Question

A recent survey showed that among 725 randomly selected subjects who completed 4 years of college, 139 smoke and 586 do not smoke. Determine a 95% confidence interval for the true proportion of the given population that smokes.

Answer 1

The 95% confidence interval for the true proportion of the given population that smokes is (0.1630, 0.2204).

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)`.

Sample of 725, 139 smoke:

This means that
`n = 725, \pi = (139)/(725) = 0.1917`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a p-value of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.1917 - 1.96\sqrt{(0.1917*0.8083)/(725)} = 0.1630`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.1917 + 1.96\sqrt{(0.1917*0.8083)/(725)} = 0.2204`

The 95% confidence interval for the true proportion of the given population that smokes is (0.1630, 0.2204).