Question

A new screening test for Lyme disease is developed for use in the general population. The sensitivity and specificity of the new test are 60% and 70%, respectively. Three hundred people are screened at a clinic during the first year the new test is implemented. Assume the true prevalence of Lyme disease among clinic attendees is 10%. The predictive value of a positive test is:

Answer 1

The predictive value of a positive test is 18.2%.

Explanation:

The population screened is 300.

The true prevalence value of the Lyme disease among clinic attendees is 10\% \Rightarrow 0.10.10% = 0.10.

The proportion that sensitivity is 60\% \Rightarrow 0.6060% = 0.60

The proportion of the specificity is 70\% \Rightarrow 0.70.70% = 0.70

Step 2 of 2

We have to calculate the probability value of the predictive value of positive test.

\begin{array}{c}\\\left( \begin{array}{l}\\{\rm{Positive Predictive }}\\\\{\rm{value}}\left( {PPV} \right)\\\end{array} \right) = \frac{{{\rm{Prevalence}} \times {\rm{Sensitivity}}}}{{\left( {{\rm{Prevalence}} \times {\rm{Sensitivity}}} \right) + \left( {{\rm{1 - Prevalence}} \times 1 - {\rm{Specificity}}} \right)}}\\\\ = \frac{{0.1 \times 0.60}}{{\left( {0.1 \times 0.60} \right) + \left( {1 - 0.1 \times 1 - 0.70} \right)}}\\\\ = \frac{{0.06}}{{0.33}}\\\\ = 0.181818\\\\ = 0.182{\rm{ }}\left( {{\rm{Round to 3 decimal place}}} \right)\\\end{array}

(Positive Predictive value, PPV) =

(Prevalence×Sensitivity)/Prevalence×Sensitivity +(1−Prevalence×1−Specificity)

= (0.1×0.60)/0.1×0.60 + (1−0.1×1−0.70) ​

= 0.06 /0.33

= 0.181818

(Positive Predictive value, PPV) = 0.182 (3decimals)

We have to convert the PPV into percentage,

\begin{array}{c}\\PPV = 0.182 \times 100\\\\ = 18.2\% \\\end{array}

PPV=0.182×100

= 18.2% ​

Therefore, the predictive positive value is 18.2%.