similar to: Zinb for Non-interger data

Displaying 20 results from an estimated 700 matches similar to: "Zinb for Non-interger data"

2011 May 23
1
Interpreting the results of the zero inflated negative binomial regression
Hi, I am new to R and has been depending mostly on the online tutotials to learn R. I have to deal with zero inflated negative binomial distribution. I am however unable to understand the following example from this link http://www.ats.ucla.edu/stat/r/dae/zinbreg.htm The result gives two blocks. *library(pscl) zinb<-zeroinfl(count ~ child + camper | persons, dist = "negbin", EM =
2010 Jun 21
1
ZINB by Newton Raphson??
Dear all.. I have a respon variable y. Predictor variable are x1, x2, x3, x4, x5 (1) What is the syntax to get paramater estimation of ZINB Model by Newton Raphson (not BFGS) (2) What syntax to plot probability of observed & predicted of ZINB Thx. Regards Krist. [[alternative HTML version deleted]]
2018 Apr 09
2
Warning en modelo ZINB
Buenas tardes, Estoy estimando un modelo binomial negativo de ceros inflados (ZINB) utilizando el comando zeroinfl() del paquete pscl. Al ejecutarlo me da el siguiente aviso: Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred ¿Sabéis que significa y si puedo usar el modelo aún con ese aviso? ¿Los coeficientes son fiables? Muchas gracias, Miriam
2010 Apr 12
1
zerinfl() vs. Stata's zinb
Hello, I am working with zero inflated models for a current project and I am getting wildly different results from R's zeroinfl(y ~ x, dist="negbin") command and Stata's zinb command. Does anyone know why this may be? I find it odd considering that zeroinfl(y ~ x, dist="poisson") gives identical to output to Stata's zip function. Thanks, --david [[alternative
2018 Apr 09
2
Warning en modelo ZINB
Muchas gracias por la respuesta. He mirado y los coeficientes no son altos pero sí tengo una gran cantidad de ceros en la variable dependiente (más del 90%). Sin embargo, al incluir otro tipo de variables independientes no me da ese aviso, dejando la misma variable dependiente. ¿Cómo podría utilizar stan/rstan de forma sencilla para diagnosticar el modelo? Muchas gracias El Lun, 9 de Abril de
2018 Apr 09
3
Warning en modelo ZINB
¿Quieres decir que para un nivel de una variable categorica todas las observaciones de la variable respuesta sean ceros? Gracias El Lun, 9 de Abril de 2018, 19:59, Carlos J. Gil Bellosta escribió: > ¿Podría ser que para algún nivel de alguna variable independiente > categórica solo hubiese ceros? En ese caso, casi seguro, aparecería ese > tipo de warning. > > El lun., 9 abr. 2018 a
2012 Jul 13
1
Vuong test
Dear All, I am using the function vuong from pscl package to compare 2 non nested models NB1 (negative binomial I ) and Zero-inflated model. NB1 <-  glm(, , family = quasipoisson), it is an object of class: "glm" "lm" zinb <- zeroinfl( dist = "negbin") is an object of class: "zeroinfl"   when applying vuong function I get the following: vuong(NB1,
2018 Apr 09
2
Warning en modelo ZINB
Hola de nuevo Carlos, he probado a quitar esa variable categórica y me sigue dando el aviso... El Lun, 9 de Abril de 2018, 20:17, Carlos J. Gil Bellosta escribió: > Si, creo que el motivo del warning puede ser ese. Es hipotético, pero > plausible. Sobre todo cuando tienes más de un 90% de ceros. > > El coeficiente de ese nivel para el modelo de la mixtura (ceros vs > binomial >
2012 Dec 10
1
Marginal effects of ZINB models
Dear all, I am modeling the incidence of recreational anglers along a stretch of coastline, and with a vary large proportion of zeros (>80%) have chosen to use a zero inflated negative binomial (ZINB) distribution. I am using the same variables for both parts of the model, can anyone help me with R code to compute overall marginal effects of each variable? My model is specified as follows:
2005 May 17
1
Vuong test
Hi, I have two questions. First, I'd like to compare a ZINB model to a negativ binomial model with the Vuong test, but I can't find how to performe it from the zicount package. Does a programm exist to do it ? Second, I'd like to know in which cases we have to use a double hurdle model instead of a zero inflated model. Many thanks, St??phanie Payet REES France R??seau
2004 Feb 10
1
generate random sample from ZINB
I want to generate 1,000 random samples of sample size=1,000 from ZINB. I know there is a rnegbin() to generate random samples from NB, and I know I can use the following process: do i=1 to 1000 n=0 do i=1 to 1000 if runi(1)>0.1 then x(i) = 0; else x(i)=rnegbin(); n=n+1; if n>1000 then stop; end; output; end; Anybody can help me out with the R code? Thanks very much ahead of time.
2010 Jun 21
0
Re ZINB by Newton Raphson??
Dear Mr.Zeileis & all. (1)     Thx for your reply. Yes, I am talk about the function zeroinfl() from the package "pscl". I want to use Newton Raphson to get parameter             estimation ZINB, so I try this: ----------------------------------------------------------------------------------------------------------------------------------         > zinb <- zeroinfl(y
2012 Oct 14
2
Poisson Regression: questions about tests of assumptions
I would like to test in R what regression fits my data best. My dependent variable is a count, and has a lot of zeros. And I would need some help to determine what model and family to use (poisson or quasipoisson, or zero-inflated poisson regression), and how to test the assumptions. 1) Poisson Regression: as far as I understand, the strong assumption is that dependent variable mean = variance.
2012 Dec 14
1
Beta-coefficients for ZINB model
Dear users, Does anyone have any idea how to generate standardised beta coefficients for a ZINB model in R to compare which explanatory variables are having the greatest impact on the dependent variable? Thanks, Jeremy [[alternative HTML version deleted]]
2016 Apr 18
1
ZINB multi-level model using MCMCglmm
Hi, I am Olga Viedma. I am running a Zero-inflated negative binomial (ZINB) multi-level model using MCMCglmm package. I have a doubt. Can I use the "Liab" outputs as fitted data, instead of the predicted values from "predict"? The liab outputs fit very well with the observed data, whereas the predicted values are so bad. Thanks in advance, Olga Viedma D . Olga
2010 Mar 03
1
Zero inflated negative binomial
Hi all, I am running the following model: > glm89.nb <- glm.nb(AvGUD ~ Year*Trt*Micro) where Year has 3 levels, Trt has 2 levels and Micro has 3 levels. However when I run it has a zero inflated negative binomial (as I have lots of zeros) I get the below error message: > Zinb <- zeroinfl(AvGUD ~ Year*Trt*Micro |1, data = AvGUD89, dist = "negbin") Error in optim(fn =
2003 Oct 29
1
One inflated Poisson or Negative Binomal regression
Hello I am interested in Poisson or (ideally) Negative Binomial regression with an inflated number of 1 responses I have seen JK Lindsey's fmr function in the gnlm library, which fits zero inflated Poisson (ZIP) or zero inflated negative binomial regression, but the help file states that for ' Poisson or related distributions the mixture involves the zero category'. I had thought
2006 Jan 24
1
non-finite finite-difference value[]
Dear R-helpers, running a zeroinflated model of the following type: zinb = zeroinfl(count=response ~., x = ~ . - response, z = ~. - response, dist = "negbin", data = t.data, trace = TRUE) generates the following message: Zero-Inflated Count Model Using logit to model zero vs non-zero Using Negative Binomial for counts dependent variable y: Y 0 1 2 3 359 52 7 3 generating
2018 Apr 18
2
Interpretar intercepto de modelo ZINB
Buenas tardes, ¿Cómo interpretarías el intercepto que da R en un modelo de ceros inflados? Por un lado en la parte de conteo tengo un intercepto de -4.2 y en la parte de ceros de 102, ambos salen significativos (***). ¿Qué me dirían? Gracias
2008 Jun 05
1
GAM hurdle models
Hello, I have been using mgcv to run GAM hurdle models, analyzing presence/absence data with GAM logistic regressions, and then analyzing the data conditional on presence (e.g. without samples with no zeros) with GAMs with a negative binomial distribution. It occurs to me that using the negative binomial distribution on data with no zeros is not right, as the negative binomial allows zeros.