A random sample of 5008 adults in a country includes 760 who do not use the Internet. Construct a 95% confidenceinterval estimate of the percentage of adults in the country who do not use the Internet. Based on the result, does itappear that the percentage of adults in the country who do not use the Internet is different from 48%, which was thepercentage in the year 2000?Construct a 95% confidence interval estimate of the percentage of adults in the country who do not use the Internet.%

Answered: A random sample of 5008 adults in a…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Assume a nationwide random survey of 800 adults asked about attitudes toward “alternative medicine” such as acupuncture, massage therapy, and herbal therapy. Among the respondents, 448 said they would use alternative medicine if traditional medicine was not producing the results they wanted. Give a 99% confidence interval for the proportion of all adults who would use alternative medicine.
A survey conducted by the U.S. department of Labor found the 48 out of 500 heads of households were unemployed. Construct a 90% confidence interval estimate of the proportion of unemployed heads of households in the population.
A Gallup poll of 1487 adults showed that 43% of the respondents have Facebook pages. Construct a 95% confidence interval for the proportion of adults who have a Facebook page.
A recent survey showed that 42% of US employers were likely to require higher employee contributions for health care coverage. Suppose the survey was based on a sample of 500 companies. Find the 99% confidence interval for the proportion of all companies likely to require higher employee contributions for health care coverage. (Round to four decimal places)
Based on a sample of 1000 adult Americans, USA Today reported that 840 of them preferred cheese on their burgers. Calculate and interpret an 88% confidence interval for the proportion of all adult Americans who prefercheese on their burgers.
If the number of smokers in a sample of 40 students selected from university A is 14 and the number of smokers in a sample of 380 students selected from university B is 84. Then the 99% confidence interval for the difference between proportions of smokers in the two universities (PB- PA) is closest to O a. (-0.1450,-0.1130) O b. (-0.3310,0.0729) O c. None of these o d. (-0.3110,0.0534) e. (-0.1430,-0.1150)
In a survey of 3038 adults, 1448 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.
In a survey of 500 students, 327 said they were planning to stay in town for the holidays. What is the point estimate for the proportion of students staying home for the holidays? Construct the 95% confidence interval for the proportion of students staying home for the holidays. Assume all requirements have been met.
If a Random Sample of 450 adults includes 155 who identify as happy, compute an 80% confidence interval for the true proportion of adults who are happy.
In the United States, 1000 residents aged 15 or older were surveyed and 870 replied that they were satisfied with the water quality. The 90% confidence interval estimate of all U.S residents satisfied with their water quality is
A study was designed to estimate the prevalence of smoking in Harris county. Of 675 people sampled, 522 were non-smokers. What is the 95% Confidence Interval estimate for the prevalence? (74.2%, 80.5%) (19.5%, 25.8%) (.19, .25)
a survey of 3080 adults, 1471 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results

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