A scientist designed a medical test for a certain disease. Among 100 patients wh

Question

A scientist designed a medical test for a certain disease. Among 100 patients who have the disease, the test will show the presence of the disease in 96 eases out of 100, and will fail to show the presence of the disease in the remaining 4 cases out of 100. Among those who do not have the disease, the test will erroneously show the presence of the disease in 3 cases out of 100. and will show there is no disease in the remaining 97 cases out of 100. What is the probability that a patient who tested positive on this test actually has the disease, if it is estimated that 20% of the population has the disease? What is the probability that a patient who tested negative on this test actually docs not have the disease, if it is estimated that 4% of the population has the disease? Reliability of Testing. A certain virus infects one in every 200 people. A test used to detect the virus in a person is positive 80% of the time when the person has the virus and 5% of the time when the person docs not have the virus. Let A be the event "the person is infected" and B be the event "the person tests positive." Using Bayes' Theorem, when a person tests positive, determine the probability that the person is infected Using Bayes' Theorem, when a person tests negative, determine the probability that a person is not infected A drug company is developing a new pregnancy-test kit for use on an outpatient basis. The company uses the pregnancy test on 100 women who arc known to be pregnant for whom 95 test results are positive. The company uses test on 100 other women who arc known to not be pregnant, of whom 99 test negative. What is the sensitivity of the test? What is the specificity of the test? the company anticipates that of the women who will use the pregnancy-test kit. 10% will actually be pregnant. What is the PV (predictive value positive) of the test?

Explanation / Answer

1. D be the event people have disease and ND be the event people dont have disease.

Let S be the success event of Test and Failure event of F

It is given that

P(S/D) =0.96

P(F/D) =0.04

P(S/ND) =0.03

P(F/ND)=0.97

a. P(D/S) = [P(S/D) P(D) ] / [P(S/D)P(D)+P(S/ND)P(ND)]

it is given P(D) =0.2

P(D/S) =(0.96 X 0.2)/ (0.96 X 0.2 +0.03 X0.8) =0.888

The probability the person will have actually have disease who was tested positive

b. It is given here P(D) =4% and P(ND)=96%

P(ND/F) =[P(F/ND) P(ND] )/ [P(F/D)P(D)+P(F/ND)P(ND)]

= (0.97 X0.96) /(0.97 X 0.96 +0.04 X 0.04) = 0.998

2. Let D be the event of infection and ND be the rate of not infection. Let S be the event of success of test and F be the event of negative result in test.

The sub questions under it an be solved using the same formulae as first question.

2-Part 1

Let P be the event of pregnancy and NP be the rate of no pregnancy. Let S be the event of success of test and F be the event of negative result in test.

Sensitivity of the test is to identify true positives. So it is calculated as Pr(T/P) X 100 =95

Specificity of the test is to identify true negatives and avoid false positive. It is calculated as Pr(N/NP) X 100 =99

Part 2.

Positive predictive value is the probability that woman who were tested positive were indeed pregnant.

it is given by formula

PPV= True Positive/ (true positive+ false positive) =95/(95+(100-99)) X100 =95/96 X 100 =98.96

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