Question

Internet service: An Internet service provider sampled 545 customers, and finds that 65 of them
experienced an interruption in high-speed service during the previous month.

(b) Construct a 99.8% confidence interval for the proportion of all customers who experienced
an interruption. Round the answers to at least three decimal places.

A 99.8% confidence interval for the proportion of all customers who experienced an interruption is
_<p<_.

Answer 1

Solution :

n =545

x = 65

= x / n = 65 / 545 = 0.119

1 - = 1 - 0.119 = 0.881

At 99.8% confidence level the z is ,

= 1 - 99.8% = 1 - 0.998 = 0.002

/ 2 = 0.002 / 2 = 0.001

Z/2 = Z0.001 = 3.090

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

= 3.090 * (((0.119 * 0.881) / 545)

= 0.043

A 99.8% confidence interval for population proportion p is ,

- E < P < + E

0.119 - 0.043 < p < 0.119+ 0.043

0.076 < p < 0.162