Question

Drug testing in the workplace. In Canada, the Supreme Court recently ruled that employees can be tested for drugs only if management has reasonable cause to administer the test. An article in Chance(Vol. 28, 2015) focused on the misclassification rates of such drug tests. A false positiveoccurs when a drug test administered to a non– drug user yields a positive result. A false negativeoccurs when a drug test administered to a drug user yields a negative result.

1. The author presented the following scenario: Suppose that 5% of a population of workers consists of drug users. The drug test has a false positive rate (i.e., the probability of a positive drug test given the worker is a non–drug user) of 5% and a false negative rate (i.e., the probability of a negative drug test given the worker is a drug user) of 5%. A worker selected from this population is drug tested and found to have a positive result. What is the probability that the worker is a non–drug user? Apply Bayes’ Rule to obtain your answer.
2. Recall that in Canada, drug tests can be administered only with probable cause. Hence, for workers that are tested, it is likely that a high proportion of them use drugs. Now consider a population of workers in which 95% are drug users. Again, assume both the false positive rate and false negative rate of the drug test are 5%. A worker selected from this population is drug tested and found to have a positive result. What is the probability that the worker is a non–drug user?

Short answer

1. The probability that an employee is not a drug user is 0.5.
2. The probability that a worker is not a drug user is 0.0028.

Step-by-step solution

Important formula

The formula for probability is$P=\frac{favourableoutcomes}{totaloutcomes}$

The Baye’s formula is

role="math" localid="1659703278889" $\begin{array}{rcl}\mathit{P}\left(\frac{B_i}{A}\right)& \mathbf{=}& \frac{\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{i}}\mathbf{\cap }\mathbf{A}\mathbf{)}}{\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}}\\ & \mathbf{=}& \frac{\mathbf{P}\mathbf{\left(}{\mathbf{B}}_{\mathbf{i}}\mathbf{\right)}\mathbf{P}\left(\frac{A}{B_i}\right)}{\mathbf{P}\mathbf{\left(}{\mathbf{B}}_{\mathbf{1}}\mathbf{\right)}\mathbf{P}\left(\frac{A}{B_1}\right)\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{B}}_{\mathbf{2}}\mathbf{\right)}\mathbf{P}\left(\frac{A}{B_2}\right)\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{B}}_{\mathbf{k}}\mathbf{\right)}\mathbf{P}\left(\frac{A}{B_k}\right)}\\ & & \end{array}$

(a)Step 2: The probability that the worker is a non–drug user.

a.

Here,

D=Drug user.

T=Drug test is positive

Apply the Baye’s formula:

$\begin{array}{rcl}P\left(D^c\right|T)& =& \frac{P(D^c\cap T)}{P\left(T\right)}\\ & =& \frac{P\left(T\right|D^c)\times P(D^c)}{P\left(T\right)}\\ & & \end{array}$

Now find the value of P(T),

$\begin{array}{rcl}P\text{(}T\text{)}& =& P\text{(}{D}^c\cap T\text{)}+P\text{(}D\cap T\text{)}\\ & =& \text{[}P\text{(}T|D^c\text{)}\times P\text{(}{D}^c\text{)]}+\text{[}P\text{(}T|D\text{)}\times P\text{(}D\text{)]}\\ & =& \text{[(}1-P\text{(}D\text{)}P\text{(}T|D^c\text{)]}+\text{[(}1-P\text{(}T|D\text{))}\times P\text{(}D\text{)]}\\ & =& \text{[(}1-0.05\text{)(}0.05\text{)]}+\text{[}0.05\text{(}1-0.05\text{)]}\\ & =& 0.095\\ & & \end{array}$

Then,

$\begin{array}{rcl}P\text{(}{D}^c|T\text{)}& =& \frac{0.05\text{(}1-0.05\text{)}}{0.095}\\ & =& 0.5\\ & & \end{array}$

Hence, there is a 0.5 chance that a worker does not utilize drugs.

The probability that the worker is a non–drug user

b.

$\begin{array}{rcl}P\text{(}T\text{)}& =& \text{[}1-0.95\text{(}0.05\text{)}+0.95\text{(}1-0.05\text{)]}\\ & =& 0.905\\ & & \end{array}$

$\begin{array}{rcl}P\left(D^c\right|T)& =& \frac{0.05(1-0.95)}{0.905}\\ & =& 0.0028\\ & & \end{array}$

Therefore,the probability that a worker is not drug user is 0.0028.