Using R Studio, complete the following:

Personal Loan Acceptance. 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 $480(=9\mathrm{.}6\%)$ 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 functions melt$()$ and cast$(),$ or function 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 $[$P$($A $$ B$)$ means $$the probability of A given B$]$:

i$.$ P$($CC $=1$ Loan $=1)($the proportion of credit card holders among the loan acceptors$)$

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

iii. P$($Loan $=1)($the proportion of loan acceptors$)$

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

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

vi$.$ P$($Loan $=0)$

e$.$ Use the quantities computed above to compute the naive Bayes probability

P$($Loan $=1$ CC $=1,$ Online $=1).$

f$.$ Compare this value with the one obtained from the pivot table in $($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 naive 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 $($e$).$