**Predicting Tire Success using TireRatings Survey Results

Project Objectives

To investigate the relationship between wet traction performance, noise level, and customer purchasing likelihood.

Question 1 & 2: Develop the logistic regression model

Step 1: Install and load required libraries

#install.packages("readxl")
#install.packages("Hmisc")
#install.packages("pscl")

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)
## Type 'citation("pROC")' for a citation.
## 
## Attaching package: 'pROC'
## The following objects are masked from 'package:stats':
## 
##     cov, smooth, var

Step 2: Import & clean the data

Class_Exercise_15_TireRatings <- read_excel("/Users/amritahimmatraopet/Desktop/R Markdown/Class Exercise 15_TireRatings.xlsx")
Class_Exercise_15_TireRatings <- subset(Class_Exercise_15_TireRatings, select = -c(Tire, Buy_Again))

Step 3: Summarize the data

head(Class_Exercise_15_TireRatings)
## # A tibble: 6 × 3
##     Wet Noise Purchase
##   <dbl> <dbl>    <dbl>
## 1   8     7.2        0
## 2   8     7.2        1
## 3   7.6   7.5        1
## 4   6.6   5.4        0
## 5   5.8   6.3        0
## 6   6.3   5.7        0
Data Description: A dexcription of some of the features presented in the table below.

Variable      | Definition
1. Wet        | Tire's wet traction performance
2. Noise      | Tire's noise level
3. Buy Again  | Whether customer wants to buy again
4. Purchase   | Likeliness for customer to purchase (1: yes and 0: no)
summary(Class_Exercise_15_TireRatings)
##       Wet            Noise          Purchase     
##  Min.   :4.300   Min.   :3.600   Min.   :0.0000  
##  1st Qu.:6.450   1st Qu.:6.000   1st Qu.:0.0000  
##  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.   :1.0000
Interpretation: The median Wet Traction is 7.75 and the median Noise Level is 7.10

Step 4: Feature selection (i.e., correlation analysis)

corr <- rcorr(as.matrix(Class_Exercise_15_TireRatings))
corr
##           Wet Noise Purchase
## Wet      1.00  0.76     0.74
## Noise    0.76  1.00     0.72
## Purchase 0.74  0.72     1.00
## 
## n= 68 
## 
## 
## P
##          Wet Noise Purchase
## Wet           0     0      
## Noise     0         0      
## Purchase  0   0

Step 5: Build the logistic regression model

model <- glm(Purchase ~ Wet + Noise, data = Class_Exercise_15_TireRatings, family = binomial)
summary(model)
## 
## Call:
## glm(formula = Purchase ~ Wet + Noise, family = binomial, data = Class_Exercise_15_TireRatings)
## 
## 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
Interpretation: All the independent variables were significant (p value < 0.05).

Question 3: Overall Model Significance

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   0.8305269
The McFadden R-Squared value is 0.705, meaning that our LR model explains about 70.5% of the variability in the outcome relative to a model with no predictors. This is considered a strong fit. 

Question 4 & 5: Predicting with new information

new_data <- data.frame(Wet = 8, Noise = 8)
probability <- predict(model, newdata = new_data, type = "response")
probability_percent <- probability * 100
probability_percent
##        1 
## 88.36964
The probability of a customer probably or definitely purchasing a particular tire again with a Wet performance rating of 8 and a Noise performance rating of 8 is 88.37%, whereas if the Wet rating was 7 and the Noise performance rating was 7, the probability is 4.06%. This explains further how the probability changes as the independent variables change.