similar to: Linear regression equation and coefficient matrix

Hello everyone, I have estimated system of three linear equations with one non-linear restrictions with nlsystemfit. I was wondering how I can calculate the R-squared (or some alternative coefficient of determination) for the whole system. This is automatically given by linear systemfit but not by nlsystemfit. I can get the values for each of the equations separately, but apparently not for

Readers, I am having difficulty understanding how to enter commands into r. I have data arranged as: 100, 200, 300 5.6, 6.7, 7.8 8.9, 9.0, 0.1 1.2, 2.3, 3.4 The data is saved in csv format and I use the command 'read.table' to import into r. The values 5.6...3.4 are a function of values 100,...300, i.e. 100,...300 are independent variables (x). The function is a power series:

Hello R users, I have a substantial question about statistics, not about R itself, but I would love to have an answer from an R user, in form of an example in R syntax. I have spent whole Sunday searching in Google and browsing the books. I've been really close to the answer but there are at least three standard errors you can talk about in the linear regression and I'm really confused.

Hi, Is it possible to do a Matrix Plot and in the cell perform a linear regression also adding to the cell the r2 and the equation. If so, how? Thanks, As hz -- View this message in context: http://r.789695.n4.nabble.com/Matrix-Plot-and-linear-regression-tp2321613p2321613.html Sent from the R help mailing list archive at Nabble.com.

Hi, Based on some modification that I did to the R Cookbook Graphs Scatterplots code, link:http://wiki.stdout.org/rcookbook/Graphs/Scatterplots%20(ggplot2) I have some questions and I will appreciate a help: - How do I change the legend title? - How can I change the for each linear regression its color and linetype? - How can I add for both the linear regression lines their equations and Rseq

Hi all, I've a linear equation of the form: 0.95=2 ( [3+ln(x/3)]^-13  + 4 [3+2ln(x/3)]^-13  + [3+4ln(x/3)]^-13 ) and I want to solve it for x, can I do this using R? Thanks in advance. Maram. [[alternative HTML version deleted]]

Sorry, I didn't mean a linear equation, of course this equation is not linear, I meant an equation in one unknown. Hi all, I've a linear equation of the form: 0.95=2 ( [3+ln(x/3)]^-13? + 4 [3+2ln(x/3)]^-13? + [3+4ln(x/3)]^-13 ) and I want to solve it for x, can I do this using R? Thanks in advance. Maram. ? ? ? ??? [[alternative HTML version deleted]] -------------- next

Hi I have a very big data.frame : str(fslu12) 'data.frame': 277200 obs. of 11 variables: $ V1 : num 304 304 304 304 304 ... $ V2 : num 300 300 300 300 300 300 300 300 300 300 ... $ V3 : num 10 10 10 10 10 10 10 10 10 10 ... $ V4 : num 79.1 52 30.6 37.3 63 79.1 52 30.6 37.3 63 ... $ V5 : num 0.005 0.00498 0.00496 0.00491 0.00484 ... $ V6 : num 0.01 0.00997 0.00988 0.00969

Hello all! I have a problem with a double exponential equation. this are my data's> structure(list(proc = c(1870.52067384719, 766.789388745793, 358.701545859122, 237.113777545511, 43.2726259059654, 148.985133316262, 92.6242882655781, 88.4521557193262, 56.6404686159112, 27.0374477259404, 34.3347291080268, 18.3226992991316, 15.2196612445747, 5.31600719692165, 16.7015717397302,

Fellow R Users: I'm not extremely familiar with lda or R programming, but a recent editorial review of a manuscript submission has prompted a crash cousre. I am on this forum hoping I could solicit some much needed advice for deriving a classification equation. I have used three basic measurements in lda to predict two groups: male and female. I have a working model, low Wilk's lambda,

Colleagues, I am trying to learn to use uniroot to solve a simple linear equation. I define the function, prove the function and a call to the function works. When I try to use uniroot to solve the equation I get an error message, Error in yfu n(x,10,20) : object 'x' not found. I hope someone can tell we how I can fix the problem

On 06/10/2020 11:00 a.m., Sorkin, John wrote: > Colleagues, > I am trying to learn to use uniroot to solve a simple linear equation. I define the function, prove the function and a call to the function works. When I try to use uniroot to solve the equation I get an error message, > Error in yfu n(x,10,20) : object 'x' not found. > > I hope someone can tell we how I can fix

Dear all, I generated a Correlograms and used the panel.ellipse (confidence ellipse and smoothed line) option. Is there a way to get instead of the smoothed line the linear regression? Thanks, As hz -- View this message in context: http://r.789695.n4.nabble.com/Correlograms-and-linear-regression-tp2331071p2331071.html Sent from the R help mailing list archive at Nabble.com.

Hi, I have an excel sheet (already imported to R) with multiple columns and I am looking for a way in R that will allow me to generate a plot for every possible pair and its linear regression line/data. Any tip/idea/script how do to so. Thanks, As hz -- View this message in context: http://r.789695.n4.nabble.com/Automated-plot-and-linear-regression-line-data-tp2328027p2328027.html Sent from

Hi - I am confronting a situation where I have a set of structural equation and one or two of my responses are multinomial. I understand that sem would not deal with the unordered response. So I am thinking of the following two ways: 1. Expanding my response to a new set of binary variables corresponding to each label of my multinomial response. Then use each of these as a separate response in my

Dear users, I am trying to show the equation (including coefficients from the model estimates) for a gam model but do not understand how to. Slide 7 from one of the authors presentations (gam-theory.pdf URL: http://people.bath.ac.uk/sw283/mgcv/) shows a general equation log{E(yi )} = ?+ ?xi + f (zi ) . What I would like to do is put my model coefficients and present the equation used. I am an

Thanks to Gabor, Duncan, and Peter. I knew the answer had something to do with solving for a and b in terms of mean and variance. I will build a function using the equations you provided Duncan and will look into using Mathomatic in the future Gabor. Appreciate the help. Peter, this was not homework but I understand your concern. I don't use listserves that often but they do open a whole

Hi So, I have a dataset and I'm trying to solve for a parameter in an equation using the dataset. Before applying more sophisticated statistical techniques, I want to tell R to solve it out for each observation. I know I have to use a loop for it (and I have done that before, but am a bit rusty), but before I do that, I'm trying to get R to solve it for just one observation, to make sure

I have a dataset X Y1 1200 1.375 4000 0.464 1333.33 0.148 444.44 0.047 148.148 0.014 49.383 0.005 16.461 0.004 I have to find a curve fit for the above dataset based on a 4-parameter logistic equation viz. Y1 = d + ((a-d)/(1+(X/cc)^b)), where X and Y1 are the values above. I need to know how to solve the above equation for values a, b, c, d. -- View this message in context:

Dear Helplist, I want to produce forecasts from a structural equation model. With the SEM package the model setup and its estimation is possible. However, I have not figured out how to obtain ex ante forecasts, i.e. applying the Gauss-Seidel algorithm to the estimated structural equations for provided values of the exogenous variables (i.e.: y_t = -inv(A)*B*x_t). Does anyone know if the there is