Question

Naive Bayes Theorem

See the dataset D in Table 1. It consists of clinical data about 14 patients. Using the data in D, determine the Naive Bayes classifier and predict the patients in Table 2. Then, compare with your ‘predicted’ ones with the ground-truth label (i.e., column ’Disease’) and report the accuracy P.

Table 1: Dataset D with clinical data of 14 patients

ID	HBP	BMI	Drink	Weight	Disease
1	“Yes”	“Normal”	“No”	“Overweight”	“Yes”
2	“No”	“Normal”	“Yes”	“Normal”	“No”
3	“No”	“Critical”	“No”	“Overweight”	“Yes”
4	“No”	“High”	“Yes”	“Overweight”	“Yes”
5	“Yes”	“Critical”	“Yes”	“Obese”	“Yes”
6	“Yes”	“High”	“Yes”	“Normal”	“Yes”
7	“No”	“High”	“No”	“Obese”	“No”
8	“Yes”	“Normal”	“Yes”	“Normal”	“Yes”
9	“Yes”	“Critical”	“No”	“Obese”	“Yes”
10	“No”	“Normal”	“No”	“Overweight”	“No”
11	“No”	“Critical”	“Yes”	“Normal”	“Yes”
12	“Yes”	“High”	“No”	“Overweight”	“No”
13	“Yes”	“Normal”	“Yes”	“Overweight”	“Yes”
14	“Yes”	“High”	“No”	“Obese”	“No”

Table 2: Test data with additional 5 patients

ID	HBP	BMI	Drink	Weight	Disease
15	“Yes”	“Normal”	“No”	“Overweight”	“Yes”
16	“No”	“Normal”	“Yes”	“Normal”	“No”
17	“No”	“Critical”	“No”	“Overweight”	“Yes”
18	“No”	“High”	“Yes”	“Overweight”	“Yes”
19	“Yes”	“Critical”	“Yes”	“Obese”	“Yes”

Answer 1

In table 1 from Disease column, P(Yes) = 9/14, P(No) = 5/14

The table 1 can be split as (For Yes and No):

Id

HBP

BMI

Drink

Weight

Disease

Id

HBP

BMI

Drink

Weight

Disease

1

Yes

Normal

No

Overweight

Yes

2

No

Normal

Yes

Normal

No

3

No

Critical

No

Overweight

Yes

7

No

High

No

Obese

No

4

No

High

Yes

Overweight

Yes

10

No

Normal

No

Overweight

No

5

Yes

Critical

Yes

Obese

Yes

12

Yes

High

No

Overweight

No

6

Yes

High

Yes

Normal

Yes

14

Yes

High

No

Obese

No

8

Yes

Normal

Yes

Normal

Yes

9

Yes

Critical

No

Obese

Yes

11

No

Critical

Yes

Normal

Yes

13

Yes

Normal

Yes

Overweight

Yes

Now, from test data Table 2,

Row 1: P(X) = (8/14)*(5/14)*(7/14)*(6/14) = 0.044.

P(Disease = Yes | X) = 0.0000344/0.044 = 0.00078

P(Disease = No | X) = 0/0.044 = 0

Predicted Row 1: Yes

Similarly,

since, the probability P(X | Disease = No) P(Disease = No) = 0, for rest all other rows the prediction for the disease will be ‘Yes’ only. Hence ID 16 is misclassified here.

So, the accuracy = (4/5)*100 = 80%