In a random sample of 150 customers of a high-speed internet provider, 63 said that their service had been interrupted one or more times in the past month.

a. Find a 95% confidence interval for the proportion of customers whose service was interrupted one or more times in the past month.

b. Find a 99% confidence interval for the proportion of customers whose service was interrupted one or more times in the past month.

c. Find the sample size needed for a 95% confidence interval to specify the proportion to within ±0.05.

d. Find the sample size needed for a 99% confidence interval to specify the proportion to within ±0.05.

Answer:

Step 1 of 5:

Suppose that, a random sample of 150 customers of a high-speed internet provider, 63 said that their service had been interrupted one or more times in the past month.

Here,

x = 63, n = 150

Step 2 of 5:

a). Here we have to find a 95% confidence interval for the proportion of customers whose service was interrupted one or more times in the past month.

Then,

= 150+ 4

= 154.

=

= 0.4221

A 95% confidence interval for ‘p’ is given by

For a 95% confidence interval:

Z - value for 95% confidence interval is

Then,

(0.3441, 0.5001)

Therefore, a 95% confidence interval for the proportion of customers whose service was interrupted one or more times in the past month is (0.3441, 0.5001).

Step 3 of 5:

b). We need to find a 99% confidence interval for the proportion of customers whose service was interrupted one or more times in the past month.

Then,

= 150+ 4

= 154.

=

= 0.4221

A 99% confidence interval for ‘p’ is given by

For a 99% confidence interval:

Z - value for 99% confidence interval is

Then,

= 0.42210.1027

(0.3194, 0.5248)

Therefore, a 99% confidence interval for the proportion of customers whose service was interrupted one or more times in the past month is (0.3194, 0.5248).