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 (□□) (round to three decimal places as needed)

Explanation:

Step1: Calculate \(z_{\alpha/2}\)

For a \(99\%\) confidence interval, \(\alpha = 1 - 0.99=0.01\), and \(\alpha/2 = 0.005\). From the standard normal distribution table, \(z_{\alpha/2}=z_{0.005} = 2.576\)

Step2: Calculate the margin of error \(E\)

Given \(n = 3000\) and \(\hat{p}=0.68\) (for the proportion of students who completed an apprenticeship or internship). The formula for the margin of error for a proportion is \(E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\)
Substitute the values: \(\hat{p}=0.68\), \(1-\hat{p}=0.32\), \(n = 3000\), \(z_{\alpha/2}=2.576\)

$$
LATEXBLOCK0
$$

Step3: Calculate the confidence interval

The confidence interval for a proportion is \(\hat{p}-ESubstitute \(\hat{p}=0.68\) and \(E = 0.02194\)
\(0.68- 0.02194=0.658\) and \(0.68 + 0.02194=0.702\)

Answer:

\((0.658,0.702)\)