Due to the COVID-19 pandemic, many college classes have been moved online. The college would like toestimate the proportion of students that have access to reliable high-speed internet.A random sample of 329 college students was selected and it was found that 230 students out of the samplehad access to reliable high-speed internet.Using a 96% confidence level, determine a confidence interval for the proportion of all students at thecollege who have access to reliable high-speed internet.O (65.75 % , 74.07 % )O (64.72 % , 75.1 % )O (63.4 %, 76.42 % )O (64.42 % , 75.4% )> Next Question

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Answered: Due to the COVID-19 pandemic, many…

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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According to the website www.collegedrinkingprevention.gov, "About 25 percent of college students report academic consequences of their drinking including missing class, falling behind, doing poorly on exams or papers, and receiving lower grades overall." A statistics student is curious about drinking habits of students at his college. He wants to estimate the mean number of alcoholic drinks consumed each week by students at his college. He plans to use a 95% confidence interval. He surveys a random sample of 66 students. The sample mean is 3-75 alcoholic drinks per week. The sample standard deviation is 3.68 drinks. Construct the 95% confidence interval to estimate the average number of alcoholic drinks consumed each week by students at this college. Your answer should be rounded to 2 decimal places.
The gas mileage for a certain model of car is known to have a standard deviation of 6 mi/gallon. A simple random sample of 36 cars of this model is chosen and found to have a mean gas mileage of 28.4 mi/gallon. Construct a 99% confidence interval for the mean gas mileage for this car model.
A graduate student wanted to estimate the average time spent studying among graduate students at her school. She randomly sampled graduate students from her school and obtained a 99% confidence interval of (17,25) hours/week. What is the margin of error
A survey by the society of Human Resource Management asked 350 job seekers why employees change jobs so frequently. The answer selected most (126 times) was "higher compensation elsewhere" as the reason for changing jobs. Then the 95% confidence interval estimate for the population proportion of job seekers who would select "higher compensation elsewhere" as the reason for changing jobs is approximately.
An internet service provider sampled 540 customers and found that 75 of them had experienced an interruption in high-speed service during the previous month. Compute a 99% confidence interval for the proportion of customers who have experienced a service interruption during the previous month.
According to the website www.collegedrinkingprevention.gov, "About 25 percent of college students report academic consequences of their drinking including missing class, falling behind, doing poorly on exams or papers, and receiving lower grades overall." A statistics student is curious about drinking habits of students at his college. He wants to estimate the mean number of alcoholic drinks consumed each week by students at his college. He plans to use a 90% confidence interval. He surveys a random sample of 63 students. The sample mean is 3.7 alcoholic drinks per week. The sample standard deviation is 3.2 drinks. Construct the 90% confidence interval to estimate the average number of alcoholic drinks consumed each week by students at this college. 1. The t value is - 2. The standard error has a value of -- 3. The margin of error has a value of 4. Construct the 90% confidence interval to estimate the average number of alcoholic drinks consumed each week by students at this college.
A recent survey of two countries in South America reports on data about foreign-born citizens. Of 500 randomly selected people in Peru, 125 reported as being born in another country; meanwhile, 75 of 800 randonly selected Argentinians reported as being foreign-born. Use this information to create a 95% confidence interval for the true difference in proportions of foreign born citizens between Peru and Argentina.
A study of 22 goats reveals that the average length of a goat’s tail is 6.3 inches. If the standarddeviation is known to be 1.2 inches find a 95% confidence interval for the mean tail length.
A random sample of ni = 262 people who live in a city were selected and 104 identified as a "dog person." A random sample of n2 = 117 people who live in a rural area were selected and 50 identified as a "dog person." Find the 99% confidence interval for the difference in the proportion of people that live in a city who identify as a "dog person" and the proportion of people that live in a rural area who identify as a "dog person." Round answers to to 4 decimal places. < pi - P2 -
A phase three clinical trial for a new medication proposed to treat depression has 1191 participants. If all these individuals are from the group of individuals who were given the medication and 759 individuals who received the medication experienced improvement in their symptoms of depression, construct a 90% confidence interval for the true proportion of individuals on the medication who experienced improvement in their symptoms of depression.
A researcher studying reaction time of drivers states that, "A 95% confidence interval for the mean time it takes for a driver to apply the brakes after seeing the brake lights on a vehicle in front of him is 1.2 to 1.8 seconds." What are the point estimate and margin of error for this interval? Point estimate = 1.5 seconds; margin of error = 0.95 seconds. Point estimate = 1.2 seconds; margin of error = 0.3 seconds. Point estimate = 1.5 seconds; margin of error = 0.6 seconds. Point estimate = 1.5 seconds; margin of error = 0.3 seconds. a O b O d
A study of 420,064 cell phone users found that 136 of them developed cancer of the brain or nervous system. Prior to this study of cell phone use, the rate of such cancer was found to be 0.0343% for those not using cell phones . Use the sample data to construct 95% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system.

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