The department of health of a certain state estimates a 10% rate of HIV
at risk | test positive | test negative | total |
HIV | 475 | 25 | 500 |
HIV free | 223 | 4277 | 4500 |
total | 698 | 4302 | 5000 |
b) of those with HIV,postive = 475/500=
0.9500 = 95%
of those with test positive, HIV are = 475/698=
0.6805 or 68.10%
option a) is correct
c)
P(disease|test positive) = P(disease and positive)/P(test positve)= =475/698= 0.6805 or 68.05%
A patient in the at risk category who test postive has a 68.1%
chance of having disease which is greater than the overall at risk
incidence rate of 10%
d)
gen. population | test positive | test negative | total |
HIV | 53 | 7 | 60 |
HIV free | 987 | 18953 | 19940 |
total | 1040 | 18960 | 20000 |
of those with HIV,% of test positive = 53/60*100=
88.3%
of those test positive,% of having HIV=53/1040*100=
5.1%
option b) is correct
e) P(disease| positive) = P(disease and positive)/P(positive) =
5.1%
the chance of patient having HIV is 5.1% ,compared to overall
incidence rate of 0.3%
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