Question

Critical Thinking: Interpreting results from a test for smoking

It is estimated that roughly half of smokers lie when asked about their smoking involvement. Pulse Co-Oximeters may be a way to get information about smoking without relying on patients’ statements. Pulse CO-oximeters use light that shines through a fingernail, and it measures carbon monoxide in blood. These devices are used by firemen and emergency departments to detect carbon monoxide poisoning, but they can also be used to identify smokers. The accompanying table lists results from people aged 18–44 when the pulse CO-oximeters is set to detect a 6% or higher level of carboxyhemoglobin (based on data from “Carbon Monoxide Test Can Be Used to Identify Smoker,” by Patrice Wendling, Internal Medicine News, Vol. 40., No. 1, and Centers for Disease Control and Prevention).

CO-Oximetry Test for Smoking

	Positive Test Results	Negative Test Results
Smoker	49	57
Non-smoker	24	370

Positive Predictive Value Find the positive predictive value of the test by finding the probability that the subject smokes, given that the test yields a positive result.

Short answer

The positive predictive value of the test is 0.671.

Step-by-step solution

Given information

The results from people aged 18-44 to detect the level of carboxyhemoglobin are based on the data from “Carbon Monoxide Test” used to identify smokers.

Describe conditional probability

A conditional probability of event is a probability obtained with the additional information that some other event has already occurred.

Notation for conditional probability is$P\left(B|A\right)$denotes the conditional probability of event B occurring, given that event A has already occurred.

Formula for conditional probability is as,

$P\left(B|A\right)=\frac{P\left(A \text{and} B\right)}{P\left(A\right)}$

Where,

$P\left(A \text{and} B\right)$denotes that both A and B occur, and denotes that event A occurs.

Summarize the given information

The test for smoking involves 500 people. Out of the 500 people, 106 are smokers and 394 are non-smokers. And, 73 people got the positive test results and 427 got the negative test results.

The above information is summarized in the following table,

	Positive Test Results	Negative Test Results	Total
Smoker	49	57	106
Non-smoker	24	370	394
Total	73	427	500

Define the events

Positive predictive value is the probability that a subject smokes,given the test yields a positive result.

Define the two events as,

A be the event that the test yields a positive result, and B be the event that the subject smokes.

Find the probability of a subject smokes, given that the test yields a positive result.

The probability of a subject smokes, given that the test yields a positive result is,$P\left(B|A\right)$.

By using conditional probability,

$P\left(\text{B|A}\right)=\frac{P\left(A \text{and} B\right)}{P\left(A\right)} ...\left(1\right)$

Find the probability that the test yields a positive result

The probability that the test yields a positive result is,

$$\begin{gathered} P\left(A\right)=\frac{\text{Number of positive} \text{test} \text{results}}{\text{Total}\text{number of surveyed subjects}} \\ =\frac{73}{500} \end{gathered}$$

Find the probability that a subject does not smoke and had negative test result

The probability that a subject does not smoke and had negative test result is,

$$\begin{gathered} P\left(A \text{and} B\right)=\frac{\text{Number of subject who smoke and had positive test result}}{\text{Total number of subjects surveyed}} \\ =\frac{49}{500} \end{gathered}$$

Find the positive predictive value 

Positive predictive value is the probability that a subject smokes, given that the test yields a positive result.

Substitute the probability of $P\left(\text{subject smoke and yield positive test result}\right)$and$P\left(\text{Positive result of test}\right)$ in equation (1),

$$\begin{gathered} P\left(B|A\right)=\frac{P\left(A \text{and} B\right)}{P\left(A\right)} \\ =\frac{\frac{49}{500}}{\frac{73}{500}} \\ =\frac{49}{73} \\ =0.6712 \end{gathered}$$

Thus, the probability that a subject smokes, given that the test yields a positive result is 0.671.