Question

stat crunch A simple random sample of size n=250 individuals who are currently employed is asked if they work at home at least once per week. Of the 250 employed individuals​ surveyed, 41 responded that they did work at home at least once per week. Construct a​ 99% confidence interval for the population proportion of employed individuals who work at home at least once per week.

Answer 1

The 99% confidence interval for the population proportion of employed individuals who work at home at least once per week is between (0.104, 0.224).

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 = 250, \pi = (41)/(250) = 0.164`

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a pvalue of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.575`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.164 - 2.575\sqrt{(0.164*0.836)/(250)} = 0.104`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.164 + 2.575\sqrt{(0.164*0.836)/(250)} = 0.224`

The 99% confidence interval for the population proportion of employed individuals who work at home at least once per week is between (0.104, 0.224).