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% (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 (0.607, 0.653) (round to three decimal places as needed.) the 99% confidence interval for the proportion of college students that feel underemployed is (0.457, 0.503) (round to three decimal places as needed.) the 99% confidence interval for the proportion of college students that prefer to work for a large company is (□,□)

Explanation:

Step1: Identify the formula for confidence interval

The formula for a confidence interval for a population proportion is $\hat{p}\pm z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$, where $\hat{p}$ is the sample proportion, $z$ is the z - score, and $n$ is the sample size. For a 99% confidence interval, the z - score $z = 2.576$.

Step2: Calculate for "Prefer to work for a large company"

Given $\hat{p}=0.14$ and $n = 3000$.
First, calculate the margin of error $E=z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=2.576\sqrt{\frac{0.14\times(1 - 0.14)}{3000}}$.

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Then, the lower limit is $\hat{p}-E=0.14 - 0.01632=0.124$ and the upper limit is $\hat{p}+E=0.14+0.01632 = 0.156$.

Answer:

$(0.124,0.156)$