/*! 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 21 The article "CSI Effect Has Juri... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The article "CSI Effect Has Juries Wanting More Evidence" (USA Today, August 5,2004 ) examines how the popularity of crime-scene investigation television shows is influencing jurors' expectations of what evidence should be produced at a trial. In a survey of 500 potential jurors, one study found that 350 were regular watchers of at least one crime-scene forensics television series. a. Assuming that it is reasonable to regard this sample of 500 potential jurors as representative of potential jurors in the United States, use the given information to construct and interpret a \(95 \%\) confidence interval for the proportion of all potential jurors who regularly watch at least one crime- scene investigation series. b. Would a \(99 \%\) confidence interval be wider or narrower than the \(95 \%\) confidence interval from Part (a)?

Short Answer

Expert verified
a) The 95% confidence interval for the proportion of potential jurors who are regular watchers of at least one crime-scene investigation series is (0.6568, 0.7432). b) A 99% confidence interval would be wider than the 95% confidence interval.

Step by step solution

01

Calculate the Proportion

First, find out the proportion \(p\) of potential jurors who are regular watchers of at least one crime-scene forensics television series. The number of watchers is 350 and the total sample size is 500. So \(p = \frac{350}{500} = 0.7\).
02

Calculate Standard Error

The standard error (\(SE\)) can be calculated with the formula \(SE = \sqrt{p(1-p)/n}\) where \(n\) is the sample size. So, \( SE = \sqrt{0.7*(1-0.7)/500} = 0.02236\).
03

Construct 95% Confidence Interval

A 95% confidence interval is calculated as \(p \pm Z*SE\) where \(Z\) is the Z-score for the desired confidence level. For a 95% confidence interval, \(Z = 1.96\). The interval then becomes \(0.7 \pm 1.96*0.02236\). The resulting interval is \((0.6568, 0.7432)\). This means we are 95% confident that the true proportion of potential jurors who consistently watch at least one crime-scene investigation series is between 65.68% and 74.32%.
04

Compare 99% Confidence Interval with 95% Confidence Interval

A 99% confidence interval would be wider than the 95% confidence interval because the higher the confidence level, the wider the interval. This is because to be more confident about our estimate, we need to allow for more possible values.

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.

Proportion Calculation
The first step in determining a confidence interval is to calculate the proportion of interest from the sample data. In the given problem, we are interested in the proportion of potential jurors who are regular watchers of crime-scene investigation TV series. Let's break down the process:
  • Numerator: The number of successful observations, in this case, jurors who regularly watch, is 350.
  • Denominator: The total sample size from which the successful observations are drawn, which is 500.
To get the proportion (\( p \)), simply divide the number of watchers by the total sample size: \( p = \frac{350}{500} = 0.7 \).
This result means that 70% of the surveyed potential jurors regularly watch at least one crime-scene forensics series. Understanding the proportion is crucial, as it forms the foundation for further calculations, like the standard error and the confidence interval.
Standard Error
The standard error is a key statistic used to measure the variability of a sample proportion. It helps us understand how much the sample proportion (\( p \)) might vary from the true population proportion. The formula for standard error when dealing with proportions is: \( SE = \sqrt{\frac{p(1-p)}{n}} \), where \( n \) is the total number in the sample.
For the juror problem, plug in the numbers:\[ SE = \sqrt{\frac{0.7(1-0.7)}{500}} = \sqrt{\frac{0.21}{500}} = 0.02236 \]
This standard error of approximately 0.02236 tells us that the proportion we calculated from the sample (0.7) might vary by this amount if we took other, similar samples from the same population.
In simpler terms, the smaller the standard error, the more precise our estimate. Small standard errors imply that the sample proportion is a good estimate of the true population proportion.
Z-score
The Z-score is an important statistical tool that helps us determine the confidence interval for our sample proportion. It indicates how many standard deviations an element is from the mean. In the context of confidence intervals, the Z-score corresponds to the desired level of confidence.
  • For a 95% confidence interval, the Z-score is approximately 1.96.
  • This means we are using 1.96 standard deviations around our sample proportion to create our confidence interval.
The formula for the confidence interval is: \( p \pm Z \times SE \). Plugging in the numbers:\[ 0.7 \pm 1.96 \times 0.02236 = (0.6568, 0.7432) \]
This calculation gives the range within which we are 95% confident that the true proportion of all potential jurors lies. Choosing a 99% confidence interval instead would require a larger Z-score (approximately 2.576), resulting in a wider interval. This wider range reflects the greater certainty we want, allowing more room for the true proportion.

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

If a hurricane was headed your way, would you evacuate? The headline of a press release issued January 21, 2009 by the survey research company International CommunicationsResearch(icrsurvey.com) states, "Thirtyone Percent of People on High-Risk Coast Will Refuse Evacuation Order, Survey of Hurricane Preparedness Finds." This headline was based on a survey of 5046 adults who live within 20 miles of the coast in high hurricane risk counties of eight southern states. In selecting the sample, care was taken to ensure that the sample would be representative of the population of coastal residents in these states. Use this information to estimate the proportion of coastal residents who would evacuate using a \(98 \%\) confidence interval. Write a few sentences interpreting the interval and the confidence level associated with the interval.

Fat content (in grams) for seven randomly selected hot dogs that were rated as very good by Consumer Reports (www.consumerreports.org) is shown below. Is it reasonable to use this data and the \(t\) confidence interval of this section to construct a confidence interval for the mean fat content of hot dogs rated as very good by Consumer Reports? Explain why or why not. \(\begin{array}{lllllll}14 & 15 & 11 & 10 & 6 & 15 & 16\end{array}\)

The article “Kids Digital Day: Almost 8 Hours" (USA Today, January 20,2010 ) summarized results from a national survey of 2002 Americans age 8 to 18 . The sample was selected in a way that was expected to result in a sample representative of Americans in this age group. a. Of those surveyed, 1321 reported owning a cell phone. Use this information to construct and interpret a \(90 \%\) confidence interval estimate of the proportion of all Americans age 8 to 18 who own a cell phone. b. Of those surveyed, 1522 reported owning an MP3 music player. Use this information to construct and interpret a \(90 \%\) confidence interval estimate of the proportion of all Americans age 8 to 18 who own an MP3 music player. c. Explain why the confidence interval from Part (b) is narrower than the confidence interval from Part (a) even though the confidence level and the sample size used to compute the two intervals was the same.

"Tongue Piercing May Speed Tooth Loss. Researchers Say" is the headline of an article that appeared in the San Luis Obispo Tribune (June 5,2002 ). The article describes a study of 52 young adults with pierced tongues. The researchers found receding gums, which can lead to tooth loss, in 18 of the participants. Construct a \(95 \%\) confidence interval for the proportion of young adults with pierced tongues who have receding gums. What assumptions must be made for use of the \(z\) confidence interval to be appropriate?

Seventy-seven students at the University of Virginia were asked to keep a diary of conversations with their mothers, recording any lies they told during these conversations (San Luis Obispo Telegram-Tribune, August 16,1995 ). It was reported that the mean number of lies per conversation was \(0.5 .\) Suppose that the standard deviation (which was not reported) was \(0.4 .\) a. Suppose that this group of 77 is a random sample from the population of students at this university. Construct a \(95 \%\) confidence interval for the mean number of lies per conversation for this population. b. The interval in Part (a) does not include \(0 .\) Does this imply that all students lie to their mothers? Explain.

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.