Question

One percent of all individuals in a certain population are carriers of a particular disease. A diagnostic test for this disease has a 93% detection rate for carriers and a 2% false positive rate. Suppose that an individual is tested. What is the probability that an individual who tests negative does not carry the disease? What is the specificity of the test?

Answer 1

Answer:

(1) The probability that an individual who tests negative does not carry the disease is 0.9709.

(2) The specificity of the test is 98%.

Step-by-step explanation:

Denote the events as follows:

X = a person carries the disease

Y = the test detected the disease.

Given:

$P(X) = 0.01\\P(Y|X)=0.93\\P(Y|X^{c})=0.02$

The probability of a person not carrying the disease is:

$P(X^{c})=1-P(X)=1-0.01=0.99$

The probability that the test does not detects the disease when the person is carrying it is:

$P(Y^{c}|X)=1-P(Y|X)=1-0.93=0.07$

The probability that the test does detects the disease when the person is not carrying it is:

$P(Y^{c}|X^{c})=1-P(Y|X^{c})=1-0.02=0.98$

(1)

Compute the probability that an individual who tests negative does not carry the disease as follows:

$P(X^{c}|Y^{c})=\frac{P(Y^{c}|X^{c})P(X^{c})}{P(Y^{c}|X^{c})P(X^{c})+P(Y^{c}|X)P(X)} \\=\frac{(0.98\times 0.99)}{(0.98\times 0.99)+(0.07\times 0.01)} \\=0.9709$

Thus, the probability that an individual who tests negative does not carry the disease is 0.9709.

(2)

By specificity it implies that how accurate the test is.

Compute the probability of negative result when the person is not a carrier as follows:

$P(Y^{c}|X^{c})=1-P(Y|X^{c})=1-0.02=0.98$

Thus, the specificity of the test is 98%.