Sequential Screening With Two Diagnostic Kits
A clinic screens for a disease that affects 3% of its patients. Two kits are available. Kit A has sensitivity 0.92 and specificity 0.95. Kit B has sensitivity 0.85 and specificity 0.98. A patient who tests positive on Kit A is automatically tested again with Kit B. The clinic wants to know how much the second positive result raises the probability of disease, and whether the order of the two kits matters. Staff also ask whether Kit B alone would be a better screening choice than the sequential rule. The clinic's statistician warns that the two kits may not be conditionally independent given disease status, because both detect similar biological markers. The clinic must decide whether to report the double-positive probability as a reliable estimate or to present it as an optimistic upper bound.
1. Assuming the two kit results are conditionally independent given disease status, calculate the probability that a patient has the disease after testing positive on Kit A and then positive on Kit B.
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Let D denote disease. The prevalence is P(D) = 0.03 and P(D′) = 0.97.
After a positive Kit A result: P(A+) = 0.03(0.92) + 0.97(0.05) = 0.0276 + 0.0485 = 0.0761. P(D | A+) = 0.0276 / 0.0761 ≈ 0.3627.
Now apply Kit B. Assuming conditional independence: P(B+ | D) = 0.85 and P(B+ | D′) = 0.02.
Numerator: P(D and A+ and B+) = 0.0276 × 0.85 = 0.02346. Denominator: 0.02346 + 0.97 × 0.05 × 0.02 = 0.02346 + 0.00097 = 0.02443.
So P(D | A+ and B+) = 0.02346 / 0.02443 ≈ 0.960, or about 96.0%.
2. Calculate the positive predictive value of Kit B alone and compare it with the sequential result. Explain why the order of the tests does not affect the final posterior if conditional independence holds.
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For Kit B alone: P(B+) = 0.03(0.85) + 0.97(0.02) = 0.0255 + 0.0194 = 0.0449. PPV = P(D | B+) = 0.0255 / 0.0449 ≈ 0.568, or about 56.8%.
The sequential rule gives about 96.0%, much higher than Kit B alone. The reason is that a double positive is a much rarer event among healthy patients, so the false-positive contribution is greatly reduced.
If conditional independence holds, the joint probability of the two results given disease status factorises. Bayes' theorem then gives P(D | A+ and B+) = P(D)P(A+ | D)P(B+ | D) / [P(D)P(A+ | D)P(B+ | D) + P(D′)P(A+ | D′)P(B+ | D′)]. The product in the numerator and denominator is symmetric in A and B, so reversing the order gives the same posterior.
3. State one reason conditional independence may fail. If the tests share the same biological marker, explain whether the calculated double-positive probability is likely to be too high or too low.
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Conditional independence may fail if both kits respond to the same biological marker or if one kit's false positives tend to occur in the same patients as the other kit's false positives. In that case, a positive on Kit A carries information about the likely result on Kit B even after disease status is known.
If the tests' errors are positively correlated, then among healthy patients a positive on A makes a positive on B more likely than the independence calculation assumes. The false-positive term P(A+ and B+ | D′) is then larger than the product used above. Since this term appears in the denominator, the true posterior P(D | A+ and B+) would be lower than 0.960. Thus the independence-based figure should be presented as optimistic, i.e. likely too high, unless the clinic has evidence that the kits err independently.