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The article "Viewers Speak Out Against Reality TV" (Associated Press, September 12,2005\()\) included the following statement: "Few people believe there's much reality in reality TV: a total of \(82 \%\) said the shows are either 'totally made up' or 'mostly distorted."" This statement was based on a survey of 1,002 randomly selected adults. Calculate and interpret the margin of error for the reported percentage.

Short Answer

Expert verified
The margin of error for the reported percentage when calculated comes out to be around \( \pm \) 2.18%, at a 95% confidence interval. This means that if the survey was repeated many times, 95% of the time, the result would lie between \(79.82 \% \) and \(84.18 \% \).

Step by step solution

01

Understanding the Provided Information

From the exercise, we know that there is a reported percentage of \(82 \% \) based on a survey of 1,002 randomly selected adults. The reported percentage or \( p \) is equal to \( 0.82 \), and the sample size or \( n \) is equal to 1,002.
02

Calculating the Margin of Error

The formula to get the margin of error (\(E\)) is given by \(E = Z \sqrt{\frac{p(1-p)}{n}}\), where \(Z\) is the z-value, which is typically \(1.96\) for a 95% confidence interval; \(p\) is the reported percentage or proportion, and \(n\) is the sample size. By plugging into the formula, we get that \(E = 1.96 \sqrt{\frac{0.82(1-0.82)}{1002}}\).
03

Interpreting the Result

The result obtained from the above calculation gives the margin of error. This means that if the survey were to be conducted many times, the true proportion of people who believe that reality TV shows are either totally made up or mostly distorted would lie within the calculated margin of error around the measured percentage of \(82 \% \) in about 95 out of 100 times (95% confidence interval).

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Key Concepts

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

Survey Sampling
Survey sampling is a critical process in research whereby a subgroup, or a sample, is chosen from a larger population to participate in a study. This selection process is crucial because it allows researchers to make inferences about the entire population based on the data collected from the sample. In order for these inferences to be valid, the sample must be representative of the population, which is often achieved by random selection.

Random sampling ensures that each member of the population has an equal chance of being included in the sample, hence avoiding bias. In our exercise, the survey of 1,002 adults is assumed to be randomly selected, which helps in giving validity to the survey results and subsequent calculations of the margin of error.
Confidence Interval
A confidence interval gives an estimated range of values which is likely to include an unknown population parameter, the estimated range being calculated from a given set of sample data. It is tied to a specified level of confidence, such as 95%, which, in simple terms, means that if the same population is sampled multiple times, we would expect the true population parameter to fall within this interval 95 out of 100 times.

The calculated margin of error is used to define the width of the confidence interval around the observed sample mean or proportion. For example, with an 82% proportion, if the margin of error is found to be 3%, then the confidence interval would be 79% to 85%, indicating where the true population proportion is believed to lie with 95% confidence.
Z-Value
The z-value is a numerical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. In the context of determining the margin of error for a confidence interval, the z-value corresponds to the number of standard deviations a data point must be from the mean to fall within a specified percentage of the data in a normal distribution.

The z-value used in calculating the margin of error is typically determined based on the desired confidence level. For a 95% confidence level, the z-value is often 1.96, which means that the true proportion is estimated to be within 1.96 standard deviations from the observed sample proportion 95% of the time. This z-value changes as the confidence level changes, with higher confidence levels requiring a higher z-value, and vice versa.
Sample Size
Sample size is the number of observations or replicates included in a statistical sample. It is a fundamental aspect of any empirical study in which the goal is to make inferences about a population from a sample. In the context of the margin of error, sample size directly affects the precision of the confidence interval estimates.

Larger sample sizes generally result in a smaller margin of error, assuming the level of variability in the population is constant, because they provide more information about the population and thus reduce the uncertainty in estimates of the population parameter. In our exercise scenario, the sample size is 1,002, which provides a balance between accuracy and the resources required to conduct the survey.

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Most popular questions from this chapter

The study "Digital Footprints"(Pew Internet \& American Life Project, www.pewinternet.org, 2007 ) reported that \(47 \%\) of Internet users have searched for information about themselves online. The \(47 \%\) figure was based on a representative sample of Internet users. Suppose that the sample size was \(n=300\) (the actual sample size was much larger). a. Use the given information to estimate the proportion of Internet users who have searched for information about themselves online. What statistic did you use? b. Use the sample data to estimate the standard error of \(\hat{p}\). c. Calculate and interpret the margin of error associated with the estimate in Part (a).

The article "Career Expert Provides DOs and DON'Ts for Job Seekers on Social Networking" (CareerBuilder.com, August 19,2009 ) included data from a survey of 2,667 hiring managers and human resource professionals. The article noted that more employers are now using social networks to screen job applicants. Of the 2,667 people who participated in the survey, 1,200 indicated that they use social networking sites such as Facebook, MySpace, and LinkedIn to research job applicants. Assume that the sample is representative of hiring managers and human resource professionals. Answer the four key questions (QSTN) to confirm that the suggested method in this situation is a confidence interval for a population proportion.

In a study of 1,710 schoolchildren in Australia (Herald Sun, October 27,1994 ), 1,060 children indicated that they normally watch TV before school in the morning. (Interestingly, only \(35 \%\) of the parents said their children watched TV before school.) Construct and interpret a \(95 \%\) confidence interval for the proportion of all Australian children who say they watch TV before school. In order for the method used to construct the interval to be valid, what assumption about the sample must be reasonable?

A random sample will be selected from the population of all adult residents of a particular city. The sample proportion \(\hat{p}\) will be used to estimate \(p,\) the proportion of all adult residents who are employed full time. For which of the following situations will the estimate tend to be closest to the actual value of \(p\) ? $$ \begin{array}{ll} \text { i. } & n=500, p=0.6 \\ \text { ii. } & n=450, p=0.7 \\ \text { iii. } & n=400, p=0.8 \end{array} $$

For each of the following choices, explain which would result in a narrower large-sample confidence interval for \(p\) : a. \(95 \%\) confidence level or \(99 \%\) confidence level b. \(n=200\) or \(n=500\)

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