[R] HOw compare 2 models in logistic regression

[ONKELINX, Thierry]

Dear Santos,

Since your models are nested you can apply a likelihood ratio test.

M0 <- glm(formula = Spend_bucket ~ Freq + Address_is_res + last_update_days_ago, data = qdataset, family = binomial)
M1 <- glm(formula = Spend_bucket ~ Freq + Address_is_res, data = qdataset, family = binomial)
anova(M0, M1, test = "Chisq")

Note that using the data argument makes you code much more readable.

Best regards,

Thierry

ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
+ 32 2 525 02 51
+ 32 54 43 61 85
Thierry.Onkelinx op inbo.be
www.inbo.be

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-----Oorspronkelijk bericht-----
Van: r-help-bounces op r-project.org [mailto:r-help-bounces op r-project.org] Namens santoshdvn
Verzonden: maandag 30 april 2012 10:23
Aan: r-help op r-project.org
Onderwerp: [R] HOw compare 2 models in logistic regression

Hi,

I have created 2 models in logistic regression and got the predictor values are significance on output. Here are 2 summaries of 2 model. HOw can we compare 2 models by what factor or coefficient and say which model is best

Please help
---------------------------------------------------------------------------------------------------------------
*MODEL 1*

glm(formula = qdataset$Spend_bucket ~ qdataset$Freq + qdataset$Address_is_res +
    qdataset$last_update_days_ago, family = binomial)

Coefficients:
                                Estimate Std. Error z value Pr(>|z|)
(Intercept)                   -1.572e+00  2.344e-01  -6.708 1.98e-11 ***
qdataset$Freq                  8.763e-01  8.229e-02  10.649  < 2e-16 ***
qdataset$Address_is_res       -8.743e-01  2.116e-01  -4.132 3.60e-05 ***
qdataset$last_update_days_ago -4.568e-04  7.625e-05  -5.991 2.09e-09 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 1228.41  on 988  degrees of freedom Residual deviance:  957.74  on 985  degrees of freedom
AIC: 965.74

Number of Fisher Scoring iterations: 5
-------------------------------------------------------------------------------------------------------------------------------
*MODEL 2*
glm(formula = qdataset$Spend_bucket ~ qdataset$Freq + qdataset$Address_is_res,
    family = binomial)

Deviance Residuals:
    Min       1Q   Median       3Q      Max
-2.5798  -0.6428  -0.5894   0.7271   2.2402

Coefficients:
                        Estimate Std. Error z value Pr(>|z|)
(Intercept)              -2.6148     0.1732 -15.094  < 2e-16 ***
qdataset$Freq             0.9526     0.0794  11.997  < 2e-16 ***
qdataset$Address_is_res  -0.7622     0.2044  -3.729 0.000192 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 1228.41  on 988  degrees of freedom Residual deviance:  995.18  on 986  degrees of freedom
AIC: 1001.2

Number of Fisher Scoring iterations: 5


--------------------------------------------------------------------------------------

Thanks,
Santosh

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