can some please detail me the createX command in bayesm package? To make things easy for you to help me, let me put forward my problem Suppose I have 3 covariates (say X matrix) and my Y has 3 categories say (1,2,3). Now from the CreateX I understand that the data matrix say 'Xa' must be of dimension n* (naxp), where 'na' is the number of variables and 'p' is

Hello everybody, This is a bug involving functions in core R package: graphics::hist.default, grDevices::nclass.FD, and base::pretty.default. It is not yet on Bugzilla. I cannot submit it myself, as I do not have an account. Could somebody else add it for me, perhaps? That would be much appreciated. Kind regards, Sietse Sietse Brouwer Summary ------- Floating point errors can cause a data

> On Oct 13, 2017, at 2:51 AM, PIKAL Petr <petr.pikal at precheza.cz> wrote: > > Hi > > You expect us to solve your problem but you ignore advice already recieved. > > Your data are unreadable, use dput(yourdata) instead. see ?dput > >> test<-read.table("clipboard", heade=T) > Error in scan(file = file, what = what, sep = sep, quote = quote,

Hi, Can Someone inform me how to generate RV's using the below CDF, by inverse technique. Thanks for your help and time. My CDF is as follows \[ F(x)=0 \ \text{if} \ x < 0\]\[ F(x)=\{\frac{x-x_i}{x_{i+1}-x_{i}}*(p_{i+1}-p_{i})\}+p_{i}\ \forall \ x_{i}\leq x < x_{i+1} \] \[ F(x)=1 \ \text{if} \ x > x_{i+1} \] Regards Murthy

Hemant's problem is that the indicators are not distributed uniformly. With a uniform distribution, categorization gives a reasonably optimal separation of cases. One approach would be to drop categorization and calculate the overall score as the mean of the standardized indicator scores. Whether this is an option I do not know. I did offer an "eyeball" set of breaks in a previous

I have the following code. > f <- cph(formula = Surv(TimeToDeath, Dead == "Yes") ~1,data=single.dat, x=T, y=T, surv=T) > for(i in c('A', 'B', 'C', 'D', 'E', 'F')){ > f <-update(f,as.formula(paste('Surv(TimeToDeath, Dead == "Yes")~',i,sep=''))) > print(summary(f, paste(i,"=1st

Another naive stats question. I'm trying to better understand what boxplots are telling me. I think what I see is the median and the boundaries of the 1st and 3rd quartiles. The whiskers represent the range of the data unless there are points which are outside "range" (default: 1.5) times the distance from the median to that quartile. Is that right? I've read the

Dear all, I did post this more or less identical mail in a follow up to another question I posted, but under another heading. I try again, but now under the correct header. upon running this code (from the Hmisc library-latex function) I believe the call to summary.formula is allright and produces wonderful tables, but the latex command results in a correct formatted table but where all the