Question

Of 150 adults selected randomly from one town, 30 of them smoke. Construct a 99% confidence interval for the true percentage of all adults in the town that smoke.

Answer 1

The 99% confidence interval for the true percentage of all adults in the town that smoke is (0.1159, 0.2841).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence interval
`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:

Of 150 adults selected randomly from one town, 30 of them smoke. This means that
`n = 150, p = (30)/(150) = 0.2`

99% confidence interval

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)} = 0.20 - 2.575\sqrt{(0.20*0.80)/(150)} = 0.1159`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.20 + 2.575\sqrt{(0.20*0.80)/(150)}{119}} = 0.2841`

The 99% confidence interval for the true percentage of all adults in the town that smoke is (0.1159, 0.2841).