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

Firlakuza [10]

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, $0.9664 \times 100 = 96.64$

6 0

2 years ago

Check all solutions to the equation. If there are no solutions, check "None."

Margaret [11]

X can be 4 or -4

As a minus by a minus is a positive

8 0

2 years ago

Josiah, the manager of a pet store, bought 72 oz of water conditioner to use in his fish tanks. His small tanks would require 2

kodGreya [7K]

If we are going to analyze the problem very well, the equation that we will be able to make would be: 2x + 6y = 72

72 is the oz of water conditioner to use in his fish tanks.

small tanks would require 2 oz of conditioner

large tanks would require 6 oz of conditioner

There is only one point which satisfy's the above equation.That is the points of (0, 12) and (36, 0)

6 0

3 years ago

People attending a football game either supported the home team or the visiting team. If 1369 of the people attending supported

Anna [14]

Answer:

37%

Step-by-step explanation:

To find the percentage of people that supported the home team, we need to find the total number of people.

This is because the percentage of people supporting the home team will be given by:

Number of people supporting home team / total number of people

The total number of people is the people supporting the home team plus the people supporting the visiting team:

1369+2331=3700

Therefore, our percentage will be:

1369/3700

Plug this into our calculator and we get:

0.37

Convert that a percentage for a final answer of 37%

7 0

2 years ago

Read 2 more answers

Two different suppliers, A and B, provide a manufacturer with the same part. All supplies of this part are kept in a large bin.

koban [17]

Answer:

The probability of selecting a non-defective part provided by supplier A is 0.807.

Step-by-step explanation:

Let A = a part is supplied by supplier A, B = a part is supplied by supplier B and D = a part is defective.

Given:

P (D|A) = 0.05, P(D|B) = 0.09

A supplies four times as many parts as B, i.e. n (A) = 4 and n (B) = 1.

Then the probability of event A and B is:

$P(A)=\frac{n(A)}{n(A)+N(B)}= \frac{4}{4+1}=0.80\\P(B)\frac{n(B)}{n(A)+N(B)}= \frac{1}{4+1}=0.20$

Compute the probability of selecting a defective product:

$P(D)=P(D|A)P(A)+P(D|B)P(B)\\=(0.05\times0.80)+(0.09\times0.20)\\=0.058$

The probability of selecting a non-defective part provided by supplier A is:

$P(A|D')=\frac{P(D'|A)P(A)}{P(D')} = \frac{(1-P(D|A))P(A)}{1-P(D)}\\=\frac{(1-0.05)\times0.80}{(1-0.058)}\\ =0.80679\\\approx0.807$

Thus, the probability of selecting a non-defective part provided by supplier A is 0.807.

5 0

3 years ago