Answered: The file UniversalBank.csv contains…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Related questions

Question

The file UniversalBank.csv contains data on 5000 customers of Universal Bank. 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. In this exercise, we focus on two predictors: Online (whether or not the customer is an active user of online banking services) and Credit Card (abbreviated CC below) (does the customer hold a credit card issued by the bank), and the outcome Personal Loan (abbreviated Loan below).

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

a. Create a pivot table for the training data with Online as a column variable, CC as a row variable, and Loan as a secondary row variable. The values inside the table should convey the count. In R, use function melt() and cast(), or function table(). (Refer to page 98 for information on a Pivot Table.)

b. Consider the task of classifying a customer who owns a bank credit card and is actively using online banking services. Looking at the pivot table, what is the probability that this customer will accept the loan offer? [This is the probability of loan acceptance (Loan = 1) conditional on having a bank credit card (CC = 1) and being an active user of online banking services (Online = 1).

c. Create two separate pivot tables for the training data. One will have Loan (rows) as a function of Online (columns) and the other will have Loan (rows) as a function of CC.

d. Compute the following quantities [Recall: P(A | B) means “the probability of A given B”]:

i. P(CC = 1 | Loan = 1) (i.e., the probability of credit card holders among the loan acceptors)

ii. P(Online = 1 | Loan = 1)

iii. P(Loan = 1) (i.e., the proportion of lean acceptors)

iv. P(CC = 1 | Loan = 0)

v. P(Online = 1 | Loan = 0)

vi. P(Loan = 0)

e. Use the quantities computed above in Part (d) to compute the naïve Bayes probability:

P(Loan = 1 | CC = 1, Online = 1)

f. Compare this value with the one obtained from the pivot table in Part (b). Which is a more accurate estimate?

g. Which of the entries in this table are needed for computing

P(Loan = 1 | CC = 1, Online = 1)?

In R, run naïve Bayes on the data. Examine the model output on training data, and find the entry that corresponds to P(Loan = 1 | CC = 1, Online = 1). Compare this to the number you obtained in Part (e) above.

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, computer-science and related others by exploring similar questions and additional content below.

Similar questions