Question

An article in the NewYork Times (February 17, 1999) about the PSA blood test for detecting prostate cancer stated that, of men who had this disease, the test fails to detect prostate cancer in 1 in 4 (so called false-negative results), and of men who did not have it, as many as two-thirds receive false-positive results. Let C (C) denote the event of having (not having) prostate cancer and let +(-) denote a positive (negative) test result.

a. Which is true: P(-|C) = 1/4 or P(C|-) = 1/4? P(CC|+) = 2/3 or P(+|CC) = 2/3?

b. What is the sensitivity of this test?

c. Of men who take the PSA test, suppose P(C) = 0.01. Find the cell probabilities in the 2 × 2 table for the joint distribution that cross classifies Y = diagnosis (+,-) with X = true disease status (C,??).

d. Using (c), find the marginal distribution for the diagnosis.

e. Using (c) and (d), find P(C|+), and interpret.

Answer 1

Solution:

a) The test fails to detect prostate cancer in 1 in 4 means P(-|C) =P(-C) / P(C) =1/4

Two-thirds receive false-positive results means P(+|)=P(+) / P(C)=2/3

So,P(-|C) = 1/4 and P(+|) = 2/3 are true.

b) Sensitivity is the proportion of patients with disease who test positive.

So,P(+|C)=1-P(-|C)=1-(1/4)=3/4

c) Of men who take the PSA test, suppose P(C) = 0.01.

We know P(C) = 0.01, its complement is P() = 1-P(C)=1-0.01=0.99. These are the total(3rd column)

From part (a), we have P(-C) = P(-|C) *P(C)=1/4*0.01=0.0025(1st row)

So, P(+C)=0.01-0.0025=0.0075(1st row)

P(-|)=1-P(+|)=1-(2/3)=1/3

P(-C) =P(-|C)* P(C)=1/3*0.99=0.33

So,P(+|)=0.99-0.33=0.66

P(+)=0.0075+0.66=0.6675 and P(-)=0.0025+0.33=0.3325

Using the above let us fill the below table and is as follows.

Status of disease	+	-	Total
C	0.0075	0.0025	0.01
	0.66	0.33	0.99
Total	0.6675	0.3325	1

d) From the above table marginal distribution for the diagnosis is P(+)=0.6675 and P(-)=0.3325