A survey is done at a company to determine how satisfied people are with their jobs, with 1 being very satisfied and 3 being very dissatisfied. Since there are so many employees, only 50 people are given the survey.A JMP Appletjmp?v Report▼ DistributionsV SatisfactionV FrequenciesLevel CountProb118 0.3600019 0.38000313 0.26000Total50 1.00000N Missing3 LevelsV DistributionsV DistributionsConceal contentsSatisfactionV Satisfactionv Confidence Intervalsv Confidence IntervalsLevel CountLevelProb Lower CI Upper CI 1-AlphaProb Lower CI Upper CI 1-Alpha0.900Count118 0.36000 0.258197 0.476177118 0.36000 0.2112930.541510.990219 0.38000 0.276014 0.4963060.90019 0.38000 0.227329 0,5607880.990313 0.26000 0.172179 0.3724610.900313 0.26000 0.1353730.440860.990TotalNote: Computed using score confidence intervals.50Total50Note: Computed using score confidence intervals.v Confidence Intervalsv Confidence IntervalsLevel CountProb Lower CI Upper CI 1-AlphaLevelCountProb Lower CI Upper CI 1-Alpha118 0.36000 0.2413870.498590.950118 0.36000 0.269655 0.4614880.850219 0.38000 0.258628 0.5184960.950219 0.38000 0.287829 0.4817220.850313 0.26000 0.158715 0.3955320.950313 0.26000 0.181529 0.3575730.850Total50Total50Note: Computed using score confidence intervals.Note: Computed using score confidence intervals. O (a)Determine the proportion of employees who chose the satisfaction rating 2 (neutral).Find the 99% confidence interval of the population proportion associated with the satisfaction rating of 2. (Round your answers to three decimal places.)O (b)The company is interested in which confidence intervals associated with 1 (very satisfied) contain the proportion 0.25, or 25%. Determine which confidence levels satisfy this. (Select all that apply.)O 85%O 90%O 95%O 99%Determine if the statement is true or false.If the same sampling type was done numerous times and a 90% confidence interval was given each time, at least 90% of these intervals would contain the true population proportion of people selecting 1 as the satisfaction rating.O TrueO False

Introduction: The 100 (1 – α) % confidence interval for the population proportion, p is: Here, n is…

Determine the proportion of employees who chose the satisfaction rating 2 (neutral). Find the 99% confidence interval of the population proportion associated with the satisfaction rating of 2. (Round your answers to three decimal places.) O (b) The company is interested in which confidence intervals associated with 1 (very satisfied) contain the proportion 0.25, or 25%. Determine which confidence levels satisfy this. (Select all that apply.) O 85% O 90% O 95% O 99% Determine if the statement is true or false. If the same sampling type was done numerous times and a 90% confidence interval was given each time, at least 90% of these intervals would contain the true population proportion of people selecting 1 as the satisfaction rating. O True O False

Determine the proportion of employees who chose the satisfaction rating 2 (neutral). Find the 99% confidence interval of the population proportion associated with the satisfaction rating of 2. (Round your answers to three decimal places.) O (b) The company is interested in which confidence intervals associated with 1 (very satisfied) contain the proportion 0.25, or 25%. Determine which confidence levels satisfy this. (Select all that apply.) O 85% O 90% O 95% O 99% Determine if the statement is true or false. If the same sampling type was done numerous times and a 90% confidence interval was given each time, at least 90% of these intervals would contain the true population proportion of people selecting 1 as the satisfaction rating. O True O False

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

You are interested in finding a 95% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 12 randomly selected physical therapy patients. Round answers to 3 decimal places where possible.
I think the answer to this last part is C
Write your answer in the p +- e notation. A study found that out of 39,700 drivers that went through a traffic intersection, 794 ran a red light. Construct a 90% confidence interval to estimate the proportion of people who ran the light. We will use estimation using confidence intervals. This time we will be using estimation of proportion. x ____________ n ____________ p hat ___________ C-level ___________
A new fertilizer was applied to the soil of 246 bean plants. 25% showed increased growth. Find the margin of error and 95% confidence interval for the percentage of all bean plants which show increased growth after application of the fertilizer. Round all answers to 2 decimal places. Margin of Error (as a percentage):Confidence Interval: % to %
Explain also how you use Z table ..
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 74 dates, the mean record high daily temperature in a certain city has a mean of 82.58° F. Assume the population standard deviation is 14.82°F. The 90% confidence interval is (Round to two decimal places as needed.) The 95% confidence interval is ( I). (Round to two decimal places as needed.) Which interval is wider? Choose the correct answer below. The 95% confidence interval The 90% confidence interval
R1
The 99% confidence interval for the proportion of college students that prefer to work for a large company is
Quantitative Analysis Determine the confidence interval of the set of data in both columns and write your answer on the space provided. Then on the blank before each number, write: A. If the standard error in column A is greater than the standard error in column B; B. If the standard error in column B is greater than the standard error in column A; C. If the values of standard error in both columns are equal; D. If the relationships cannot be determined from the information given. Column A Column B Determine the confidence interval using 95% confidence level: 2-3. x=34 4-5. * = 36 0=2 0=5 n = 140 n = 110 Determine the confidence interval using 90% confidence level: 7-8. = 14 9-10. = 13 s = 1.45 s = 2.01 n = 15 n = 15 Find the 95% confidence interval of the proportion of x: 12-13. 14-15.x = 12 n = 39 n = 140 1. 6. 11.
View both photos attached and answer the question below. Estimate the net sale in future promotions, 95% confidence intervals for the population means were constructed. See output below. Explain and compare the confidence interval for net sales for Promotional and Regular customers. Make sure to clearly describe the confidence intervals in the context of the problem.
The figure to the right shows the results of a survey in which 2000 college graduates from the year 2016 were asked questions about employment. Construct 99% confidence intervals for the population proportion of college students who gave each response. Employment College students' responses to questions about employment Expect to stay at first employer for 3 or more years Completed an apprenticeship or intemship 70% 66% Employed in field of study 61% Feel underemployed 46% Prefer to work for a large company 15% The 99% confidence interval for the proportion of college students that expect to stay at their first employer for 3 or more years is (Round to three decimal places as needed.)
The figure to the right shows the results of a survey in which 2000 college graduates from the year 2016 were asked questions about employment. Construct 99% confidence intervals for the population proportion of college students who gave each response. Employment College students' responses to questions about employment Expect to stay at first employer for 3 or more years Completed an apprenticeship or internship 70% 66% Employed in field of study 61% Feel underemployed 46% Prefer to work for a large company 15% The 9970 Conuenit initivaliui int proportion un cuiityt diuutid inal txptul iu diay al ineli idi empivyti iui j'ui muit years is (0.01, 0.121). (Round to three decimal places as needed.) The 99% confidence interval for the proportion of college students that have completed an apprenticeship or internship is (J.). (Round to three decimal places as needed.)