similar to: How to Write a Model in R that has the Log taken of the Intercept

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'm trying to create a linear model for a dataset that has a breakpoint e.g. # dummy dataset x <- 1:20 y <- c(1:10,seq(10.5,15,0.5)) plot(x,y) I've modelled this using the following formula: temp <- lm(y ~ x*(x<=10)+x*(x>10)) I want to be able to omit the intercept (i.e. force the line through zero) from the first of these segments (x<=10) so that I'm only

Hi, I have been using lm in R to do a linear regression and find the slope coefficients and value for R-squared. The R-squared value reported by R (R^2 = 0.9558) is very different than the R-squared value when I use the same equation in Exce (R^2 = 0.328). I manually computed R-squared and the Excel value is correct. I show my code for the determination of R^2 in R. When I do not set 0 as the

Not sure what you are trying to do here. The immediate issue is that you are getting 'y' on the RHS, because that is the 1st column in Dataset. So "for (i in 2:3)" might be closer to intention. However, a 0/1 regresson with no intercept implies that the mean for the "0" group is zero, and with two regressors that the mean is zero for the (0,0) group. Looking at the

I have two related variables, each with 16 points (x and Y). I am given variance and the y-intercept. I know how to create a regression line and find the residuals, but here is my problem. I have to make a loop that uses the seq() function, so that it changes the slope value of the y=mx + B equation ranging from 0-5 in increments of 0.01. The loop also needs to calculate the negative log

It's easy to run a linear regression on a simple model without an intercept just by doing this: lm(y ~ x1 + x2 -1) Is there a similar trick to suppress the intercept when your model is in a large dataframe and you don't want to write out the names of individual columns? -- View this message in context:

Hello, I am using the lm to fit a linear model to data, I was wondering if there is a way to display the equation on a plot using the extracted lm coefficients? I am using the plot() function to create the plot/.png. Example: lm_mod <- lm(data1~data2) intercept = 48.54 slope = 0.4856 I want to display equation on a plot as y=0.4856x+48.54 Any thoughts or suggestions are appreciated.

Hi all this is a silly question since i should know the answer. lm(y~x) perfroms linear regression with the intercept included. How do i estimate the equation without the intercept? cheers

Dear list members, this is not an R question and forgive me for using the list for irrelevant questions, but this is the only place I know where I can find some good statisticians and I need an expert opinion. There is this power law kinetic model of the form: M=kt^n where t is the time, M is the fraction of drug released, k is the rate constant and n is an exponent related to the mechanism of

Hi There, I have a set of data (xi,yi).I want to fit them with the equation y=mx. note: in the above equation, there is no intercept. I don't know how to use common software such as R , matlab, sas, or spss to do this kind of regression. Does anyone know how to do this? I know it is easy to use least square method to do this by programming. But I want to find if there exists some common

Dear list, I have a question regarding the meaning of intercept term in a two-way anova model without interaction term. for example (let's assume there is no interaction between factor1 and factor2) : > df         val        factor1 factor2 1  48.61533       A      t1 2 171.13535       B      t1 3  65.96884       C      t1 4  63.71222       A      t2 5  80.22049       B      t2 6 

Hi I have a follow up question, relating to subsetting to list items. After using the list and min(sapply()) method to adjust the length of the variables, I specify a dynamic regression equation using the variables in the list. My list looks like this: Dcr<- list(Dcre1=DCred1,Dcre2=DCred2,Dcre3=DCred3,Dbobc1=DBoBC1,Dbobc2=DBoBC2,Dbobc3=DBoBC3,...) By specifying the list items with names, I

Hi, I'm a new user of R and I'm trying to make a linear model from this kind of dataset x [1] 16.87 19.93 25.85 20.94 17.06 19.49 19.93 25.45 27.74 20.15 25.81 21.06 17.17 20.03 25.50 27.79 20.44 16.88 19.93 25.79 z<-x-10 y [1] 0.80 1.27 2.22 1.32 0.90 1.18 1.84 2.41 2.97 1.25 2.07 1.41 1.14 1.66 2.59 3.51 1.53 0.81 1.26 2.30 plot(x,y) I want to be able to force the line of

Hi, I'm trying to estimate the parameters of a state space model of the following form measurement eq: z_t = a + b*y_t + eps_t transition eq y_t+h = (I -exp(-hL))theta + exp(-hL)y_t+ eta_{t+h}. The problem is that the distribution of the innovations of the transition equation depend on the previous value of the state variable. To be exact: y_t|y_{t-1} ~N(mu, Q_t) where Q is a diagonal

Hi there, I am trying to use linear regression to solve the following equation - y <- c(0.2525, 0.3448, 0.2358, 0.3696, 0.2708, 0.1667, 0.2941, 0.2333, 0.1500, 0.3077, 0.3462, 0.1667, 0.2500, 0.3214, 0.1364) x2 <- c(0.368, 0.537, 0.379, 0.472, 0.401, 0.361, 0.644, 0.444, 0.440, 0.676, 0.679, 0.622, 0.450, 0.379, 0.620) x1 <- 1-x2 # equation lmFit <- lm(y ~ x1 + x2) lmFit Call:

Dear advanced statisticians, *******Objectif******** I try to set up linear models with mean as intercept: Answer: y Variable: x, as factor of two modalities: x(1), x(2). I would like to have a model as: y = mean(y)+A(i)+residuals, with i in (1,2) and A(1) coefficient for x(1) and A(2) coefficient for x(2). *******Trials in R******* ## Firstly: I write in R: >Model<-lm(y~x,Data)

Dear friends, I have written the following lines in R console wich already exist in pdf file systemfit: data( "GrunfeldGreene" ) library( "plm" ) GGPanel <- plm.data( GrunfeldGreene, c( "firm", "year" ) ) greeneSur <- systemfit( invest ~ value + capital, method = "SUR", + data = GGPanel ) greenSur I have obtained the following incomplete

Hallo, yesterday I was puzzled when I discovered that I probabliy miss something in the interepretation of intercept in two-way lm models. I thought that the intercept, using the default contr.treatment contrasts, represents the mean of the group of observations having zero in all column of the model.matrix. It turns out not to be case To be more more clear I am attaching a short example:

Dear R-list, I apologize for my many emails but I think I know how to desctribe my problem differently and more clearly. My question is how to compare linear models with intercept and those without intercept using maximizing adjusted R^2 strategy. Now I do it like the following: > library(leaps) > n=20 > x=matrix(rnorm(n*3),ncol=3) > b=c(1,2,0) > intercept=1 >

Dear R-Users, How can I put a pre-defined regression model into to an object of class lm in order to use the predict.lm function. A simplified example: I would normally run a regression analysis on a dataset, > germany<-lm(RENT~AGE1, in.mi01) > summary(germany) Call: lm(formula = RENT ~ AGE1, data = in.mi01) Residuals: Min 1Q Median 3Q Max