Question

the figure to the right shows the results of a survey in which 3000 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 70% completed an apprenticeship or internship 68% employed in field of study 63% feel underemployed 48% prefer to work for a large company 14% the 99% confidence interval for the proportion of college students that expect to stay at their first employer for 3 or more years is (0.678, 0.722) (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 (0.658, 0.702) (round to three decimal places as needed ) the 99% confidence interval for the proportion of college students that are employed in their field of study is (□.□) (round to three decimal places as needed.)

Explanation:

Step1: Calculate the margin of error

The formula for the margin of error \(E\) for a proportion is \(E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\). For a 99% confidence interval, \(\alpha=1 - 0.99 = 0.01\), so \(\alpha/2=0.005\). From the standard normal distribution table, \(z_{\alpha/2}=z_{0.005} = 2.576\). Here, \(\hat{p}=0.63\) (the sample proportion) and \(n = 3000\) (sample size).

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Step2: Calculate the confidence interval

The confidence interval for a proportion is \(\hat{p}-E

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Answer:

\((0.607,0.653)\)