Question

A department of education reported that in 2007, 68% of students enrolled in college or a trade school within 12 months of graduating from high school. In 2013, a random sample of 160 individuals who graduated from high school 12 months prior was selected. From this sample, 106 students were found to be enrolled in college or a trade school. Complete parts a through c below. a. Construct a 90% 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 2013. and an upper limit of A 90% confidence interval to estimate the actual proportion has a lower limit of (Round to three decimal places as needed.)

Answer 1

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = x / n = 106 / 160 = 0.663

1 - $\hat p$ = 1 - 0.663 = 0.337

Z$\alpha$/2 = Z_0.05 = 1.645

Margin of error = E = Z_{$\alpha$ / 2} * (($\hat p$ * (1 - $\hat p$ ))$\sqrt$ / n)

= 1.645 ($\sqrt$((0.663 * 0.337) / 160)

= 0.061

A 90% confidence interval for population proportion p is ,

$\hat p$ ± E  

= 0.663 ± 0.061

= ( 0.602, 0.724 )

lower limit = 0.602

upper limit = 0.724