For a random sample of 96 smokers, the mean amount of short-term absenteeism from work was 2.15 hours per month, and the sample standard deviation was 2.09hours per month. For an independent random sample of 206 employees who had never smoked, the mean amount of absenteeism was 1.69 hours per month, and thesample standard deviation was 1.91 hours per month. Find a 99% confidence interval for the difference between the two population means.O a. -0.19< ux-uy < 111O b. 019 < px-uy <1.11O ch 0.29< HX-Hy<211Q d. 0.29< jx-Hy < 2.11230428145740-46402

Answered: Image /qna-images/answer/a3d58fb1-f91d-4527-89f3-48939bccc1ef.jpg

Answered: For a random sample of 96 smokers, the…

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 study is conducted to find the mean life of a certain manufacturer's car batteries. A random sample of 34 batteries is selected and a mean of 47.5 months with a standard deviation of 7.3 months is found. Find a 95% confidence interval for the mean battery life.
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The monthly advertising expenditure of the company is normally distributed with astandard deviation of $200. If a sample of 36 randomly selected months yields a meanadvertising expenditure of $1360 monthly, what is a 94% confidence interval for the mean ofthe company’s monthly advertising expenditure?
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In a random sample of 17 people, the mean commute time to work was 30.2 minutes and the standard deviation was 7.1 minutes. Assume the population is normally distributed and use a t- distribution to construct a 95% confidence interval for the population mean y. What is the margin of error of y? Interpret the results.
A study of a local high school tried to determine the mean number of text messages that each student sent per day. The study surveyed a random sample of 114 students in the high school and found a mean of 195 messages sent per day with a standard deviation of 73 messages. Determine a 95% confidence interval for the mean, rounding all values to the nearest whole number.
A study was conducted to estimate hospital costs for car accident victims who woreseat belts. Suppose that 46 randomly selected cases had an average of $15150and a standard deviation of $8500. If the average hospital costs for those that didnot wear seatbelts was $19000, construct a 90% confidence interval to see if theaverage costs are significantly different.
A sports news station wanted to know whether people who live in the North or the South are bigger sports fans. For its study, 121 randomly selected Southerners were surveyed and found to watch a mean of 4.6 hours of sports per week. In the North, 176 randomly selected people were surveyed and found to watch a mean of 3.2 hours of sports per week. Find a 90% confidence interval for the true difference between the mean numbers of hours of sports watched per week for the two regions if the South has a population standard deviation of 1.6 hours per week and the North has a population standard deviation of 1.4 hours per week. Let Population 1 be people who live in the South and Population 2 be people who live in the North. Round the endpoints of the interval to one decimal place, if necessary. Lower endpoint: Upper endpoint:
A sample of 12 suburban hotels showed that the mean revenue is $1117. It is known that the standard deviation is $121.8. Construct the 90% confidence interval for the true average revenue of all suburban hotels.
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