To develop a logistic regression model to predict the likelihood of a customer purchasing a tire again based on its Wet and Noise performance ratings
###install.packages("readxl")
###install.packages("Hmisc")
###install.packages("pscl")
if(!require(pROC)) install.packages("pROC")
## Loading required package: pROC
## Type 'citation("pROC")' for a citation.
##
## Attaching package: 'pROC'
## The following objects are masked from 'package:stats':
##
## cov, smooth, var
library(readxl)
library(Hmisc)
##
## Attaching package: 'Hmisc'
## The following objects are masked from 'package:base':
##
## format.pval, units
library(pscl)
## Classes and Methods for R originally developed in the
## Political Science Computational Laboratory
## Department of Political Science
## Stanford University (2002-2015),
## by and under the direction of Simon Jackman.
## hurdle and zeroinfl functions by Achim Zeileis.
library(pROC)
tirerating_df <- read_excel("~/Downloads/RMarkdown/Class Exercise 15.xlsx")
tirerating_df
## # A tibble: 68 × 5
## Tire Wet Noise Buy_Again Purchase
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 BFGoodrich g-Force Super Sport A/S 8 7.2 6.1 0
## 2 BFGoodrich g-Force Super Sport A/S H&V 8 7.2 6.6 1
## 3 BFGoodrich g-Force T/A KDWS 7.6 7.5 6.9 1
## 4 Bridgestone B381 6.6 5.4 6.6 0
## 5 Bridgestone Insignia SE200 5.8 6.3 4 0
## 6 Bridgestone Insignia SE200-02 6.3 5.7 4.5 0
## 7 Bridgestone Potenza G 019 Grid 7.7 5.2 5 0
## 8 Bridgestone Potenza RE92 5 6.2 2.5 0
## 9 Bridgestone Potenza RE92A 5.6 6.4 2.7 0
## 10 Bridgestone Potenza RE960AS Pole Position 8.8 8.5 8.1 1
## # ℹ 58 more rows
tr_df <- subset(tirerating_df, select = -c(Tire))
summary(tr_df)
## Wet Noise Buy_Again Purchase
## Min. :4.300 Min. :3.600 Min. :1.400 Min. :0.0000
## 1st Qu.:6.450 1st Qu.:6.000 1st Qu.:3.850 1st Qu.:0.0000
## Median :7.750 Median :7.100 Median :6.150 Median :0.0000
## Mean :7.315 Mean :6.903 Mean :5.657 Mean :0.4412
## 3rd Qu.:8.225 3rd Qu.:7.925 3rd Qu.:7.400 3rd Qu.:1.0000
## Max. :9.200 Max. :8.900 Max. :8.900 Max. :1.0000
corr <- rcorr(as.matrix(tr_df))
corr
## Wet Noise Buy_Again Purchase
## Wet 1.00 0.76 0.91 0.74
## Noise 0.76 1.00 0.83 0.72
## Buy_Again 0.91 0.83 1.00 0.83
## Purchase 0.74 0.72 0.83 1.00
##
## n= 68
##
##
## P
## Wet Noise Buy_Again Purchase
## Wet 0 0 0
## Noise 0 0 0
## Buy_Again 0 0 0
## Purchase 0 0 0
model <- glm(Purchase ~ Wet + Noise, data = tr_df, family = binomial)
summary(model)
##
## Call:
## glm(formula = Purchase ~ Wet + Noise, family = binomial, data = tr_df)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -39.4982 12.4779 -3.165 0.00155 **
## Wet 3.3745 1.2641 2.670 0.00760 **
## Noise 1.8163 0.8312 2.185 0.02887 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 93.325 on 67 degrees of freedom
## Residual deviance: 27.530 on 65 degrees of freedom
## AIC: 33.53
##
## Number of Fisher Scoring iterations: 8
pR2(model)
## fitting null model for pseudo-r2
## llh llhNull G2 McFadden r2ML r2CU
## -13.7649516 -46.6623284 65.7947536 0.7050093 0.6199946 0.8305269
###Question 4&5: predicting with new information
new_data1 <- data.frame(Wet = 8, Noise = 8) #buy again
new_data2 <- data.frame(Wet = 7, Noise = 7) # not buy again
prob1 <- predict(model, new_data1, type = "response") #probability that customers will buy again
prob1 *100 #88.37% that customers will buy again
## 1
## 88.36964
prob2 <- predict(model, new_data2, type = "response") #probability that customers will not buy again
prob2 *100 #4.06% that customers will not buy again
## 1
## 4.058753