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.
the 99% 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.)

Explanation:

Step1: Identify the formula for confidence interval

The formula for a confidence interval for a population proportion is $\hat{p}\pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$, where $\hat{p}$ is the sample proportion, $n$ is the sample size, and $z_{\alpha/2}$ is the critical value.
For a 99% confidence interval, $\alpha=1 - 0.99 = 0.01$, so $\alpha/2=0.005$. From the standard normal distribution table, $z_{0.005}=2.576$.
Here, $\hat{p}=0.7$ (70% expressed as a decimal) and $n = 3000$.

Step2: Calculate the margin of error

First, calculate $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.7\times(1 - 0.7)}{3000}}=\sqrt{\frac{0.7\times0.3}{3000}}=\sqrt{\frac{0.21}{3000}}\approx\sqrt{7\times10^{-5}}\approx0.00837$.
Then, the margin of error $E=z_{\alpha/2}\times\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=2.576\times0.00837\approx0.0216$.

Step3: Calculate the confidence interval

The lower limit is $\hat{p}-E=0.7 - 0.0216 = 0.6784$ and the upper limit is $\hat{p}+E=0.7+0.0216 = 0.7216$.

Answer:

$(0.678,0.722)$