/*! 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 25 The article "Viewers Speak Out A... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 1002 randomly selected adults. Compute and interpret a bound on the error of estimation for the reported percentage.

Short Answer

Expert verified
The estimated margin of error is 2.2%. This means we can be 95% confident that the true proportion of adults who believe most reality TV shows are either 'totally made up' or 'mostly distorted' is between 79.8% and 84.2%.

Step by step solution

01

Understand the Concept of Error Margin

Margin of Error is calculated using the formula \(E = Z * \sqrt{{(p*(1-p))/n}}\) where \(E\) is the margin of error, \(Z\) is the Z-score (for a 95% confidence level, Z=1.96), \(p\) is the sample proportion (in this case 0.82), and \(n\) is the size of the sample (in this case 1002).
02

Substitute Corresponding Values into the Formula

Now, we will substitute corresponding values into the formula. Therefore, the formula becomes \(E = 1.96 * \sqrt{{(0.82*(1-0.82))/1002}}\).
03

Calculation

By calculating, we get \(E\) approximately equal to 0.022 or 2.2%.
04

Interpret the Result

The margin of error of 2.2% implies that if the survey were conducted 100 times, the results would be within 2.2% of the reported percentage 95 out of 100 times. This means that we can be 95% confident that the true proportion of adults who believe most reality TV shows are either 'totally made up' or 'mostly distorted' is between 79.8% and 84.2%.

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.

Error Estimation
Understanding the margin of error starts with grasping the concept of error estimation. It is a statistical measurement of the potential discrepancy between the observed sample proportion and the true population proportion. The formula for estimating the margin of error is typically expressed as \(E = Z * \sqrt{{(p*(1-p))/n}}\), where \(E\) represents the margin of error.

When applying error estimation, it is important to consider the sample size and the variability in the data. Larger sample sizes and lower variability both contribute to a smaller margin of error, indicating more precise estimates. For the given example, a sample size of 1002 adults provides a reasonably small margin of error, suggesting that the survey results are a fairly accurate reflection of the entire adult population's perceptions of reality TV.
Confidence Level
The confidence level is integral to understanding statistical results. It indicates the degree of certainty that the true parameter lies within the margin of error of the estimated value. Commonly used confidence levels include 90%, 95%, and 99%, with higher levels offering more reliability at the expense of a wider margin of error.

The Z-score in the error estimation formula corresponds to the desired confidence level. For a 95% confidence level, the Z-score is 1.96, meaning that if the same survey were to be repeated many times, the proportion of respondents who believe reality TV to be fabricated would fall between the computed range of error 95 times out of 100. It's a balance between confidence and precision; as confidence level increases, so does the margin of error.
Sample Proportion
The sample proportion, represented by \(p\) in our margin of error formula, is the percentage of survey respondents that exhibit the characteristic being measured—in this case, the belief that reality TV is not genuine. It is calculated by dividing the number of respondents who agree with the statement by the total number of respondents. With a sample proportion of 0.82, it suggests that a significant majority—82%—of the selected adults share this skepticism towards reality TV.

This sample proportion is a snapshot of the larger population's opinions. But, because it is based on a sample rather than the entire population, we need the margin of error to account for the fact that different samples could yield slightly different proportions.
Survey Sampling
Survey sampling involves selecting a subset of the population to represent the whole. The key to effective sampling is ensuring that the sample is randomly selected, which helps to avoid bias and ensures that the results are generalizable to the entire population. The quality of survey sampling can significantly influence the margin of error; a well-designed, random sampling method will lead to a more accurate estimation of the population parameter.

In the exercise mentioned, a sample of 1002 randomly selected adults was chosen. This sizeable random sample helps to provide a reliable representation of the adult population's opinion on reality TV. When interpreting survey results, it's crucial to consider the sampling method used to assess the credibility of the findings.

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

An Associated Press article on potential violent behavior reported the results of a survey of 750 workers who were employed full time iSan Luls Obispo Tribune. September \(7,\) 1999), Of those surveyed, 125 indicated that they were so angered by a coworker during the past year that they felt like hitting the coworker (but didn't). Assuming that it is reasonable to regard this sample of 750 as a random sample from the population of full-time workers, use this information to construct and interpret \(90 \%\) confidence interval estimate of \(p,\) the true proportion of full-time workers so angered in the last year that they wanted to hit a colleague.

The formula used to compute a large-sample confidence interval for \(p\) is $$ \hat{p} \pm(z \text { critical value }) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(95 \%\) d. \(80 \%\) b. \(90 \%\) e. \(85 \%\) c. \(99 \%\)

In an AP-AOL sports poll (Associated Press, December 18,2005\(), 394\) of 1000 randomly selected U.S. adults indicared that they considered themselves to be baseball fans. Of the 394 baseball fans, 272 stated that they thought the designated hitter rule should either be expanded to both baseball leagues or eliminated. a. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults who consider themselves to be bascball fans. b. Construct a \(95 \%\) confidence interval for the proportion of those who consider themselves to be baseball fans who think the designated hitter rule should be expanded to both leagues or eliminated. c. Explain why the confidence intervals of Parts (a) and (b) are not the same width even though they both have a confidence level of \(95 \%\).

The authors of the paper "Deception and Design . The Impact of Communication Technology on Lying Behavior" (Proceedings of Computer Human Interaction [2004]\()\) asked 30 students in an upper division communications course at a large university to keep a journal for 7 days, recording each social interaction and whether or not they told any lies during that interaction. A lie was defined as "any time you intentionally try to mislead someone." The paper reported that the mean number of lies per day for the 30 students was 1.58 and the standard deviation of number of lies per day was 1.02 a. What assumption must be made in order for the \(t\) confidence interval of this section to be an appropriate method for estimating \(\mu,\) the mean number of lies per day for all students at this university? b. Would you recommend using the \(t\) confidence interval to construct an estimate of \(\mu\) as defined in Part (a)? Explain why or why not.

The two intervals (114.4,115.6) and \((114.1,\) 115.9 ) are confidence intervals (computed using the same sample data) for \(\mu=\) true average resonance frequency (in hertz) for all tennis rackets of a certain type. a. What is the value of the sample mean resonance frequency? b. The confidence level for one of these intervals is \(90 \%\) and for the other it is \(99 \%\). Which is which, and how can you tell?

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.