4, A certain blood test for some disease gives ‘positive\' result for 90% of pat

Question

4, A certain blood test for some disease gives ‘positive' result for 90% of patient whoare infected with the disease (detection rate Pr(PID)-90%), but also a 'positive' result for 5% of people who are NOT infected (false positive rate Pr(P ) =5% ten thousands (prevalence rate of Pr(D) =0.01%). o). Ihe infection rate among the population is one in a) What is the probability that a person who tested positive' is indeed infected with this disease? Define an'accuracy, measure as: Pr(DIP)×Pr(DIP), what will be its value for this test? The company can invest X number of dollars into R&D; and to improve either the detection rate to 99% OR to reduce the false positive rate to 4% (but not both), what should the company do to improve the test's accuracy the most? b) c)

Explanation / Answer

a)

Pr(D|P) = Pr(D and P)/Pr(P) = 0.90 * 0.01/100 / ( 0.90 * 0.01/100 + (1 - 0.01/100) * 0.05)

= 0.001796

b)

accuracy = Pr (D|P) * Pr(D'|P')

Pr(D'|P') = Pr (D' and P')/Pr(P')

= (1-0.01/100)*0.95 / ( (1-0.01/100)*0.95 + 0.01/100 * 0.10)

= 0.99998947274

hence

accuracy = 0.001796 *0.99998947274 = 0.00179598109

c)

if detection rate 99 %

then

accuracy = 0.99 * 0.01/100 / ( 0.99 * 0.01/100 + (1 - 0.01/100) * 0.05) * (1-0.01/100)*0.95 / ( (1-0.01/100)*0.95 + 0.01/100 * 0.01)

=0.0019762825

if Pr (P|D') = 4%

then accuracy = 0.90 * 0.01/100 / ( 0.90 * 0.01/100 + (1 - 0.01/100) * 0.04) * (1-0.01/100)*0.96 / ( (1-0.01/100)*0.96+ 0.01/100 * 0.10)

=0.00224514948

since in 2nd accuracy is more ,

we should reduce the false positive rate to 4 %

Please rate

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