/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 Lyme disease is the leading tick... [FREE SOLUTION] | 91Ó°ÊÓ

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Lyme disease is the leading tick-borne disease in the United States and Europe. Diagnosis of the disease is difficult and is aided by a test that detects particular antibodies in the blood. The article "Laboratory Considerations in the Diagnosis and Management of Lyme Borreliosis" (American Journal of Clinical Pathology [1993]: \(168-174\) ) used the following notation: + 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 were reported in the article:\(\begin{aligned} P(L) &=0.00207 \\ P\left(L^{C}\right) &=0.99793 \\ P(+\mid L) &=0.937 \\ P(-\mid L) &=0.063 \\ P\left(+\mid L^{C}\right) &=0.03 \\ P\left(-\mid L^{C}\right) &=0.97 \end{aligned}\) 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 the table constructed in Part (b) to calculate this probability.

Short Answer

Expert verified
In the context of this problem, for every 1000 people, 2 people are likely to have Lyme disease. If a person tests positive for Lyme disease, there is only approximately a 6.25% chance that they actually have the disease.

Step by step solution

01

Interpretation of the probabilities

Here's what each probability means in the context of this problem: 1. \(P(L) = 0.00207\): Out of the total population, around \(0.207\)% of people have Lyme disease. 2. \(P(L^{C}) = 0.99793\): Out of the total population, approximately \(99.793\)% do not have Lyme disease. 3. \(P(+ | L) = 0.937\): Given a person has the Lyme disease, there is a \(93.7\)% probability that he/she will test positive in the blood test. 4. \(P(- | L) = 0.063\): Given a person has the Lyme disease, there is a \(6.3\)% probability that his/her test will be a false negative. 5. \(P(+ | L^{C}) = 0.03\): Among those who do not have Lyme disease, around \(3\)% will have a false positive result, which means the test result will say they have Lyme disease even though they don't. 6. \(P(- | L^{C}) = 0.97\): Among those who do not have Lyme disease, \(97\)% will have a correct negative result, which means the test result will correctly say they don't have Lyme disease.
02

Construct a hypothetical 1000-person table

To create the hypothetical table, multiply the probabilities by 1000. Here is the probability table: - The number of people with Lyme disease = \(P(L) \times 1000 = 2.07\), round it to 2.- The number of people without Lyme disease = \(P(L^{C}) \times 1000 = 997.93\), round it to 998 (to total 1000).- The number of people with Lyme disease who test positive = \(P(+ | L) \times 2 = 1.874\), round it to 2.- The number of people with Lyme who test negative = \(P(- | L) \times 2 = 0.126\), round it to 0. - The number of people without Lyme disease but test positive = \(P(+ | L^{C}) \times 998 = 29.94\), round it to 30. - The number of people without Lyme disease and test negative = \(P(- | L^{C}) \times 998 = 969.06\), round it to 968 (to keep total 998).
03

Calculate the likelihood of having Lyme disease given a positive test.

The objective is to find the probability that a person has Lyme disease, given the test result is positive, denoted as \(P(L | +)\). This can be calculated by dividing the number of true positives by the total number of positives. This is \(P(L | +) = \frac{\text{Number of people with disease who test positive}}{\text{Total number of people who test positive}} = \frac{2} {2+30} \approx 0.0625\).

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Key Concepts

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

Bayes' Theorem
Bayes' Theorem is a fundamental concept in the field of statistics, helping us calculate the probability of a hypothesis based on prior knowledge and new evidence. It's particularly useful in scenarios where you need to update the probability of a condition as more data becomes available.

Bayes' Theorem is expressed as:\[P(A|B) = \frac{P(B|A) \, P(A)}{P(B)}\]Where:
  • \(P(A|B)\) is the probability of event A given event B has occurred.
  • \(P(B|A)\) is the probability of event B given event A is true.
  • \(P(A)\) is the probability of event A on its own.
  • \(P(B)\) is the probability of event B on its own.
In the context of disease diagnosis, Bayes' Theorem allows us to calculate how likely it is for a person to have a disease given a positive test result by using known probabilities of various outcomes. For instance, in the Lyme disease example, it was used to determine \(P(L|+)\), the probability of having Lyme disease given a positive test result.
Statistical Interpretation
Statistical interpretation involves translating numerical data into meaningful insights. This practice is crucial in understanding probabilities, especially in medical testing.

For Lyme disease testing:
  • \(P(L) = 0.00207\): tells us that Lyme disease is quite rare, with only 0.207% of people likely having it in this context.
  • \(P(+|L) = 0.937\): indicates the test's sensitivity, meaning it correctly identifies 93.7% of people with the disease.
  • \(P(-|L^C) = 0.97\): shows the test's specificity, correctly indicating negatives 97% of the time for those without the disease.
Proper statistical interpretation helps make informed decisions, such as interpreting the implications of a positive result and understanding the chances of false positives or negatives. It allows us to comprehend both the capabilities and limitations of diagnostic tests.
Disease Diagnosis Statistics
Disease diagnosis statistics are crucial in evaluating the effectiveness of medical tests. These statistics help in understanding test outcomes, informing patient care, and shaping medical guidelines.

Key statistical terms in this area include:
  • **Prevalence**: The proportion of a population found to have a condition. In the Lyme disease example, the prevalence is very low (about 0.207%).
  • **Sensitivity**: This measures how effectively a test identifies true positives. For Lyme disease, the sensitivity is 93.7%.
  • **Specificity**: This measures the accuracy of identifying true negatives. The specificity in our case is 97%.
  • **Positive Predictive Value (PPV)**: The probability that subjects with a positive screening test truly have the disease, calculated as \(P(L|+) = \frac{2}{2+30} \approx 0.0625\) in the example. This shows 6.25% of the positive test results are true cases.
These metrics help determine the reliability of tests in various contexts and assist healthcare providers in making data-driven decisions. By evaluating these statistics, physicians can better understand the likelihood of a patient's test results being accurate, enabling improved treatment strategies. Recognizing these statistics ensures both doctors and patients make well-informed decisions.

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Most popular questions from this chapter

According to The Chronicle for Higher Education (Aug. 26, 2011), there were 787,325 Associate degrees awarded by U.S. community colleges in the \(2008-2009\) academic year. A total of 488,142 of these degrees were awarded to women. a. If a person who received an Associate degree in 2008 2009 is selected at random, what is the probability that the selected person will be female? b. What is the probability that the selected person will be male?

The article "Checks Halt over 200,000 Gun Sales" (San Luis Obispo Tribune, June 5,2000 ) reported that required background checks blocked 204,000 gun sales in \(1999 .\) The article also indicated that state and local police reject a higher percentage of would-be gun buyers than does the FBI, stating, "The FBI performed 4.5 million of the 8.6 million checks, compared with 4.1 million by state and local agencies. The rejection rate among state and local agencies was \(3 \%,\) compared with \(1.8 \%\) for the FBI." a. Use the given information to estimate \(P(F), P(S)\), \(P(R \mid F),\) and \(P(R \mid S),\) where \(F=\) event that a randomly selected gun purchase background check is performed by the \(\mathrm{FBI}, S=\) event that a randomly selected gun purchase background check is performed by a state or local agency, and \(R=\) event that a randomly selected gun purchase background check results in a blocked sale. b. Use the probabilities from Part (a) to create a "hypothetical \(1000 "\) table. Use the table to calculate \(P(S \mid R),\) and write a sentence interpreting this value in the context of this problem.

Consider a chance experiment that consists of selecting a student at random from a high school with 3,000 students. a. In the context of this chance experiment, give an example of two events that would be mutually exclusive. b. In the context of this chance experiment, give an example of two events that would not be mutually exclusive.

A study of how people are using online services for medical consulting is described in the paper "Internet Based Consultation to Transfer Knowledge for Patients Requiring Specialized Care" (British Medical Journal [2003]: \(696-699)\). Patients using a particular online site could request one or both (or neither) of two services: specialist opinion and assessment of pathology results. The paper reported that \(98.7 \%\) of those using the service requested a specialist opinion, \(35.4 \%\) requested the assessment of pathology results, and \(34.7 \%\) requested both a specialist opinion and assessment of pathology results. Consider the following two events: \(S=\) event that a specialist opinion is requested \(A=\) event that an assessment of pathology results is requested a. What are the values of \(P(S), P(A)\), and \(P(S \cap A)\) ? b. Use the given probability information to set up a "hypothetical 1000 " table with columns corresponding to \(S\) and not \(S\) and rows corresponding to \(A\) and \(\operatorname{not} A .\) c. Use the table to find the following probabilities: i. the probability that a request is for neither a specialist opinion nor assessment of pathology results. ii. the probability that a request is for a specialist opinion or an assessment of pathology results.

A large cable company reports that \(42 \%\) of its customers subscribe to its Internet service, \(32 \%\) subscribe to its phone service, and \(51 \%\) subscribe to its Internet service or its phone service (or both). a. Use the given probability information to set up a "hypothetical \(1000 "\) table. b. Use the table to find the following: i. the probability that a randomly selected customer subscribes to both the Internet service and the phone service. ii. the probability that a randomly selected customer subscribes to exactly one of the two services.

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