similar to: based on mean and std

Dear all, I would like to ask your help concerning an error message I get. I have the following struct str(CRagentInTime[[1]]) List of 2 $ timelag: int 0 $ CRagent:List of 50 ..$ :List of 3 .. ..$ CRmap: num [1:256, 1:256] NA NA NA NA NA NA NA NA NA NA ... .. ..$ xy : num [1:2] 10 177 .. ..$ sr : num [1:49] -94.9 -92.8 -79.5 -97.6 -78.4 ... and I wanted to select all the sr fields

http://r.789695.n4.nabble.com/file/n3567636/ecdfs.jpg ecdfs.jpg http://r.789695.n4.nabble.com/file/n3567636/ecdf_curve.gif ecdf_curve.gif Hello, I have generated a plot of two empirical CDFs (attachment 1). As a result, they are stepwise when plotted. The following code was used: > plot(ecdf(mut), do.points=FALSE, verticals=TRUE, xlim=range(mut, non), > col="red") >

Hi to all, maybe the last question was not clear enough. I did not found any hints how to decide whether it should use lower.tail or not. As it is an extra R-feature ( written in http://finzi.psych.upenn.edu/R/Rhelp02a/archive/66250.html ) I do not find anything about it in any statistical books of me. Regards Carmen

Hi R community! I am analyzing the data set "motorins" in the package "faraway" by using the generalized additive model. it shows the following error. Can some one suggest me the right way? library(faraway) data(motorins) motori <- motorins[motorins$Zone==1,] library(mgcv) >amgam <- gam(log(Payment) ~ offset(log(Insured))+ s(as.numeric(Kilometres)) + s(Bonus) + Make +

Hi All,  In order to compare two different logistic regressions, I think I need to compare them based on their BIC values, but I am not sure if the smaller BIC would mean a better model or the reverse is true? Thanks a lot,Andra [[alternative HTML version deleted]]

Let's say you want to compare "one observation" with a sample, how would you use R to get a p-value for that single observation itself? To clarify what I'm asking: We know you use a one-sample t test to compare an actual sample to a hypothetical value, and a Wilcoxon test if it's not normally distributed, in R either "t.test( )" or "wilcox.test( )".

Dear all, I would like for a given vector to calculate the pdf and the cdf and then plot these two graphs. Is there in R any toolkit that can help me do that fast? Regards Alex

Hello all, I would like to plot the emperical CDF, normal CDF and pareto CDF in the same graph and I amusing the following codes. "z" is a vector and I just need the part when z between 1.6 and 3. plot(ecdf(z), do.points=FALSE, verticals=TRUE, xlim=c(1.6,3),ylim=c(1-sum(z>1.6)/length(z), 1)) x <- seq(1.6, 3, 0.1) lines(x,pgpd(x, 1.544,0.4373,-0.2398), col="red") y

Under platform x86_64-pc-linux-gnu arch x86_64 os linux-gnu system x86_64, linux-gnu status major 2 minor 13.1 year 2011 month 07

Hi, all: Does anybody know how to avoid the intercept term in cv.glmnet coefficient? When I say "avoid", it does not mean using coef()[-1] to omit the printout of intercept, it means no intercept at all when doing the analysis. Thanks. [[alternative HTML version deleted]]

Hi all, I am using the glmnet R package to run LASSO with binary logistic regression. I have over 290 samples with outcome data (0 for alive, 1 for dead) and over 230 predictor variables. I currently using LASSO to reduce the number of predictor variables. I am using the cv.glmnet function to do 10-fold cross validation on a sequence of lambda values which I let glmnet determine. I then take

This practical problem in maximum likelihood estimation must be encountered quite a bit. What do you do when a data point has a probability that comes out in numerical evaluation to zero? In calculating the log likelihood you then have a log(0) problem. Here is a simple example (probit) which illustrates the problem: x<-c(1,2,3,4,100) ntrials<-100 yes<-round(ntrials*pnorm((x-3)/1))

Hi, I am trying to do parameter estimation with optim, but I can't get it to work quite right-- I have an equation X = Y where X is a gaussian, Y is a multinomial distribution, and I am trying to estimate the probabilities of Y( the mean and sd of X are known ), Theta1, Theta2, Theta3, and Theta4; I do not know how I can specify the constraint that Theta1 + Theta2 + Theta3 + Theta4 = 1 in

Hi, Does anyone know what the distribution for the product of two correlated normal? Say I have X~N(a, \sigma1^2) and Y~N(b, \sigma2^2), and the \rou(X,Y) is not equal to 0, I want to know the pdf or cdf of XY. Thanks a lot in advance. yu [[alternative HTML version deleted]]

In the R source, nmath/pnorm.c contains the code for a rational function approximation for the normal cdf. These constants are listed: const double a[5] = { 2.2352520354606839287, 161.02823106855587881, 1067.6894854603709582, 18154.981253343561249, 0.065682337918207449113 }; The source file cites a paper by Cody (1969) and states that these

Dear Experts, I need to calculate the cdf of the standard normal distribution, i.e. H(x) = 1/sqrt(2*pi) integral(exp(-z^2/2) dz), where z is b/w -infi to infi. I know there should be a way to do it in R, but did not know to do it. I'd appreciate any help you could offer. Charlie Liu Graduate student intern at EPA/ECO

Dear all, do you know any easy way based on a vector input how to plot easily cdf and pdf. I would like to thank you in advance for your help Regards Alex

Hi, I'm trying to find a way to take evaluate the Normal CDF over an interval and return the result on the log scale. This works, but I think it isn't numerically stable: log(pnorm(a, mean = x, sd = y) - pnorm(b, mean = x, sd = y)) Does anyone know of a single function that does the above? Or knows of a way to make it more stable? I'd really appreciate any suggestions!

Hello! I have a problem with integrate() in my function nctspa(). Integrate produces an error message "evaluation of function gave a result of wrong length". I don't know what that means. Could anyone suggest me what is wrong with my function? These are the examples of function calls that work OK: nctspa(a=1:10,n=5) nctspa(a=1:10, n=5, mu=2, theta=3, renorm=0) This does not work:

Hi, I want to generate random samples from truncated normal say Normal(0,1)Indicator((0,1),(2,4)). It has more than one intervals. In the library msm, it seems to me that the 'lower' and 'upper' arguments can only be a number. I tried rtnorm(1,mean=0,sd=1, lower=c(0,2),upper=c(1,4)) and it didn't work. Can you tell me how I can do truncated normal at more than one intervals?