Complete Question:
Air traffic controllers are required to undergo random drug testing. A urine test detects the presence of amphetamines and barbiturates. The sensitivity of the test is known to be .96 and the specificity is .93. Based on past drug testing of air traffic controllers, the FAA reports that the probability of drug use at a given time is .007.
a.What is the probability that a randomly selected air traffic controller will test positive? (If the test comes back positive, the individual will have to undergo a second test that is more accurate, but more expensive than the urine test.)
b.Find the probability that a person truly uses drugs, given that his test is positive, i.e.,find the positive predictive value.21
c. Find the probability that the individual is not a drug user given that the test result is negative i.e. negative predictive valued.
d. How does the positive predictive value change if the prevalence rate increases to 15%?
e. d. How does the negative predictive value change if the prevalence rate increases to 15%?
Answer:
answer for a = 0.07623
answer for b = 0.088
answer for c = 0.9997
answer for d = them positive predictive value increases with increase prevalence rate
answer for e = the negative predictive value decreases with increase prevalence rate
Step-by-step explanation:
Tree Diagram for Air traffic controllers
Sensitivity = 0.96 = P (+T|D) = probability of the test being positive or the probability of drug use at that time
Specificity = 0.93 = P (-T|U) = probability of the test being negative or the probability of no drug
Use at that time.
Note that this symbol | stands for OR and this symbol stands for ^ AND
P (D) = probability of drug use at a given time = 0.007
P (U) = probability of no drug use at a given time =1 - 0.007 = 0.993
The Tree diagram is shown below(The first uploaded image)
Where P (-T | D) is the probability that the test will be negative or the probability that there is drug use at that time = 1 - P (+T | D) = 0.04
And P (+T | U) is the probability that the test I positive or the probability that there is no drug use at that time = 1 - P (-T | U) = 0.07
We have that:
P (+T ^ D) = 0.96 *0.007 = 0.00672
P (-T ^ D) = 0.04 * 0.007 = 0.0028
P (+T ^ U) = 0.07 * 0.993 = 0.6951
P (-T ^ U) = 0.93 * 0.993 = 0.92349
Contingency Table for Air traffic :
The table is shown on the below (i.e the second uploaded image)
Considering question a
P (+T) = probability of text being positive = P (+T ^ D) + P (+T ^ U) = 0.00672 + 0.6951 = 0.7623
Or from the table P(+T) = 0.7623/1 = 0.7623
Considering question b
From the question we can see that we are looking for
P(D | +T) = P(+T ^ D) ÷ P(+T)
= 0.00672 ÷ 0.7623 = 0.088
Considering Question C
From the question we can see that we are looking for
P(U | -T) = P(-T ^ U) ÷ P(-T)
Where P(-T) = P (-T ^ D) + P (-T ^ U) = 0.0028 + 0.92349 = 0.92377
Hence,
P(U | -T) = 0.92349 ÷ 0.92377 = 0.9997
Considering Question d
increase in the prevalence rate simply means that the probability of the drug use increased from 0.007 to 0.15
Now we have that
P (+T ^ D) = 0.96 *0.15 = 0.144
P (-T ^ D) = 0.04 * 0.15 = 0.006
P (+T ^ U) = 0.07 * 0.85 = 0.0595
P (-T ^ U) = 0.93 * 0.85 = 0.7905
For this change the positive predictive value will be
P(D | +T) = P(+T ^ D) ÷ P(+T)
Where P(+T) = P (+T ^ D) + P (+T ^ U) = 0.144 + 0.0595 = 1.4995
P(D | +T) = P(+T ^ D) ÷ P(+T) = 0.144 ÷ 1.4995
= 0.7076
From this value we see that as the prevalence rate increase the (considering 15%) the positive predictive value increases from 0.088 to 0.7076
Considering Question e
From the question we can see that we are looking for
P(U | -T) at an increase prevalence of 15%
P(U | -T) = P(-T ^ U) ÷ P(-T)
Where P(-T) = P (-T ^ D) + P (-T ^ U) = 0.006 + 0.7905 = 0.7950
Hence,
P(U | -T) = 0.7905 ÷ 0.7950 = 0.992
In light of the answer above compared to the previous answer we see that the negative predictive value decreases with increase prevalence rate
