Question

The American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 44 who smoke. Step 2 of 2 : Suppose a sample of 632 Americans over 44 is drawn. Of these people, 462 don't smoke. Using the data, construct the 90% confidence interval for the population proportion of Americans over 44 who smoke. Round your answers to three decimal places.

Answer 1

The 90% confidence interval for the population proportion of Americans over 44 who smoke is (0.24, 0.298)

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:

632 American, 462 don't smoke, 632 - 462 = 170 smoke.

Proportion who smoke, so

`n = 632, \pi = (170)/(632) = 0.269`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.269 - 1.645\sqrt{(0.269*0.731)/(632)} = 0.24`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.269 + 1.645\sqrt{(0.269*0.731)/(632)} = 0.298`

The 90% confidence interval for the population proportion of Americans over 44 who smoke is (0.24, 0.298)