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

2008 Jul 16

4

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

1

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

1

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

1

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

1

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

1

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

4

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

2

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

2

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

2

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

2

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

3

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

1

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

1

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

1

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

2

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

1

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

1

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

3

min() return factor class values

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2011 Oct 06

1

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