A 2018 poll of 2214 randomly selected U.S. adults found that 45.62% planned to watch at least a "fair amount" of a particular sporting event in 2018. In 2014, 51% of U.S. adults reported planning to watch at least a "fair amount."a. Does this sample give evidence that the proportion of U.S. adults who planned to watch the 2018 sporting event was less than the proportion who planned to do so in 2014? Use a 0.05 significance level.b. After conducting the hypothesis test, a further question one might ask is what proportion of all U.S. adults planned to watch at least a "fair amount" of the sporting event in 2018. Use the sample data to construct a 90% confidence intervalfor the population proportion. How does your confidence interval support your hypothesis test conclusion?a. State the null and alternative hypotheses. Let p be the proportion of U.S. adults that planned to watch at least a "fair amount" of the sporting event.Ho: pHa: P(Туреcimals. Do not round.)%3D>

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Answered: A 2018 poll of 2214 randomly selected…

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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A scientific study was conducted to determine the hours of sleep that college students get per night. The study surveyed 20 college students and obtained the following data:8, 8.3, 7.8, 7.7, 8.6, 8.1, 7.7, 8.5, 8.1, 8.2, 8.2, 7.9, 8.5, 7.9, 8.3, 8.2, 7.8, 7.8, 7.8, 7.7Assuming the population standard deviation is σ=0.31σ=0.31, construct a 86% confidence interval for the average hours of sleep per night that college students get.xˉ=x̄=α2=α2=zα2=zα2=Margin of Error: E=E=We are 86% confident that college students get an average of between hours and hours of sleep per night. Submit QuestionQuestion
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