A surprising number of young adults (ages 19 to 25) still live in their parents' homes. A random sample by the National Institutes of Health included 2253 men and2629 women in this age group. The survey found that 986 of the men and 923 of the women lived with their parents. Construct and interpret a 99% confidenceinterval for the difference in the true proportions of men and women aged 19 to 25 who live in their parents' homes.What conditions are met? (select all that apply) (Remember you check BOTH populations.)O 923 > 30O 10(923) < all women aged 19 to 25O 986 > 30O 2253 > 30O a random sampleV 2629 > 30O 986 > 10O 10(2629) < all women aged 19 to 25a random sampleO 1267 > 10O 1706 > 10O 923 > 10O 10(986) < all men aged 19 to 25O 10(2253) < all men aged 19 to 25O O

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Answered: A surprising number of young 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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A doctor would like to estimate the mean difference in the number of hours of sleep for seniors in high school and seniors in college. To do so, he selects a random sample of 50 high school seniors from all high schools in his county. He also selects a random sample of 50 seniors from the colleges in his county. He would like to construct a 95% confidence interval for the true mean difference in the number of hours of sleep for seniors in high school and seniors in college. Are the conditions for inference met? Yes, all three conditions for inference are met. No, the random condition is not met for both samples. No, the 10% condition is not met for both samples. No, the Normal/large sample condition is not met for both samples.
57 out of 70 students said they were comfortable with “working in a workplace with a lot of activity,” and 38 out of 71 said they were comfortable with “doing group projects.” Find a 90% confidence interval for the difference between these two proportions.
In a sample of 1901 patients who underwent a certain type of surgery, 22% experiencedcomplications. Find a 90% confidence interval for the proportion of all those undergoingthis surgery who experience complications
The Nielsen company surveyed 257 owners of android phones and found that 217 of them planned to get another android as their next phone. What is the lower bound for the 95% confidence interval for the proportion of android users who plan to get another android.
A medical study showed that 57 of 300 persons failed to recover from a given disease. Find a 95% confidence interval for the mortality rate of this disease.
In a survey of 1,626,773 U.S. adults, 49,311 personally identified as lesbian, gay, bisexual, or transgender. Construct a 95% confidence interval for the population portion of U.S. adults who personally identify as gay, lesbian,bisexual or transgender.
An internet service provider sampled 540 customers and found that 55 of them experienced an interruption in their service during the previous month. Construct a 95% confidence interval for the proportion of all the customers who have experienced a service interruption. Interpret the results.
Out of 100 people sampled 48 had kids. Based on this construct a 90% confidence interval for the true population proportion of people with kids.
A marketing survey involves product recognition in New York and California. Of 558 New Yorkers surveyed, 183 knew the product while 186 out of 614 Californians knew the product. Construct a 98% confidence interval for the difference between the two population proportions. Interpret the results.
A random survey of 36 airline passengers indicates that twelve have never flown before. Find a 95% confidence interval for the true proportion of airline passengers taking their first flight. Interpret
A surprising number of young adults (ages 19 to 25) still live in their parents' homes. A random sample by the National Institutes of Health included 2253 men and 2629 women in this age group. The survey found that 986 of the men and 923 of the women lived with their parents. Construct and interpret a 99% confidence interval for the difference in the true proportions of men and women aged 19 to 25 who live in their parents' homes. What conditions are met? (select all that apply) (Remember you check BOTH populations.) O 923 > 30 O 10(923) 30 O 2253 > 30 O a random sample V 2629 > 30 O 986 > 10 O 10(2629) 10 O 1706 > 10 O 923 > 10 O 10(986) < all men aged 19 to 25 O 10(2253) < all men aged 19 to 25 O O
A genealogist randomly selected 203 Americans and asked about their heritage. Of those randomly selected 6.25% of them claimed to have French heritage. Determine a 99% confidence interval for the percentage of Americans who claim to have French heritage.

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