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)

  1. Load the data from the file store_sales.csv and name it store_sales.
  2. Split the data into training (70%) and test (30%) sets. Use set.seed(2025) to ensure reproducibility.
# a) Load data
# Your code here
store_sales <- read.csv("store_sales.csv")
# b) Split data into training and test sets
set.seed(2025)
# Your code here
sample_size <- floor(0.7 * nrow(store_sales))
train_indices <- sample(1:nrow(store_sales), sample_size)

train_data <- store_sales[train_indices, ]
test_data <- store_sales[-train_indices, ]

2. Logistic Regression (5 points)

  1. Fit a logistic regression model to predict High_Sales using all other variables as predictors. Please use the training dataset.
# Your code here
model <- glm(High_Sales ~ ., data = train_data, family = "binomial")
  1. 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?
# Your code here
summary(model)
## 
## Call:
## glm(formula = High_Sales ~ ., family = "binomial", data = train_data)
## 
## 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

Comments:Price coefficient = -0.179454, p-value = 7.06e-12 (significant). Odds ratio = 0.836. Each $1 price increase reduces the odds of high sales by 16.4%

  1. Generate predicted probabilities on the testing dataset. Then, obtain the predicted classes using 0.6 as the cutoff value (threshold).
# Your code here
pred_probs <- predict(model, newdata = test_data, type = "response")

pred_classes <- ifelse(pred_probs >= 0.6, 1, 0)
  1. Create a confusion matrix using the testing dataset and report Misclassification Rate (MR).
# Your code here
conf_matrix <- table(Predicted = pred_classes, Actual = test_data$High_Sales)
conf_matrix
##          Actual
## Predicted  0  1
##         0 63 16
##         1  3 38
misclass_rate <- mean(pred_classes != test_data$High_Sales)
misclass_rate
## [1] 0.1583333

Comments: Confusion matrix: 63 True Negatives, 38 True Positives, 16 False Negatives, 3 False Positives. Misclassification Rate = 15.8%. Model is 84.2% accurate

  1. 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?
# Your code here
order_index <- order(pred_probs, decreasing = TRUE)
actual_sorted <- test_data$High_Sales[order_index]

tpr <- cumsum(actual_sorted == 1) / sum(actual_sorted == 1)
fpr <- cumsum(actual_sorted == 0) / sum(actual_sorted == 0)

plot(fpr, tpr, type = "l", main = "ROC Curve")
abline(0, 1, lty = 2)

sum(diff(fpr) * (tpr[-1] + tpr[-length(tpr)]) / 2)
## [1] 0.9374299

Comments: ROC curve is well above the diagonal line, showing good performance. AUC = 0.937 (above 0.8 means good predictive performance)

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)

  1. Load the data from the file insurance.csv and name it insurance.
  2. 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.
# Your code here
insurance <- read.csv("insurance.csv")

# b) Split data into training and test sets
set.seed(2025)
# Your code here
train_size <- round(0.7 * nrow(insurance))
train_indices <- sample(1:nrow(insurance), train_size)
train_data <- insurance[train_indices, ]
test_data <- insurance[-train_indices, ]

2. Regression Tree (4 points)

  1. 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.
# Your code here
library(rpart)
library(rpart.plot)

tree_model <- rpart(charges ~ ., data = train_data)

rpart.plot(tree_model)

  1. 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?
# Your code here
large_tree <- rpart(charges ~ ., data = train_data, cp = 0.001)

plotcp(large_tree)

Comments: I choose cp = 0.01 because that’s where the error line flattens out.This prevents overfitting while keeping good accuracy.

  1. Refit the regression tree using the cp selected in question (b). Then obtain the predictions on the testing data.
# Your code here
final_tree <- rpart(charges ~ ., data = train_data, cp = 0.01)
predictions <- predict(final_tree, newdata = test_data)


# d) 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?

#```{r}
# Your code here
pred_initial <- predict(tree_model, newdata = test_data)
mse_initial <- mean((test_data$charges - pred_initial)^2)
mse_initial
## [1] 30205107
pred_pruned <- predict(final_tree, newdata = test_data)
mse_pruned <- mean((test_data$charges - pred_pruned)^2)
mse_pruned
## [1] 30205107

Comments: - Initial tree MSE = 30205107 - Pruned tree MSE = 30205107 Both models have the same MSE, so either model is acceptable.The pruned tree is slightly preferred because it’s simpler (fewer splits) while maintaining the same accuracy.

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.