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7.1 Personal Loan Acceptance. Universal Bank is a relatively young bank growing rapidly in terms of overall customer acquisition. The majority of these customers are liability customers (depositors) with varying sizes of relationship with the bank. The customer base of asset customers (borrowers) is quite small, and the bank is interested in expanding this base rapidly to bring in more loan business. In particular, it wants to explore ways of converting its liability customers to personal loan customers (while retaining them as depositors).

A campaign that the bank ran last year for liability customers showed a healthy conversion rate of over 9% success. This has encouraged the retail marketing department to devise smarter campaigns with better target marketing. The goal of our analysis is to model the previous campaign's customer behavior to analyze what combination of factors make a customer more likely to accept a personal loan. This will serve as the basis for the design of a new campaign.

The file UniversalBank.xls contains data on 5000 customers. The data include customer demographic information (age, income, etc.), the customer's relationship with the bank (mortgage, securities account, etc.), and the customer response to the last personal loan campaign (Personal Loan). Among these 5000 customers, only 480(= 9.6%) accepted the personal loan that was offered to them in the earlier campaign.

Partition the data into training (60%) and validation (40%) sets.

  1. Perform a k-NN classification with all predictors except ID and ZIP code using k = 1. Remember to transform categorical predictors with more than two categories into dummy variables first. Specify the success class as 1 (loan acceptance), and use the default cutoff value of 0.5. How would this customer be classified?

  2. What is a choice of k that balances between overfitting and ignoring the predictor information?

  3. Show the classification matrix for the validation data that results from using the best k.

  4. Classify the customer using the best k.

  5. Repartition the data, this time into training, validation, and test sets (50% : 30% : 20%). Apply the k-NN method with the k chosen above. Compare the classification matrix of the test set with that of the training and validation sets. Comment on the differences and their reason.

7.2 Predicting Housing Median Prices. The file BostonHousing.xls contains information on over 500 census tracts in Boston, where for each tract 14 variables are recorded. The last column (CAT.MEDV) was derived from MEDV, such that it obtains the value 1 if MEDV>30 and 0 otherwise. Consider the goal of predicting the median value (MEDV) of a tract, given the information in the first 13 columns.

Partition the data into training (60%) and validation (40%) sets.

  1. Perform a k-NN prediction with all 13 predictors (ignore the CAT.MEDV column), trying values of k from 1 to 5. Make sure to normalize the data (click "normalize input data"). What is the best k chosen? What does it mean?

  2. Predict the MEDV for a tract with the following information, using the best k: (Copy this table with the column names to a new worksheet and then in "Score new data" choose "from worksheet.")


  3. Why is the error of the training data zero?

  4. Why is the validation data error overly optimistic compared to the error rate when applying this k-NN predictor to new data?

  5. If the purpose is to predict MEDV for several thousands of new tracts, what would be the disadvantage of using k-NN prediction? List the operations that the algorithm goes through in order to produce each prediction.


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