Incidence of a rare disease. Only 1 in 1000 adults is afflicted with a rare disease for which a diagnostic test has been developed. The test is such that when an individual actually has the disease, a positive result will occur 99% of the time, whereas an individual without the disease will show a positive test result only 2% of the time (the sensitivity of this test is 99% and the specificity is 98%; in conrast, the Sept. 22, 2012 issue of The Lancet reports that the first at-home HIV test has a sensitivity of only 92% and a specificity of 99.98%). If a randomly selected individual is tested and the result is positive, what is the probability that the individual has the disease? To use Bayes' theorem, let A1 = individual has the disease, A2 = individual does not have the disease, and B = positive test result. Then P(A1) = , P(A2) = , P(B|A1) = , and P(B|A2) = . The tree diagram for this problem is below. Next to each branch corresponding to a positive test result, the (Answers to choose from in drop down menu are: subtraction, division, multiplication, addition) rule yields the recorded probabilities. Therefore, P(B) = 0.00099 + 0.01998 = (entered to five decimal places), from which we haveP(A1|B) = P(A1 ∩ B)P(B) = 0.000990.02097 = (rounded to three decimal places)This result seems counterintuitive; the diagnostic test appears so accurate we expect someone with a positive test result to be highly likely to have the disease, whereas the computed conditional probability is only 0.047. However, the rarity of the disease implies that most positive test results arise from errors rather than from diseased individuals. The probability of having the disease has (Options for drop down menu are: Increased, Decreased) by a multiplicative factor of 47 (from prior ? to posterior 0.047); but to get a further increase in the posterior probability, a diagnostic test with much smaller error rates is needed. Example 2.31Incidence of a rare disease. Only 1 in 1000 adults is afflicted with a rare disease for which a diagnostic test has been developed. The test is such that when an individual actually has the disease, a positive result willoccur 99% of the time, whereas an individual without the disease will show a positive test result only 2% of the time (the sensitivity of this test is 99% and the specificity is 98%; in conrast, the Sept. 22, 2012 issue ofThe Lancet reports that the first at-home HIV test has a sensitivity of only 92% and a specificity of 99.98%). If a randomly selected individual is tested and the result is positive, what is the probability that the individualhas the disease?To use Bayes' theorem, let A, = individual has the disease, A, = individual does not have the disease, and B = positive test result. Then P(A,) = |P(B|A,) =and P(B|A,) =The tree diagram for this problem is below., P(A,) =P(A¡ N B) = .0009999B = +Test.01.001B' = -TestA, = Has diseaseP(A2 N B) = .01998.999A2 = Doesn't have disease02B = +Test98B' = -TestTree diagram for the rare-disease problemNext to each branch corresponding to a positive test result, the --Select---rule yields the recorded probabilities. Therefore, P(B) = 0.00099 + 0.01998 =which we have(entered to five decimal places), fromP(A,\B) =P(A, N B)0.00099(rounded to three decimal places)%3DP(B)0.02097This result seems counterintuitive; the diagnostic test appears so accurate we expect someone with a positive test result to be highly likely to have the disease, whereas the computed conditional probability is only0.047. However, the rarity of the disease implies that most positive test results arise from errors rather than from diseased individuals. The probability of having the disease has --Select--- v by a multiplicative factor of47 (from priorto posterior 0.047); but to get a further increase in the posterior probability, a diagnostic test with much smaller error rates is needed.

Answered: Incidence of a rare disease. Only 1 in…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

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Incidence of a rare disease. Only 1 in 1000 adults is afflicted with a rare disease for which a diagnostic test has been developed. The test is such that when an individual actually has the disease, a positive result will occur 99% of the time, whereas an individual without the disease will show a positive test result only 2% of the time (the sensitivity of this test is 99% and the specificity is 98%; in conrast, the Sept. 22, 2012 issue of The Lancet reports that the first at-home HIV test has a sensitivity of only 92% and a specificity of 99.98%). If a randomly selected individual is tested and the result is positive, what is the probability that the individual has the disease?
To use Bayes' theorem, let A₁ = individual has the disease, A₂ = individual does not have the disease, and B = positive test result. Then P(A₁) = , P(A₂) = , P(B|A₁) = , and P(B|A₂) = . The tree diagram for this problem is below.

Next to each branch corresponding to a positive test result, the (Answers to choose from in drop down menu are: subtraction, division, multiplication, addition) rule yields the recorded probabilities. Therefore, P(B) = 0.00099 + 0.01998 = (entered to five decimal places), from which we have

P(A₁|B) =

P(A₁ ∩ B)

P(B)

0.00099

0.02097

(rounded to three decimal places)

This result seems counterintuitive; the diagnostic test appears so accurate we expect someone with a positive test result to be highly likely to have the disease, whereas the computed conditional probability is only 0.047. However, the rarity of the disease implies that most positive test results arise from errors rather than from diseased individuals. The probability of having the disease has (Options for drop down menu are: Increased, Decreased) by a multiplicative factor of 47 (from prior ? to posterior 0.047); but to get a further increase in the posterior probability, a diagnostic test with much smaller error rates is needed.

Example 2.31
Incidence of a rare disease. Only 1 in 1000 adults is afflicted with a rare disease for which a diagnostic test has been developed. The test is such that when an individual actually has the disease, a positive result will
occur 99% of the time, whereas an individual without the disease will show a positive test result only 2% of the time (the sensitivity of this test is 99% and the specificity is 98%; in conrast, the Sept. 22, 2012 issue of
The Lancet reports that the first at-home HIV test has a sensitivity of only 92% and a specificity of 99.98%). If a randomly selected individual is tested and the result is positive, what is the probability that the individual
has the disease?
To use Bayes' theorem, let A, = individual has the disease, A, = individual does not have the disease, and B = positive test result. Then P(A,) = |
P(B|A,) =
and P(B|A,) =
The tree diagram for this problem is below.
, P(A,) =
P(A¡ N B) = .00099
99
B = +Test
.01
.001
B' = -Test
A, = Has disease
P(A2 N B) = .01998
.999
A2 = Doesn't have disease
02
B = +Test
98
B' = -Test
Tree diagram for the rare-disease problem
Next to each branch corresponding to a positive test result, the --Select---
rule yields the recorded probabilities. Therefore, P(B) = 0.00099 + 0.01998 =
which we have
(entered to five decimal places), from
P(A,\B) =
P(A, N B)
0.00099
(rounded to three decimal places)
%3D
P(B)
0.02097
This result seems counterintuitive; the diagnostic test appears so accurate we expect someone with a positive test result to be highly likely to have the disease, whereas the computed conditional probability is only
0.047. However, the rarity of the disease implies that most positive test results arise from errors rather than from diseased individuals. The probability of having the disease has --Select--- v by a multiplicative factor of
47 (from prior
to posterior 0.047); but to get a further increase in the posterior probability, a diagnostic test with much smaller error rates is needed.

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