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Difficulty: HardDiagnostic and Screening Test Performance

A gastroenterology research group is evaluating a novel point-of-care salivary assay for celiac disease. During initial clinical validation at a tertiary care celiac referral center, the assay demonstrated a sensitivity of 90%90\% and a specificity of 90%90\%. Based on the specific patient population at this referral center, the positive predictive value of the assay was calculated to be 90%90\%. Following regulatory approval, a large regional primary care network implements the assay for routine screening of asymptomatic adults. Epidemiological data confirms that the true prevalence of celiac disease in this primary care network is exactly one-fifth (15\frac{1}{5}) of the prevalence observed at the tertiary care referral center. If a patient in the primary care network receives a positive result on this salivary assay, what is the probability that the patient actually has celiac disease?

  1. A
    10%10\%
  2. B
    18%18\%
  3. 50%50\%Answer
  4. D
    69%69\%
  5. E
    90%90\%

Answer

50%50\%
The question asks for the probability that a patient has the disease given a positive test result, which is the definition of positive predictive value (PPV). First, we must find the initial prevalence (pp) at the referral center. Using the formula PPV=Sensitivity×p(Sensitivity×p)+((1−Specificity)×(1−p))PPV = \frac{Sensitivity \times p}{(Sensitivity \times p) + ((1 - Specificity) \times (1 - p))}, we substitute the known values: 0.90=0.90p0.90p+0.10(1−p)0.90 = \frac{0.90p}{0.90p + 0.10(1 - p)}. Solving this algebraic equation yields 0.81p+0.09=0.90p0.81p + 0.09 = 0.90p, which simplifies to p=0.50p = 0.50 (a 50%50\% prevalence). Next, we determine the prevalence in the primary care network, which is stated to be one-fifth of the referral center's prevalence: 0.50×15=0.100.50 \times \frac{1}{5} = 0.10 (10%10\%). Finally, we calculate the new PPV using this 10%10\% prevalence: PPV=0.90×0.10(0.90×0.10)+(0.10×0.90)=0.090.09+0.09=0.090.18=0.50PPV = \frac{0.90 \times 0.10}{(0.90 \times 0.10) + (0.10 \times 0.90)} = \frac{0.09}{0.09 + 0.09} = \frac{0.09}{0.18} = 0.50. Therefore, there is a 50%50\% probability that a patient with a positive test in the primary care setting actually has celiac disease.

Step-by-Step Solution

1
Calculate the true prevalence of the disease at the tertiary care referral center.
The prevalence is 50%50\%.
Using Bayes' theorem and the known PPV (90%90\%), Sensitivity (90%90\%), and Specificity (90%90\%), we solve for prevalence (pp): 0.90=0.90p0.90p+0.10(1−p)0.90 = \frac{0.90p}{0.90p + 0.10(1 - p)}. This simplifies to p=0.50p = 0.50.
2
Determine the disease prevalence in the new primary care network.
The new prevalence is 10%10\%.
The prompt states the prevalence in the primary care setting is one-fifth (15\frac{1}{5}) of the referral center's prevalence. 0.50×0.20=0.100.50 \times 0.20 = 0.10.
3
Calculate the new positive predictive value (PPV) for the primary care network.
The new PPV is 50%50\%.
Apply the PPV formula with the new prevalence (10%10\%): PPV=0.90×0.10(0.90×0.10)+(0.10×0.90)=0.090.09+0.09=0.090.18=0.50PPV = \frac{0.90 \times 0.10}{(0.90 \times 0.10) + (0.10 \times 0.90)} = \frac{0.09}{0.09 + 0.09} = \frac{0.09}{0.18} = 0.50.

Key Concept

The positive predictive value (PPV) of a diagnostic test is highly dependent on the disease prevalence in the tested population. To find a new PPV when moving a test to a different population, the original prevalence must first be mathematically derived if not explicitly provided.
Estimated Time:2m 30s
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Diagnostic and Screening Test Performance

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