Testing for HIV An article in the New York Times some time a

Testing for HIV. An article in the New York Times some time ago reported that college students are beginning to routinely ask to be tested for the AIDS virus. The standard test for the HIV virus is the Elias test, which tests for the HIV antibodies. It is estimated that this has a 99.8% sensitivity and a 99.8% specificity. A 99.8% sensitivity means that, in a large-scale screening test, for every 1000 people tested who have the virus we can expert 998 people to test positive and 2 to have a false negative test. A 99.8% specificity means that, in a large-scale screening test, for every 1000 people tested who do not have the virus we can expert 998 people to have a negative test and 2 to have a false positive test. (a) The New York Times article remarks that it is estimated that about 2 in every 1000 college students have the HIV virus. Assume that a large group of randomly chosen college students, say 100,000, are tested by the Elias test. If a student tests positive, what is the chance that this student has the HIV virus? (b) What would this probably be for a population at high risk, where 5% of the population has the HIV virus?

Solution

Let D is student has the HIV virus; ND is student has no virus

P(D) = 2/1000 = 0.002; P(ND) = 998/1000 = 0.998

P(+ve| D) =0.998 ;P(-ve|D) = 0.002; P(-ve|ND) = 0.998 ; P(+ve|ND) = 0.002

If a student tests positive, the chance that this student has the HIV virus is P(D|+ve)

P(D|+ve)= P( D and +ve) /P(+ve)

P(+ve) = P(D)* P(+ve|D) + P(ND) *P(+ve|ND) = 0.002*0.998 + 0.998 *0.002 = 0.00392

P( D and +ve)=P(D)* P(+ve|D)=0.002*0.998 =0.001996

P(D|+ve)= P( D and +ve) /P(+ve) =0.5

(b) If P(D) = 5% =0.05

then

P(+ve) = P(D)* P(+ve|D) + P(ND) *P(+ve|ND) = 0.05*0.998 + 0.95 *0.002 = 0.0518

P( D and +ve)=P(D)* P(+ve|D)=0.05*0.998 =0.0499

P(D|+ve)= P( D and +ve) /P(+ve) =0.96332