similar to: AIC, AICc, and K

Displaying 20 results from an estimated 800 matches similar to: "AIC, AICc, and K"

2008 Jul 16
Likelihood ratio test between glm and glmer fits
Dear list, I am fitting a logistic multi-level regression model and need to test the difference between the ordinary logistic regression from a glm() fit and the mixed effects fit from glmer(), basically I want to do a likelihood ratio test between the two fits. The data are like this: My outcome is a (1,0) for health status, I have several (1,0) dummy variables RURAL, SMOKE, DRINK, EMPLOYED,
2013 May 21
Calculating AIC for the whole model in VAR
Hello! I am using package "VAR". I've fitted my model: mymodel<-VAR(mydata,myp,type="const") I can extract the Log Liklihood for THE WHOLE MODEL: logLik(mymodel) How could I calculate (other than manually) the corresponding Akaike Information Criterion (AIC)? I tried AIC - but it does not take mymodel: AIC(mymodel) # numeric(0) Thank you! -- Dimitri Liakhovitski
2006 Dec 12
Calculating AICc using conditional logistic regression
I have a case-control study that I'm analysing using the conditional logistic regression function clogit from the survival package. I would like to calculate the AICc of the models I fit using clogit. I have a variety of scripts that can calculate AICc for models with a logLik method, but clogit does not appear to use this method. Is there a way I can calculate AICc from clogit in R? Many
2004 Jul 16
Does AIC() applied to a nls() object use the correct number of estimated parameters?
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$Run == 1, ] > ## using a selfStart model > fm1DNase1 <- nls( density ~ SSlogis( log(conc), Asym, xmid, scal ), DNase1 ) Using AIC() function: >
2017 Jun 08
stepAIC() that can use new extractAIC() function implementing AICc
I would like test AICc as a criteria for model selection for a glm using stepAIC() from MASS package. Based on various information available in WEB, stepAIC() use extractAIC() to get the criteria used for model selection. I have created a new extractAIC() function (and extractAIC.glm() and extractAIC.lm() ones) that use a new parameter criteria that can be AIC, BIC or AICc. It works as
2010 Feb 16
nls.lm & AIC
Hi there, I'm a PhD student investigating growth patterns in fish. I've been using the minpack.lm package to fit extended von Bertalanffy growth models that include explanatory covariates (temperature and density). I found the nls.lm comand a powerful tool to fit models with a lot of parameters. However, in order to select the best model over the possible candidates (without covariates,
2016 Apr 07
Contenido de un objeto/modelo ARIMA
Buenos días, Os cuento: Cargo la librería "Forecast" y ejecuto su función Arima(...) sobre una serie temporal: mimodelo <- Arima(miST$miserie, ...); Ahora si ejecuto las siguientes sentencias, voy obteniendo los resultados contenidos en "mimodelo", pero algunos de ellos no sé lo que son: mimodelo[[1]] obtengo los coeficientes del modelo ARIMA mimodelo[[2]] obtengo el
2011 Sep 04
AICc function with gls
Hi I get the following error when I try and get the AICc for a gls regression using qpcR: > AICc(gls1) Loading required package: nlme Error in n/(n - p - 1) : 'n' is missing My gls is like this: > gls1 Generalized least squares fit by REML Model: thercarnmax ~ therherbmax Data: NULL Log-restricted-likelihood: 2.328125 Coefficients: (Intercept) therherbmax 1.6441405
2009 Apr 15
AICs from lmer different with summary and anova
Dear R Helpers, I have noticed that when I use lmer to analyse data, the summary function gives different values for the AIC, BIC and log-likelihood compared with the anova function. Here is a sample program #make some data set.seed(1); datx=data.frame(array(runif(720),c(240,3),dimnames=list(NULL,c('x1','x2','y' )))) id=rep(1:120,2); datx=cbind(id,datx) #give x1 a
2006 Jul 12
AICc vs AIC for model selection
Hi, I am using 'best.arima' function from forecast package to obtain point forecast for a time series data set. The documentation says it utilizes AIC value to select best ARIMA model. But in my case the sample size very small - 26 observations (demand data). Is it the right to use AIC value for model selection in this case. Should I use AICc instead of AIC. If so how can I modify
2013 Mar 29
Error message in dredge function (MuMIn package) used with binary GLM
Hi all, I'm having trouble with the model generating 'dredge' function in the MuMIn 'Multi-model Inference' package. Here's the script: globalmodel<- glm(TB~lat+protocol+tested+ streams+goats+hay+cattle+deer, family="binomial") chat<- deviance(globalmodel)/59 #There we 59 residual degrees of freedom in this global model. models<-
2009 Feb 25
indexing model names for AICc table
hi folks, I'm trying to build a table that contains information about a series of General Linear Models in order to calculate Akaike weights and other measures to compare all models in the series. i have an issue with indexing models and extracting the information (loglikehood, AIC's, etc.) that I need to compile them into the table. Below is some sample code that illustrates my
2003 Jun 22
Using weighted.mean() in aggregate()
Dear R users, I have a question on using weighted.mean() while aggregating a data frame. I have a data frame with columns Sub, Length and Slope: > x[1:5,] Sub Length Slope 1 2 351.547 0.0025284969 2 2 343.738 0.0025859390 3 1 696.659 0.0015948968 4 2 5442.338 0.0026132544 5 1 209.483 0.0005304225 and I would like to calculate the weighted.mean of Slope, using Length
2011 Oct 25
difficulties with MuMIn model generation with coxph
Hi All, I'm having trouble with the automatized model generation (dredge) function in the MuMIn package. I'm trying to use it to automatically generate subsets of models from a global cox proportional hazards model, and rank them based on AICc. These seems like it's possible, and the Mumin documentation says that coxph is supported. However, when I run the code (see below), it gives
2011 Aug 27
Degrees of freedom in the Ljung-Box test
Dear list members, I have 982 quotations of a given stock index and I want to run a Ljung-Box test on these data to test for autocorrelation. Later on I will estimate 8 coefficients. I do not know how many degrees of freedom should I assume in the formula for Ljung-Box test. Could anyone tell me please? Below the formula: Box.test(x, lag = ????, type = c("Ljung-Box"), fitdf = 0)
2012 Jun 26
Ljung-Box test (Box.test)
I fit a simple linear model y = bX to a data set today, and that produced 24 residuals (I have 24 data points, one for each year from 1984-2007). I would like to test the time-independence of the residuals of my model, and I was recommended by my supervisor to use the Ljung-Box test. The Box.test function in R takes 4 arguments:  x a numeric vector or univariate time series. lag the statistic
2018 Jan 18
Time-dependent coefficients in a Cox model with categorical variants
First, as others have said please obey the mailing list rules and turn of First, as others have said please obey the mailing list rules and turn off html, not everyone uses an html email client. Here is your code, formatted and with line numbers added. I also fixed one error: "y" should be "status". 1. fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) 2. p
2013 Apr 24
pglm package: fitted values and residuals
I'm using the package pglm and I'have estimated a "random probit model". I need to save in a vector the fitted values and the residuals of the model but I can not do it. I tried with the command fitted.values using the following procedure without results: library(pglm) m1_S<-pglm(Feed ~ Cons_PC_1 + imp_gen_1 + LGDP_PC_1 + lnEI_1 +
2007 Jan 09
min() return factor class values
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2011 Oct 06
anova.rq {quantreg) - Why do different level of nesting changes the P values?!
Hello dear R help members. I am trying to understand the anova.rq, and I am finding something which I can not explain (is it a bug?!): The example is for when we have 3 nested models. I run the anova once on the two models, and again on the three models. I expect that the p.value for the comparison of model 1 and model 2 would remain the same, whether or not I add a third model to be compared