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

Required conditions: n1>30 n2>30 n1*p1^>=10 n1*(1-p1^)>=10 n2*p2>=10 n2*p2^>=10

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 random sample of 1200 high school freshmen was taken and it was found that 559 of them were reading at or above their grade level. identify the population proportion that is being estimated, create a 95% confidence interval for the proportion, and interpret the confidence interval
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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.
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
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.
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Out of 300 people sampled, 279 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids.
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A doctor claims that the mean number of hours of sleep that seniors in high school get per night differs from the mean number of hours of sleep college seniors get per night. To investigate, 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 constructs 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. The resulting interval is (0.57, 1.25). Based upon the interval, can the doctor conclude that mean number of hours of sleep that seniors in high school get per night differs from the mean number of hours of sleep college seniors get per night? yes because 1 is in the confidence interval yes because 0 is not in the confidence interval no because 1 is in the confidence interval no because 0 is not in the confidence interval
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.

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