similar to: linear regression and testing the slope

Dear R users, I would like to draw some histograms as seen in the page whose address I wrote below. I searched through the web a lot and I found a page which describes how I can do it for older versions of R. For newer versions they recommend to install the package R.basics in R.clusters but this does not exist. The address of the web page is http://www1.maths.lth.se/help/R/plot.histogram/

Dear R users, I am a master student in Mathematics and I am writing my thesis in statistics. I need to use R and unfortunately I do not have any experience with a computer program. Could you please help me about chi squared goodness of fit test with R? In R-help website I saw a message about how to do that but I do not know how to cut the data into bins and calculate the expected numbers in each

Dear R users, First of all, thank you for your help about goodness of fit tests. I have another set of data and i am sure it is incomplete. I searched for incomplete data analysis with R but I could not find any suggestion, or method. Lets say (totally hypothetical situation) from a distribution you can generate a continuous random sample between the numbers 0 and 100. but I have data from 20 to

Hi All, I am interested in aggregating a data frame based on 2 categories--mean effect size (r) for each 'id's' 'mod1'. The 'with' function works well when aggregating on one category (e.g., based on 'id' below) but doesnt work if I try 2 categories. How can this be accomplished? # sample data id<-c(1,1,1,rep(4:12)) n<-c(10,20,13,22,28,12,12,36,19,12,

I've created a short program to print a table of learning curve factors. However, I cannot figure out how to format the table to: 1) Get rid of the [1]s in the first column and replace it with the values of N. 2) Line up the first row with the factors (decimal fractions). Thanks for any help. The complete program and output is as follows: > Lc<-seq(0.70,0.95,0.05) #Specify learning

Dear r-helpers, xx <- c(0.000, 0.210, 0.714, 0.514, 1.000, 0.190, 0.590, 0.152) yy <- c(0.000, 0.265, 0.256, 0.521, 0.538, 0.761, 0.821, 1.000) aa <- c(19, 19, 19, 21, 19, 21, 21, 21) x0 <- xx[c(1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 5, 6, 6, 7)] y0 <- yy[c(1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 5, 6, 6, 7)] x1 <- xx[c(2, 3, 3, 4, 6, 4, 5, 5, 6, 7, 7, 7, 8, 8)] y1 <- yy[c(2, 3, 3, 4, 6, 4, 5,

Dear all R-users, I wanted to fit a Garch(1,1) model to a dataset by: >garch1 = garch(na.omit(dat)) But I got a warning message while executing, which is: >Warning message: >NaNs produced in: sqrt(pred$e) The garch parameters that I got are: > garch1 Call: garch(x = na.omit(dat)) Coefficient(s): a0 a1 b1 1.212e-04 1.001e+00 1.111e-14 Can any one

Hi, I'm regressing various body dimensions upon body mass using the 'lmodel2' function, as I'm keen to obtain both OLS and RMA slope values. I also wish to create a plot of the regressions, with the 95% confidence interval of both the slope and intercept. I know how to plot 95% ci bands of the OLS slope using lm with the 'predict' function and 'matlines'. Does

Greetings, I would like to use the "boot" function to generate a bootstrap confidence interval for the slope in a SLR that has a zero intercept. My attempt to do this is shown below. Is this the correct implementation of the boot function to solve this problem? In particular, should I be doing anything with the residuals in the "bs" function (e.g., using weighted residuals)?

Dear R users, Using R lme function, I found that both fixed and random effects of variable A on variable B are significant. Now, I'd like to analyze what variables are predicting differences in the slope. In other words, I'd like to know what variables (e.g., variable C) are predicting individual differences in the effects of A on B. I have many data points for A and B for each

Dear All, I fear that this is a really easy question but I do seem to go around in circles.. I have 2 points on a plot and would like to calculate the slope of the line drawn through these 2 points. that cant be so hard?! Thank you in advance, Katrin -- Katrin Fleischer Vrije Universiteit Amsterdam Faculty of Earth and Life Sciences Subdepartment Hydrolgy and Geo-Environmental Sciences Room

library(ROCR) n <- 1000 fitglm <- function(iteration,intercept,sigma,tau,beta){ x <- rnorm(n,0,sigma) ystar <- intercept+beta*x z <- rbinom(n,1,plogis(ystar)) xerr <- x + rnorm(n,0,tau) model<-glm(z ~ xerr, family=binomial(logit)) *int*<-coef(model)[1] *slope*<-coef(model)[2] # when add error you are suppose to get slightly bias slope. However when I change

Hello. I want to fit a smoother spline (or an equivalent local regression method) to a series of data in which the initial value of the 1st derivative (slope) is constrained to a specific value. Is it possible to do this? If so, how? Bill Shipley [[alternative HTML version deleted]]

0down votefavorite <https://stats.stackexchange.com/questions/296569/how-to-obtain-random-slope-random-intercept-plots-for-categorical-response-varia#> I'm trying to generate a random slope random intercept plot after ordinal regression using the clmmfunction from the ordinal package in R. I have aggression levels which are categorical with six levels. Earlier, I made random intercept

Hi all I am not an experienced user of R. I am trying to perform DIF analysis using Lordif and I get the follow error: > GroupDIF <- lordif(resp.data=Resp, group=Group, criterion="R2", pseudo.R2="McFadden", R2.change=0.02) Iteration: 500, Log-Lik: -137340.437, Max-Change: 0.00119 EM cycles terminated after 500 iterations. (mirt) | Iteration: 1, 14 items flagged for DIF

Hello, I have a linear model with different slopes for different treatment groups. I need to pairwise compare the different slope estimates for the different treatment groups. Is there a package that does pairwise comparisons of slope coefficients, making the appropriate adjustments in the P values? Thanks, John. -- ========================================== John Janmaat Department of

Hello All I am currently looking at spatial data - Chorophyll A concentration in sea water over a wide geographic area. These data are used to determine the location of ocean fronts and hence where tuna are located. A front is identified by a steep gradient in the change in chloroA concentration. I have been looking at these data qualitatively using persp, contour, and image but would like to

hello i have a mixed effect model which gives slope and intercept terms for 6 groups (diagnosis (3 levels) by risk group(2 levels)). the fixed part of the model is -- brain volume ~ Diagnosis + Risk Group + (Risk Group * age : Diagnosis) - 1 thus allowing risk group age/slope terms to vary within diagnosis and omitting a nonsignificant diagnosis by risk group intercept (age was centered)

Dear R-Users, I am new to R, so please excuse the ignorance. I have data: x (2.14, 2.41, 1.09, 0.17, 8.18) y (3.81, 5.13, 0.63, 0.75, 6.35) I would like to use simple linear regression and test 2 things: 1) slope of line of best fit is statistically different from 1 2) y-intercept is statically different from 0 Could anyone provide me with the R terminology to do this? Thanks so much, Dan

I am learning ggplot2, and need your help. When I try > p <- ggplot(mtcars, aes(x = wt, y=mpg)) + geom_point() > p + geom_abline(slope=5) (from http://had.co.nz/ggplot2/geom_abline.html) the slope of the abline does not change, but this works: > p + geom_abline(intercept=20) In order to have slope work, I have to use > p + geom_abline(aes(slope=5)) Is it a bug, or is there