Question

In a recent year, a study found that 77% of adults ages 18–29 had internet access at home. A researcher wanted to estimate the proportion of undergraduate college students (18–29 years) with access, so she randomly sampled 187 undergraduates and found that 165 had access. Estimate the true proportion with 99% confidence. Round intermediate answers to five decimal places. Round your final answer to one decimal place.

Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 165 / 187 = 0.882

1 - = 1 - 0.882 = 0.118

Z/2
= Z_{0.005} = 2.576

Margin of error = E = Z_{
/ 2} * ((
* (1 -
))
/ n)

= 2.576 (((0.882 * 0.118) / 187)

= 0.061

A 99% confidence interval for population proportion p is ,

± E

= 0.882 ± 0.061

= ( 0.821, 0.943 )

= ( 82.1%, 94.3% )

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