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Alenkinab [10]
3 years ago
9

he American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 45 w

ho smoke. Step 2 of 2: Suppose a sample of 830 Americans over 45 is drawn. Of these people, 631 don't smoke. Using the data, construct the 85% confidence interval for the population proportion of Americans over 45 who smoke. Round your answers to three decimal places.
Mathematics
1 answer:
Sonja [21]3 years ago
7 0

Answer:

The 85% confidence interval is (0.2187, 0.2613).

Step-by-step explanation:

We have a large sample of size n = 830 Americans over 45. We have observed that 631 people don't smoke. Let p be the population proportion of Americans over 45 who smoke. An point estimate of p is \hat{p} = 199/830 = 0.24 and an estimate of the standard deviation of \hat{p} is \sqrt{\hat{p}(1-\hat{p})/n} = \sqrt{(0.24)(0.76)/830} = 0.0148. Therefore, because we have a large sample, the 85% confidence interval for the population proportion is given by \hat{p}\pm z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n} = 0.24\pm z_{0.15/2}0.0148 = 0.24\pm z_{0.075}0.0148 where z_{0.075} is the 7.5th quantile of the standard normal distribution, i.e., -1.4395. Therefore the 85% confidence interval is 0.24\pm (1.4395)(0.0148) or equivalently (0.2187, 0.2613).

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