A Roper Starch survey asked employees ages 25 to 34 whether they would prefer better health insurance or a raise in salary. Of the 500 employees surveyed, 340 said they would prefer better health insurance. Using this sample information, develop a 98% confidence interval for the true proportion of employees who would prefer better health insurance.

A Roper Starch survey asked employees ages 25 to 34 whether they would prefer better health insurance or a raise in salary. Of the 500 employees surveyed, 340 said they would prefer better health insurance. Using this sample information, develop a 98% confidence interval for the true proportion of employees who would prefer better health insurance.

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

See similar textbooks

Similar questions

Example: You would like to construct a 90% confidence interval for the proportion of graduating seniors that go directly on to graduate school. You randomly sample 386 graduating seniors and find that 73 will be enrolling in graduate school in the upcoming semester. 1. What is your point estimate? 2. Construct a 90% confidence interval for the true proportion of graduating seniors that go directly on to graduate school? 3.How many graduating seniors will you have to sample to have a margin of error within 2%?
Repairs: A sample of six records for repairs of a component showed the following costs: 95, 87, 20, 75, 90, 96. Construct a 90% confidence interval for the mean cost of a repair for this type of component. Use tables and technology. Round your answers to at least one decimal place.
help 1
On April 15th, 1912, the R.M.S. Titanic sank into the North Atlantic Ocean, resulting in the loss of more than 1,500 passengers and crew. Below is a table which summarizes the total number of passengers and the number of survivors for the crew, first-, second-, and third-class. Find the proportion of survivors by type of service (crew, first, second, third). Test the hypothesis at the 99% confidence level that the proportion of first-class survivors is the same as third class (.25).What do you calculate for the test-statistic? What is the critical statistic? What is the conclusion of your hypothesis test? Passengers survivedCrew 885 212First 325 202Second 285 118Third 706 178Total 2201 710
See attachment
A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective. Step 2 of 2: Suppose a sample of 542 floppy disks is drawn. Of these disks, 37 were defective. Using the data, construct the 95% confidence interval for the population proportion of disks which are defective. Round your answers to three decimal places.
A college administrator is interested in determining the proportion of students who receive some sort of financial aid. Instead of examining the records for all students, the administrator randomly draws 120 students and finds that 48 of them are receiving financial aid. Use a 90% confidence interval to estimate the true proportion of students who receive financial aid.
The National Association of Colleges and Employers sponsors the Graduating Student and Alumni Survey.Part of the survey gauges student optimism in landing a job after graduation. According to one year’s surveyresults, published in American Demographics, among the 1218 respondents, 733 said that they expecteddifficulty finding a job. Construct a 90% confidence interval for the proportion of students who expecteddifficulty finding a job. Interpret your result.
A political scientist claims that 38% of first-year college students characterize themselves as being “moderate” or “middle of the road” as far as their political affiliation is concerned. Believing this claimed value is too high, you survey a random sample of 400 first-year college students and find that 120 characterize themselves as being “moderate” or “middle of the road.” Based on this information, what will the test statistic be? Choose the answer below that is closest to what you calculate, and try not to do a lot of rounding until you get to the very end of your calculations. 1. -0.3 2. -1.2 3. -2.6 4. -3.3 5. None of the other answer options are correct because the test statistic should be positive, not negative.
17
An online retailer has identified the maximum willingness of 10 consumers to pay for the same product. The 10 prices are: $10.00, $11.50, $9.50, $10.20, $11.20, $10.50, $9.80, $10.10, $9.90, $11.00 The retailer seeks a 95% confidence interval for the mean price consumers are willing to pay. The sample mean is calculated as $10.37, and the sample standard deviation is $0.66. Also consider the following t-values: Degrees of freedom 8 a. (9.99, 10.75) 9 10 90% confidence 1.86 1.833 1.812 95% confidence 2.306 2.262 2.228 99% confidence 3.355 3.250 3.169 Selecting the correct t-value, calculate the limits of the confidence interval.
In a study by Sallie Mae and Gallup, 323 of the 569respondents said that they paid for their education by taking out student loans. Find a 95% confidence interval for the proportion of students who paid for their education by taking out student loans.