/*! 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

A Gallup survey found that \(46 \%\) of women and \(37 \%\) of men experience pain on a daily basis (San Luis Obispo Tribune, April 6,2000 ). Suppose that this information is representative of U.S. adults. If a U.S. adult is selected at random, are the events selected adult is male and selected adult experiences pain on a daily basis independent or dependent? Explain.

A single-elimination tournament with four players is to be held. A total of three games will be played. In Game 1 , the players seeded (rated) first and fourth play. In Game 2 , the players seeded second and third play. In Game \(3,\) the winners of Games 1 and 2 play, with the winner of Game 3 declared the tournament winner. Suppose that the following probabilities are known: \(P(\) Seed 1 defeats Seed 4\()=0.8\) \(P(\) Seed 1 defeats \(\operatorname{Seed} 2)=0.6\) \(P(\) Seed 1 defeats \(\operatorname{Seed} 3)=0.7\) \(P(\) Seed 2 defeats \(\operatorname{Seed} 3)=0.6\) \(P(\) Seed 2 defeats Seed 4\()=0.7\) \(P(\) Seed 3 defeats Seed 4) \(=0.6\) a. How would you use random digits to simulate Game 1 of this tournament? b. How would you use random digits to simulate Game 2 of this tournament? c. How would you use random digits to simulate the third game in the tournament? (This will depend on the outcomes of Games 1 and \(2 .\) ) d. Simulate one complete tournament, giving an explanation for each step in the process. e. Simulate 10 tournaments, and use the resulting information to estimate the probability that the first seed wins the tournament. f. Ask four classmates for their simulation results. Along with your own results, this should give you information on 50 simulated tournaments. Use this information to estimate the probability that the first seed wins the tournament. g. Why do the estimated probabilities from Parts (e) and (f) differ? Which do you think is a better estimate of the actual probability? Explain.

In an article that appears on the website of the American Statistical Association (www.amstat.org), Carlton Gunn, a public defender in Seattle, Washington, wrote about how he uses statistics in his work as an attorney. He states: I personally have used statistics in trying to challenge the reliability of drug testing results. Suppose the chance of a mistake in the taking and processing of a urine sample for a drug test is just 1 in \(100 .\) And your client has a "dirty" (i.e., positive) test result. Only a 1 in 100 chance that it could be wrong? Not necessarily. If the vast majority of all tests given - say 99 in 100 - are truly clean, then you get one false dirty and one true dirty in every 100 tests, so that half of the dirty tests are false. Define the following events as \(T D=\) event that the test result is dirty \(T C=\) event that the test result is clean \(D=\) event that the person tested is actually dirty \(C=\) event that the person tested is actually clean a. Using the information in the quote, what are the values of i. \(P(T D \mid D)\) iii. \(P(C)\) ii. \(P(T D \mid C)\) iv. \(P(D)\) b. Use the probabilities from Part (a) to construct a "hypothetical 1000 " table. c. What is the value of \(P(T D)\) ? d. Use the table to calculate the probability that a person is clean given that the test result is dirty, \(P(C \mid T D)\). Is this value consistent with the argument given in the quote? Explain.

Roulette is a game of chance that involves spinning a wheel that is divided into 38 equal segments, as shown in the accompanying picture. A metal ball is tossed into the wheel as it is spinning, and the ball eventually lands in one of the 38 segments. Each segment has an associated color. Two segments are green. Half of the other 36 segments are red, and the others are black. When a balanced roulette wheel is spun, the ball is equally likely to land in any one of the 38 segments. a. When a balanced roulette wheel is spun, what is the probability that the ball lands in a red segment? b. In the roulette wheel shown, black and red segments alternate. Suppose instead that all red segments were grouped together and that all black segments were together. Does this increase the probability that the ball will land in a red segment? Explain. c. Suppose that you watch 1000 spins of a roulette wheel and note the color that results from each spin. What would be an indication that the wheel was not balanced?

Suppose you want to estimate the probability that a randomly selected customer at a particular grocery store will pay by credit card. Over the past 3 months, 80,500 payments were made, and 37,100 of them were by credit card. What is the estimated probability that a randomly selected customer will pay by credit card?

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