similar to: Intervalos de confianza de la varianza de los residuos en un modelo no lineal.-

??Gracias Javier,? 2017-04-01 12:07 GMT-03:00 <javier.ruben.marcuzzi en gmail.com>: > Mi duda es la siguiente, la varianza residual en su modelo, es homogénea o > heterogénea, ?La varianza es homogénea, común a todas las observaciones. Digamos que el modelo es el siguiente, y=f(x, betas)+e con f alguna función no lineal cuyos parámetros son betas y con e~N(0, sigma2). Uno suele

Hi! When checking validity of a model for a large number of experimental data I thought it to be interesting to check the information provided by the summary method programmatically. Still I could not find out which method to use to get to those data. Example (not my real world data, but to show the point): [BEGIN] > DNase1 <- subset(DNase, Run == 1) > fm1DNase1 <- nls(density ~

I'm wondering whether AIC scores extracted from nls() objects using AIC() are based on the correct number of estimated parameters. Using the example under nls() documentation: > data( DNase ) > DNase1 <- DNase[ DNase$Run == 1, ] > ## using a selfStart model > fm1DNase1 <- nls( density ~ SSlogis( log(conc), Asym, xmid, scal ), DNase1 ) Using AIC() function: >

Hi, I am trying to run the following nonlinear regression model. > nreg <- nls(y ~ exp(-b*x), data = mydf, start = list(b = 0), alg = "default", trace = TRUE) OUTPUT: 24619327 : 0 24593178 : 0.0001166910 24555219 : 0.0005019005 24521810 : 0.001341571 24500774 : 0.002705402 24490713 : 0.004401078 24486658 : 0.00607728 24485115 : 0.007484372

Hello! I work with : R : Copyright 2006, The R Foundation for Statistical Computing Version 2.3.1 (2006-06-01) On Windows XP Professional (Version 2002) SP2. At this moment I use the function "nls" combined with a selfStar model (SSmicmen, related to Michaelis-Menten equation, and provided by the "stats" package). When I realise the following operation (cf. p 59 of the

Hi all, I try to write a simple function in a script. The script is as follows yo<-function(Xdata) { n<-length(Xdata[,1]) Lgm<-nls(formula=LgmFormula, data=Xdata, start=list(a=1500,b=0.1),weights=Xdata$Qe) return(Lgm) } After the execution of the script, when I call the function yo on data called NC60.DATA I get an error. #yo(NC60.DATA) Erreur dans eval(expr, envir, enclos)

Dear all, I encounter some discrepancies when comparing the deviance of a weighted and unweigthed model with the AIC values. A general example (from 'nls'): DNase1 <- subset(DNase, Run == 1) fm1DNase1 <- nls(density ~ SSlogis(log(conc), Asym, xmid, scal), DNase1) This is the unweighted fit, in the code of 'nls' one can see that 'nls' generates a vector

Dear All, I encounter some discrepancies when comparing the deviance of a weighted and unweigthed model with the AIC values. A general example (from 'nls'): DNase1 <- subset(DNase, Run == 1) fm1DNase1 <- nls(density ~ SSlogis(log(conc), Asym, xmid, scal), DNase1) Now for a weighted fit: fm2DNase1 <- nls(density ~ SSlogis(log(conc), Asym, xmid, scal),

Agradezco la ayuda recibida me vino muy bien,,, el problema de instalar el paquete lubridate es que no está en los repositorios de Debian,,, al usar RStudio creo que se debe instalar en ~/R/x86_64-pc-linux-gnu-library previa compilación de fuente bajada de r-cran,,, cosa que será otro pedido de ayuda en el futuro si sale mal,,, Saludos Horacio El día 30 de diciembre de 2016, 7:05, Carlos Ortega

Thanks Adaikalavan, however the problem remains. Considering AIC() as applied to the linear model in AIC() help documentation: > data(swiss) > lm1 <- lm(Fertility ~ . , data = swiss) > AIC(lm1) [1] 326.0716 Clearly this includes the estimation of the residual standard error as an estimated parameter, as this gives the correct score: > -2*logLik(lm1) + 2*(length(coef(lm1))+1)

Dear all, I am studying a bit the various support functions that exist for extracting information from fitted model objects. From the help files it is not completely clear to me whether the number returned by nobs() should be the same as the "nobs" attribute of the object returned by logLik(). If so, then there is a slight inconsistency in the methods for 'nls' objects with

Dear list, I have encountered a problem with predict.nls (Windows XP, R.2.5.0), but I am not sure if it is a bug... On the nls man page, an example is: DNase1 <- subset(DNase, Run == 1) fm2DNase1 <- nls(density ~ 1/(1 + exp((xmid - log(conc))/scal)), data = DNase1, start = list(xmid = 0, scal = 1)) alg = "plinear", trace =

Buenas, esta pregunta es un poco elemental, pero estoy haciendo mis primeras experiencias en R. Yo tengo un CSV con el siguiente formato... fecha;Gastos;media móvil;Holt Winter 31/08/02;2498,5;;2498,5 29/09/02;2250,93;2320,63;2424,229 31/10/02;2212,46;2097,87;2360,6983 30/11/02;1830,22;2092,78;2201,55481 ,,,,,,, los guardo en un dataframe con gastos=read.csv2("indice_gastos.csv"),

Hello, How do I specify the formula and random effects without a startup object ? I thought it would be a mixture of nls and lme. after trying very hard, I ask for help on using nlme. Can someone hint me to some examples? I constructed a try using the example from nls: #variables are density, conc and Run #all works fine with nls DNase1 <- subset(DNase, Run == 1 ) fm2DNase1 <- nls(

Hello I followed the example in page 59, chapter 11 of the 'Introduction to R' manual. I entered my own x,y data. I used the least squares. My function has 5 parameters: p[1], p[2], p[3], p[4], p[5]. I plotted the x-y data. Then I used lines(spline(xfit,yfit)) to overlay best curves on the data while changing the parameters. My question is how do I calculate the residual sum of squares.

Hola, paso el mini programita q estoy viendo, lo q me llama la atencion es una parte donde se definen las funciones. Probe primero meter adentro del boots la estadistica a estimar usando directamente gini(varible, pesos) pero no me dejo. Vi q en el ej del manual de boots, siempre define antes la funcion, entonces probe definir antes una funcion haciendo grini<-function(x) {gini(variable,

I am trying to calculate robust standard errors for a logit model. I installed the package "car" and tried using hccm.default, but that required an lm object. Is there some way to do a similar operation for a glm object? x <- hccm.default(glm(winner ~ racebl + racehi + raceas + inchi + incmed + edhs + edcol + edba + agec1 + agec4 + sex + margin + regla + regbay + regsc +

What am I failing to understand here? The script below works fine if the dataset being used is DNase1 <- DNase[ DNase$Run == 1, ] per the example given in help(nlrob). Obviously, I am trying to understand how to use nls and nlrob to fit curves to data using R. #package=DAAG attach(codling) plot(pobs~dose) #next command returns 'step factor reduced below min factor

I?m new to R so maybe this issue has been asked before and I still could not read the complete set of past messages sent to the list. I found a weird behabiour that I will explain with a simple example. Lets consider the following block of commands: > x <- diag(c(1,4,10)) > x [,1] [,2] [,3] [1,] 1 0 0 [2,] 0 4 0 [3,] 0 0 10 > invx <- x^-1 > invx

Dear WizaRds, A run of nls produces the following concise summary: > summary(cs.wt) Formula: 0 ~ wt.MM(conc, time, A1, a1, A2, a2) Parameters: Estimate Std. Error t value Pr(>|t|) A1 4.814e+02 2.240e+01 21.495 0.0296 * a1 7.401e-01 7.435e-02 9.956 0.0637 . A2 1.613e+02 1.738e+01 9.280 0.0683 . a2 1.770e-02 7.324e-03 2.417 0.2497