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Chapter 6: Problem 46

The Australian newspaper The Mercury (May 30 , 1995) reported that, based on a survey of 600 reformed and current smokers, \(11.3 \%\) of those who had attempted to quit smoking in the previous 2 years had used a nicotine aid (such as a nicotine patch). It also reported that \(62 \%\) of those who quit smoking without a nicotine aid be gan smoking again within 2 weeks and \(60 \%\) of those who used a nicotine aid began smoking again within 2 weeks. If a smoker who is trying to quit smoking is selected at random, are the events selected smoker who is trying to quit uses a nicotine aid and selected smoker who has attempted to quit begins smoking again within 2 weeks inde pendent or dependent events? Justify your answer using the given information.

Short Answer

Expert verified
The events 'selected smoker who is trying to quit uses a nicotine aid' and 'selected smoker who has attempted to quit begins smoking again within 2 weeks' are dependent events.

Step by step solution

01

Interpret the Data

The exercise gives us the following information: 11.3% of smokers who have attempted to quit in the last two years used a nicotine aid, 62% of those who quit without a nicotine aid started smoking again within two weeks, and 60% of those who used a nicotine aid began smoking again within two weeks.
02

Determine the Probability of A and B

Let A be the event that smoker uses a nicotine aid and B be the event that the smoker starts smoking again within two weeks. From the given data, the probability that a randomly selected smoker uses a nicotine aid \( (P(A)) \) is 11.3% or 0.113. The point is to determine the probability of both A and B happening \( (P(A and B)) \). This is the same as the probability of a smoker who used a nicotine aid starting smoking again within two weeks which is 60% or 0.6 of the 11.3% that used a nicotine aid. Therefore, \( P(A and B) = 0.113 * 0.6 = 0.0678 \)
03

Determine Independence

Two events A and B are independent if and only if \( P(A and B) = P(A)P(B) \). If this equality doesn't hold, then the events are dependent. We still need to calculate \( P(B) \), the probability that a smoker starts smoking again within two weeks. This does include those who did use and those who did not use a nicotine aid. Using total probability theorem, this can be written as \( P(B) = P(B|A)P(A) + P(B|A')P(A') \), where \( A' \) stands for event A not happening. \( P(B|A) \) is 0.6 (or 60%), \( P(A) \) is 0.113 (or 11.3%), \( P(B|A') \) is 0.62 (or 62%) and \( P(A') \) is 1 - 0.113 = 0.887. Putting these values into equation yields \( P(B) = 0.6*0.113 + 0.62 * 0.887 = 0.6201 \).
04

Check for Independence

Now compare \( P(A and B) \) with \( P(A)P(B) \). \( P(A and B) = 0.0678 \) and \( P(A)P(B) = 0.113 * 0.6201 = 0.0701 \). The two values are not equal, so the events A and B are not independent but dependent.

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Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics concerned with analyzing random phenomena. The central objects of probability theory are random variables, which provide a quantitative description of the outcomes of uncertain events, and events, which are certain outcomes or combinations of outcomes. Probability is quantified as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. The more likely an event is to occur, the closer its probability is to 1. Understanding probability helps us make informed decisions in the face of uncertainty and is widely applied across various fields such as finance, science, and engineering.

In the context of the provided exercise, we consider events related to smoking cessation. Here, probability theory helps us to analyze the likelihood of a smoker who is trying to quit, using a nicotine aid or relapsing within a specified time frame. The problem involves understanding and calculating the probability of independent and dependent events, which is essential for determining the relationship between specific behaviors and outcomes in such health-related studies.
Nicotine Aid and Smoking Relapse
In public health, a significant area of study is the examination of factors that influence smoking relapse among individuals attempting to quit. Nicotine replacement therapies (NRTs), such as nicotine patches, gums, and lozenges, are widely used to support smoking cessation. They work by delivering controlled amounts of nicotine to the body, which helps to reduce the cravings and withdrawal symptoms that many people experience when they stop smoking.However, the success of NRTs and their impact on relapse rates is an area of ongoing research. The exercise addresses this by considering the probability of a smoker using a nicotine aid and the probability of them relapsing within two weeks. These probabilities are crucial for understanding the dependency between the use of nicotine aids and relapse rates. The findings can guide future strategies on how to improve the effectiveness of NRTs and support individuals who are trying to quit smoking.
Total Probability Theorem
The total probability theorem is a fundamental rule in probability theory, which relates the probability of an event to the probabilities of several mutually exclusive events that cover the entire sample space. In mathematical terms, if we have events \( B_1, B_2, \ldots, B_n \) that are mutually exclusive and exhaustive, and an event \( A \) whose probability we wish to find, the theorem states that:\[ P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2) + \ldots + P(A|B_n)P(B_n) \]where \( P(A|B_i) \) is the conditional probability of event \( A \) occurring given that event \( B_i \) has occurred, and \( P(B_i) \) is the probability of event \( B_i \) occurring. This rule is particularly useful when dealing with complex probability problems where events depend on several other events. As demonstrated in the exercise, the total probability theorem allows us to calculate the overall probability of relapse (\( P(B) \) by considering both the group of smokers who used a nicotine aid (\( P(A) \) and those who did not (\( P(A') \) providing a comprehensive picture of risk that is integral for both statistical analysis and informed public health planning.

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Most popular questions from this chapter

Information from a poll of registered voters in Cedar Rapids, Iowa, to assess voter support for a new school tax was the basis for the following statements (Cedar Rapids Gazette, August 28,1999 ): The poll showed 51 percent of the respondents in the Cedar Rapids school district are in favor of the tax. The approval rating rises to 56 percent for those with children in public schools. It falls to 45 percent for those with no children in public schools. The older the respondent, the less favorable the view of the proposed tax: 36 percent of those over age 56 said they would vote for the tax compared with 72 percent of 18- to 25 -year-olds. Suppose that a registered voter from Cedar Rapids is selected at random, and define the following events: \(F=\) event that the selected individual favors the school \(\operatorname{tax}, C=\) event that the selected individual has children in the public schools, \(O=\) event that the selected individual is over 56 years old, and \(Y=\) event that the selected individual is \(18-25\) years old. a. Use the given information to estimate the values of the following probabilities: i. \(P(F)\) ii. \(P(F \mid C)\) iii. \(P\left(F \mid C^{C}\right)\) iv. \(P(F \mid O)\) v. \(P(F \mid Y)\) b. Are \(F\) and \(C\) independent? Justify your answer. c. Are \(F\) and \(O\) independent? Justify your answer.

The general addition rule for three events states that $$ \begin{aligned} P(A \text { or } B \text { or } C)=& P(A)+P(B)+P(C) \\ &-P(A \text { and } B)-P(A \text { and } C) \\ &-P(B \text { and } C)+P(A \text { and } B \text { and } C) \end{aligned} $$ A new magazine publishes columns entitled "Art" (A), "Books" (B), and "Cinema" (C). Suppose that \(14 \%\) of all subscribers read A, \(23 \%\) read \(\mathrm{B}, 37 \%\) read \(\mathrm{C}, 8 \%\) read \(\mathrm{A}\) and \(\mathrm{B}, 9 \%\) read \(\mathrm{A}\) and \(\mathrm{C}, 13 \%\) read \(\mathrm{B}\) and \(\mathrm{C}\), and \(5 \%\) read all three columns. What is the probability that a randomly selected subscriber reads at least one of these three columns?

The USA Today article referenced in Exercise \(6.37\) also gave information on seat belt usage by age, which is summarized in the following table of counts: $$ \begin{array}{lcc} & \begin{array}{c} \text { Does Not Use } \\ \text { Seat Belt } \\ \text { Regularly } \end{array} & \begin{array}{c} \text { Uses } \\ \text { Seat Belt } \\ \text { Regularly } \end{array} \\ \hline 18-24 & 59 & 41 \\ 25-34 & 73 & 27 \\ 35-44 & 74 & 26 \\ 45-54 & 70 & 30 \\ 55-64 & 70 & 30 \\ 65 \text { and older } & 82 & 18 \\ \hline \end{array} $$ Consider the following events: \(S=\) event that a randomly selected individual uses a seat belt regularly, \(A_{1}=\) event that a randomly selected individual is in age group \(18-24\), and \(A_{6}=\) event that a randomly selected individual is in age group 65 and older. a. Convert the counts to proportions and then use them to compute the following probabilities: i. \(P\left(A_{1}\right)\) ii. \(P\left(A_{1} \cap S\right)\) iii. \(P\left(A_{1} \mid S\right)\) iv. \(P\left(\right.\) not \(\left.A_{1}\right) \quad\) v. \(P\left(S \mid A_{1}\right) \quad\) vi. \(P\left(S \mid A_{6}\right)\) b. Using the probabilities \(P\left(S \mid A_{1}\right)\) and \(P\left(S \mid A_{6}\right)\) computed in Part (a), comment on how \(18-24\) -year-olds and seniors differ with respect to seat belt usage.

Insurance status - covered (C) or not covered (N) \- is determined for each individual arriving for treatment at a hospital's emergency room. Consider the chance experiment in which this determination is made for two randomly selected patients. The simple events are \(O_{1}=(\mathrm{C}, \mathrm{C})\) \(O_{2}=(\mathrm{C}, \mathrm{N}), O_{3}=(\mathrm{N}, \mathrm{C})\), and \(O_{4}=(\mathrm{N}, \mathrm{N}) .\) Suppose that probabilities are \(P\left(O_{1}\right)=.81, P\left(O_{2}\right)=.09, P\left(O_{3}\right)=.09\), and \(P\left(O_{4}\right)=.01\). a. What outcomes are contained in \(A\), the event that at most one patient is covered, and what is \(P(A)\) ? b. What outcomes are contained in \(B\), the event that the two patients have the same status with respect to coverage, and what is \(P(B)\) ?

USA Today (June 6,2000 ) gave information on seat belt usage by gender. The proportions in the following table are based on a survey of a large number of adult men and women in the United States: $$ \begin{array}{l|cc} \hline & \text { Male } & \text { Female } \\ \hline \text { Uses Seat Belts Regularly } & .10 & .175 \\ \begin{array}{l} \text { Does Not Use Seat Belts } \\ \text { Regularly } \end{array} & .40 & .325 \\ \hline \end{array} $$ Assume that these proportions are representative of adults in the United States and that a U.S. adult is selected at random. a. What is the probability that the selected adult regularly uses a seat belt? b. What is the probability that the selected adult regularly uses a seat belt given that the individual selected is male? c. What is the probability that the selected adult does not use a seat belt regularly given that the selected individual is female? d. What is the probability that the selected individual is female given that the selected individual does not use a seat belt regularly? e. Are the probabilities from Parts (c) and (d) equal? Write a couple of sentences explaining why this is so.
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