/*! 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 65 In an article that appears on th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In an article that appears on the web site 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, and \(C=\) event that the person tested is actually clean. a. Using the information in the quote, what are the values of \mathbf{i} . ~ \(P(T D \mid D)\) iii. \(P(C)\) ii. \(P(T D \mid C) \quad\) iv. \(P(D)\) b. Use the law of total probability to find \(P(T D)\). c. Use Bayes' rule to evaluate \(P(C \mid T D)\). Is this value consistent with the argument given in the quote? Explain.

Short Answer

Expert verified
The values of i. \(P(T D | D) = 1\), ii. \(P(T D | C) = 0.01\), iii. \(P(C) = 0.99\), and iv. \(P(D) = 0.01\). The total probability \(P(T D) = 0.02\). The Bayesian probability \(P(C | T D) = 0.495\). This result is consistent with the argument in the quote indicating that about half of the positive tests could be false positives.

Step by step solution

01

Assign Values

From the statement, we know that:\n- \(P(T D | D) = 1\) (if person is dirty, the test is always dirty), \n- \(P(T D | C) = 0.01\) (there is a 1% chance that a clean person will test dirty), \n- \(P(C) = 0.99\) (99% of people tested are clean), \n- \(P(D) = 0.01\) (1% of people tested are dirty).
02

Calculate Total Probability

We can use the law of total probability to calculate \(P(T D)\). The total probability law states that the probability of event A is equal to the sum of the probability of A intersecting with each event Bi in a partition of the sample space.\nSo, \(P(T D) = P(T D | D) * P(D) + P(T D | C) * P(C) = 1 * 0.01 + 0.01 * 0.99 = 0.02.\)
03

Use Bayes' Rule

Bayes' Rule is used to reverse conditional probabilities. The formula is: \n\(P(A | B) = P(B | A) * P(A) / P(B)\)\nWe need to find \(P(C | T D)\), the probability that a person is clean given a dirty test.\nSo, \(P(C | T D) = P(T D | C) * P(C) / P(T D) = 0.01 * 0.99 / 0.02 = 0.495\)\nThis indicates that there is a 49.5% chance that a person is clean even if the test is dirty. This result is consistent with the public defender's statement that 'half of the dirty tests are false'.
04

Interpretation

The public defender's statistical argument checks out mathematically. For a test that only makes a mistake 1% of the time, if 99% of the people are clean, then about 50% of the positive tests could be false positives. It is important to consider the base rates, in this case the fact that a very large majority of the tested people are clean.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Law of Total Probability
The Law of Total Probability is a crucial concept in probability theory. It helps us determine the probability of a certain event when there are multiple ways for it to occur. Imagine you're trying to find out how likely it is for a test to show a dirty result, considering both scenarios: whether the person is actually dirty (has used drugs) or clean.This law tells us that the probability of any event (say "a dirty test result") is the sum of the probabilities of that event intersecting with each possible outcome. Here's how it works in our example:
  • Probability that a person is dirty and gets a dirty test: this is 1%.
  • Probability that a person is clean but still gets a dirty test result: this is 0.99% of the time.
We calculate the total probability by adding these two paths: \[P(T D) = P(T D \mid D) \cdot P(D) + P(T D \mid C) \cdot P(C)\]Using the specific probabilities given in the exercise:\[P(T D) = (1 \cdot 0.01) + (0.01 \cdot 0.99) = 0.02\]This calculation shows that there is a 2% chance of getting a dirty test result overall.
Bayes' Rule
Bayes' Rule is a powerful tool for updating the probability of a hypothesis as more evidence or information becomes available. It's used to "flip" conditional probabilities. That means it's highly useful when you need to determine the likelihood of an initial assumption or cause given an outcome.Let's say you have a dirty test result and you want to know, given this dirtiness, how likely the person is actually clean. The formula for Bayes' Rule is:\[P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}\]In our example, it helps to find:
  • The probability of a clean person, given they've received a dirty test result (\(P(C \mid T D)\)).
Plugging in the values from the exercise:\[P(C \mid T D) = \frac{P(T D \mid C) \cdot P(C)}{P(T D)} = \frac{0.01 \cdot 0.99}{0.02} = 0.495\]This means there is approximately a 49.5% chance that a person is clean even though the test result shows dirty, perfectly aligning with the public defender’s argument.
False Positive Rate
Understanding the False Positive Rate is vital, particularly in testing scenarios. A false positive occurs when a test incorrectly indicates the presence of a condition (such as drug use) when it isn't there. In our given example, it's when a clean person returns a dirty test. The false positive rate can mislead interpretations of test accuracy, especially if a test is widely used among a majority group that doesn’t possess the condition, just like our exercise where 99% of people tested are actually clean. Why is this crucial?
  • It teaches us that even tests with a low error rate can produce a lot of false positives if the base rate (actual occurrence of condition within the population) is significantly low.
  • In this scenario, the false positive rate is a key factor that resulted in about half the positive test results being wrong.
Highlighting the importance of reviewing not just the error rate of tests, but the broader context of testing populations to truly understand the test's effectiveness.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Three friends (A, B, and C) will participate in a round-robin tournament in which each one plays both of the others. Suppose that \(P(\) A beats \(B)=.7, P(\) A beats \(C)=.8\), \(P(\mathrm{~B}\) beats \(\mathrm{C})=.6\), and that the outcomes of the three matches are independent of one another. a. What is the probability that A wins both her matches and that \(\mathrm{B}\) beats \(\mathrm{C}\) ? b. What is the probability that \(A\) wins both her matches? c. What is the probability that A loses both her matches? d. What is the probability that each person wins one match? (Hint: There are two different ways for this to happen.)

The article "Doctors Misdiagnose More Women, Blacks" (San Luis Obispo Tribune, April 20, 2000) gave the following information, which is based on a large study of more than 10,000 patients treated in emergency rooms in the eastern and midwestern United States: 1\. Doctors misdiagnosed heart attacks in \(2.1 \%\) of all patients. 2\. Doctors misdiagnosed heart attacks in \(4.3 \%\) of black patients. 3\. Doctors misdiagnosed heart attacks in \(7 \%\) of women under 55 years old. Use the following event definitions: \(M=\) event that a heart attack is misdiagnosed, \(B=\) event that a patient is black, and \(W=\) event that a patient is a woman under 55 years old. Translate each of the three statements into probability notation.

Many cities regulate the number of taxi licenses, and there is a great deal of competition for both new and existing licenses. Suppose that a city has decided to sell 10 new licenses for \(\$ 25,000\) each. A lottery will be held to determine who gets the licenses, and no one may request more than three licenses. Twenty individuals and taxi companies have entered the lottery. Six of the 20 entries are requests for 3 licenses, 9 are requests for 2 licenses, and the rest are requests for a single license. The city will select requests at random, filling as many of the requests as possible. For example, the city might fill requests for \(2,3,1\), and 3 licenses and then select a request for \(3 .\) Because there is only one license left, the last request selected would receive a license, but only one. a. An individual has put in a request for a single license. Use simulation to approximate the probability that the request will be granted. Perform at least 20 simulated lotteries (more is better!). b. Do you think that this is a fair way of distributing licenses? Can you propose an alternative procedure for distribution?

Approximately \(30 \%\) of the calls to an airline reservation phone line result in a reservation being made. a. Suppose that an operator handles 10 calls. What is the probability that none of the 10 calls result in a reservation? b. What assumption did you make to calculate the probability in Part (a)? c. What is the probability that at least one call results in a reservation being made?

The Cedar Rapids Gazette (November 20, 1999) reported the following information on compliance with child restraint laws for cities in Iowa: $$ \begin{array}{lcc} & \begin{array}{c} \text { Number of } \\ \text { Children } \\ \text { Observed } \end{array} & \begin{array}{c} \text { Number } \\ \text { Properly } \\ \text { City } \end{array} & \text { Restrained } \\ \hline \text { Cedar Falls } & 210 & 173 \\ \text { Cedar Rapids } & 231 & 206 \\ \text { Dubuque } & 182 & 135 \\ \text { Iowa City (city) } & 175 & 140 \\ \text { Iowa City (interstate) } & 63 & 47 \\ \hline \end{array} $$ a. Use the information provided to estimate the following probabilities: i. The probability that a randomly selected child is properly restrained given that the child is observed in Dubuque. ii. The probability that a randomly selected child is properly restrained given that the child is observed in a city that has "Cedar" in its name. b. Suppose that you are observing children in the Iowa City area. Use a tree diagram to illustrate the possible outcomes of an observation that considers both the location of the observation (city or interstate) and whether the child observed was properly restrained.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.