1. Getting a good night's sleep is essential for teenage students to perform optimally during school.According to the National Sleep Foundation, high school students need at least 8 hours of sleep foroptimal performance at school the following day. Based on a sleep study of 459 students, a 95%Confidence Interval for the true mean sleeping time (in hours) was calculated to be (7, 7.28) hours.a. Based on the Confidence Interval given, can you say with high degree of confidence that thetrue mean of sleeping time is below 8 hours? In a sentence or two explain why or why not.b. If we were to increase our sample size, what would we expect to happen to the confidenceinterval? Get wider/ get narrower/ stay the same?c. If we were to increase our level of confidence, what would we expect to happen to the confidenceinterval? Get wider/ get narrower/ stay the same?

The sample size, i.e., the sampled no. o students, n=459. The 95% CI for the true mean sleeping time…

Answered: 1. Getting a good night's sleep is…

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 of 420,098 cell phone users found that 0.0324% 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 0.0335% for those not using cell phones. Complete parts (a) and (b). a. Use the sample data to construct a 95% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system Do not round until the final answer. Then round to three decimal places as needed.] R
1. Getting a good night's sleep is essential for teenage students to perform optimally during school. According to the National Sleep Foundation, high school students need at least 8 hours of sleep for optimal performance at school the following day. Based on a sleep study of 459 students, a 95% Confidence Interval for the true mean sleeping time (in hours) was calculated to be (7, 7.28) hours. a. Researchers want to test if students, on average, are getting more than the required amount of sleep to perform optimally at school. State the null and alternative hypothesis in proper notation. b. What conclusion can you make about the null and alternative hypothesis based on the confi- dence interval ?
Use Table V in Appendix A to determine the t-percentile that is required to construct each of the following two-sided confidence intervals: (Express your answers to 3 decimal places.) Confidence level = 95 %, degrees of freedom = 11 , Confidence level = 95% , degrees of freedom = 14 Confidence level = 99 %, degrees of freedom = 12 Confidence level = 99% , degrees of freedom = 10
A Gallup poll of 1487 adults showed that 43% of the respondents have Facebook pages. Construct a 95% confidence interval for the proportion of adults who have a Facebook page.
In a survey funded by the UW school of medicine, 750 of 1000 adult Seattle residents said they did not believe they could come down with a sexually transmitted infection (STI). Construct a 95% confidence interval estimage of the proportion of adult Seattle residents who don't believe they can contract an STI. (Use a z score of 1.96 for your computations.) (.728, .772) (.723, .777) (.718, .782) (.713, .878) (.665, .835)
5. A researcher constructed a 95% confidence interval for the true mean number of years students take to earn a bachelor's degree. His results were (4.319, 4.841). a. What is the point estimate for the average number of years to complete a bachelor's degree? b. Does his interval support or contradict the widely held belief that it takes 4 years to complete a bachelor's degree? Why or why not?
At a large high school, an assistant principal wants to know how intelligent the students at the school are. An intelligence test is administered to an SRS of 50 students, out of over 2000 students. The mean score was 115 and the standard deviation was 15. Construct a 99% confidence interval for the mean score of all students in the school. A) (109.32, 120.69) B) (119.58, 123.50) C) (110.87, 129.41) D) Can not be determined from the information given.

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