Question

The figure to the right shows the results of a survey in which 2500 college graduates from the year 2016 were asked questions about employment. Construct 95​% confidence intervals for the population proportion of college students who gave each response. A table labeled "Employment, College students' responses to questions about employment" consists of five rows containing the following information from top to bottom, with row listed first and information listed second: Expect to stay at first employer for 3 or more years, 72 percent; Completed an apprenticeship or internship, 67 percent; Employed in field of study, 63 percent; Feel underemployed, 48 percent; Prefer to work for a large company, 15 percent. 72%67%63%48%15% The 9595​% confidence interval for the proportion of college students that expect to stay at their first employer for 3 or more years is left parenthesis nothing comma nothing right parenthesism,m. ​(Round to three decimal places as​ needed.) please breakdown the problem. Thank you

Answer 1

We are asked to find 95% confidence interval estimate for population proportion ( P )of college students that expect to stay at their first employer for 3 or more years

We are given = 0.72 and m = 2500

Lower bound = - E

Upper bound = + E

E is margin of error =

z is critical value follows standard normal distribution , we can find its value using z score table.

We are given confidence level = 0.95

Therefore α = 1 - 0.95 = 0.05 , 1 - (α/2) = 0.975

So we have to find z score corresponding to area 0.9750 on z score table

So z = 1.96

E =

E = 0.0176

Lower bound = - E = 0.72 - 0.0176 = 0.702

Upper bound = + E = 0.72 +0.0176 = 0.738

Therefore 95% confidence interval estimate for population proportion ( P )of college students that expect to stay at their first employer for 3 or more years ( 0.702,0.738 )