Question

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

Answer 1

The correct answer is "0.3410, 0.4990".

Explanation:

Given values are:

`n=150`

`p=(63)/(150)`

`=0.42`

At 95% confidence interval,

C = 95%

z = 1.96

As we know,

⇒
`E=z\sqrt{(p(1-p))/(n) }`

By substituting the values, we get

`=1.96\sqrt{(0.42* 0.58)/(150) }`

`=1.96\sqrt{(0.2436)/(150) }`

`=0.0790`

hence,

The confidence interval will be:

=
`p \pm E`

=
`0.42 \pm 0.079`

=
`(0.3410,0.4990)`