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drek231 [11]
2 years ago
11

A data mining routine has been applied to a transaction dataset and has classified 88 records as fraudulent (30 correctly so) an

d 952 as non fraudulent (920 correctly so). Construct the classification matrix and calculate the error rate.Suppose that this routine has an adjustable cutoff (threshold) mechanism by which you can alter the proportion of records classified as fraudulent. Describe how moving the cutoff up or down would affect the following:a. The classification error rate for records that are truly fraudulentb. The classification error rate for records that are truly non fraudulent

Mathematics
1 answer:
Firlakuza [10]2 years ago
6 0

Answer:

The classification matrix is attached below

Part a

The classification error rate for the records those are truly fraudulent is 65.91%.

Part b

The classification error rate for records that are truly non-fraudulent is 96.64%

Step-by-step explanation:

The classification matrix is obtained as shown below:

The transaction dataset has 30 fraudulent correctly classified records out of 88 records, that is, 30 records are correctly predicted given that an instance is negative.

Also, there would be 88 - 30 = 58 non-fraudulent incorrectly classified records, that is, 58 records are incorrectly predicted given that an instance is positive.

The transaction dataset has 920 non-fraudulent correctly classified records out of 952 records, that is, 920 records are correctly predicted given that an instance is positive.

Also, there would be 952 - 920 = 32 fraudulent incorrectly classified records, that is, 32 records incorrectly predicted given that an instance is negative.

That is,

                                                                            Predicted value

                           Active value                 Fraudulent       Non-fraudulent

                              Fraudlent                         30                       58

                          non-fraudulent                   32                     920

The classification matrix is obtained by using the information related to the transaction data, which is classified into fraudulent records and non-fraudulent records.

The error rate is obtained as shown below:

The error rate is obtained by taking the ratio of \left( {b + c} \right)(b+c) and the total number of records.

The classification matrix is, shown above

The total number of records is, 30 + 58 + 32 + 920 = 1,040

The error rate is,

\begin{array}{c}\\{\rm{Error}}\,{\rm{rate}} = \frac{{b + c}}{{{\rm{Total}}}}\\\\ = \frac{{58 + 32}}{{1,040}}\\\\ = \frac{{90}}{{1,040}}\\\\ = 0.0865\\\end{array}  

The percentage is 0.0865 \times 100 = 8.65

(a)

The classification error rate for the records those are truly fraudulent is obtained by taking the rate ratio of b and \left( {a + b} \right)(a+b) .

The classification error rate for the records those are truly fraudulent is obtained as shown below:

The classification matrix is, shown above and in the attachment

The error rate for truly fraudulent is,

\begin{array}{c}\\FP = \frac{b}{{a + b}}\\\\ = \frac{{58}}{{30 + 58}}\\\\ = \frac{{58}}{{88}}\\\\ = 0.6591\\\end{array}  

The percentage is, 0.6591 \times 100 = 65.91

(b)

The classification error rate for records that are truly non-fraudulent is obtained by taking the ratio of d and \left( {c + d} \right)(c+d) .

The classification error rate for records that are truly non-fraudulent is obtained as shown below:

The classification matrix is, shown in the attachment

The error rate for truly non-fraudulent is,

\begin{array}{c}\\TP = \frac{d}{{c + d}}\\\\ = \frac{{920}}{{32 + 920}}\\\\ = \frac{{920}}{{952}}\\\\ = 0.9664\\\end{array}

The percentage is, 0.9664 \times 100 = 96.64

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Cost of tickets

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Students = $2

Total tickets sold = 280

Total revenue = $1010

Let

x = number of adults tickets

y = number of teachers tickets

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x + y + z = 280

6x + 4y + 2z = 1010

If the number of adult tickets sold was twice the number of teacher tickets

x = 2y

Substitute x=2y into the equations

x + y + z = 280

6x + 4y + 2z = 1010

2y + y + z = 280

6(2y) + 4y + 2z = 1010

3y + z = 280

12y + 4y + 2z = 1010

3y + z = 280 (1)

16y + 2z = 1010 (2)

Multiply (1) by 2

6y + 2z = 560 (3)

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Subtract (3) from (2)

16y - 6y = 1010 - 560

10y = 450

Divide both sides by 10

y = 450/10

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y = 45

Substitute y=45 into (1)

3y + z = 280

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z = 280 - 135

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z = 145

Substitute the values of y and z into

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x + 45 + 145 = 280

x + 190 = 280

x = 280 - 190

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x = 90

Therefore,

number of adults tickets sold = x = 90

number of teachers tickets = y = 45

number of students tickets = z = 145

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