The figure to the right shows the results of a survey in which 1500 college graduates from the year 2016 were asked questions about employment. Employment College students' responses to questions about employment Construct 99% confidence intervals for the population proportion of college students who gave each response. Expect to stay at first employer for 3 or more years 73% Completed an appranticeship or intemship 67% Employed in field of study 61% Feel underemployed 53% Prefer to work for a large company 17% stay at their first émployer for 3 or more years is (0.700.0.760 ). The 99% confidence interval for the proportion of college students that expect (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 ( !D (Round to three decimal places as needed.)

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%?

You are interested in finding a 98% confidence interval for the average commute that non-residential students have to their college. The data below show the number of commute miles for 10 randomly selected non-residential college students. Round answers to 3 decimal places where possible.   22 19 10 26 24 27 18 21 21 24   b. With 98% confidence the population mean commute for non-residential college students is between  and   miles. c. If many groups of 10 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About  percent of these confidence intervals will contain the true population mean number of commute miles and about  percent will not contain the true population mean number of  commute miles.

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. Employment College students' responses to questions about employment Construct 95% confidence intervals for the population proportion of college students who gave each response. Expect to stay at first employer for 3 or more years 70% 67% Completed an apprenticeship or intemship 62% Employed in field of study Feel underemployed 54% Prefer to work for a large company 17% The 95% 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.)

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

Your analysis of the raw data indicates that 16 of the 30 students reported driving no more than 20 miles per day. Calculate a 98% confidence interval for the sample proportion.  What Z Score was used?

Treating the 2015 Tobacco rate data as a sample, construct a 95% confidence interval for the mean tobacco rate for males and for females. Interpret each interval. Does it appear that males use tobacco at a higher rate than females? Justify your answer, using the confidence intervals in your explanation. Construct the same confidence interval for males at the 99% confidence level. Compare to the male 95% confidence interval from the previous question. What happened to the interval when the confidence level increased? Explain why. Prevalence of smoking any tobacco product among persons aged >= 15 years Male 2015 Prevalence of smoking any tobacco product among persons aged >= 15 years Female 2015 51.2 7.6     37.2 27.8         29.5 18.4 52.3 1.5 16.7 13.1 35.5 34.8 46.5 0.4     48.8 7.6 39.8 0.7 13.1 0.9 46.2 10.6 26.5 20     17.7 1     30.5 17.1 47.2 30     19.3 11.3 29.3 3.1 42.4 28.2 36 4.5         22.2 3.5 44.1 2.8…

(Help me with part 4) se the sample data and confidence level given below to complete parts​ (a) through​ (d).   A research institute poll asked respondents if they felt vulnerable to identity theft. In the​ poll, n=919 and x=594 who said​ "yes." Use a 90% confidence level.   LOADING... Click the icon to view a table of z scores.       Question content area bottom Part 1 ​a) Find the best point estimate of the population proportion p.   0.6460.646 ​(Round to three decimal places as​ needed.) Part 2 ​b) Identify the value of the margin of error E.   E=0.0260.026 ​(Round to three decimal places as​ needed.) Part 3 ​c) Construct the confidence interval.   0.6200.620<p<0.6720.672 ​(Round to three decimal places as​ needed.) Part 4 ​d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below.     A. One has 90​% confidence that the sample proportion is equal to the population proportion.   B. There is a 90​% chance that…

Where did they get the 0.21 from? Please explain