similar to: Formatting data for bootstrapping for confidence intervals

Hello, We are a group of PhD students working in the field of toxicology. Several of us have small data sets with N=10-15. Our research is mainly about the association between an exposure and an effect, so preferrably we would like to use linear regression models. However, most of the time our data do not fulfill the model assumptions for linear models ( no normality of y-varible achieved even

I apologize for this post. I am new to R (two days) and I have tried and tried to calculated confidence intervals for medians. Can someone help me? Here is my data: institution1 0.21 0.16 0.32 0.69 1.15 0.9 0.87 0.87 0.73 The first four observations compose group 1 and observations 5 through 9 compose group 2. I would like to create a bootstrapped 90% confidence interval on the difference of

Hi, I have the following problem. It has two parts. 1. I need to calculate the stationary probabilities of a Markov chain, eg if the transition matrix is P, I need x such that xP = x in other words, the left eigenvectors of P which have an eigenvalue of one. Currently I am using eigen(t(P)) and then pick out the vectors I need. However, this seems to be an overkill (I only need a single

Hi all, Running simulations, I'm generating market response to 2 factors X&Y.. There is no closed form for the market response.. The results are store in a matrix Z(X <- seq(.02,.98,.02), Y <- seq(.01,.19,.01)).. For optmization purpose I need to approximate the values for any factor X in 0,02-0,98 and Y in 0,01-0,19 How can I do it ? For one factor : Xn-1 < x <= Xn

Hi, Sorry for repeated question. I performed logistic regression using lrm and penalized it with pentrace function. I wanted to get confidence intervals of odds ratio of each predictor and summary(MyModel) gave them. I also tried to get bootstrapping standard errors in the logistic regression. bootcov function in rms package provided them. Then, I found that the confidence intervals provided by

Hi to all, how could I to rotate automatically a data sheet which was imported by read.xls? x1 x2 x3 .... xn y1 1 4 7 ... xn/y1 y2 2 5 8 .... xn/y2 y3 3 6 9 ....xn/y2 yn ... ... ... Xn/Yn to y1 y2 y3 .... yn x1 1 2 3 ..... Yn/x1 x2 4 5 6 .... Yn/x2 x3 7 8 9 .... Yn/x2 xn ... ... ... ..... Yn/xn Kind regards Knut

Dear all, I am interested to calculate confidence interval for fitted values in general for non-linear regressions. Lets say we have y=f(x1,x2,..xN) where f() is a non-linear regression. I would like to calculate a confidence interval for new prediction f(a1,..,aN). I am aware of techniques for calculating confidence intervals for coeffiecients in specific non-linear regressions and with them

The programming language mosml comes with foldr that 'accumulates' a function f over a list [x1,x2,...,xn] with initial value b as follows foldr f b [x1,x2,...,xn] = f(x1,...,f(xn-1,f(xn,b))...) Observe that "list" should have same elements so in R terminology it would perhaps be appropriate to say that the accumulation takes place over a 'vector'. I wonder if R

Can someone offer some advice on how to properly evaluate a SYMSXP from a .Call ? I have the following in R: variable xn, with an attribute "mu" which references the variable mu in the global environment. I know "references" is a loose term; mu was defined in this fashion as a way to implement deferred binding: foo <- function(x,mu) { attr(x,"mu") <-

Hi, I have written a small C++ function and compile it. However in R I can't see the function I have defined in C++. I have read some web-pages about Rcpp and C++ but it is a bit confusion for me. Anyway, This is the C++-code: #include <Rcpp.h> using namespace Rcpp; // [[Rcpp::export]] List compute_values_cpp(int totalPoints = 1e5, double angle_increment = 0.01, int radius =

Dear all, The documentation for `all.equal.numeric` says Numerical comparisons for ?scale = NULL? (the default) are done by first computing the mean absolute difference of the two numerical vectors. If this is smaller than ?tolerance? or not finite, absolute differences are used, otherwise relative differences scaled by the mean absolute difference. But the actual behaviour

Hi I need advice regarding constraint optimization with large number of variables. I need to solve the following problem max f(x1,...,xn) x1,..xn x1=g1(x1,...,xn) . . xn=gn(x1,...,xn) I am using Rdonlp2 package which works well until 40 variables in my case. I need to solve this problem with over 300 variables. In this case Rdonlp2 is very very slowly. I know

Hi guys, I'm trying to use the the integral function to estimate the area under a PDF and a crossing curve. first I stated the function with several vectors in it: fn=function(a,b,F,mu,alpha,xi) { x<-vector() fs<-function(x) { c <- (mu+(alpha*(1-(1-F)^xi)/xi)) tmp <- (1 + (xi * (x - mu))/alpha) ((as.numeric(tmp > 0) * (tmp^(-1/xi - 1) *

.Call("compute_values_cpp") Also, if you were passing arguments to the C++ function you would need to declare the function differently. Do a search on "Rcpp calling C++ functions from R" HTH, Eric On Sun, Dec 3, 2017 at 3:06 AM, Martin M?ller Skarbiniks Pedersen < traxplayer at gmail.com> wrote: > Hi, > > I have written a small C++ function and compile it.

Hello, Can anyone give me some suggestion in term of calculating the sum below. Is there a function in R that can help doing it faster? x1, x2, ...xn where xi can be 0 or 1. I want to calculate the following: sum{ beta[a+sum(xi), b+n-sum(xi) ]* [ (1-x1)dnorm(0,1)+x1dnorm(2,1) ]* [ (1-x2)dnorm(0,1)+x2dnorm(2,1) ]* ...* [ (1-xn)dnorm(0,1)+xndnorm(2,1) ] } The sum in the beginning is over all

Hi, there could you help me coding this problme? I am just starting to leard the R. So I really need help Question is from Ross, Simulation, 4th Edition. ch3 14. with x1=23, x2=66 Xn=3*Xn-1+5*Xn-2 mod(100) n>=3 we will call the sequence Un=Xn/100 n>=1 find the first 14 values thank you sophia

Dear all, I considered an ordinary ridge regression problem. I followed three different ways: 1. estimate beta without any standardization 2. estimate standardized beta (standardizing X and y) and then again convert back 3. estimate beta using lm.ridge() function X<-matrix(c(1,2,9,3,2,4,7,2,3,5,9,1),4,3) y<-t(as.matrix(cbind(2,3,4,5))) n<-nrow(X) p<-ncol(X) #Without

Dear all, For an ordinary ridge regression problem, I followed three different approaches: 1. estimate beta without any standardization 2. estimate standardized beta (standardizing X and y) and then again convert back 3. estimate beta using lm.ridge() function X<-matrix(c(1,2,9,3,2,4,7,2,3,5,9,1),4,3) y<-as.matrix(c(2,3,4,5)) n<-nrow(X) p<-ncol(X) #Without standardization

Hello folks, Any ideas how to do this? data.frame is a data frame with column names "x1",...,"xn" y is a response variable of length dim(data.frame)[1] I want to write a function function(y, data.frame){ lm(y~x1+...+xn) } This would be easy if n was always the same. If n is arbitrary how could I feed the x1+...+xn terms into lm(response~terms)? Thanks Richard -- Dr.

Dear Jon, thank you for raising the issue, >>>>> Jon Clayden <jon.clayden at gmail.com> >>>>> on Tue, 28 Jul 2015 12:14:48 +0100 writes: > Sorry; minor clarification. The actual test criterion in the example I > gave is of course abs((0.1-0.102)/0.1) < 0.01, not abs(0.1) < 0.01. In > any case, this does not match (my reading of) the docs,