Rapid antigen testing in two prevalence groups
A community clinic evaluates a new rapid antigen test for a respiratory virus. In a validation study, the test had sensitivity 0.85 and specificity 0.92. The clinic plans to use the test in two groups: Group A has prevalence 4% (routine screening), and Group B has prevalence 20% (symptomatic contacts). The clinic has 2,500 tests and expects 1,000 people in Group A and 300 people in Group B during the next month. A nurse claims that because the test is '85% accurate,' a positive result means an 85% chance of infection in both groups. The clinic director asks for a quantitative comparison of the positive predictive value (PPV) in the two groups and for the expected number of false positives in Group A if all 1,000 people are tested. Assume test performance is the same in both groups and that prevalence estimates are correct.
1. Calculate PPV in Group A and Group B. Show the arithmetic.
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Use PPV = (Se × p) / [(Se × p) + (1 − Sp)(1 − p)].
Group A: Se × p = 0.85 × 0.04 = 0.034; false-positive term = 0.08 × 0.96 = 0.0768. PPV = 0.034 / (0.034 + 0.0768) = 0.034 / 0.1108 ≈ 0.307, or 30.7%.
Group B: 0.85 × 0.20 = 0.17; 0.08 × 0.80 = 0.064. PPV = 0.17 / 0.234 ≈ 0.727, or 72.7%.
The nurse's 85% claim confuses sensitivity with PPV; PPV depends strongly on prevalence.
2. If all 1,000 Group A people are tested, how many true positives and false positives are expected? Interpret.
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True positives = 1000 × 0.04 × 0.85 = 34. False positives = 1000 × 0.96 × 0.08 = 76.8 ≈ 77. Expected total positives = 34 + 77 = 111, so PPV ≈ 34/111 = 30.6%.
Most positive results in Group A would be false positives; a positive test should trigger confirmatory testing, especially in low-prevalence screening. This is why the nurse's statement is wrong even though sensitivity is 85%.
3. The clinic has 2,500 tests and expects 1,000 Group A plus 300 Group B people. If it pools 5 Group A samples per pool and resolves positive pools individually, what is the expected number of tests for Group A? Compare with individual testing.
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Pool size 5: 1000/5 = 200 pools. Pool positivity = 1 − (1 − 0.04)^5 = 1 − 0.96^5 ≈ 1 − 0.8154 = 0.1846. Expected positive pools = 200 × 0.1846 ≈ 36.9. Resolve each positive pool with 5 individual tests: 36.9 × 5 ≈ 184.6. Total Group A tests ≈ 200 + 184.6 = 384.6, about 385.
Add Group B's 300 individual tests: total ≈ 685, well below 2,500. Individual testing would require 1,300 tests. Pooling saves tests but can dilute low-positive samples and change sensitivity; the calculation assumes pooling does not change test performance.