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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.

Short Answer

Expert verified
The outcomes \(L\) and \(F\) are not independent as the product of their individual probabilities \(0.29\) does not equal the probability of both \(0.31\).

Step by step solution

01

Recall the concept of independence

Two events \(A\) and \(B\) are independent if the probability of both events occurring is the product of the probabilities of each event occurring on its own. Mathematically, this can be represented as \(P(A \cap B) = P(A) \times P(B)\). This is the mathematical condition for independence.
02

Substituting the given values

The values provided in this exercise are \(P(L) = 0.58\), \(P(F) = 0.50\), and \(P(L \cap F) = 0.31\). We substitute these values into the condition for independence.
03

Compute the product of \(P(L)\) and \(P(F)\)

We calculate the product of \(P(L)\) and \(P(F)\) which gives \(0.58 \times 0.50 = 0.29\).
04

Compare the product with \(P(L \cap F)\)

Next, we compare the product \(0.29\) with the given probability \(P(L \cap F) = 0.31\). Since \(0.31\) does not equal \(0.29\), \(L\) and \(F\) are not independent.

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Key Concepts

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

Event Independence
Event independence is a fundamental concept in probability theory where two events, A and B, are considered independent if the occurrence of one does not affect the probability of the occurrence of the other. In simple terms, knowing that event A has occurred gives us no information about whether event B has happened or will happen, and vice versa.

Mathematically, event independence is defined by the formula: \[ P(A \cap B) = P(A) \times P(B) \]. If this equation holds true, then events A and B are independent. To demonstrate with a real-world example, think of flipping a coin and rolling a die simultaneously. The result of the coin flip (getting heads or tails) does not affect the outcome of the die roll, making these two events independent.

Applying this to the textbook problem, if the probability of a randomly selected adult learning something new about health (event L) is independent of the probability of being female (event F), then we should expect the combined probability of both learning something new and being female (P(L and F)) to be the product of their individual probabilities. However, the solution indicates a disparity between the calculated product and the given probability, suggesting that these events are not independent.
Probability Theory

Understanding Probability

Probability theory is a branch of mathematics that deals with quantifying the likelihood of events occurring. It forms the backbone of statistical analysis and is widely applied in fields like science, finance, and philosophy. The probability of an event A is usually expressed as a number between 0 and 1, where 0 indicates that the event will definitely not occur, and 1 indicates that the event will definitely occur. Events with probabilities closer to 1 are more likely to occur than events with probabilities closer to 0.

When dealing with multiple events, probability theory provides a structure to determine the composite likelihood of various outcomes. It's crucial in providing the rules for how probabilities interact, such as in the case of independent events. However, the exercises of probability are not limited to the theory; they include analyzing data to estimate probabilities of real-world events, as in the scenario detailed in the original exercise where probabilities were estimated from data collected in a survey.
Statistical Analysis

Importance of Statistical Analysis

Statistical analysis is an essential process that involves collecting, reviewing, and interpreting data to discover patterns and trends. This informs decision-making and conclusions in research and real-world applications. Throughout statistical analysis, probability theory plays a crucial role, especially when making inferences about a population based on a sample.

For instance, in the original exercise, the Centers for Disease Control (CDC) conducted a survey to infer the behaviors of adult Americans. By using probabilities, the CDC was able to make estimates about the larger population's behaviors based on the survey results. Statistical analysis also helps to determine the relationship between different factors, such as gender and the likelihood of learning about health topics through media, which is at the core of the original exercise. By understanding the analysis and the resultant probabilities, researchers can draw conclusions about the dependency or inter-relationship of events, such as whether learning outcomes are independent of gender among TV viewers, a crucial insight that impacts health communication strategies.

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Most popular questions from this chapter

An article in the New york Times (March 2, 1994) reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P(\) survival \()=.01\) for people who suffer a cardiac arrest in New York City (The artide attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings. Similar studies in smaller cities showed higher survival rates.) a. Give a relative frequency interpretation of the given probability. b. The research that was the basis for the New York Times article was a study of 2329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2329 cardiac arrest sufferers do you think survived? Explain.

\quad\( + Approximately \)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 results in a reservation? b. What assumption did you make in order to calculate the probability in Part (a)? c. What is the probability that at least one call results in a reservation being made?

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.

Five hundred first-year students at a state university were classified according to both high school grade point average (GPA) and whether they were on academic probation at the end of their first semester. The data are summarized in the accompanying table. \begin{tabular}{lcccc} & \multicolumn{3}{c} { High School GPA } & \\ \cline { 2 - 4 } Probation & 2.5 to & 3.0 to & 3.5 and & \\ & \(<3.0\) & \(<3.5\) & Above & Total \\ \hline Yes & 50 & 55 & 30 & 135 \\ No & 45 & 135 & 185 & 365 \\ Total & 95 & 190 & 215 & 500 \\ \hline \end{tabular} a. Construct a table of the estimated probabilities for each GPA-probation combination by dividing the number of students in each of the six cells of the table by \(500 .\) b. Use the table constructed in Part (a) to approximate the probability that a randomly selected first-year student at this university will be on academic probation at the end of the first semester. c. What is the estimated probability that a randomly selected first-year student at this university had a high school GPA of 3.5 or above? d. Are the two outcomes selected student has a GPA of 3.5 or above and selected student is on academic probation at the end of the first semester independent outcomes? How can you tell? e. Estimate the proportion of first-year students with high school GPAs between 2.5 and 3.0 who are on academic probation at the end of the first semester.

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, nine are requests for 2 licenses, and the rest are requests for a single license. The city will select requests at random, filling as much of the request 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 who wishes to be an independent driver 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?

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