Question

Out of 100 people sampled, 42 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three places.

Answer 1

The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).

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 z-score that has a p-value of
`1 - (\alpha)/(2)`.

Out of 100 people sampled, 42 had kids.

This means that
`n = 100, \pi = (42)/(100) = 0.42`

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a p-value 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)} = 0.42 - 2.575\sqrt{(0.42*0.58)/(100)} = 0.293`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.42 + 2.575\sqrt{(0.42*0.58)/(100)} = 0.547`

The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).