similar to: Don't understand removing constant on 1-way ANOVA

Hello, I would like to perform a multivariate regression analysis to model the relationship between m responses Y1, ... Ym and a single set of predictor variables X1, ..., Xr. Each response is assumed to follow its own regression model, and the error terms in each model can be correlated. Based on my readings of the R help archives and R documentation, the function lm() should be able to

Dear list members, I am trying to estimate parameters of the AR(1)-GARCH(1,1) model. I have one additional dummy variable for the AR(1) part. First I wanted to do it using garchFit function (everything would be then estimated in one step) however in the fGarch library I didn't find a way to include an additional variable. That would be the formula but, as said, I think it is impossible to add

Dear All, I'm trying to use the ols function in the Design library (version 2.1.1) of R to estimate parameters of a linear model, and then use the contrast function in the same library to test various contrasts. As a simple example, suppose I have three factors: feature (3 levels), group (2 levels), and patient (3 levels). Patient is coded as a non-unique identifier and is

Dear list, I'm currently trying to use the rms package to get predicted ordinal responses from a conditional ratio model. As you will see below, my model seems to fit well to the data, however, I'm having trouble getting predicted mean (or fitted) ordinal response values using the predict function. I have a feeling I'm missing something simple, however I haven't been able to

Hi all, I have a very easy questions (I hope). I had measure a property of plants, growing in three different substrates (A, B and C). The rest of the conditions remained constant. There was very high variation on the results. I want to do address, whether there is any difference in the response (my measurement) from substrate to substrate?

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

Petr, It works perfectly! But I still have a question; I have fit the following data; x,y,z 1,10,11 2,11,15 3,12,21 4,13,29 5,14,39 6,15,51 7,16,65 8,17,81 9,18,99 10,19,119 >dat.lm <- lm(z~I(x^2)+y, data=dat) >dat.lm Call: lm(formula = z ~ I(x^2) + y, data = dat) Coefficients: (Intercept) I(x^2) y 1.841e-14 1.000e+00 1.000e+00 How do I create the

Hi, I want to know what is there returned values of 'lm'. 'class' and 'lm' does not show that the returned value has the variable coefficients, etc. I am wondering what is the command to show the detailed information. If possible, I aslo want the lower level information. For example, I want to show that 'coefficients' is a named list and it has 2 elements.

Dear R users, how can I extrapolate values listed in the summary of an lm model but not directly available between object values such as the the standard errors of the calculated parameters? for example I got a model: mod <- lm(Crd ~ 1 + Week, data=data) and its summary: > summary(mod) Call: lm(formula = Crd ~ 1 + Week, data = data, model = TRUE, y = TRUE) Residuals: Min

Hi, I thought that 'coefficients' is a named list, but I can not refer to its element by something like r$coefficients$y. I used str() to check r. It says the following. Can somebody let me know what it means? ..- attr(*, "names")= chr [1:2] "(Intercept)" "y" $ Rscript lm.R > x=1:10 > y=1:10 > r=lm(x~y) > class(r) [1] "lm" >

Dear all, The following is a example that I run and hope to get a linear model. However, I find the lm() can not give correct coefficients for the linear model. I hope it's just my own mistake. Please help. TIA. Regards, Jinsong > x [1] 3.760216 3.997288 3.208872 3.985417 3.265704 3.497505 2.923540 3.193937 [9] 3.102787 3.419574 3.169374 2.928510 3.153821 3.100385 3.768770 3.610583

In my post at https://stat.ethz.ch/pipermail/r-help/2011-October/292019.html I included an undefined term "ej". The problem code should be as follows. It seems like a simple linear programming problem, but for some reason my code is not finding the solution. obj <- c(rep(0,3),1) col1 <-c(1,0,0,1,0,0,1,-2.330078923,0) col2 <-c(0,1,0,0,1,0,1,-2.057855981,0) col3

> On Mar 5, 2018, at 2:27 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > David: > > I believe your response on SO is incorrect. This is a standard OFAT (one factor at a time) design, so that assuming additivity (no interactions), the effects of drugA and drugB can be determined via the model you rejected: >> three groups, no drugA/no drugB, yes drugA/no drugB,

But of course the whole point of additivity is to decompose the combined effect as the sum of individual effects. "Mislead" is a subjective judgment, so no comment. The explanation I provided is standard. I used it for decades when I taught in industry. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into

Hi everybody, I'm trying to fit a Garch(1,1) process to the DAX returns. My data consists of about 2300 10day-logreturns in chronologically descending order (see attachment). But if I use the garch function I get a very high alpha_1 and a quite low beta, which doesn't make that much sense. I think I am missing something, but have no idea what it might be. I'd appreciate it a lot

> On Mar 5, 2018, at 3:04 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > But of course the whole point of additivity is to decompose the combined effect as the sum of individual effects. Agreed. Furthermore your encoding of the treatment assignments has the advantage that the default treatment contrast for A+B will have a statistical estimate associated with it. That was a

I've recently come across the following results reported from the lm() function when applied to a particular type of admittedly difficult data. When working with small data sets (for instance 3 points) with the same response for different predicting variable, the resulting slope estimate is a reasonable approximation of the expected 0.0, but the p-value of that slope estimate is a surprising

I'd like to model the relationship between m responses Y1, ..., Ym and a single set of predictor variables X1, ..., Xr. Each response is assumed to follow its own regression model, and the error terms in each model can be correlated. My understanding is that although lm() handles vector Y's on the left-hand side of the model formula, it really just fits m separate lm models. What should

Dear helpers, I have a 2x2 mixed design with two "groups" (between-subjects) and two presentation-types (within-subjects). The difference between groups is in the order of manipulations: group.CD having first a block of present.type.C and then a block of present.type.D, each block containing 31 trials. group.DC having first a block of present.type.D and then a block of present.type.C,

Hello, In a book (David W. Stockburger, "Multivariate Statistics: Concepts, Models, and Applications", chapter 12 "Contrasts, Special and Otherwise", available online at http://www.psychstat.smsu.edu/multibook) I've found some examples of doing analysis of variance on a contrast basis. I attach my solution (in R, the book uses SPSS) to this problem. Am I computing the