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Chapter 6: Problem 2

The paper "Predictors of Complementary Therapy Use among Asthma Patients Results of a Primary Care Survey"" (Health and Social Care in the Community 12008 h \(155-164\) ) included the accompanying rable. The table summarizes the responses given by 1077 asthma patients to two questions: Question 1: Do conventional asthma medications usually help your asthma symptoms? Question 2: Do you use complementary therapies (such as herbs, acupuncture, aroma therapy) in the treatment of your asthma? \begin{tabular}{l|cc} & Doesn't Use Complementary Therapies & Does Use Complementary Theraples \\ \hline Comventional Medica- & 816 & 131 \\ \multicolumn{1}{c|} { tions Usually Help } \\ Conventional Medi- & & \\ cations Usually Do Not Help & 103 & 27 \\ \hline \end{tabular} From this information, we can estimate that the proportion who use complementary therapies is \(\frac{131+27}{1077}=\frac{158}{1077}=.147 .\) For those who report that conventional medications usually help, the proportion who use complementary therapies is \(\frac{131}{947}=.138\) and for those who report that conventional medications usually do not help, the proportion who use comple- mentary therapies is \(\frac{27}{130}=.208\) If one of these 1077 patients is selected at random, are the outcomes selected patient reports that conventional medications wually help and selected patient uses complementary therapies independent or dependent? Explain.

Short Answer

Expert verified
The events are not independent. The probability of a patient using complementary therapies given that conventional medicines work (\(0.138\)) is not equal to the overall probability of a patient using complementary therapies (\(0.147\)). This means that the usage of complementary therapies is slightly less likely among those who report that conventional medicines usually work, suggesting that there exists a relationship between the two events.

Step by step solution

01

Calculate the Overall Probability of Using Complementary Therapies

First, calculate the probability of an individual using complementary therapies regardless of the effectiveness of conventional medicine. This is calculated by dividing the total number of cases where complementary therapies are used by the total number of cases. From the data, this would be \(\frac{131+27}{1077} = 0.147\)
02

Calculate The Probability of Using Complementary Therapies Given Conventional Medicines Work

Next, calculate the probability that an individual uses complementary therapies, given that they stated that conventional medicines usually help their symptoms. This can be calculated by dividing the number of cases where both conditions are met by the total number of cases where conventional medicines help. So, this probability would be \(\frac{131}{947} = 0.138\)
03

Compare Probabilities

Finally, compare the probabilities calculated in steps 1 and 2. If they are equal, the two events are independent; if they are not equal, the events are dependent. Comparing \(0.147\) and \(0.138\), it is seen that the probabilities are not equal.

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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 that deals with calculating the likelihood of events occurring. While it may sound complex, it can be broken down into simple parts that make it easier to grasp.
In this context, we use it to determine how likely it is for asthma patients to use complementary therapies like herbs and acupuncture.
To find these probabilities, we look at various conditions and events, such as whether conventional medicines help. When discussing probability, it's essential to understand the following concepts:
  • **Sample Space**: This is the total number of possible outcomes. For our problem, the sample space is all 1077 asthma patients.
  • **Event**: An event is any specific outcome or a combination of outcomes you鈥檙e looking for. Here, events can include patients using complementary therapies or reporting that conventional medicines help.
  • **Probability of an Event**: This value is calculated by dividing the number of successful outcomes by the sample space. For example, the probability of using complementary therapies is calculated by dividing the number of patients using these therapies by the total number of patients, which gives us 0.147.
Understanding probability theory helps in assessing health behaviors and outcomes, like in this study of asthma patients.
Independence and Dependence
In probability theory, independence and dependence help determine the relationship between two events.
In the context of our exercise, these events include whether conventional asthma medications help and whether complementary therapies are used by patients.

Independent Events

Two events are independent if the occurrence of one does not affect the probability of the other occurring. For instance, if using complementary therapies is independent of conventional meds effectiveness, then the probability of using these therapies would equate when conventional medicines help or don鈥檛 help.

Dependent Events

Events are dependent if the occurrence of one event does affect the probability of the other. When assessed, the probabilities differ, indicating a relationship or dependency.
In this exercise, we conclude dependence because the probabilities of using complementary therapies (0.138 when meds work vs 0.147 overall) are not equal. This suggests that patients鈥� choices in therapy likely depend on how effective they find conventional medications.
Contingency Tables
Contingency tables are a powerful tool in statistics. They help summarize and analyze the relationship between two categorical variables.
In our exercise, the contingency table provides a snapshot of asthma patients, showing whether they use complementary therapies and if they feel conventional meds help. A basic contingency table includes rows and columns, with each cell presenting frequencies or counts of observations. For instance:
  • **Rows** might represent if conventional meds help or not.
  • **Columns** can show whether complementary therapies are used.
  • **Cells** display the count of patients for each category combination.
Contingency tables make it easy to calculate conditional probabilities and observe why probabilities are similar or not, aiding in evaluating independence or dependence in this example. By breaking down data in a straightforward manner, these tables simplify complex statistical relationships.

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Most popular questions from this chapter

Many fire stations handle emergency calls for medical assistance as well as calls requesting firefighting equipment. A particular station says that the probability that an incoming call is for medical assistance is .85. This can be expressed as \(P(\) call is for medical assistance \()=.85 .\) a. Give a relative frequency interpretation of the given probability. b. What is the probability that a call is not for medical assistance? c. Assuming that successive calls are independent of one another (i.e., knowing that one call is for medical assistance doesn't influence our assessment of the probability that the next call will be for medical assistance), calculate the probability that both of two successive calls will be for medical assistance. d. Still assuming independence, calculate the probability that for two successive calls, the first is for medical assistance and the second is not for medical assistance. e. Still assuming independence, calculate the probability that exactly one of the next two calls will be for medical assistance. (Hint: There are two different possibilities that you should consider. The one call for medical assistance might be the first call, or it might be the second call.) f. Do you think it is reasonable to assume that the requests made in successive calls are independent? Explain.

Consider a system consisting of four components, as pictured in the following diagram: Components 1 and 2 form a series subsystem, as do Components 3 and 4\. The two subsystems are connected in parallel. Suppose that \(P(1\) works \()=.9\), \(P(2\) works \()=, 9, P(3\) works \()=.9,\) and \(P(4\) works \()=.9\) and that these four outcomes are independent (the four components work independently of one another). a. The \(1-2\) subsystem works only if both components work. What is the probability of this happening? b. What is the probability that the \(1-2\) subsystem doesn't work? that the \(3-4\) subsystem doesn't work? c. The system won't work if the \(1-2\) subsystem doesn't work and if the \(3-4\) subsystem also doesn't work. What is the probability that the system won't work? that it will work? d. How would the probability of the system working change if a \(5-6\) subsystem was added in parallel with the other two subsystems? e. How would the probability that the system works change if there were three components in series in each of the two subsystems?

Is ultrasound a reliable method for determining the gender of an unborn baby? Consider the following data on 1000 births, which are consistent with summary values that appeared in the online journal of Statistics Education ("New Approaches to Learning Probability in the First Statistics Course" 120011 ): \begin{tabular}{l|cc} & Ultrasound Predicted Female & Utrasound Predicted Male \\ \hline Actual Gender Is Female & 432 & 48 \\ Actual Gender is Male & 130 & 390 \\ \hline \end{tabular} Do you think that a prediction that a baby is male and a prediction that a baby is female are equally reliable? Explain, using the information in the table to calculate estimates of any probabilities that are relevant to your conclusion.

Three friends \((A, B,\) and \(C)\) will participate in a round-robin tournament in which each one plays both of the others. Suppose that \(P(\mathrm{~A}\) beats \(\mathrm{B})=.7, P(\mathrm{~A}\) beats C) \(=.8\), and \(P(B\) beats \(C)=.6\) 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. Calculate the probability of each separately, and then add.)

The report "TV Drama/Comedy Viewers and Health Information" (www.cdc.gov/healthmarketing) describes the results of a large survey involving approximately 3500 people that was conducted for the Centers for Disease Control. The sample was selected in a way that the Centers for Disease Control believed would result in a sample that was representative of adult Americans, One question on the survey asked respondents if they had learned something new about a health issue or disease from a TV show in the previous 6 months. Consider the following outcomes: \(L=\) outcome that a randomly selected adult American reports learning something new about a health issue or disease from a TV show in the previous 6 months and \(F=\) outcome that a randomly selected adult American is female Data from the survey were used to estimate the following probabilities: $$ P(L)=.58 \quad P(F)=.50 \quad P(L \text { and } F)=.31 $$ Are the outcomes \(L\) and \(F\) independent? Use probabilities to justify your answer.
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