Question

The American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 46 who smoke.

Step 1 of 2: Suppose a sample of 2172 Americans over 46 is drawn. Of these people, 1738 don't smoke. Using the data, estimate the proportion of Americans over 46 who smoke. Enter your answer as a fraction or a decimal number rounded to three decimal places.
The American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 46 who smoke.
Step 2 of 2: Suppose a sample of 2172 Americans over 46 is drawn. Of these people, 1738 don't smoke. Using the data, construct the 98 % confidence interval for the population proportion of Americans over 46 who smoke. Round your answers to three decimal places

Answer 1

1.) .800

2.).779 to .821

Explanation:

We must assume that the sample that we are given to work with is representative of the population, in which case we just need to find the proportion of people in the sample who don't smoke. This is
`(1738)/(2172)` = .8002, or .800 rounded to three decimal places.

Part 2 will build off the answer in step 1. The middle of the confidence interval will be the estimate gained from the sample, or .800. The critical value of 98% equals 2.326 when using the normal model. The standard deviation can be calculated using the formula
`\sqrt{((proportion)(1-proportion))/(n) }`. Plugging in the numbers you get
`\sqrt{((.8)(.2))/(2172) }`= .009.

This means the confidence interval is .800±.021