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 at inbo.be
www.inbo.be
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-----Oorspronkelijk bericht-----
Van: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
Namens santoshdvn
Verzonden: maandag 30 april 2012 10:23
Aan: r-help at 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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