Statistics HELP

Question

According to a study, 77% of adults ages 18-29 years had broadband Internet access at home in 2011. A researcher wanted to estimate the proportion of undergraduate college students (18-23) with access, so she randomly sampled 182 undergraduates and found that 159 had access. Estimate the true proportion with 99% confidence. Round your answers to three decimal places.

Answer 1

n = 182 undergraduates

x = 159 selected

p = 159/182 = 0.87

z = 2.58 (This is from the z-table for critical values)

Use the one proportion confidence interval.

p ± z√p(1-p)/n = 0.87 ± 2.58√(0.87)(1-0.87)/182

= 0.87 ± 2.58√(0.87)(0.13)/182

= (-0.806, 0.934)

We are 99% confident that the true proportion lies between -0.806 and 0.934.