Question

An Internet service provider sampled 473 customers, and finds that 110 of them experienced an interruption in high-speed service during the previous month. What is the upper bound for the 99% confidence interval for the proportion of customers who experienced an interruption? Round to three decimal places (for example: 0.419). Write only a number as your answer.

Answer 1

Solution :

Given that,

n = 473

x = 110

= x / n =110 /473 = 0.233

1 - = 1 - 0.233 = 0.767

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01

Z = Z_0.01 = 2.326

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

= 2.326* (((0.233 * 0.767) / 473) = 0.043

A 99 % confidence interval for population proportion p is ,

+ E

0.233 + 0.043

0.276

The 99% confidence interval for the population proportion p is :upper bound =  0.276