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Lyme disease is transmitted by infected ticks. Several tests are available for people with symptoms of Lyme disease. One of these tests is the EIA/IFA test. The paper "Lyme Disease Testing by Large Commercial Laboratories in the United States" (Clinical Infectious Disease [2014]: \(676-681\) ) found that \(11.4 \%\) of those tested actually had Lyme disease. Consider the following events: \(+\) represents a positive result on the blood test \- represents a negative result on the blood test \(L\) represents the event that the patient actually has Lyme disease \(L^{C}\) represents the event that the patient actually does not have Lyme disease The following probabilities are based on percentages given in the paper: $$ \begin{array}{r} P(L)=0.114 \\ P\left(L^{C}\right)=0.886 \end{array} $$ $$ \begin{array}{c} P(+\mid L)=0.933 \\ P(-\mid L)=0.067 \\ P\left(+\mid L^{C}\right)=0.039 \\ P\left(-\mid L^{C}\right)=0.961 \end{array} $$ a. For each of the given probabilities, write a sentence giving an interpretation of the probability in the context of this problem. b. Use the given probabilities to construct a hypothetical 1000 table with columns corresponding to whether or not a person has Lyme disease and rows corresponding to whether the blood test is positive or negative. c. Notice the form of the known conditional probabilities; for example, \(P(+\mid L)\) is the probability of a positive test given that a person selected at random from the population actually has Lyme disease. Of more interest is the probability that a person has Lyme disease, given that the test result is positive. Use information from the table constructed in Part (b) to calculate this probability.

Step by step solution

01

Interpret Probabilities in Context:

Each of the given probabilities refers to a specific event related to Lyme disease testing. They are: 1. \(P(L) = 0.114\): 11.4% of those tested actually have Lyme disease. 2. \(P\left(L^{C}\right) = 0.886\): 88.6% of those tested actually do not have Lyme disease. 3. \(P(+\mid L) = 0.933\): 93.3% of those who have Lyme disease will test positive. 4. \(P(-\mid L) = 0.067\): 6.7% of those who have Lyme disease will test negative. 5. \(P\left(+\mid L^{C}\right) = 0.039\): 3.9% of those who do not have Lyme disease will test positive. 6. \(P\left(-\mid L^{C}\right) = 0.961\): 96.1% of those who do not have Lyme disease will test negative.

02

Create the 1000 Table:

We will create a hypothetical table with 1000 people, representing the different possible outcomes of the test. - \(P(L) = 0.114\) of 1000 people have Lyme disease, which gives us 114 people with Lyme disease. - \(P\left(L^{C}\right) = 0.886\) of 1000 people do not have Lyme disease, which gives us 886 people without Lyme disease. - \(P(+\mid L) = 0.933\) of 114 people with Lyme disease, which gives us 106 people who test positive and have Lyme disease. - \(P(-\mid L) = 0.067\) of 114 people with Lyme disease, which gives us 8 people who test negative and have Lyme disease. - \(P\left(+\mid L^{C}\right) = 0.039\) of 886 people without Lyme disease, which gives us 35 people who test positive and do not have Lyme disease. - \(P\left(-\mid L^{C}\right) = 0.961\) of 886 people without Lyme disease, which gives us 851 people who test negative and do not have Lyme disease. Now, arrange these numbers in a table: | | Lyme Disease | No Lyme Disease | Total | | --- | --- | --- | --- | | Positive Test | 106 | 35 | 141 | | Negative Test | 8 | 851 | 859 | | Total | 114 | 886 | 1000 |

03

Calculate the Desired Probability:

We want to find the probability that a person has Lyme disease given that they tested positive. This is the probability \(P(L \mid +)\). Using the numbers from our table, we can calculate this probability: $$ P(L \mid +)=\frac{\text{Number of people with Lyme Disease and a Positive Test}}{\text{Total Number of Positive tests}}=\frac{106}{141} \approx 0.7518 $$ So, the probability that a person has Lyme disease given they tested positive is approximately 75.18%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lyme Disease Testing

Understanding Lyme disease testing is essential for interpreting the probability exercises often encountered in statistics. Lyme disease, a condition transmitted by infected ticks, can be diagnosed with various tests – one being the EIA/IFA test.
To determine the effectiveness of such tests, consider the probability of a positive or negative test result in relation to the actual presence of the disease. This is where conditional probabilities come to play, helping us predict the likelihood of a particular outcome.

Probability Interpretation

Probability interpretation involves making sense of the likelihood of certain events. When discussing medical tests such as those for Lyme disease, this becomes crucial. For example, a probability of 0.933 or 93.3% for testing positive given the presence of Lyme disease (\(P(+\mid L) = 0.933\)) means that the EIA/IFA test is quite sensitive to detecting the disease when it is indeed present.

Conversely, a probability like 0.039 or 3.9% for testing positive when Lyme disease is not present (\(P(+\mid L^{C}) = 0.039\)) indicates a relatively low chance of a false positive. Here, probability serves as a guide to understanding the reliability and effectiveness of medical testing procedures, influencing patient diagnosis and treatment pathways.

Hypothetical 1000 Table

A hypothetical 1000 table is a visual aid to simplify complex conditional probabilities. By projecting recorded probabilities onto a cohort of 1000 people, we can better visualize and analyze the data.
In the context of the Lyme disease exercise, 114 people are expected to actually have the disease, and among them, 106 will likely get a positive test result due to the test's sensitivity. Similarly, of the 886 without the disease, about 35 might receive false positives. This table helps us easily calculate vital probabilities like the chance of truly having Lyme disease after a positive test result. Employing this table serves as an invaluable tool in translating statistical probabilities into more understandable figures.