#Logistic Regression: TireRatings Problem
##Question 1&2 #Step 1: Load Data
library(readxl)
library(Hmisc)## Warning: package 'Hmisc' was built under R version 4.3.3##
## Attaching package: 'Hmisc'## The following objects are masked from 'package:base':
##
## format.pval, unitslibrary(pscl)## Warning: package 'pscl' was built under R version 4.3.3## 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)## Warning: package 'pROC' was built under R version 4.3.3## Type 'citation("pROC")' for a citation.##
## Attaching package: 'pROC'## The following objects are masked from 'package:stats':
##
## cov, smooth, var#Step 2: Import Data
TireRating_data <- read_excel(file.choose())#Step 3: Clean Data
tire_data <- subset(TireRating_data, select = -c(Buy_Again))#Step 4: Summarize Data
head(tire_data)## # A tibble: 6 × 4
## Tire Wet Noise Purchase
## <chr> <dbl> <dbl> <dbl>
## 1 BFGoodrich g-Force Super Sport A/S 8 7.2 0
## 2 BFGoodrich g-Force Super Sport A/S H&V 8 7.2 1
## 3 BFGoodrich g-Force T/A KDWS 7.6 7.5 1
## 4 Bridgestone B381 6.6 5.4 0
## 5 Bridgestone Insignia SE200 5.8 6.3 0
## 6 Bridgestone Insignia SE200-02 6.3 5.7 0summary(tire_data)## Tire Wet Noise Purchase
## Length:68 Min. :4.300 Min. :3.600 Min. :0.0000
## Class :character 1st Qu.:6.450 1st Qu.:6.000 1st Qu.:0.0000
## Mode :character Median :7.750 Median :7.100 Median :0.0000
## Mean :7.315 Mean :6.903 Mean :0.4412
## 3rd Qu.:8.225 3rd Qu.:7.925 3rd Qu.:1.0000
## Max. :9.200 Max. :8.900 Max. :1.0000Interpretation: The Median Wet preformance is 7.75, with a median Noise Preformance of 7.10,
resulting in people purchasing tires based on these variables
Interpretation: All the predictors are significant with the target vairable
#Step 6: Logistic Regression
model <- glm(Purchase ~ Wet + Noise, data = tire_data, family = binomial)
summary(model)##
## Call:
## glm(formula = Purchase ~ Wet + Noise, family = binomial, data = tire_data)
##
## 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##Question 3: Overall Model Significance
null_model <- glm(Purchase ~ 1, data = tire_data, family = binomial)#Likelihood Ratio Test
anova(null_model, model, test = "Chisq")## Analysis of Deviance Table
##
## Model 1: Purchase ~ 1
## Model 2: Purchase ~ Wet + Noise
## Resid. Df Resid. Dev Df Deviance Pr(>Chi)
## 1 67 93.325
## 2 65 27.530 2 65.795 5.162e-15 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Interpretation: The factors of Wet Preformance and Noise Preformance
Do significantly predict the likelihood of purchasing the type of tire
#Pseudo R-Squared
pR2(model)## fitting null model for pseudo-r2## llh llhNull G2 McFadden r2ML r2CU
## -13.7649516 -46.6623284 65.7947536 0.7050093 0.6199946 0.8305269Interpretation: A McFadden R-Squared of 0.7050 means that our LR model
explains about 70.50% of the variability in the outcome, relative to a model
with no predictors. This is considered an Average fit, as the value of 0.7050
is in between 0.5 and 0.74
#Area Under the Curve
roc_curve <- roc(tire_data$Purchase, fitted(model))## Setting levels: control = 0, case = 1## Setting direction: controls < casesplot(roc_curve)
auc(roc_curve)## Area under the curve: 0.9741Interpretation: An AUC score of 0.88 indicates that the LR model has a high
level of accuracy in predicting purchasing the tires
##Question 4&5: Predicting with new Info
new_data1 <- data.frame(Wet = 7, Noise = 7)
new_data2 <- data.frame(Wet = 8, Noise = 8)#Predict Probability
prob1 <- predict(model, new_data1, type = "response")
prob1 * 100 # 4.06% probability of purchasing this specific tire## 1
## 4.058753prob2 <- predict(model, new_data2, type = "response")
prob2 * 100 # 88.37% probability of purchasing this specific tire## 1
## 88.36964