Instructions

This is the R portion of your final exam.

Follow the instructions carefully and write your R code in the provided chunks. You will be graded on the correctness of your code, the quality of your analysis, and your interpretation of the results.

Submission: Please make sure the RMD is knittable and submit the RMD file along with the generated HTML report.

Troubleshooting: If you find errors in your code that prevent the RMD file from knitting, please comment them out (add # before the code). I will give you partial credit based on your logic.

Good luck!

Part I: Logistic Regression

0. Background

Context: You have been hired by a retail consulting firm to analyze the sales performance of a company selling child car seats. The company wants to identify the key drivers of high sales performance to optimize their marketing and store layout strategies.

They have provided you with a dataset (store_sales.csv) containing data from 400 different store locations. Your goal is to build classification models to predict whether a store will have “High Sales” (Yes) or not.

Data Dictionary:

High_Sales (Target): Factor with levels Yes and No. Indicates if the store sold more than 8,000 units.
CompPrice: Price charged by the nearest competitor at each location.
Income: Community income level (in thousands of dollars).
Advertising: Local advertising budget for the company at each location.
Population: Population size of the region (in thousands).
Price: Price charged for the car seats at each site.
ShelveLoc: A factor indicating the quality of the shelving location for the car seats at the site (Good, Bad, or Medium).
Age: Average age of the local population.
Education: Education level at each location.
Urban: Factor (Yes/No) indicating if the store is in an urban location.
US: Factor (Yes/No) indicating if the store is in the US.

1. Data Preparation (0.5 point)

Load the data from the file store_sales.csv and name it store_sales.
Split the data into training (70%) and test (30%) sets. Use set.seed(2025) to ensure reproducibility.

# a) Load data
store_sales <- read.csv("store_sales.csv")
str(store_sales)

## 'data.frame':    400 obs. of  11 variables:
##  $ CompPrice  : int  138 111 113 117 141 124 115 136 132 132 ...
##  $ Income     : int  73 48 35 100 64 113 105 81 110 113 ...
##  $ Advertising: int  11 16 10 4 3 13 0 15 0 0 ...
##  $ Population : int  276 260 269 466 340 501 45 425 108 131 ...
##  $ Price      : int  120 83 80 97 128 72 108 120 124 124 ...
##  $ ShelveLoc  : chr  "Bad" "Good" "Medium" "Medium" ...
##  $ Age        : int  42 65 59 55 38 78 71 67 76 76 ...
##  $ Education  : int  17 10 12 14 13 16 15 10 10 17 ...
##  $ Urban      : chr  "Yes" "Yes" "Yes" "Yes" ...
##  $ US         : chr  "Yes" "Yes" "Yes" "Yes" ...
##  $ High_Sales : int  1 1 1 0 0 1 0 1 0 0 ...

# b) Split data into training and test sets
set.seed(2025)
n <- nrow(store_sales)
train_idx <- sample(1:n, size = round(0.7 * n))
store_train <- store_sales[train_idx, ]
store_test  <- store_sales[-train_idx, ]

2. Logistic Regression (5 points)

Fit a logistic regression model to predict High_Sales using all other variables as predictors. Please use the training dataset.

library(stats)
store_train$High_Sales <- as.factor(store_train$High_Sales)
store_test$High_Sales  <- as.factor(store_test$High_Sales)

logit_model <- glm(High_Sales ~ ., data = store_train, family = binomial)
summary(logit_model)

## 
## Call:
## glm(formula = High_Sales ~ ., family = binomial, data = store_train)
## 
## Coefficients:
##                  Estimate Std. Error z value Pr(>|z|)    
## (Intercept)     -6.787364   3.390753  -2.002  0.04531 *  
## CompPrice        0.201653   0.034124   5.909 3.43e-09 ***
## Income           0.034553   0.010828   3.191  0.00142 ** 
## Advertising      0.371516   0.075899   4.895 9.84e-07 ***
## Population      -0.002566   0.001927  -1.332  0.18296    
## Price           -0.179454   0.026173  -6.856 7.06e-12 ***
## ShelveLocGood    9.338531   1.407041   6.637 3.20e-11 ***
## ShelveLocMedium  4.270635   0.891648   4.790 1.67e-06 ***
## Age             -0.087756   0.019324  -4.541 5.59e-06 ***
## Education       -0.133422   0.103935  -1.284  0.19924    
## UrbanYes         0.140172   0.594386   0.236  0.81357    
## USYes           -1.626954   0.830909  -1.958  0.05023 .  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 375.21  on 279  degrees of freedom
## Residual deviance: 106.23  on 268  degrees of freedom
## AIC: 130.23
## 
## Number of Fisher Scoring iterations: 7

Use the summary() function to examine your fitted model. What is the estimated coefficient for Price? Is it statistically significant? Please interpret the number using the odds ratio?

summary(logit_model)$coefficients["Price", ]

##      Estimate    Std. Error       z value      Pr(>|z|) 
## -1.794535e-01  2.617334e-02 -6.856347e+00  7.064380e-12

exp(coef(logit_model)["Price"])

##     Price 
## 0.8357268

Comments:

This means higher prices are associated with lower odds of high sales, holding other variables constant.

Generate predicted probabilities on the testing dataset. Then, obtain the predicted classes using 0.6 as the cutoff value (threshold).

test_prob <- predict(logit_model, newdata = store_test, type = "response")
test_pred_class <- ifelse(test_prob >= 0.6, "Yes", "No")
test_pred_class <- factor(test_pred_class, levels = levels(store_test$High_Sales))

Create a confusion matrix using the testing dataset and report Misclassification Rate (MR).

cm_test <- table(Predicted = test_pred_class, Actual = store_test$High_Sales)
cm_test

##          Actual
## Predicted 0 1
##         0 0 0
##         1 0 0

MR_test <- mean(test_pred_class != store_test$High_Sales)
MR_test

## [1] NA

Comments:

MR represents the proportion of misclassified observations in the test set which there was an NA with the MR_Test.

Draw an ROC curve and calculate the AUC (Area Under Curve) for this logistic regression model on the testing dataset. Do you think the prediction performance is acceptable?

library(pROC)

## Type 'citation("pROC")' for a citation.

## 
## Attaching package: 'pROC'

## The following objects are masked from 'package:stats':
## 
##     cov, smooth, var

roc_obj <- roc(store_test$High_Sales, test_prob)

## Setting levels: control = 0, case = 1

## Setting direction: controls < cases

plot(roc_obj, col = "blue", main = "ROC Curve – Logistic Regression (Test)")

auc(roc_obj)

## Area under the curve: 0.9374

Comments:

Based on the computed AUC this model has a 0.9374 which is very strong ability to distinguish between high sales and non-high sales.

Part II: Regression Tree

0. Background

You have been hired by a health insurance company to improve their pricing strategy. They want to understand which factors contribute most to high individual medical costs.

Data Dictionary:

charges (Target): Individual medical costs billed by health insurance.
age: Age of primary beneficiary.
sex: Insurance contractor gender (female, male).
bmi: Body mass index (providing an understanding of body weights that are relatively high or low relative to height).
children: Number of children covered by health insurance / Number of dependents.
smoker: Smoking status (yes, no).
region: The beneficiary’s residential area in the US (northeast, southeast, southwest, northwest).

1. Data Preparation (0.5 point)

Load the data from the file insurance.csv and name it insurance.
Split the data into training (70%) and test (30%) sets. Use set.seed(2025) to ensure reproducibility. Hint: use round() function to retain only the integer part of a number.

insurance <- read.csv("insurance.csv")
str(insurance)

## 'data.frame':    1338 obs. of  7 variables:
##  $ age     : int  19 18 28 33 32 31 46 37 37 60 ...
##  $ sex     : chr  "female" "male" "male" "male" ...
##  $ bmi     : num  27.9 33.8 33 22.7 28.9 ...
##  $ children: int  0 1 3 0 0 0 1 3 2 0 ...
##  $ smoker  : chr  "yes" "no" "no" "no" ...
##  $ region  : chr  "southwest" "southeast" "southeast" "northwest" ...
##  $ charges : num  16885 1726 4449 21984 3867 ...

insurance$sex    <- as.factor(insurance$sex)
insurance$smoker <- as.factor(insurance$smoker)
insurance$region <- as.factor(insurance$region)



# b) Split data into training and test sets
set.seed(2025)
n2 <- nrow(insurance)
train_idx2 <- sample(1:n2, size = round(0.7 * n2))
ins_train <- insurance[train_idx2, ]
ins_test  <- insurance[-train_idx2, ]

2. Regression Tree (4 points)

Fit a regression tree based on the training data, using the charges as response variable, and all other variables as predictors. Then visualize the tree using rpart.plot.

library(rpart)
library(rpart.plot)

tree_model <- rpart(charges ~ ., data = ins_train, method = "anova")
rpart.plot(tree_model, main = "Initial Regression Tree for Charges")

Pruning: Fit a large regression tree with cp = 0.001, and then use plotcp() function to view the complexity parameter plot. Based on this plot, what value of cp you would you choose to prune the tree, and why?

big_tree <- rpart(
  charges ~ .,
  data = ins_train,
  method = "anova",
  control = rpart.control(cp = 0.001)
)

plotcp(big_tree)

printcp(big_tree)

## 
## Regression tree:
## rpart(formula = charges ~ ., data = ins_train, method = "anova", 
##     control = rpart.control(cp = 0.001))
## 
## Variables actually used in tree construction:
## [1] age      bmi      children region   smoker  
## 
## Root node error: 1.3144e+11/937 = 140282620
## 
## n= 937 
## 
##           CP nsplit rel error  xerror     xstd
## 1  0.6365601      0   1.00000 1.00299 0.063864
## 2  0.1314088      1   0.36344 0.36608 0.022202
## 3  0.0669267      2   0.23203 0.23279 0.017152
## 4  0.0103270      3   0.16510 0.16991 0.015611
## 5  0.0100514      4   0.15478 0.16636 0.015667
## 6  0.0074812      5   0.14473 0.15817 0.015599
## 7  0.0059708      6   0.13724 0.15142 0.015665
## 8  0.0030658      7   0.13127 0.14455 0.015318
## 9  0.0022378      8   0.12821 0.14392 0.015363
## 10 0.0019148     10   0.12373 0.14334 0.015288
## 11 0.0016684     12   0.11990 0.14223 0.015084
## 12 0.0015433     13   0.11823 0.14340 0.015226
## 13 0.0013601     14   0.11669 0.14353 0.015322
## 14 0.0011696     15   0.11533 0.14525 0.015444
## 15 0.0011223     16   0.11416 0.14710 0.015615
## 16 0.0010000     17   0.11304 0.14551 0.015438

Comments:

I chose cp = 0.001 because it corresponds to the smallest cross-validated error and represents a good trade-off between model complexity and predictive performance.

Refit the regression tree using the cp selected in question (b). Then obtain the predictions on the testing data.

best_cp <- big_tree$cptable[which.min(big_tree$cptable[,"xerror"]), "CP"]

pruned_tree <- prune(big_tree, cp = best_cp)

pred_initial <- predict(tree_model, newdata = ins_test)
pred_pruned  <- predict(pruned_tree, newdata = ins_test)

Obtain the out-of-sample MSE (Mean Squared Error) for both the initial tree model in (a) and the pruned tree model in (c). Which model is preferred for this data, and why?

MSE_initial <- mean((ins_test$charges - pred_initial)^2)
MSE_pruned  <- mean((ins_test$charges - pred_pruned)^2)

MSE_initial

## [1] 30205107

MSE_pruned

## [1] 28961204

Comments:

The pruned tree has a lower MSE, which suggests that pruning reduced overfitting and improved predictive performance.

End of Exam. Please double-check that your RMD file knits successfully. Submit both the RMD and the generated HTML report.

Reminder: If a specific chunk causes an error, comment it out to allow the file to knit. Failure to submit an HTML report may result in a point deduction.

Final Exam: Coding Part

Instructions

Part I: Logistic Regression

0. Background

1. Data Preparation (0.5 point)

2. Logistic Regression (5 points)

Part II: Regression Tree

0. Background

1. Data Preparation (0.5 point)

2. Regression Tree (4 points)