search for: interceptable

Hi there, I trying to solve this problem for the whole day not going anywhere, so I really hope maybe somebody can help me in this community... I've got an object coefficient2 which I want to plot in differerent ways, with colors and labels added to the points, but somehow there seems to be a problem if a value is NA within the independent variable, resulting in false labels and false colors

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 >

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 I am using R 2.2.0 under SuSE 10 I want to use lm() to get the slope and intercept for several daatasets and store them in a database. So far so good - but how do I extract the slope and the intercept from the result from lm()? my code looks like this: lmNNDens <- lm(log(DensNN$MeanNN) ~ log(DensNN$MeanDensity)) anovaLM <- anova(lmNNDens) Results$slope[No] <- ???lmNNDens???

Hello, I am trying to fit my Elisa results (absorbance readings) to a standard curve. To create the standard curve model, I performed a 4-parameter logistic fit using the 'drc' package (ExpectedConc~Absorbance). This gave me the following: > FourP A 'drc' model. Call: drm(formula = Response ~ Expected, data = SC, fct = LL.4()) Coefficients: b:(Intercept) c:(Intercept)

Hi all, I hope this question is not too trivial. I can't find an explanation anywhere (Stats and R books, R-archives) so now I have to turn to the R-list. Question: If you have a factorial design with two factors (say A and B with two levels each). What does the intercept coefficient with treatment.contrasts represent?? Here is an example without interaction where A has two levels A1 and

Hi, I am not a statistics expert, so I have this question. A linear model gives me the following summary: Call: lm(formula = N ~ N_alt) Residuals: Min 1Q Median 3Q Max -110.30 -35.80 -22.77 38.07 122.76 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 13.5177 229.0764 0.059 0.9535 N_alt 0.2832 0.1501 1.886 0.0739

Dear List, I've just tried to specify a GAM without an intercept -- I've got one of the (rare) cases where it is appropriate for E(y) -> 0 as X ->0. Naively running a GAM with the "-1" appended to the formula and the calling "predict.gam", I see that the model isn't behaving as expected. I don't understand why this would be. Google turns up this old

Hi, I know that +1 is used to specify an intercept in a R model. An example of this would be: y~x+1 If I want to have a model where the log of the intercept is taken, the equation y~x+log(1) will not take the log of the intercept. Any suggestions on how to take the log of the intercept will be appreciated. Thanks [[alternative HTML version deleted]]

Below is a SIMEX object that was generated with the "simex" function from the "simex" package applied to a logistic regression fit. From this mountain of information I would like to extract all of the values summarized in this line: .. ..$ variance.jackknife: num [1:5, 1:4] 1.684 1.144 0.85 0.624 0.519 ... Can someone suggest how to go about doing this? I can extract the

I want to do a logistic regression without an intercept term. This option is absent from glm, though present in some of the inner functions glm uses. I gather glm is the standard way to do logistic regression in R. Hoping it would be passed in, I said > r <- glm(brain.cancer~epilepsy+other.cancer, c3, > family=binomial(link="logit"), intercept=FALSE) which produced

Dear all, I have the following output generated by linear regression. Since there is only one regression intercept and one slope for one set of data, what is the meaning of std. error for intercept and that of slope? Thanks in advance. Sincerely, Minghua > data(thuesen) > attach(thuesen) > lm(short.velocity~blood.glucose) Call: lm(formula = short.velocity ~ blood.glucose)

Hi everybody, I am trying to fit a GAM model without intercept using library mgcv. However, the result has nothing to do with the observed data. In fact the predicted points are far from the predicted points obtained from the model with intercept. For example: #First I generate some simulated data: library(mgcv) x<-seq(0,10,length=100) y<-x^2+rnorm(100) #then I fit a gam model with

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

Dear r-helpers, I can run the examples on the mcsamp help page. For example: **************************************** > M1 <- lmer (y1 ~ x + (1|group)) > (M1.sim <- mcsamp (M1)) fit using lmer, 3 chains, each with 1000 iterations (first 500 discarded) n.sims = 1500 iterations saved mean sd 2.5% 25% 50% 75% 97.5% Rhat n.eff beta.(Intercept)

Dear all, is there anyone who can help me extracting p-values from an Anova object from the car library? I can't seem to locate the p-values using str(result) or str(summary(result)) in the example below > A <- factor( rep(1:2,each=3) ) > B <- factor( rep(1:3,times=2) ) > idata <- data.frame(A,B) > fit <- lm( cbind(a1_b1,a1_b2,a1_b3,a2_b1,a2_b2,a2_b3) ? sex,

Hi! I'm failing to understand the value of the intercept value in a multiple linear regression with categorical values. Taking the "warpbreaks" data set as an example, when I do: > lm(breaks ~ wool, data=warpbreaks) Call: lm(formula = breaks ~ wool, data = warpbreaks) Coefficients: (Intercept) woolB 31.037 -5.778 I'm able to understand that the value of

Hi, I've a list containing parameters (intercepts & coefficients) of 12 regressions fitted > coeff [[1]] (Intercept) anno -427017.1740 217.0588 [[2]] (Intercept) anno -39625.82146 21.78025 ..... [[12]] (Intercept) anno 257605.0343 -129.7646 I want create a data frame with two columns (intercept and anno)using data in these list. Any help

Hi I have the following code which I would like to simplify. Id does linear regressions and returns the r-squares, and the coefficients. It runs slow, as it is doing the regressions for each - is it possible to get the values in a dataframe which looks as follow: expert | xx | seeds | r.squared | slope | intercept Thanks in advance, Rainer library(reshape) rsqs <- as.data.frame(

Hello R experts I have two linear regressions for sexes (Male, Female, Unknown). All have a good correlation between body length (response variable) and head length (explanatory variable). I know it is not recommended, but for a good practical reason (the purpose of study is to find a single conversion factor from head length to body length), the regressions need to go through the origin (0