Question

a countrys education department reported that in 2015, 68.0% of students enrolled in college or a trade school within 12 months of graduating high school. in 2017, a random sample of 180 individuals who graduated from high school 12 months prior was selected. from this sample, 105 students were found to be enrolled in college or a trade school. complete parts a through c

a. construct a 95% confidence interval to estimate the actual proportion of students enrolled in college or a trade school within 12 months of graduating from high school in 2017.

the confidence interval has a lower limit of \\( \square \\) and an upper limit of \\( \square \\).
(round to three decimal places as needed.)

Explanation:

Step1: Calculate sample proportion

Sample proportion $\hat{p}=\frac{x}{n}$, where $x = 105$ (number of successes) and $n=180$ (sample size).
$\hat{p}=\frac{105}{180}\approx0.583$

Step2: Find $z -$ value

For a $95\%$ confidence interval, the $z -$ value $z_{\alpha/2}=1.96$ (from standard normal distribution table).

Step3: Calculate margin of error

Margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
Substitute $\hat{p}=0.583$, $n = 180$, and $z_{\alpha/2}=1.96$
$E=1.96\sqrt{\frac{0.583\times(1 - 0.583)}{180}}$
First calculate inside the square - root: $0.583\times(1 - 0.583)=0.583\times0.417 = 0.243$
$\sqrt{\frac{0.243}{180}}=\sqrt{0.00135}\approx0.037$
$E=1.96\times0.037\approx0.073$

Step4: Calculate confidence interval limits

Lower limit $=\hat{p}-E$
$=0.583 - 0.073=0.510$
Upper limit $=\hat{p}+E$
$=0.583+0.073 = 0.656$

Answer:

The confidence interval has a lower limit of $0.510$ and an upper limit of $0.656$.