Question

In a sample of 500 adults, 345 had children. Construct a 99% confidence interval for the true population proportion of adults with children.

Answer 1

The 99% confidence interval for the true population proportion of adults with children is (0.6367, 0.7433).

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)`.

In a sample of 500 adults, 345 had children.

This means that
`n = 500, \pi = (345)/(500) = 0.69`

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)} = 0.69 - 2.575\sqrt{(0.69*0.31)/(500)} = 0.6367`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.69 + 2.575\sqrt{(0.69*0.31)/(500)} = 0.7433`

The 99% confidence interval for the true population proportion of adults with children is (0.6367, 0.7433).