Question

A test for diabetes classifies 99% of people with the disease as diabetic and 10% of those who don't have the disease as diabetic. It is known that 12% of the population is diabetic.

a) what are the false positive and false negative rates?

b) what is the probability that someone classified as diabetic does in fact have the disease?

i) solve the problem by drawing up a contingency table and

ii) solve the problem using conditional probability and the law of total probability

Answer 1

A test for diabetes classifies 99% of people with the disease as diabetic and 10% of those who don't have the disease as diabetic. It is known that 12% of the population is diabetic.

P(Positive | Disease) = 99%

P(Negative | Disease)= 1 - 99% = 1% ................using P(A|B) = 1- P(A' |B)

P(Positive | No Disease) = 10%

P(negative | No Disease) = 1- 10% = 90%

P(Disease) = 12%

Therefore P(No disease) = 1- 12% = 88%

Before drawing the contingency table we can create a tree diagram

							Intersecting
	Disease			Test
	P(D)	12%		P(+ve|D)	99%		P(D and +ve)	0.1188
				P(-ve|D)	1%		P(-ve and D)	0.0012
	P(D')	88%		P(+ve|D')	10%		P(D' and +ve)	0.088
				P(-ve|D')	90%		P(D' and -ve)	0.792

i) solve the problem by drawing up a contingency table and

	Disease	No disease	Total
Positive	11.88%	8.8%  FP	20.68%
Negative	0.12%  FN	79.2%	79.32%
total	12%	88%	100%

a) what are the false positive and false negative rates?

b) what is the probability that someone classified as diabetic does in fact have the disease?

P(disease | Positive) = (0.1188 / 0.2068)

= 0.5745

ii) solve the problem using conditional probability and the law of total probability

Therefore

a) what are the false positive and false negative rates?

False positive = People who test positive but do not have the disease.

= P(Positve and no disease)

= P(Positive | No Disease) *P(No disease)

= 10% * 88%

False positive = 0.088 = 8.8%

false negative = P(Negative and disease )

= P(Negative | diease) * P(Disease)

= (1 - P(positive | disease)] * P(Disease)

= 1% * 12%

False negative = 0.0012 = 0.12%

b) what is the probability that someone classified as diabetic does in fact have the disease?

P(Disease | Positive) = P(Positive and Disease) / P(Positive )

P(Positive and Disease) = P(Positive | Disease) * P(Disease)

= 99% * 12%

= 0.1188

P( Positve) = P(Positive and Disease) + P(Positive and No Disease)

= 0.2068

P(Disease | Positive) = 0.1188 / 0.2068

P(Disease | Positive) = 0.5745