Caries Screening: Positive Predictive Value Across Prevalences
A school-based dental screening program uses a visual-tactile caries screening test. In School A, 500 children are screened, and 12% are known from a full dental examination to have active caries. The screening test has a sensitivity of 88% and a specificity of 92%. In School B, the same test is used on 500 children, but the true caries prevalence is only 4%. The program director is concerned that positive screening results may be less trustworthy in School B, even though the test itself has not changed. The director also wants to know what specificity would be required in School A to achieve a positive predictive value of at least 80% while keeping sensitivity at 88%. Assume the full examination is a perfect gold standard and that all 500 children in each school are tested.
1. Calculate the positive predictive value and negative predictive value in School A. Use expected counts and show the 2×2 table.
Read the worked answer
Prevalence = 12%, so among 500 children:
- True caries: 500 × 0.12 = 60.
- No caries: 500 × 0.88 = 440.
Apply sensitivity 88% and specificity 92%:
- True positives: 60 × 0.88 = 52.8.
- False negatives: 60 × 0.12 = 7.2.
- False positives: 440 × (1 − 0.92) = 440 × 0.08 = 35.2.
- True negatives: 440 × 0.92 = 404.8.
Positive predictive value (PPV):
PPV = TP ÷ (TP + FP) = 52.8 ÷ (52.8 + 35.2) = 52.8 ÷ 88 = 0.60 = 60%.
Negative predictive value (NPV):
NPV = TN ÷ (TN + FN) = 404.8 ÷ (404.8 + 7.2) = 404.8 ÷ 412 = 0.9825 ≈ 98.3%.
So in School A, 60% of screen positives are true cases, while 98.3% of screen negatives are truly disease-free. Fractional counts are expected values, not actual children.
2. Calculate the positive predictive value in School B and explain why it differs from School A.
Read the worked answer
In School B, prevalence is 4%:
- True caries: 500 × 0.04 = 20.
- No caries: 500 × 0.96 = 480.
- True positives: 20 × 0.88 = 17.6.
- False positives: 480 × 0.08 = 38.4.
PPV = 17.6 ÷ (17.6 + 38.4) = 17.6 ÷ 56 = 0.314 ≈ 31.4%.
NPV = (480 × 0.92) ÷ [(480 × 0.92) + (20 × 0.12)] = 441.6 ÷ (441.6 + 2.4) = 0.9946 ≈ 99.5%.
The test's sensitivity and specificity are unchanged, but PPV falls from 60% to about 31% because the disease is rarer. With only 20 true cases, the same 8% false-positive rate among 480 healthy children produces 38.4 false positives-more than twice the number of true positives. In a low-prevalence population, a positive screen is much more likely to be a false alarm, so confirmatory examination is essential.
3. What specificity would be required in School A to achieve a PPV of at least 80% while keeping sensitivity at 88%?
Read the worked answer
In School A, true positives remain 52.8. For PPV ≥ 0.80:
TP ÷ (TP + FP) ≥ 0.80
52.8 ÷ (52.8 + FP) ≥ 0.80
52.8 ≥ 0.80 × (52.8 + FP)
52.8 ≥ 42.24 + 0.80FP
10.56 ≥ 0.80FP
FP ≤ 13.2.
The 440 healthy children can contribute at most 13.2 false positives, so the false-positive rate must be:
13.2 ÷ 440 = 0.03 = 3%.
Therefore specificity must be at least:
1 − 0.03 = 0.97 = 97%.
At 97% specificity, FP = 13.2 and PPV = 52.8 ÷ (52.8 + 13.2) = 52.8 ÷ 66 = 0.80 exactly. Improving specificity from 92% to 97% would cut false positives by 62.5% and make positives far more trustworthy.