Home / Questions / 349482
Accepted
Juan Luis
Asked: 2020-04-26 12:05:30 +0800 CST 2020-04-26 12:05:30 +0800 CST

How to compare nested logit models using ANOVA in R?

  • 772

I have two binomial logit models :

  • Model A:
modelo_logit_viv <- glm( SAP ~sexo + edad + peso + niv_est + enf_cron +  sit_lab +
                           frec_act_fis   + GHQ_12 + ingreso_eq +
                           n_dormitorios + cont_indus +    delincuencia, # variables de estudio
                         data = datos_modelo, family = binomial(link = "logit"),na.action = "na.omit")
  • Model B (nested model of A):
modelo_logit <- glm(SAP ~ sexo + edad + peso + niv_est + enf_cron  +  sit_lab + 
                       frec_act_fis   + ingreso_eq + GHQ_12, 
                       data = modelo_logit_viv$model, family = binomial(link = "logit"))

c(edad, peso, ingreso_eq, GHQ_12)They are continuous variables, the rest are categorical variables (factor).

I want to analyze whether the characteristics of the dwelling (study variables) have an influence on the self-perceived health status (SAP). All variables are significant in both models. However, I want to perform an analysis of variance (ANOVA) between these two models to check that model B is better than model B . So I run:

anova(modelo_logit,modelo_logit_viv)

And I get the following table:

  Resid. Df Resid. Dev Df Deviance
1     16805      15439            
2     16802      15420  3   18.644

Can the interpretation of this table tell me if there is an influence between housing characteristics and health status?

Another way to ask this question is: Is the 18.644F test statistic that I should compare with Snedecor's F table to determine whether or not to accept the null hypothesis that there is a difference in means between the two models?

If not, how can I compare these two models with an ANOVA in R?

r

1 Answers

  1. Best Answer

    The deviation term that we see when executing anova(modelo_logit,modelo_logit_viv) is the difference between the sum of the prediction errors , so if modelo_logitloses better than modelo_logit_viv, the deviation should decrease, as it is the case.

    This suggests that the study variables indeed have explanatory power or some association with the dependent variable SAP.

    We can check this using the Chi-square test that is implemented in anova(). An example with dataframe iris:

    data = iris
data$Species=ifelse(data$Species=="versicolor",1,0)
full_model = glm(Species ~ .,data=data,family=binomial)
red_model = glm(Species ~ Sepal.Width,data=data,family=binomial)

anova(red_model,full_model,test="Chisq")
Analysis of Deviance Table

Model 1: Species ~ Sepal.Width
Model 2: Species ~ Sepal.Length + Sepal.Width + Petal.Length + Petal.Width
  Resid. Df Resid. Dev Df Deviance Pr(>Chi)  
1       148     151.93                       
2       145     145.07  3   6.8562  0.07662

    On the other hand, if the study variables included in the model are significant under the Wald test (we can see this by running summary(modelo_logit_viv)) the likelihood ratio should be similar to , so we also calculate McFadden's pseudo R^2:

    modelo_nulo <- glm(SAP~1, family = "binomial",data = datos_modelo)

pseudo_R_viv_signif <- 1 - logLik(modelo_logit_viv)/logLik(modelo_nulo)
pseudo_R_viv_signif  # 0.49

pseudo_R <- 1 - logLik(modelo_logit)/logLik(modelo_nulo) # 0.48
pseudo_R

    We see that the values ​​are very similar c(0.49,0.48)and the Chi-square test in my model is also significant. We can conclude then that the study variables are relevant when explaining the state of health ( SAP)

    • 0