search for: intervall

Hi, I would like to add confidence intervales to an ANCOVA with 2 covariates when using xyplot. What would be a good way of accomplishing this? --8<---------------cut here---------------start------------->8--- rm(list = ls(all = TRUE)) rm(list = c(ls())) library(lattice) ## 1. generate data random <- rnorm(200) y <- abs(random) x1.cont <- abs(random) x2.fac <-

Hi >i have this data > X [1] 5.79 1579.52 2323.70 68.85 426.07 110.29 108.29 1067.60 17.05 22.66 [11] 21.02 175.88 139.07 144.12 20.46 43.40 194.90 47.30 7.74 0.40 [21] 82.85 9.88 89.29 215.10 1.75 0.79 15.93 3.91 0.27 0.69 [31] 100.58 27.80 13.95 53.24 0.96 4.15 0.19 0.78 8.01 31.75 [41] 7.35 6.50

Dear R-Users, I am resending this message just to reminder my question regarding the calculation of a bootstrap confidence intervals for a GAM plot. I am trying to apply a bootstrap to a GAM in order to calculate the 95% confidence intervals for a smooth curve obtained by the ?plot.gam? function of the mgcv package. Nonetheless, I am getting some difficulties in transposing the results for

Dear R, I am using this R version:R version 2.11.1 (2010-05-31)(Cran Mirror Berlin) It seems to me, that R constructs a wrong confidence intervall if you try to get a one sided t-test. If the true mean is 1 and my alternative hypothesis (H1) says that mu is smaller ("less")than zero the conf. intervall should reach +∞ and not -∞ if it is constructed for the H0 saying that mu is greater or equal to zero. > rnorm(20,1)->n &...

...plot.Map function by defining the foreground colour: jet.colors = colorRampPalette(c("#00007F", "blue", "#007FFF", "cyan", "#7FFF7F", "yellow", "#FF7F00", "red", "#7F0000")) n=100 cols <- jet.colors(n) Intervalls = cut(as.numeric(Vector),n, na.rm=TRUE) fgs <-cols[Intervalls] ## is there some option/possibility here to choose colours differently ?? plot.Map(Shape.map, fg=fgs, ol=NA, bty="n", xlab="", ylab="", xaxt="n", yaxt="n") My problem is the foll...

Hi everybody, I am sorry that I am kind of spamming this forum, but I have searched for some input everywhere and cant really find a nice solution for my problem. Data looks like: price 2011-11-01 08:00:00 0.000000000 2011-11-01 08:00:00 0.000000000 2011-11-01 08:02:00 0.000000000 2011-11-01 08:03:00 -0.017033339 2011-11-01 08:13:00 0.000690001 2011-11-01

Hello, I am a little bit confused regarding the density values obtained from the function kde2d() from the package MASS because the are not in the intervall [0,1] as I would expect them to be. Here is an example: x <- c(0.0036,0.0088,0.0042,0.0022,-0.0013,0.0007,0.0028,-0.0028,0.0019,0.0026,-0.0029,-0.0081,-0.0024,0.0090,0.0088,0.0038,0.0022,0.0068,0.0089,-0.0015,-0.0062,0.0066) y <- c(0.00614,0.00194,0.00264,0.00064,0.00344,0.00314,-0.00006,-0....

Hello I recently switched from apcupsd to NUT2.2.2 without any troubles. I tried to figure out how I can configure the self-test intervall. The command upsrw lists only the variables battery.charge.low battery.runtime.low ups.delay.shutdown ups.delay.start The command usbhid-ups -D and the listing under http://obsvermes.org/cgi-bin/nut/upsstats.cgi?host=apcsmart at localhost&treemode show no hint on appropriate variable. I s...

Hi List, I have start and end times of events structure(list(start = c("15:00", "15:00", "15:00", "11:00", "14:00", "14:00", "15:00", "12:00", "12:00", "12:00", "12:00", "12:00", "12:00", "12:00", "12:00", "12:00", "12:00",

...w to calculate odds ratio from logistic model in R. Thank you very much. Guoya Stefano <stecalza at tiscalinet.it <https://stat.ethz.ch/mailman/listinfo/r-help> > writes: >Hi all. > >A simple question. >Is there a function to compute the Odds Ratio and its confidence intervall, from >a logistic model (glm(.......,family=binomial....). I've written my own, but >certainly someone did a better job. I show a simple function to do this in my introductory notes available from: http://www.myatt.demon.co.uk basically it is: lreg.or <- funct...

...No problem so far. But I have one doubt: is it also reccmended (for keeping maildata in sync) to run "doveadm replicator replicate '*'" continously on both nodes? Or is it enough on only one node? Or should I run "doveadm sync -A tcp:anotherhost.example.com" in regular intervalls? Perhaps once a day on both nodes? Thanks Urban Loesch

...egories. Therefore, the Goodman method seems like a good idea. I have found the R function "MultinomCI" {DescTools} to calculate these CIs. However, I have not found a way to calculate in advance the required sample size for the Goodman method at a fixed confidence level and length of the intervall. I would be very grateful for any suggestion. Many thanks in advance! Claudia [[alternative HTML version deleted]]

Hi all. A simple question. Is there a function to compute the Odds Ratio and its confidence intervall, from a logistic model (glm(.......,family=binomial....). I've written my own, but certainly someone did a better job. Thank you in advance, Stefano *********************************************** Stefano Calza Istituto di Statistica Medica e Biometria Universit? degli Studi di Milano Via G...

Hi I'm using the "image" function to produce a plot and I want to define the breaks using "cut" and the colors using "heat.colors". > image(interp(mat2[,2],mat2[,1],mat2[,3]),breaks=cut(mat2[,3],30),col=heat.colors(29)) Error in image.default(interp(mat2[, 2], mat2[, 1], mat2[, 3]), breaks = cut(mat2[, : must have one more break than colour The

Hi, I am using the reldist package and having problems determining the p-value for the entropy value from the reldist function. I am able to properly determine the entropy value, but cannot figure out what function to use to find the p-value. I have tried using rpy, rpluy (which provides p-values for the polarization values) and investing the results from reldist(). Thus, far I cannot find the

...have patience with me. :clap: I am trying to evalute the "internal response" for a couple of different cluster methods with the help of the AdjustedRandIndex, which is included in the mclust package. However, I do get a bit puzzled when I get a negative value as the value should be in intervall of [0,1], am I correct? Have I done something wrong? Or should I interpret the negative value as the two methods do not group the cases to the same cluster? I have for example used k-means and k-median, and choosen five clusters. Each respondent has been labelled as either beloning to cluster 1,2...

...the coefficient of a quantile regression granger-causes Y in a quantile range. I have nearly computed everything but I am searching for an estimator of the density of the distribution at several points of the distribution. As the quantreg-package of Roger Koenker is also able to compute confidence intervalls for quantile regression (which also contain data concerning the estimated density) I wanted to ask wether someone could tell me if it is possible to "extract" the density of the underlying distribution by using the quantreg package. I hope my question is not to confusing, thank you very,...

hello, i managed to get CI's for my curve - but now I need the intervall for a certain y point (y_tenth) of the curve.. can anyone help me with this? #####data: por<-data.frame(list(structure(list(run = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L), .Label = c("1", "3", "4"), class = "factor"), press = c(15,...

Hi I have a plist t which contains size measurements of individual plants, identified by the field "plate". It contains, among other, a field "year" indicating the year in which the individual was measured and the "height". The number of measurements range from 1 to 4 measurements in different years. My problem is that I would need the LAST measurement. I only

...7 variables: $ (0.3,0.325] :'data.frame': 29 obs. of 7 variables: $ (0.325,0.35]:'data.frame': 29 obs. of 7 variables: $ (0.35,0.375]:'data.frame': 17 obs. of 7 variables: And I would like to get a data.frame gathering the median value of each variable for each intervall. as.data.frame(t(sapply(rowsplit, mean)) works well but I would prefer to use median but with sapply(rowsplit, median), I get the following message "Error in median.default(X[[1L]], ...) : need numeric data". Any idea ? Thanks in advance for your help, Have a nice week-end, Ptit Bleu....