similar to: tolerance intervals

How can I get prediction intervals from a mixed-effects model? Consider the following example: library(nlme) fm3 <- lme(distance ~ age*Sex, data = Orthodont, random = ~ 1) df3.1 <- with(Orthodont, data.frame(age=seq(5, 20, 5), Subject=rep(Subject[1], 4), Sex=rep(Sex[1], 4))) predict(fm3, df3.1, interval='prediction') # M01 M01

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

Hello, I wonder if anyone has a function for calculating confidence intervals for ratios. I have mortality rates calculated for several groups from sample data -- but these can be treated as ratios?). I am entertaining the idea of bootstrapping them. Any help would be greatly appreciated. TIA, Marwan ---------------------------------------------------------------------------- ------------

Hello Bernardo, --------- If I understood your problem this script solve your problem: q<-0.15 + c(-.1,0,.1) h<-10 + c(-.1,0,.1) 5*q*h [1] 2.475 7.500 12.625 --------- OK, this solves the simple example. But what if the example is not that simple. E.g. P = 5 * q/h Here, to get the maximum tolerances for P, we need to divide the maximum value for q by the minimum value for h, and

Hi folks, I'm trying to estimate bias-corrected percentile (BCP) confidence intervals on a vector from a simple for loop used for resampling. I am attempting to follow steps in Manly, B. 1998. Randomization, bootstrap and monte carlo methods in biology. 2nd edition., p. 48. PDF of the approach/steps should be available here: https://wyocoopunit.box.com/s/9vm4vgmbx5h7um809bvg6u7wr392v6i9 If

Hello. I'm trying to find a way in R to calculate between and within subject variances and confidence intervals for some analytical method development data. I've found a reference to a method in Burdick, R. K. & Graybill, F. A. 1992, Confidence Intervals on variance components, CRC Press. This example is for Balanced Data confidence interval calculation from Pg 62. The data are

Hello, I have a question regarding bootstrap confidence intervals. Suppose we have a data set consisting of single measurements, and that the measurements are independent but the distribution is unknown. If we want a confidence interval for the population mean, when should a bootstrap confidence interval be preferred over the elementary t interval? I was hoping the answer would be

This is a follow-up mail of "the effect of blocking on the size of confidence intervals - analysis using aov". In both mails I pursue the idea of using blocking factors in order to reduce the width of confidence intervals. My dataset comprises, a quantitative response variable, namely: "response", and three categorical eplanatory variables, namely: "method",

Dear list, How is the tolerance for a model parameter in an lm() call extracted? I did not see a solution in the documentation for lm(), or predict(), nor in the archives using 'tolerance' as the search string. I also checked into the nlme package, though nothing popped out at me. Sincerely, KeithC.

On 2019-12-27 04:34, Duncan Murdoch wrote: > On 26/12/2019 11:14 p.m., Spencer Graves wrote: >> Hello, All: >> >> >> ? ????? The default "simulate" method for lm and glm seems to ignore the >> sampling variance of the parameter estimates;? see the trivial lm and >> glm examples below.? Both these examples estimate a mean with formula = >>

Full_Name: Robert W. Baer, Ph.D. Version: 1.6.2 OS: Windows 2000 Submission from: (NULL) (198.209.172.106) Problem: prop.test() does not seem to produce appropriate confidence intervals for the case where the vector length of x and n is one. (I am not certain about higher vector lengths.) As an example, I include x=6 and n=42 which has a mean proportion of 0.115. When I calculate the 95% CI

Dear R people, I notice that the confidence intervals of a very small sample (e.g. n=6) derived from the one-sample wilcox.test are just the maximum and minimum values of the sample. This only occurs when the required confidence level is higher than 0.93. Example: > sample <- c(1.22, 0.89, 1.14, 0.98, 1.37, 1.06) > summary(sample) Min. 1st Qu. Median Mean 3rd Qu. Max.

Dear List I?m using the ?lmomRFA? package to fit different distributions to my data sample. To calculate the error bounds I used: regsimq(?) and sitequantbounds(?) So my questions are: Are error bounds and confidence intervals the same thing? And: Does regsimq(? boundprob = c(0.05, 0.95)) calculate the 90 or the 95% confidence interval? If error bounds and confidence intervals are

Hi Everyone, This example code results in R 'crashing'; that is the R application closes with no warnings or error messages. #----------------------- myD <- read.table(stdin(), header=TRUE, nrows=20) Broth Salt pH Temp N Y Growth 1 310 9.0 2.92 10 90.0 NA 0 2 615 6.0 7.82 30 1.0 2 1 3 217 2.0 7.34 10 7.0 8

Greetings, I would like to calculate tolerancelimits for a series of predicted values from nonlinear regression models. I've been using the tolerance package but the self-starting functions are not in the derivitive tables. When I spellout the functions and supply starting values I repeatedly get an error message regardless of the starting values I use (see output below). Does anyone

R-Experts, Here below my R code working without error message but I don't get the results I am expecting. Here is the result I get: [1] "All values of t are equal to 0.28611928397257 \n Cannot calculate confidence intervals" NULL If someone knows how to solve my problem, really appreciate. Best, S ######################################################### # Difference in Spearman

Dear R-Experts, Thanks to Prof. Bonnett, I have got an R script working to calculate confidence intervals around the semipartial correlation coefficients. Now, I would like to calculate BCa bootstrap CIs using the boot library and the boot.ci(results, type="all") function. How could I modify my R script (here below reproducible example) to get the BCa bootstrap CIs ? CIsemipartcorr

I'm using R 2.15.1 on a 64-bit machine with Windows 7 Home Premium. Sample problem (screwy subscripted syntax is a relic of edited down a more complex script): > N <- 25 > s <- rlnorm(N, 0, 1) > require("boot") Loading required package: boot > v <- NULL # hold sample variance estimates > i <- 1 > v[i] <- var(s) # get sample variance >

For a given linear regression, I wish to find the 2-tailed t-dist probability that Y-hat <= newly observed values. I generate prediction intervals in predict() for plotting, but when I calculate my t-dist probabilities, they don't agree. I have researched the issues with variance of individual predictions and been advised to use the variance formula below (in the code). I presume my

I believe the problem is here: cor1 <- cor(x1, y1, method="spearman") cor2 <- cor(x2, y2, method="spearman") The x's and y's are not looked for in data (i.e. NSE) but in the environment where the function was defined, which is standard evaluation. Change the above to: cor1 <- with(d, cor(x1, y1, method="spearman")) cor2 <- with(d, cor(x2, y2,