search for: 0.025

Dear R help, I have some readings in three dimensions (x, y, z) and an amplitude for each. I'd like to visualize the data using ggplot, using tile plots, as I have some additional point data I would like to eventually overlay on the tile plots. I would like to subset the data by sections, slices if you will, in the z dimension, and plot the data for that slice. I can do all of this, but am

Hello, All: I am unable to manually replicate predict.lm, specifically comparing se.fit with (fit[,3]-fit[,2]): I think their ratio should be 2*qnorm((1-level)/2), and that's not what I'm getting. Consider the following slight modification of the first example in help('predict.lm'): set.seed(1) x <- rnorm(15) y <- x + rnorm(15) predict(lm(y ~ x)) new <-

Doh! Thanks very much. sg On 11/6/23 5:17 PM, John Fox wrote: > Dear Spencer, > > You need the t distribution with correct df, not the standard-normal > distribution: > > > pt(-z.confInt/2, df=13) > ??? 1???? 2???? 3???? 4???? 5???? 6???? 7???? 8???? 9??? 10??? 11 > 0.025 0.025 0.025 0.025 0.025 0.025 0.025 0.025 0.025 0.025 0.025 > ?? 12??? 13 > 0.025

I have a file: x y z 0.025 0.025 1.65775 0.025 0.050 1.62602 0.025 0.075 1.63683 0.025 0.100 1.91847 0.025 0.125 2.00913 0.025 0.150 1.82222 0.025 0.175 1.70901 0.025 0.200 1.39759 0.025 0.225 1.39089 0.025 0.250 1.04762 If I read the file like this: data<-read.table("file.dat") How do I access the vectors x,y,z that are inside the dataframe data? I studied Venables and

Hi guys, I got a problem when I was trying to use lattice to do some plot. Below is one working example which can generate one curve for A and one curve for B in each subplot. However, I would like to just show the points for B, not connecting the dots. As for A, I still want a curve (dots are connected). Is there any people having any suggestions. Data are attached. Thanks a lot, --Jerry

I am trying to use t.test on the following data: date type INTERVAL nCASES MTF SDF MTO SDO nFST MF nOBS MO MB BIASCV BIASEV ME MAE RMSE CRCF 2001-06-15 avn GE1.00 4385 0.246 0.300 1.502 0.556 1367 1.373 4385 1.502 1.471 0.285 0.164 -1.256 1.266 1.399 0.056 2001-06-15 avn

Hi, I want to compute the quantiles of Chi^2 distributions with different degrees of freedom like x<-cbind(0.005, 0.010, 0.025, 0.05, 0.1, 0.5, 0.9, 0.95, 0.975, 0.99, 0.995) df<-rbind(1:100) m<-qchisq(x,df) and hoped to get back a length(df) times length(x) matrix with the quantiles. Since this does not work, I use x<-c(0.005, 0.010, 0.025, 0.05, 0.1, 0.5, 0.9, 0.95, 0.975,

A simple question - using the following fishers test it appears that the P value is significant, but the CI includes 1. Is this result correct? > data.50p10min <- matrix(c(16,15, 8, 24),nrow=2) > fisher.test(data.50p10min) Fisher's Exact Test for Count Data data: data.50p10min p-value = 0.03941 alternative hypothesis: true odds ratio is not equal to 1 95

Good morning, suppose the following: m <- round(matrix(rnorm(16), ncol=4), 3) a <- rev(c(0.01, 0.025, 0.05, 0.1)) rownames(m) <- a colnames(m) <- c("0.25,0.75", "0.4,0.6", "0.1,0.9", "0.4,0.9") m 0.25,0.75 0.4,0.6 0.1,0.9 0.4,0.9 0.1 1.034 -0.119 -1.213 0.619 0.05 0.035 1.074 0.525 1.671 0.025 -1.687 0.960

Hello everyone, With th bar histogram (number of occurrences) hist<-c(24,7,4,1,2,1,1) of seven equally spaces classes ]1-4], ]5-8], ]9-12], ]13-16], ]17-20], ]21-24], ]25-28], I obtained shape=0.8276 and rate=0.1448. I would like to know how to build the continuous pdf of a this gamma distribution knowing these two estimated parameters such that I will be able to predict the pdf of any

Hi, Please see the email exchanges below. I am having trouble generating output that is large enough for our needs, specifically when using the GaussRF function. However, when I wrote Dr. Schlather (the author of the GaussRF function), he indicated that there is also a limit imposed by R itself. Is this something that we can overcome? Thank you very much for any assistance you may provde.

What's the current best wisdom on how to construct confidence bounds on something like a normal probability plot? I recall having read a suggestion to Monte Carlo something like 201 simulated lines with the same number of points, then sort the order statistics, and plot the 6th and 196th of these. [I use 201 not 200 because quantile(1:201, c(0.025, 0.975)) = 6 and 196 while

Hello, I hesitate to call this a bug, because I could have forgotten something important, but the MarkowitzPortfolio example in fPortfolio does not work for me. Here's my code: > library(fPortfolio) > >xmpPortfolio("\nStart: Load monthly data set of returns > ") > data(berndtInvest) > # Exclude Date, Market and Interest Rate columns from data

OK, Let's try this again! Here is the reproducible script; it is long because I had to copy the panel dataset here. My question is related to systemfit; I don't know how to get the result for the entire panel. #Reproducible script Empdata<- read.csv("/Users/ngwinuiazenui/Documents/UPLOADemp.csv") View(Empdata) install.packages("systemfit")

I have a data.frame with three columns + the row number Sample MidPoint Count [1,] 1 0.025 1 [2,] 1 0.075 3 [3,] 1 0.125 15 [4,] 1 0.175 12 [5,] 1 0.225 5 [6,] 1 0.275 8 When I export the data to a csv the column names are shifted because the row numbers are being exported as well. Sample MidPoint

Sadly you failed to set your email program to send plain text and the data is corrupted at my end. I also think you need to reduce the size of the data set... the intent here is to increase your understanding, not debug your particular analysis. I will say that I am having a very challenging time understanding what you are trying to accomplish though. What are the equations that you think need

Hi everyone, I'm working on a script trying to use foreach %dopar% but without success, so I manage to run the code with foreach %do% and looks like this: The code is part of a MCMC model for projects valuation, returning the most important results (VPN, TIR, EVA, etc.) of the simulation. foreach (simx = NsimT, .combine=cbind, .inorder=FALSE, .verbose=TRUE) %do% { MCPVMPA = MCVAMPA[simx]

Hi, I have a question about the computation of confidence intervals in the zyp package, in particular using the functions zyp.sen and confint.zyp, or zyp.yuepilon. (1) I'm a bit confused about the confidence intervals given by zyp.sen and confint.zyp. When I request a certain confidence interval in the function, the R output seems to deliver another confidence interval, e.g. when I set

Hi All, I have a dataframe, divs, that looks like this: > divs date ticker dividend 263 20010322 ADBE 0.025 264 20010628 ADBE 0.025 265 20010927 ADBE 0.025 4308 20010212 ED 0.550 4309 20010514 ED 0.410 5416 20010330 GE 0.137 5417 20010629 GE 0.137 5418 20010928

Hello all, Consider the following problem: There is a matrix of probabilities: > set.seed(1) > probs <- array(abs(rnorm(25, sd = 0.33)), dim = c(5,5), dimnames = list(1:5, letters[1:5])) > probs a b c d e 1 0.21 0.27 0.50 0.0148 0.303 2 0.06 0.16 0.13 0.0053 0.258 3 0.28 0.24 0.21 0.3115 0.025 4 0.53 0.19 0.73 0.2710 0.656 5 0.11 0.10 0.37 0.1960