QuestionThe American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 40 who smoke. Step 2 of 2: Suppose a sample of 1548 Americans over 40 is drawn. Of these people, 1100 don't smoke. Using the data, construct the $99 \%$ confidence interval for the population proportion of Americans over 40 who smoke. Round your answers to three decimal places.

Studdy Solution
Construct the confidence interval:
$$\text{Margin of Error (ME)} = z^* \times SE$$
$$ME = 2.576 \times 0.011$$
$$ME \approx 0.028$$
Calculate the confidence interval:
$$\text{Lower bound} = \hat{p} - ME = 0.289 - 0.028$$
$$\text{Upper bound} = \hat{p} + ME = 0.289 + 0.028$$
$$\text{Confidence Interval} = (0.261, 0.317)$$
The 99% confidence interval for the population proportion of Americans over 40 who smoke is:
$$\boxed{(0.261, 0.317)}$$