Two screening tests for colorectal cancer are used in series: every person positive on faecal occult blood test (specificity 90%) then undergoes colonoscopy (specificity 95%) as the second test. The final classification of disease requires both tests to be positive. What is the net specificity of this sequential strategy?

• A 56%
• B 85.5%
• C 90%
• D 99.5% ✓

Explanation

When two tests must both be positive, a non-diseased person is labelled positive only if both tests falsely turn positive: probability = 0.10 x 0.05 = 0.005. Net specificity = 1 - 0.005 = 0.995, that is 99.5%. Sequential testing raises overall specificity at the cost of sensitivity, since net sensitivity would be the product of the two sensitivities. Option B results from averaging the specificities instead of combining them multiplicatively.

Reference: Park's Textbook of Preventive and Social Medicine, 27th ed.

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