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Chapter 19: Problem 60

Enzyme immunoassay tests are used to screen blood specimens for the presence of antibodies to HIV, the virus that causes AIDS. Antibodies indicate the presence of the virus. The test is quite accurate but is not always correct. Here are approximate probabilities of positive and negative test results when the blood tested does and does not actually contain antibodies to HIV: \({ }^{12}\) $$ \begin{array}{c|cc} \hline & {\text { Test Result }} \\ & \text { Positive } & \text { Negative } \\ \hline \text { Antibodies present } & 0.9985 & 0.0015 \\ \hline \text { Antibodies absent } & 0.0060 & 0.9940 \\ \hline \end{array} $$ Suppose that \(1 \%\) of a large population carries antibodies to HIV in their blood. (a) Draw a tree diagram for selecting a person from this population (outcomes: antibodies present or absent) and testing his or her blood (outcomes: test positive or negative). (b) What is the probability that the test is positive for a randomly chosen person from this population?

Short Answer

Expert verified
The probability of a positive test for a random person is approximately 0.0159.

Step by step solution

01

Define the Probabilities

First, identify the given probabilities for the presence and absence of antibodies. The probability of having antibodies is 0.01, as 1% of the population carries them. Hence, the probability of not having antibodies is 0.99. Additionally, note the test result probabilities: the test is positive with antibodies present (0.9985) and negative without antibodies (0.9940), while the false results are 0.0015 and 0.0060, respectively.
02

Construct a Tree Diagram

Draw a tree diagram with two initial branches for 'Antibodies Present' and 'Antibodies Absent'. From each, further split into 'Test Positive' and 'Test Negative'. Label branches with the corresponding probabilities: - 'Antibodies Present': 0.01, - 'Test Positive': 0.9985, - 'Test Negative': 0.0015. - 'Antibodies Absent': 0.99, - 'Test Positive': 0.0060, - 'Test Negative': 0.9940.
03

Calculate the Probability of a Positive Test

To find the overall probability of a positive test, add probabilities from both scenarios where the test is positive. Calculate the probability of testing positive given antibodies are present as 0.01 (probability of antibodies) * 0.9985 (probability of a positive test) and for antibodies absent as 0.99 * 0.0060. Consequently, the total probability of a positive test is:\[ P( ext{Positive}) = (0.01 \cdot 0.9985) + (0.99 \cdot 0.0060) \]
04

Solve the Expression

Substitute and compute the values:- Calculate \(0.01 \cdot 0.9985 = 0.009985\).- Calculate \(0.99 \cdot 0.0060 = 0.00594\).Combine these probabilities to find:\[ P( ext{Positive}) = 0.009985 + 0.00594 = 0.015925 \]
05

Final Result

Thus, the probability that a randomly chosen person from this population tests positive is approximately 0.015925.

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Key Concepts

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

Conditional Probability
Understanding conditional probability is essential for solving problems like the one presented in the exercise concerning HIV testing. Conditional probability helps you calculate the probability of an event happening, given that another event has already occurred. For instance, in the exercise, we are interested in the probability of the test being positive depending on whether antibodies are present or absent.

In the simplest terms, conditional probability can be expressed as:
  • Probability of event A given event B is: \( P(A|B) = \frac{P(A \cap B)}{P(B)} \).

In our case:
  • \( P(\text{Positive | Antibodies present}) = 0.9985 \)
  • \( P(\text{Positive | Antibodies absent}) = 0.0060 \).

These probabilities show how likely the test is to give a positive result based on whether antibodies are present or not. These are crucial in determining the reliability of tests and understanding potential outcomes.
False Positives and Negatives
When dealing with diagnostic tests like enzyme immunoassays for HIV testing, false positives and negatives are critical concepts. These refer to incorrect test results, where a false positive means the test indicates infection when the individual is not infected, and a false negative means the test fails to detect infection when the individual is infected.

In the given exercise:
  • False Positive: Occurs with a probability of 0.0060 (Test is positive when antibodies are absent).
  • False Negative: Occurs with a probability of 0.0015 (Test is negative when antibodies are present).

These rates help in evaluating test performance. A low false positive rate is important to avoid unnecessary stress and treatments, while a low false negative rate is crucial to ensure infections are not overlooked. Understanding these probabilities helps in the design of more reliable tests and better health outcomes.
Bayesian Statistics
Bayesian Statistics provides a powerful framework for handling the kind of probabilistic inference required in the exercise. It allows you to update the probability estimate for a hypothesis as more evidence or information becomes available. This process is embodied in Bayes' Theorem, which blends prior knowledge with new evidence to get an updated probability estimate.

Bayes' Theorem is given by:
  • \( P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} \)
In this scenario:
  • \( P(\text{Antibodies | Positive}) \) describes the probability that antibodies are present given a positive test result.
  • This is calculated by considering both the likelihood of testing positive when antibodies are present and the nature of positive results amongst antibody absent tests.

Applying Bayesian Statistics allows more accurate post-test evaluations by integrating the base rate of infection (1% in the population) with test sensitivity and specificity. It shifts focus from the test result to the probability of the presence of antibodies, given the result, facilitating better-informed decisions in medical testing and treatment.

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Most popular questions from this chapter

According to the National Survey of Student Engagement (NSSE), the average amount of time that first-year college students spent preparing for class (studying, reading, writing, doing homework or lab work, analyzing data, rehearsing, and other academic activities) in 2013 was \(14.25\) hours per week. Your college wonders if the average \(\mu\) for its first-year students in 2013 differed from the national average. A random sample of 500 students who were first-year students in 2013 claims to have spent an average of \(x^{-} \bar{x}=13.4\) hours per week on homework in their first year. What are the null and alternative hypotheses for a comparison of first-year students at your college with national first-year students in 2013 ? (a) \(H_{0}: \mathrm{x}^{-} \bar{x}=14.25, H_{a}: \mathrm{x}^{-} \bar{x} \neq 14.25\) (b) \(H_{0}: x^{-} \bar{x}=13.4, H_{a}: x^{-} \bar{x}>13.4\) (c) \(H_{0}: \mu=14.25, H_{a}: \mu \neq 14.25\) (d) \(H_{0}: \mu=13.4, H_{a}: \mu>13.4\)

A randomly chosen subject arrives for a study of exercise and fitness. Describe a sample space for each of the following. (In some cases, you may have some freedom in your choice of \(S\).) (a) The subject is either female or male. (b) After 10 minutes on an exercise bicycle, you ask the subject to rate his or her effort on the rate of perceived exertion (RPE) scale. RPE ranges in wholenumber steps from 6 (no exertion at all) to 20 (maximal exertion). (c) You measure \(\mathrm{VO}_{2}\) max, the maximum volume of oxygen consumed per minute during exercise. \(\mathrm{VO}_{2}\) is generally between \(2.5\) and \(6.1\) liters per minute. (d) You measure the maximum heart rate (beats per minute). Internet search sites compete for users because they sell advertising space on their sites and can charge more if they are heavily used. Choose an Internet search attempt at random. Here is the probability distribution for the site the search uses: \({ }^{1}\). $$ \begin{array}{l|ccccc} \hline \text { Site } & \text { Google } & \text { Microsoft } & \text { Yahoo } & \text { Ask Network } & \text { Others } \\ \hline \text { Probability } & 0.64 & 0.21 & 0.12 & 0.02 & ? \\ \hline \end{array} $$ Use this information to answer Questions \(19.2\) through \(19.4 .\)

The addition rule for probabilities, \(P(A\) or \(B)=P(A)+P(B)\), is not always true. Give (in words) an example of real-world events \(A\) and \(B\) for which this rule is not true.

Infants weighing less than 1500 grams at birth are classed as "very low birth weight." Low birth weight carries many risks. One study followed 113 male infants with very low birth weight to adulthood. At age 20, the mean IQ score for these men was \(x^{-} \bar{x}=\) 87.6. \({ }^{5}\) IQ scores vary Normally with standard deviation \(\sigma=15\). Give a \(95 \%\) confidence interval for the mean IQ score at age 20 for all very-low-birth-weight males.

The Environmental Protection Agency (EPA) fuel economy ratings say that the 2016 Toyota Prius hybrid car gets 50 miles per gallon (mpg) on the highway. Deborah wonders whether the actual long-term average highway mileage \(\mu\) of her new Prius is less than \(50 \mathrm{mpg}\). She keeps careful records of gas mileage for 3000 miles of highway driving. Her result is \(x^{-} \bar{x}=49.2 \mathrm{mpg}\). What are her null and alternative hypotheses? (a) \(H_{0}: \mu=50, H_{a}: \mu<50\) (b) \(H_{0}: \mu=50, H_{a}: \mu>50\) (c) \(H_{0}: \mathrm{x}^{-} \bar{x}=50, H_{a}: \mathrm{x}^{-} \bar{x}<50\) (d) \(H_{0}: \mathrm{x}^{-} \bar{x}=50, H_{a}: \mathrm{x}^{-} \bar{x}>50\)
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