The principal of a school is interested in estimating the average income perfamily of her students. She selects a random sample of students and collectsinformation about their family income. A 95 percent confidence intervalcomputed from this data for the mean income per family is ($35,095,$45,005).Which of the following provides the best interpretation of this confidenceinterval?There is a 95% probability that the families of all the students in this school have anincome of between $35,095 and $45,005.95 percent of the students in her school are from families whose income is between$35,095 and $45,005.There is a 95% probability that the mean of another sample with the same size will fallbetween $35,095 and $45,005.There is a 95% probability that the mean income per family in the school is between$35,095 and $45,005.None of theseIf we were to take another sample of the same size and compute a 95 percentO confidence interval, we would have a 95% chance of getting the interval ($35,095,$45,005).

Answered: The principal of a school is interested…

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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Eric randomly surveyed 150 adults from a certain city and asked which team in a contest they were rooting for, either North High School or South High School. From the results of his survey, Eric obtained the following 95% confidence interval for the proportion of all adults in the city rooting for North High, (0.52,0.68). Interpret this confidence interval. A.We are 95% sure that between 52% and 68% of all adults in this city will root for North High School.B.We are 95% sure that between 52% and 68% of adults in this sample will root for North High School.C.There's a 95% chance that between 52% and 98% of all adults in this city will root for North High School.D.Between 52% and 68% of adults in this city will root for North High School 95% of the time.
Which of the following confidence intervals would have a wider width for the same set of data? A 90% interval. A 99% interval. Group of answer choices The 90% interval. The 99% interval.
Find the confidence level and α for a. a 90% confidence interval. b. a 99% confidence interval.
You plan to conduct a survey to find what proportion of the workforce has two or more jobs. You decide on the 90% confidence level and a margin of error of 2%. A pilot survey reveals that 5 of the 47 sampled hold two or more jobs. (Use Z-Distribution Table.) How many in the workforce should be interviewed to meet your requirements? (Round Z Score to 2 decimal places. Round up your answer to the nearest whole number.)
A college professor would like to estimate the proportion of students who pull an "all-nighter," meaning they study all night for an upcoming exam. She selects a random sample of 100 students from her large college and finds that the 99% confidence interval for the true proportion of students who pull an all-nighter is 0.48 to 0.62. Which of these statements is a correct interpretation of the confidence level? O Approximately 99% of all students have pulled an all-nighter. O There is a 99% probability that the true proportion of all students who pull an all-nighter is between 0.48 and 0.62. If many random samples of size 100 are selected from all students at this college, approximately 99% of the intervals would capture the true proportion who pull an all-nighter. C Approximately 99% of the sample proportions of students who pull an all-nighter, based on random samples of size 100 from the population of all students at this college, will fall between 0.48 and 0.62.
Tonya wants to estimate what proportion of her school's seniors plan to attend the prom. She interviews an SRS of 50 of the 750 seniors in her school and finds that 36 plan to go to the prom. Tonya then calculates a 90% confidence interval of (0.616, 0.824). Interpret Tonya's interval in context. 90% of the seniors at Tonya's high school are between 61.6% and 82.4% likely to attend prom. We can be 90% confident that the sample proportion is captured by the confidence interval. We are 90% confident that the interval from 0.616 to 0.824 captures the true proportion of all seniors at Tonya's high school who plan to attend the prom. O If Tonya took many samples, 90% of them would find that 72.4% of the seniors in the sample plan on attending prom. The percent of seniors at Tonya's high school who plan to attend the prom is certain to be between 61.6% and 82.4%. terms of use contact us help careers privacy policy about us
n 8.2 In 2013, the Gallup poll asked 1,039 American adults if they believe there was a conspiracy in the assassination of President Kennedy, and found that 634 believe there was a conspiracy ("Gallup news service," 2013). Estimate the proportion of American's who believe in this conspiracy using a 98% confidence interval.
If two people used the same data and one person made a 90% confidence interval and the other person made a 95% confidence interval for p, which interval would be more narrow? is it; a. The 90% Confidence interval b. The 95% Confidence interval c. We cannot tell from the information alone.
Debbie and Ethan are going to construct confidence intervals from the same simple random sample. Debbie's confidence interval will have a 98% confidence interval and Ethan's will have a 95% confidence interval. A. Which confidence interval, if there is a difference will have a larger margin of error? B. Which confidence interval is more likely to cover the population mean, if there is a difference I am using a Ti84 and need to know how to get the answer.
We randomly sample 218 individuals and find that 97 have a college degree. a.)Construct a 95% confidence interval. b.) Describe our interpretation of the confidence level.
Online News In a Harris poll of 2036 adults, 40% said that they prefer to get their news online. Construct a 95% confidence interval estimate of the percentage of all adults who say that they prefer to get their news online. Can we safely say that fewer than 50% of adults prefer to get their news online?
Four students are instructed to calculate a confidence interval based on the same sample mean (xbar: 0.76) and the same sample standard deviation s. However, each student is given a different sample size and confidence level. The following are the intervals obtained by the four students. Which of the following must be incorrect? A. (0.09, 1.43) B. (-1.11, 2.89) C. (-0.57, 2.09) D. (-2.22, 3.74)

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