Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Online News In a Harris poll of 2036 adults, 40% said that they prefer to get their news online. Construct a 95% confidence interval estimate of the percentageof all adults who say that they prefer to get their news online. Can we safely say that fewer than 50% of adults prefer to get their news online?

Short Answer

Expert verified

The 95% confidence interval estimate of the percentage of all adults who say that they prefer to get their news online is 37.9%< p< 42.1.

Yes, you can safely say that fewer than 50% of adults prefer to get their news online.

Step by step solution

01

Given information

The number of adults is n=2036.

The percentage of adults who prefer to get their news online is 40%.

The level of confidence is 95%.

02

Compute the confidence interval

From the given information, the following points can be drawn:

The proportion of adults who prefer to get their news online isp^=0.40.

The level of confidence is 95%, which implies that the level of significance is 0.05.

From the Z table, the two-tailed critical value obtained at 0.05 level of significance is 1.96.

The margin of error is computed as follows:

role="math" localid="1648195038482" E=zα2p^(1-p^)n=1.96×0.40×1-0.402036=0.0213

Therefore, the margin of error is 0.213.

The 95% confidence interval is computed as follows:

role="math" localid="1648195006699" (p^-E,p^+E)=0.40-0.0213,0.40+0.0213=0.3787,0.42130.379,0.421

Therefore, the confidence interval estimate of the percentage of all adults who say that they prefer to get their news online is37.9% <p <42.1%.

03

Provide the conclusion

Since the actual percentage for the population of adults is within the confidence limits of 37.9% and 42.1%,it can be said that fewer than 50% of adults prefer to get their news online.

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 Vaia!

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

Determining Sample Size. In Exercises 31–38, use the given data to find the minimum sample size required to estimate a population proportion or percentage.

Astrology

A sociologist plans to conduct a survey to estimate the percentage of adults who believe in astrology. How many people must be surveyed if we want a confidence level of 99% and a margin of error of four percentage points?

a. Assume that nothing is known about the percentage to be estimated.

b. Use the information from a previous Harris survey in which 26% of respondents said that they believed in astrology.

Normality Requirement What does it mean when we say that the confidence interval methodsof this section are robust against departures from normality?

Constructing and Interpreting Confidence Intervals. In Exercises 13–16, use the given sample data and confidence level. In each case, (a) find the best point estimate of the population proportion p; (b) identify the value of the margin of error E; (c) construct the confidence interval; (d) write a statement that correctly interprets the confidence interval.

Eliquis The drug Eliquis (apixaban) is used to help prevent blood clots in certain patients. In clinical trials, among 5924 patients treated with Eliquis, 153 developed the adverse reaction of nausea (based on data from Bristol-Myers Squibb Co.). Construct a 99% confidence interval for the proportion of adverse reactions.

In Exercises 9–16, assume that each sample is a simplerandom sample obtained from a population with a normal distribution.

Insomnia Treatment A clinical trial was conducted to test the effectiveness of the drug zopiclone for treating insomnia in older subjects. After treatment with zopiclone, 16 subjects had a mean wake time of 98.9 min and a standard deviation of 42.3 min (based on data from “Cognitive Behavioral Therapy vs Zopiclone for Treatment of Chronic Primary Insomnia in Older Adults,” by Sivertsen et al.,Journal of the American Medical Association,Vol. 295, No. 24). Assume that the 16 sample values appear to be from a normally distributed population and construct a 98% confidence interval estimate of the standard deviation of the wake timesfor a population with zopiclone treatments. Does the result indicate whether the treatment is effective?

In Exercises 5–8, use the relatively small number of given bootstrap samples to construct the confidence interval. Freshman 15: Here is a sample of amounts of weight change (kg) of college students in their freshman year (from Data Set 6 “Freshman 15” in Appendix B): 11, 3, 0, -2, where -2 represents a loss of 2 kg and positive values represent weight gained. Here are ten bootstrap samples: {11, 11, 11, 0}, {11, -2, 0, 11}, {11, -2, 3, 0}, {3, -2, 0, 11}, {0, 0, 0, 3}, {3, -2, 3, -2}, {11, 3, -2, 0}, { -2, 3, -2, 3}, { -2, 0, -2, 3}, {3, 11, 11, 11}. a. Using only the ten given bootstrap samples, construct an 80% confidence interval estimate of the mean weight change for the population. b. Using only the ten given bootstrap samples, construct an 80% confidence interval estimate of the standard deviation of the weight changes for the population.

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.

Sign-up for free