similar to: simple fit

Hi Mark, I don't know wether you recived a sufficient reply or not, so here are my comments to your problem. Supressing the constant term in a regression model will probably lead to a violation of the classical assumptions for this model. From the OLS normal equations (in matrix notation) (1) (X'X)b=X'y and the definition of the OLS residuals (2) e = y-Xb you get - by

Hi, I am trying to draw a perpendicular line from a point to two points. Mathematically I know how to do it, but to program it, I encounter some problem and hope can get help. Thanks. I have points, A, B and C. I calculate the slope and intercept for line drawn between A and B. I am trying to check whether I can draw a perpendicular line from C to line AB and get the x,y value for the point D

Just to inform: I posted that before in R-sig-ecology but as it might be interesting also for other useRs, I post it also to the general r-user list: Hello Alexandre, thank you very much. I also found another way to extract summarizing information from lm results over e.g. 1000 repeated model runs: results2 <- t(as.data.frame(results)) summary(results2) Although some questions popped up in

Hello. Consider the response-variable of data.frame df is constant, so analytically perfect fit of a linear model is expected. Fitting a regression line using lm result in residuals, slope and std.errors not exactly zero, which is acceptable in some way, but errorneous. But if you use summary.lm it shows inacceptable error propagation in the calculation of the t value and the corresponding

Hi group, I have this type of data x(predictor), y(response), factor (grouping x into many groups, with 6-20 obs/group) I want to fit a linear regression with one common intercept. 'factor' should only modify the slopes, not the intercept. The intercept is expected to be >0. If I use y~ x + factor, I get a different intercept for each factor level, but one slope only if I use y~ x *

Hi *, is there a way to obtain some kind of p-value for a model fitted with RMA using the lmodel2 package? I know that p-values are discussed and criticized a lot and as you can image from my question I'm not very much of a statistican (only writing my bachelor thesis). As fare as I understood the confidence interval statistic correctly, a coefficient is regarded as statistically significant

I want to fit a linear model with fixed slope e.g. y = x + b (instead of general: y = a*x + b) Is it possible to do with lm()? Regards, Ryszard -------------------------------------------------------------------------- Confidentiality Notice: This message is private and may ...{{dropped:11}}

Hi R-Help! My question: I have lifetime/failure data of machines with different stress levels and i think an weibull/extreme value distribution would fit this data. So I did: model1 <- survreg(Surv(lfailure)~stress,data=steel,dist="extreme") (where lfailure=log(failure)) Now I would like to do a likelihood ratio test to test the hypothesis H0: location parameters of the

Hi R users I'm trying to fit a model y=ax^b. I know if I made ln(y)=ln(a)+bln(x) this is a linear regression. But I obtain differente results with nls() and lm() My commands are: nls(CV ~a*Est^b, data=limiares, start =list(a=100,b=0), trace = TRUE) for nonlinear regression and : lm(ln_CV~ln_Est, data=limiares) for linear regression Nonlinear

Dear R-help, Here's a simple example of nonlinear curve fitting where nls seems to get the answer wrong on a very simple exponential fit (my R version 2.7.2). Look at this code below for a very basic curve fit using nls to fit to (a) a logarithmic and (b) an exponential curve. I did the fits using self-start functions and I compared the results with a more simple fit using a straight lm()

Hello, I am working on a simple non-parametric (Theil) regression function and and am following Hollander and Wolfe 1999 text. I would like some help making my function faster. I have compared with pre-packaged version from "MBLM", which isnt very fast either, but it appears mine is faster with N = 1000 (see results below). I plan on running this function repeatedly, and I generally

I am not certain how nlme works so I followed an example from the web ( http://www.menne-biomed.de/gastempt/gastempt1.html). I was able to successfully reproduce the example. However, when I modified my the example to use my data and with my formula, I get a set of errors having to do with the log() function. I get 10 of them (all exactly the same) and there are 10 levels in my factor variable.

Dear Bert, Thank you so much for your kind and valuable feedback. I tried finding the starting values using the approach you mentioned, then did the following to fit the nonlinear regression model: nlregmod2 <- nls(y ~ theta1 - theta2*exp(-theta3*x), start = list(theta1 = 0.37, theta2 = exp(-1.8), theta3 =

Dear List: I am having difficulties with the fitted values at different levels of a multilevel model. My data set is a series of student test scores over time with a total of 7,280 observations, 1,720 students nested witin 60 schools. The data set is not balanced. The model was fit using eg.model.1<-lme(math~year, random=~year|schoolid/childid, data=single). When I call the random

Hello, I would like to fit data with the following formula : y=V*(1+alpha*(x-25)) where y and x are my data, V is a constant and alpha is the slope I'm looking for. How to translate this into R-language ? At the moment, I only know : lm(y ~ x) Sorry for such a basic question. I thought I could find the solution in a post but I have to confess that, up to know, I'm not able to understand

Hi Folks, I'm seeking confirmation of something which is probably true but which I have not managed to find in the documentation. I have a binary response y={0.1} and a variable x and have fitted a probit response to the data with f <- glm( y~x, family=binomial(link=probit) ) and then, with a specified set of x-value X I have used the predict.glm function as p <- predict( f, X,

Oh, sorry; I changed signs in the model, fitting theta0 + theta1*exp(theta2*x) So for theta0 - theta1*exp(-theta2*x) use theta1= -.exp(-1.8) and theta2 = +.055 as starting values. -- Bert On Sun, Aug 20, 2023 at 11:50?AM Paul Bernal <paulbernal07 at gmail.com> wrote: > Dear Bert, > > Thank you so much for your kind and valuable feedback. I tried finding the > starting

We need help.... We are doing a project for a statistical class in and we are looking at world record times in different running events over time. We are trying to fit the data with a negative exponential but we just cant seem to get a function that works properly. we have on our x-axis the date and on the y-axis the time(in seconds). So as you can imagine, the times have decreased and appear to

Hello, I have a simple xyplot with rlm lines. I would like to add the a and b coefficients (y=ax+b) of the rlm calculation in each panel. I know I can do it 'outside' the xyplot command but I would like to do all at the same time. I found some posts with the same question, but no answer. Is it impossible ? Thanks in advance for your help. Ptit Bleu. x11(15,12) xyplot(df1$col2 ~

Dear Bert, Thank you for your extremely valuable feedback. Now, I just want to understand why the signs for those starting values, given the following: > #Fiting intermediate model to get starting values > intermediatemod <- lm(log(y - .37) ~ x, data=mod14data2_random) > summary(intermediatemod) Call: lm(formula = log(y - 0.37) ~ x, data = mod14data2_random) Residuals: Min