Question

An Internet service provider sampled 545 customers, and finds that 73 of them experienced an interruption in high-speed service during the previous month. Construct a

99.5% confidence interval for the proportion of all customers who experienced an interruption. Round the answers to at least three decimal places.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 73 / 545 = 0.134

1 - = 1 - 0.134 = 0.866

Z/2 = Z0.0025 = 2.81

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.81 (((0.134 * 0.866) /545 )

= 0.041

A 99.5% confidence interval for population proportion p is ,

± E

= 0.134  ± 0.041

= ( 0.093, 0.175 )