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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鈥� 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鈥�44 when the pulse CO-oximeter is set to detect a 6% or higher level of carboxyhemoglobin (based on data from 鈥淐arbon Monoxide Test Can Be Used to Identify Smoker,鈥� by Patrice Wendling, Internal Medicine News, Vol. 40., No. 1, and Centers for Disease Control and Prevention).

CO-Oximetry Test for Smoking

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

Analyzing the Results

Sensitivity Find the sensitivity of the test by finding the probability of a true positive, given that the subject actually smokes.

Step by step solution

01

Given information

The given table summarizesthe test results for smoking.

CO-Oximetry test for smoking

	Positive test result	Negative test result
Smoker	49	57
Non-smoker	24	370

02

Define sensitivity of a test

Sensitivity of a test is the probability of getting true positive result, given that the subject is a smoker.

Let event A denotes that the subject is a smoker and event B denotes that a subject has true positive results.

In this case, the event A and B is same as the event B.

Then, sensitivity is defined as,

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

03

Find the probabilities

Total number of subjects are 500.

Using the given information,

$$\begin{gathered} P\left(A\right)=\frac{\text{Number of subjects who are smokers}}{\text{Total number of subjects}} \\ =\frac{49+57}{500} \\ =\frac{106}{500} \end{gathered}$$

$$\begin{gathered} P\left(\text{B}\right)=\frac{\text{Number of subjects who are smoker and test result is positive}}{\text{Total number of subjects}} \\ =\frac{49}{500} \end{gathered}$$

04

Calculate the sensitivity

Substitute the values in equation (1),

$$\begin{gathered} P\left(B\left|A\right.\right)=\frac{\frac{49}{500}}{\frac{106}{500}} \\ =0.462 \end{gathered}$$

Therefore, theprobability of a true positive result (sensitivity of the test), given that the subject is a smoker, is 0.462.

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