This report details the exploratory data analysis and modeling of the Boston housing dataset, focusing on predicting median home values using linear regression and classification and regression trees (CART). The analysis involved data splitting, exploratory analysis, model fitting through various regression techniques, and cross-validation, revealing that regression trees performed best in terms of mean squared error across trials. Additionally, a comparison of logistic regression and CART models highlighted the effectiveness of the CART model based on misclassification rates.

1. Executive Summary:
Goal and background:
This report involves the Exploratory Data Analysis, Linear Regression model fitting and Cross validation
using K-fold CV and Classification and Regression Trees of the Boston Housing Data. Two trails were
conducted by randomly sampling and selecting 80% of the data from the Boston Housing data for training.
Rest of the data was used for testing and validation.
Approach:
The entire Boston Housing data contains 506 observations with 14 variables. The objective is to arrive at
a model that will successfully predict the response variable medv(median value of the owner-occupied
homes in $1000’s) using the other 13 predictor variables. The training data was initially split by taking 80%
of the full data. For the exploratory data analysis, a summary statistic table was generated to analyze the
training data. Further, pairwise correlation plots were obtained in order to find collinearity between the
response and the predictor variables and also for the multicollinearity between the predictor variables.
Then, box-plots were created to identify the presence of potential outliers that might affect the model.
Post EDA, a generalized linear model is generated using the training data, using all of the predictor
variables in-order to later verify the positive effect to step-wise regression and CART. The AIC of the model
is noted. Then, step-wise regression models are created in ‘forward’, ‘backward’ and ‘both’ directions and
the model with the least AIC is chosen for further analysis. The MSE of the model is calculated by testing
the observed model with 20% of the testing data. This process is carried out for two trials with different
testing and training sets.
After step-wise regression models are generated, cross-validation is performed with 5-fold CV on the
entire data. The MSE for this model is also calculated for comparison. Post this, a classification, regression
tree is created from the training data and the model is checked with the testing data.
Key observations and results:
All the steps explained in the approach is carried out for two trials by generating different training and
testing data sets.
The Mean-Square Errors are calculated for the linear regression model with all variables, stepwise
regression, K-fold CV and the regression tree. From the results, we see that the Regression trees provide
the best models in both trials. Also, the large differences in the MSEs of two models indicate that in both
cases, the data is randomly selected and hence point out the veracity of the data.
Technique used MSE (testing Trial 1) MSE (testing Trial 2)
LM (all variables–training MSE) 19.73 23.92
Linear Regression (Stepwise) 31.49 15.05
K-fold validation 23.04 23.22
Regression Tree 20.50 10.76
Comparison of MSE’s of various types of Regression and Cross-validations

2. Executive summary – problem2
Goal and Background:
The goal of this problem is to build logistic regression, CART on a single dataset and compare the results.
Approach and Results:
First, a logistic regression model is built. Best model is obtained by using AIC selection criterion. To test
the goodness of model by best AIC , we have performed a 5 fold validation on the full data set. The AUC
and misclassification rates of k- fold validation and out of sample prediction are noted and compared.
The procedure is repeated for two iterations to take account of sampling selection biases.
The following are the results for two iterations:
Iteration1 Iteration2
5 fold misclassification rate 0.3415372 0.3416461
Out of sample misclassification
rate for AIC best model
0.3299632 0.3455882
AUC of the k-fold 0.8780538 0.8780538
AUC of AIC model on out of
sample
0.8983785 0.873988
Because, 5- fold validation takes 5 random samples and summarizes the results based on these 5
samples, the estimates from 5-fold should be more reliable than the estimates given by BIC out of
sample.
Then, CART model is built on the data. The performance of CART and logistic regression model are
compared using misclassification rate. The following are the results:
Iteration1 Iteration2
Misclassification rate tree 0.2757353 0.3143382
Misclassification rate logit 0.3299632 0.3455882
Following are the classification tables of two methods:
Output of tree:
Predicted
Truth 0 1
0 662 287
1 13 216
Output of logistic:
Predicted
Truth 0 1
0 559 350
1 9 130

3. On the basis of misclassification rate, for the two iteration CART model is better. We can also use ROC as
another performance criterion.
1. Boston Housing data. Random sample a training data set that contains 80% of the original data
points. (You may stay with the same data set from HW2.)
(i) Start with exploratory data analysis. Repeat linear regression as in HW2. Conduct some residual
diagnosis.
The Original Boston dataset contains 506 records and has 14 variables. All the variables are numeric and
there are no missing values in the dataset. A brief description of the variables is given below
Variable Description
CRIM per capita crime rate by town
ZN Proportion of residential land zoned for lots over 25,000 sq.ft.
INDUS proportion of non-retail business acres per town
CHAS Charles River dummy variable (= 1 if tract bounds river; 0 otherwise)
NOX nitric oxides concentration (parts per 10 million)
RM average number of rooms per dwelling
AGE proportion of owner-occupied units built prior to 1940
DIS weighted distances to five Boston employment centers
RAD index of accessibility to radial highways
TAX full-value property-tax rate per $10,000
PTRATIO pupil-teacher ratio by town
BLACK 1000(Bk - 0.63)^2 where Bk is the proportion of blacks by town
LSTAT % lower status of the population
MEDV Median value of owner-occupied homes in $1000's
Table 1.1 Variable description – Boston data
The Boston housing data was randomly sampled twice as follows:
Dataset No. of Observations No. of Variables
Training Data 404 14
Testing Data 102 14
Table 1.2 – Split of training and testing data
The Training data was used to build a suitable regression model and carry out further studies
We started with the Exploratory Data Analysis of the Training Data which is presented next.
1. Exploratory Data Analysis
1.1 Summary Statistics

4. Trial 1 Trial 2
Variable Min Max Median Mean Min Max Median Mean
CRIM 0.01 73.53 0.27 3.62 0.001 73.53 0.261 3.371
ZN 0.00 100.00 0.00 11.14 0.00 100.00 0.00 11.33
INDUS 1.21 27.74 9.90 11.32 0.46 27.74 9.69 11.07
CHAS 0.00 1.00 0.00 0.07 0.00 1.00 0.00 0.07
NOX 0.39 0.87 0.54 0.56 0.39 0.87 0.54 0.56
RM 3.56 8.78 6.21 6.28 3.56 8.78 6.23 6.31
AGE 2.90 100.00 77.95 68.68 2.90 100.00 76.6 68.55
DIS 1.13 12.12 3.17 3.77 1.13 12.13 3.22 3.80
RAD 1.00 24.00 5.00 9.63 1.00 24.00 5.00 9.58
TAX 187.00 711.00 330.00 410.20 187.00 711.00 330.00 407.60
PTRATIO 12.60 22.000 19.10 18.54 12.60 22.00 19.10 18.48
BLACK 0.32 396.900 391.38 356.68 0.32 396.90 391.70 358.42
LSTAT 1.73 36.980 11.64 12.75 1.73 36.98 10.93 12.49
MEDV 5.00 50.000 20.90 22.27 5.00 50.00 21.40 22.86
Table 1.3 Summary statistics
As stated above, the sample data contained 80% observations. There is one dependent variable namely
MEDV and other independent variables. All the variables are numeric in nature.
1.2 Pairwise Correlation
A pairwise correlation matrix was obtained as below. A lot of variables showed high correlation. This
was confirmed from the correlation matrix obtained for this data sample. Few of the highly correlated
values are:
TRIAL 1 TRIAL 2
dis:indus = -0.710 dis:indus = -0.702
dis:age = -0.748 dis:age = -0.753
dis:nox = -0.769 dis:nox = -0.771
tax:indus = 0.706 tax:indus = 0.716
tax:rad = 0.912 tax:rad = 0.909
chas:age = 0.092 chas:age = 0.061
chas:dis = -0.095 chas:dis = -0.087
chas:nox = 0.091 chas:nox = 0.065
Table 1.4 Correlation between variables
This would be taken care of when selecting the variables for the final model.
1.3 Outliers

5. The box plots for entire data are given below. There were many variables that had a number of outliers.
The prominent ones are black, zn and crim
Fig 1.1 Similar boxplots for Trail 1 and Trail 2 for training data.
2. Linear Regression
Linear Regression was performed on the given set of variables. The medv was kept as the response
variable and rest all the variables were treated as predictor variables to obtain a rough linear model. The
relevant statistics are as follows:
a) The significance factor showed indus and age to be least significant predictors (no stars only
blanks)
b) The R2
= 0.759 and R2
Adj = 0.751 (trial 1), R2
= 0.744 and R2
Adj = 0.736 (trial 2) showed that the
rough model is a very good fit for the sample data. But this is not conclusive.
c) MSE = 19.72 (trial 1), MSE = 23.89 (trial 2)
d) AIC criterion = 2381.285 (trial 1), 2458.628 (trial 2)
e) BIC criterion = 2441.306 (trial 1), 2518.649 (trial 2)
The next step would be the Variable selection. We will use this to refine our rough model to obtain a
better fit.
3. Variable Selection
Two techniques were employed to select the correct variables for the regression model

6. a) Best Subsets Analysis – Since the number of variables were few, the best subset analysis was
employed to check the most feasible subset. 14 comparisons with 2 best fits were done. The
results were checked on the BIC and Adj R2
criteria. Below are the plots for the same.
Similar plots were obtained for Trial 1 and Trial 2.
Fig 1.2:BIC criterion
Fig 1.3: Adjusted R2
criterion
Both the criteria indicated towards removal of few variables for the best fit.

7. b) Forward and Stepwise Regression – Both forward and stepwise regression were carried out
on the sample dataset. Both the techniques revealed the same model i.e. with same
predictor variables. The final model obtained was:
medv ~ lstat + rm + ptratio + dis + nox + chas + black + zn + rad + crim +
tax
The estimated coefficients are as below:
Trial 1 Trial 2
(Intercept) 34.03 (Intercept) 37.51
lstat -0.53 lstat -0.53
rm 3.92 rm 3.68
ptratio -0.91 ptratio -0.95
dis -1.35 dis -1.54
nox -15.84 nox -17.75
chas 2.64 chas 3.05
black 0.01 Black 0.01
zn 0.04 zn 0.04
rad 0.27 rad 0.30
crim -0.09 crim -0.11
tax -0.01 tax -0.011
Table 1.5 Estimated coefficients for stepwise-regression
The AIC value for this model was 1228.91 (trial 1) 1306.61 (trial 2) which was much smaller as
compared to our previous rough model and hence hinted that this model is far better than the
model with all variables
The MSE is 19.73 (trial 1), 23.92 (trial 2) which is almost the same as the previous value
4. Residual diagnostics
Residual diagnostics was carried out on the newly build model. The results are as given below

8. Fig 1.4 Residuals vs fitted
The residuals showed a decent fit here. There were few outliers but a majority of data had an even
spread.
Fig 1.5 QQ plot
Again, the QQ plot showed some evident outliers but overall the fit was good as the sample points
coincided with the line.

9. Fig 1.6 Residuals vs leverage
For most points the Cook’s distance is very less and hence it’s a good fit.
The plots are similar for both trial 1 and trial 2
(ii) Test the out-of-sample performance. Using final linear model built from (i) on the 80% of original
data, test with the remaining 20% testing data. (Try predict() function in R.) Report out-of-sample
model MSE etc.
The out-of-sample performance was measured by fitting the linear model obtained above to the 20%
test dataset. The MSE for the test dataset is 31.4860 (trial 1), 15.05 (trial 2).
(iii) Cross validation. Use 5-fold cross validation. (Try cv.glm() function in R on the ORIGINAL 100%
data.) Does (iii) yield similar answer as (ii)?
A 5-fold cross validation was applied on the entire BOSTON dataset. The MSE calculated after the cross
validation was 23.0353 (trial 1), 23.85 (trial 2). This is different to the model obtained by step-wise
regression.
(iv) Fit a regression tree (CART) on the same data; repeat the above steps (i), (ii).

10. Fig 1.7 Classification tree
The out-of-sample performance for the regression tree was measured using the in-sample fit. The MSE
in this case is 20.504 (trial 1), 10.762 (trial 2). This is remarkably close to the value obtained with in
training sample MSE using the linear regression. Thus the regression tree gives the best fitting model for
the testing dataset.
(v) What do you find comparing CART to the linear regression model fits from HW2?
A comparison chart for the 3 methods employed on Boston data is given below
Technique used MSE (testing Trial 1) MSE (testing Trial 2)
Linear Regression (Stepwise) 31.49 15.05
K-fold validation 23.04 23.22
Regression Tree 20.50 10.76
Table 1.6 Comparison of MSE
By applying the three different techniques we found that the Mean Square Error in prediction is least
when we fit the data using Regression tree. The simple linear regression on the other hand had higher
error as compared to the Tree or K-fold validation.
But the above result comes with a caveat. The error may change with different random samples and the
linear regression may perform better with a different sample. Hence we can say that the above results
are indicative and not conclusive. However, all three techniques provide better models than generating
a linear model with all variables.

11. 2(i)
Exploratory data analysis
Below plot shows the distribution of dependent variable DLRSN in the data. As can be seen number of
1’s is close to 750 and number of 0’s is ~4500
Fig 2.1.1 histogram of dependent variable DLRSN

12. Correlation plot for all the variables. As can be seen (R3, R8) and (R9, R10) show high correlation greater
than .6 indicating a possible multi-collinearity among the independent variables. Dependent variable
DLRSN is most linearly correlated to R5 and R6 as observed in the plot.
Figure 2.1.2 Correlation plot for all the variables in bankruptcy data

13. General Linear Regression methods and comparison
Below is the comparison summary of general regression methods on bankruptcy data. With AIC criteria
we can observe that Log-Log method performed well out of the three models.
Model Intercept R1 R2 R3 R4 R5 R6 R7 R8 R9 R10 AIC
Logistic -2.56 0.207 0.584 -0.496
-
0.0817
-
0.0461 0.25 -0.47 -0.289 0.384 -1.63 2433.4
Probit -1.41 0.096 0.32 -0.258
-
0.0129 -0.024 0.0152 -0.204 -0.142 0.201 -0.896 2439.6
Log-
Log -2.54 0.166 0.448 -0.387 -0.141 -0.01 0.152 -0.41 -0.257 0.31 -1.287 2455.2
Fig 2.2.3 Comparison of general linear regression models
2(ii)
Logistic Regression Analysis- feature selection with step wise approach
In Sample Data
 Model selected from step wise regression with AIC criterion is as follows
DLRSN= -2.6 + 0.11*R1 + 0 .62*R2 - 0.45*R3 - 0.16*R4 + 0.26*R6 - 0.41*R7 - 0.36*R8 + 0.32*R9-
1.56*R10
AIC value = 2368.6 which is less than full model AIC of 2370
Mis classification rate using p=1/16 is .331
Mean Residual deviance 2412.2
Confusion matrix is as follows
Predicted
TRUE 0 1
0 2343 1388
1 52 565
Fig 2.2.1Confusion matrix on in sample training data – AIC Model

14. ROC Curve
Fig 2.2.2 AUC curve on in sample training data – AIC model
AUC = .883
 Model selected from step wise regression with BIC criterion is as follows
DLRSN= -2.58 + .62*R2 -0.44*R3 + 0.21*R6 -0.35*R7 -0.35*R8
+ .41*R9 - 1.62*R10
BIC value = 2369.2 which is less than full model BIC of 2440
Mis classification rate using p=1/16 is .332
Mean Residual deviance 2412.2

15. Confusion matrix is as follows
Predicted
TRUE 0 1
0 2339 1392
1 50 567
Fig 2.2.3 Confusion matrix on in sample training data – BIC model
Fig 2.2.4 AUC curve on in sample training data – BIC model
AUC = .882

16. 2 (ii)Logistic Regression - testing model performance with out of sample testing data
Out Sample Data
 Model selected from step wise regression with AIC criterion is as follows
DLRSN= -2.6 + 0.11*R1 + 0 .62*R2 - 0.45*R3 - 0.16*R4 + 0.26*R6 - 0.41*R7 - 0.36*R8 + 0.32*R9-
1.56*R10
Mis classification rate using p=1/16 is .356
Confusion matrix is as follows
Predicted
TRUE 0 1
0 558 371
1 17 142
2.3.1 Confusion Matrix on out of sample data – AIC model

17. Fig 2.3.2 ROC curve on testing out sample data – AIC Model
AUC = 0.859
 Model selected from step wise regression with BIC criterion is as follows
DLRSN= -2.58 + .62*R2 -0.44*R3 + 0.21*R6 -0.35*R7 -0.35*R8
+ .41*R9 - 1.62*R10
Mis classification rate using p=1/16 is 0.361
Confusion matrix is as follows
Predicted
TRUE 0 1
0 553 376
1 17 142
2.3.3 Confusion matrix on testing out sample table – BIC Model
Fig 2.3.4 ROC curve on testing out sample data – BIC model

18. AUC = .857
2(iv)
Identifying optimal cut off probability based on cost function and search grid
Below plot describes the variation of cost with cut off probability
Fig 2.4.1 cut off probability vs cost
We can observe that at cutoff probability of .10 out cost is minimal. This is close to the cut off
probability of .16 we used for our misclassification table.

19. 2 (v) 5 fold cross validation with the model DLRSN ~ R1 + R2 + R3 + R6 + R7 + R8 + R9 + R10 on the full
dataset gives the following results:
Misclassification rate : 0.3415372
The initial cost from step iii is 0.3299632. k- fold validation gives more cost
AUC for k-fold classification rate :0.8770093 and AUC based on part (iii) is 0.8983785. k-fold validation
gives less AUC compared to single out of sample values.
k- fold estimates should be more reliable because they are estimated k-times on full data
Fig-2.v.a shows the ROC of k-fold
2(vi) By using asymmetric cost of 15:1 a classification tree is fitted and it is shown in Fig.2.vi.a

20. 2 vii)
Comparing CART and logistic regression model:
(i) Comparing on the basis of misclassification rate:
Misclassification rate by obtained by tree is 0.2757353 whereas the misclassification rate
obtained by logistic regression is 0.3299632. Hence, on the basis of misclassification rate,
logistic regression is better.
Output of tree:
Predicted
Truth 0 1
0 662 287
1 13 126
Output of logit:
Predicted
True 0 1
0 599 350
1 9 130
(ii) Comparing AUCs:

21. AUC of CART model is 0.8251 whereas AUC of logistic model is 0.8983.
Hence, on the basis of AUCs , logistic regression is better
2 viii) Following is the summary of results for another run
Result Iteration 1 Iteration 2
AIC of best model 2502.027 2491.085
5 fold misclassification rate 0.3415372 0.3416461
Out of sample misclassification
rate for AIC best model
0.3299632 0.3455882
AUC of the k-fold 0.8780538 0.8780538
AUC of AIC model on out of
sample
0.8983785 0.873988
Misclassification rate tree 0.2757353 0.3143382
Misclassification rate logit 0.3299632 0.3455882
AUC of tree 0.8251 0.8231813
AUC of logit 0.87 0.873988
The change in results of k-fold validation is negligible where as we can observe considerable variation on
out of sample results. Hence, k-fold validation is more robust.
On the basis of misclassification rate, CART is better model in two iterations.