Question

Now let’s see what is implied by the study of Wuhan patients that the WSJ article describes. Here’s another quote:

“A February study of about 1,000 patients in Wuhan, China, who were hospitalized with suspected coronavirus there, where the pandemic began, found that about 60% tested positive using lab tests similar to those available in the U.S. But, almost 90% showed tell-tale signs of the virus in CT scans of their chests, the article, published in the journal Radiology, found, suggesting many patients in the group were testing negative despite active coronavirus infections.”

Here, the population is “patients hospitalized for something that seems like coronavirus”. In that population, 90% of people were infected (if we take the CT scans as definitive). But 60% of this population tested positive. Note that 60% is not the sensitivity of the test, because this includes people who were not infected and who tested positive.

e. Let’s “back out” the sensitivity of the test, instead of assuming a value for sensitivity. Using 0.9 as the prevalence of infection with Covid-19 in the tested population; using 0.99 as the specificity of the test (same as in the previous parts of this problem); and using 0.6 as the fraction of the population who tested positive, calculate the implied sensitivity of the test. HINT: Use the law of total probability.    

f. Continued. In this situation (in the situation of the Wuhan study), what is the probability that a patient who tested negative actually was not infected?

g. Suppose the goal was for the probability to be at least 0.75 that a patient who tested negative actually was not infected, in the conditions of the Wuhan study. What is the minimum value of sensitivity that would allow this goal to be achieved?

Answer 1

e.

P(Infected) = 0.9

P(Not infected) = 1 - P(Infected) = 1 - 0.9 = 0.1

Specificity = 0.99 => P(Negative | Not infected) = 0.99

P(Positive | Not infected) = 1 - P(Negative | Not infected) = 1 - 0.99 = 0.01

P(Positive) = 0.6

Sensitivity = P(Positive | Infected)

By law of total probability,

P(Positive) = P(Positive | Infected) * P(Infected) + P(Positive | Not Infected) * P(Not Infected)

0.6 = Sensitivity * 0.9 + 0.01 * 0.1

Sensitivity = (0.6 - 0.01 * 0.1) / 0.9 = 0.6655556

f.

P(Negative) = 1 - P(Positive) = 1 - 0.6 = 0.4

Probability that a patient who tested negative actually was not infected = P(Not infected | Negative)

= P(Negative | Not infected) * P(Not infected) / P(Negative) (Bayes Theorem)

= 0.99 * 0.1 / 0.4

= 0.2475

g.

Let k be the sensitivity. Then  P(Positive | Infected) = k

P(Not infected | Negative) 0.75

1 - P(Infected | Negative) 0.75

P(Infected | Negative) 0.25

P(Negative | Infected) * P(Infected) / P(Negative) 0.25   (Bayes Theorem)

P(Negative | Infected) * 0.9 / 0.4   0.25  

P(Negative | Infected) 0.1111

1 - P(Positive | Infected) 0.1111

P(Positive | Infected)   1 - 0.1111 = 0.8889

Thus, the minimum value of sensitivity that would allow this goal to be achieved is 0.8889