Question

A survey (random sample) of 700 office workers investigated telephone answering practices, including how often each office worker was able to answer incoming telephone calls and how often incoming telephone calls went directly to voice mail. A total of 280 office worker indicated that they never need voice mail and are able to take every telephone call. What is the 92% confidence interval for the actual proportion of all office workers who are able to take every telephone call

Answer 1

The 92% confidence interval for the actual proportion of all office workers who are able to take every telephone call is (0.3676, 0.4324).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

`n = 700, \pi = (280)/(700) = 0.4`

92% confidence level

So
`\alpha = 0.08`, z is the value of Z that has a pvalue of
`1 - (0.08)/(2) = 0.96`, so
`Z = 1.75`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.4 - 1.75\sqrt{(0.4*0.6)/(700)} = 0.3676`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.4 + 1.75\sqrt{(0.4*0.6)/(700)} = 0.4324`

The 92% confidence interval for the actual proportion of all office workers who are able to take every telephone call is (0.3676, 0.4324).