Tests for diseases can sometimes be wrong there are two type

Tests for diseases can sometimes be wrong, there are two types of errors that can occur; indicating a disease is present when it isn’t, or failing to indicate a disease is present when it is. The table below gives some results for one such test:

(a) What is the probability that the disease is present given that the test is positive?

(b) What is the probability that the disease is not present given that the test is negative?

(c) Without making any changes to the test, what can be done to improve the accuracy of the testing procedure to determine that a patient either has or doesn’t have the disease?

Tpositive test T- negative test D+ disease present D-= no disease present D+ D- Total T+ 25 1439 T 18 78 96 Total 43 92 135

Solution

(a)
P(D+ | T+)=25/39= 0.6410256
(b)
P(D- | T-)=78/96= 0.8125
(c)
Consider the test as a pair of two individual tests. Accept the overall test as positive if both the individual tests are positive. Accept the overall test as -ive if both the individual tests are -ive. Discard the test results if they are not the same and continue till you get a air of test results such that either both are +ve or both are -ve.
The sample space is { (T+,T+), (T-,T-) }

P((T+,T+) | D+)=P(T+ | D+) * P(T+ | D+)=25/43*25/43
P((T+,T+) | D-)=P(T+ | D-) * P(T+ | D-)=14/92*14/92

So, P(D+ | (T+,T+))=P((T+,T+) | D+)*P(D+) / (P((T+,T+) | D+)*P(D+) + P((T+,T+) | D-)*P(D-))
= (25/43*25/43*43/135) / (25/43*25/43*43/135 + 14/92*14/92*92/135 )=0.8721636

Clearly, P(D+ | (T+,T+)) > P(D+ | T+)
Hence, the test is more accurate when it gives positive.

Similarly,
P(D- | (T-,T-))=P((T-,T-) | D-)*P(D-) / (P((T-,T-) | D-)*P(D-) + P((T-,T-) | D+)*P(D+))
=(78/92*78/92*92/135) / (78/92*78/92*92/135 + 18/43*18/43*43/135)= 0.8977146

Again, P(D- | (T-,T-)) > P(D- | T-)

Hence, the performance of the test is improved thus. The new test is more accurate.