#** Predicting Tire Purchases at Tire Rack**
Project Objective
To investigate survey ratings from Tire Rack customers. ##Question 1 & 2: Develop the Model and assess Predictor Significance
###Step 1: Install and Load required Libraries
#Install.packages for the following if not done so already
library(readxl)#allows us to import excel files
library(Hmisc)#allows us to call the correlation function##
## Attaching package: 'Hmisc'## The following objects are masked from 'package:base':
##
## format.pval, unitslibrary(pscl)#allows us to call the pseudo R-square package to evaluate our model ## 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)#allows us to run the area under the curve (AUC) package to get the plot and AUC score ## Type 'citation("pROC")' for a citation.##
## Attaching package: 'pROC'## The following objects are masked from 'package:stats':
##
## cov, smooth, var###Step 2 & 3: Import and Clean the data
ratings_df <- read_excel("Class Exercise 15_TireRatings (1).xlsx")
rat_df <- subset(ratings_df, select = -c(Tire)) #drop irrelevant column###Step 4: Summarize the data
head(rat_df)## # A tibble: 6 × 4
## Wet Noise Buy_Again Purchase
## <dbl> <dbl> <dbl> <dbl>
## 1 8 7.2 6.1 0
## 2 8 7.2 6.6 1
## 3 7.6 7.5 6.9 1
## 4 6.6 5.4 6.6 0
## 5 5.8 6.3 4 0
## 6 6.3 5.7 4.5 0summary(rat_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###Step 5:Feature Selection(correlation analysis)
corr <- rcorr(as.matrix(rat_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#Interpretation: All the variables are significant with the target variable ###Step 6:Build the logistic Regression
model <- glm(Purchase ~ Wet + Noise, data = rat_df, family = binomial)
summary(model)##
## Call:
## glm(formula = Purchase ~ Wet + Noise, family = binomial, data = rat_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: 8Interpretation:All The independent Variables were significant p-value < 0.05
###Question 3: Overall model Significance
#Fit a null model
null_model <- glm(Purchase ~ 1, data = rat_df, family = binomial)###Perform likelyhood 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 ' ' 1#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##Area under the curve #Compute ROC curve and AUC score
roc_curve <- roc(rat_df$Purchase, fitted(model))## Setting levels: control = 0, case = 1## Setting direction: controls < casesplot(roc_curve)
auc(roc_curve)## Area under the curve: 0.9741```Interpretation: An AUC score closer to 1 indicates that the LR model has a high level or accuracy in predicting student retention. In this case, the score was 0.97.
##Question 4 & 5: Predicting with new Inormation
new_data1 <- data.frame(Wet = 8, Noise = 8)
new_data2 <- data.frame(Wet = 7, Noise = 7)#predict the probability
prob1 <- predict(model, newdata = new_data1, type ="response")
prob1 * 100## 1
## 88.36964prob2 <- predict(model, newdata = new_data2, type ="response")
prob2 * 100## 1
## 4.058753