CR2. The USA Today claims that 44% of adults who access the Internet read the international newsonline. You want to check the accuracy of their claim by surveying a random sample of 120adults who access the Internet and asking them if they read the international news online. Fifty-two adults respond "yes."dinelh boa• Verify that the sampling distribution of p canapproximated by the normal distribution.lemonUse a 95% confidence interval to test the newspaper's claim. Use mathematics to explain youranswer.n i bgguonda olion A Dd ltoo lo u 85odralon lo noitol b eaignialealeo narlwnodte lon w.anim .lei sonpilnoboingIf we were to increase the value of n from 120 to 200, how would the width of the confidenceinterval be effected?Statistics Unit 6 Assessment

a. It is claimed that 44% of adults, who access the Internet, read the international news online.

Answered: The USA Today claims that 44% of adults…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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You measure 49 texbook's weights, and find they have a mean weight of 85 ounces. assume the population standard deviation is 3.6 ounces. Based on this, construct a 95% confidence interval for the true population mean textbook weight. round your answer to 2 decimals
The National Center for Education Statistics surveyed 4400 college graduates about the lengths of time required to earn their bachelor's degrees. The mean is 5.15 years, and the standard deviation is 1.68 years. Based enthuse sample data, construct the 99% confidence interval for the mean time required by all college graduates. left left: right:
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In a poll, 1100 adults in a region were asked about their online vs. in-store clothes shopping. One finding was that 44% of respondents never clothes-shop online. Find and interpret a 95% confidence interval for the proportion of all adults in the region who never clothes-shop online. Click here to view page 1 of the table of areas under the standard normal curve. Click here to view page 2 of the table of areas under the standard normal curve. The 95% confidence interval is from to (Round to three decimal places as needed.)
A researcher wants to estimate the mean amount of money spentonline by consumers in a particular market. She samples 16 consumers and finds that they spend on average $854 online with a standard deviation of $152. Construct a 99% confidence interval estimate of the mean amount of money spent online. Write a statement that correctly interprets the confidence interval.
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A random sample of 64 shoppers showed that they spend an average of $27.31 per visit at the Sunday Mornings Bookstore. The standard deviation of the population is $2.14. Find the 90% confidence interval of the true mean. Please include all work showing how you found your answer.
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A poll asked whether states should be allowed to conduct random drug tests on elected officials. Of 21,327 respondents, 56% said "yes." a. Determine the margin of error for a 99% confidence interval. b. Without doing any calculations, indicate whether the margin of error is larger or smaller for a 90% confidence interval. Explain your answer. Click here to view page 1 of the table of areas under the standard normal curve. Click here to view page 2 of the table of areas under the standard normal curve. a. The margin of error for a 99% confidence interval is (Round to three decimal places as needed.) b. Without doing any calculations, indicate whether the margin of error is larger or smaller for a 90% confidence interval. A. The margin of error will be smaller for a 90% confidence interval. B. The margin of error will be about the same for a 90% confidence interval. C. The margin of error will be larger for a 90% confidence interval. D. There is insufficient information to determine the…

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