Question

A study was conducted to estimate the sensitivity and specificity of a new

procedure for detecting the presence of a kidney disease among patients

suffering from hypertension. Among the 54 hypertensive patients who

had the kidney disease, the procedure identified the disease for 45 subjects. Among the 83 hypertensive patients who did not have the kidney

disease, the procedure identified the disease for 24 subjects. Consider

a patient chosen from a certain hypertensive population in which the

prevalence of this kidney disease is 8%. If the new procedure identifies

the presence of the kidney disease for this patient, what is the probability

that patient truly has the disease? Assume that the sensitivity and specificity of the procedure remain the same as in the study mentioned above.

Answer 1

Solution

Back-up Theory

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then

Conditional Probability of B given A, denoted by P(B/A) = P(B ∩ A)/P(A)…..................................................…..............….(1)

(1) => P(B ∩ A) = P(B/A) x P(A) or P(A ∩ B) = P(A/B) x P(B) ……………………......................................................……..(1a)

P(B) = {P(B/A) x P(A)} + {P(B/A^C) x P(A^C)}………….........……………………................................................……………….(2)

P(A/B) = P(B/A) x {P(A)/P(B)}……………………………..….....................................................................……………..…….(2a)

Now, to work out the solution,

Let A represent the event that the patient has the kidney disease and B represent the event that the

procedure identifies the disease

Then, trivially, A^C represents the event that the patient does not have the kidney disease and B^C

represents the event that the procedure does not identify the disease

With the above definitions, the given statements translate into probability language as follows:

Among the 54 hypertensive patients who had the kidney disease, the procedure identified the disease

for 45 subjects. => P(B/A) = 45/54 = 5/6 …………………………………………………….................................………… . (3)

Among the 83 hypertensive patients who did not have the kidney disease, the procedure identified

the disease for 24 subjects. => P(B/A^C) = 24/84 = 2/7 ……………………………………….............................………… . (4)

Consider a patient chosen from a certain hypertensive population in which the prevalence of this

kidney disease is 8%. => P(A) = 0.08…………………………………………………………...............................…………. (5)

and hence P(A^C) = 0.92…………………………………………………………..............................…..……………………. (5a)

Now, vide (2),

P(B) = {(5/6) x (0.08)} + {(2/7) x (0.92)} [from (3), (5), (4) and (5a)]

= 0.3295 …………………………………………………………………………………………..............................………….. (6)

If the new procedure identifies the presence of the kidney disease for a patient, probability that patient truly has the disease = P(A/B)

= P(B/A) x {P(A)/P(B)} [vide (2a)]

= (5/6) x 0.08/0.3295 [vide (3), (5) and (6)]

= 0.2023 Answer

DONE