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Chapter 8: Problem 57

The 2003 Statistical Abstract of the United States reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of .30 a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of \(.02 ?\) Use \(95 \%\) confidence. b. Assume that the study uses your sample size recommendation in part (a) and finds 520 smokers. What is the point estimate of the proportion of smokers in the population? c. What is the \(95 \%\) confidence interval for the proportion of smokers in the population?

Short Answer

Expert verified
The required sample size to estimate the proportion of smokers in the population with a margin of error of 0.02 and 95% confidence level is 2,207. The point estimate of the proportion of smokers in the population is 23.55%. The 95% confidence interval for the proportion of smokers in the population is approximately (0.2143, 0.2567) or (21.43%, 25.67%).

Step by step solution

01

Part a: Calculate sample size

In order to calculate the sample size needed to estimate the proportion of smokers in the population with a margin of error of 0.02 at 95% confidence level, we need to use the following formula for sample size calculation: \( n = \frac{Z^2 \times p \times (1-p)}{E^2} \) Where: - \(n\) = sample size - \(Z\) = Z-score, which corresponds to the 95% confidence level (1.96) - \(p\) = the preliminary proportion of smokers (0.30) - \(E\) = the margin of error (0.02)
02

Calculate the sample needed using the formula

\( n = \frac{1.96^2 \times 0.30 \times (1-0.30)}{0.02^2} \) \( n = \frac{3.8416 \times 0.30 \times 0.70}{0.0004} \) \( n = 2206.79 \) To have a whole number as the sample size, we need to round up the result: \( n = 2207 \) Therefore, the sample size required to estimate the proportion of smokers in the population with a margin of error of 0.02 and 95% confidence level is 2,207.
03

Part b: Point estimate of the proportion of smokers

The study used the recommended sample size calculated in part (a) which was 2,207, and it found 520 smokers. In order to find the point estimate of the proportion of smokers, we need to divide the number of smokers by the total sample size:
04

Calculate the point estimate

\( \hat{p} = \frac{520}{2207} \) \( \hat{p} = 0.2355 \) The point estimate of the proportion of smokers in the population is 0.2355, or 23.55%.
05

Part c: 95% confidence interval for the proportion of smokers

To calculate the 95% confidence interval for the proportion of smokers in the population, we will use the following formula: Confidence Interval = \( \hat{p} \pm Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \) Where: - \( \hat{p} \) = point estimate of the proportion of smokers (0.2355) - \( Z \) = Z-score, which corresponds to the 95% confidence level (1.96) - \( n \) = sample size (2207)
06

Calculate the confidence interval

Confidence Interval = \( 0.2355 \pm 1.96 \times \sqrt{\frac{0.2355 \times (1-0.2355)}{2207}} \) Confidence Interval = \( 0.2355 \pm 1.96 \times \sqrt{\frac{0.2355 \times 0.7645}{2207}} \) Confidence Interval = \( 0.2355 \pm 0.0212 \) Therefore, the 95% confidence interval for the proportion of smokers in the population is approximately (0.2143, 0.2567) or (21.43%, 25.67%).

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Most popular questions from this chapter

A well-known bank credit card firm wishes to estimate the proportion of credit card holders who carry a nonzero balance at the end of the month and incur an interest charge. Assume that the desired margin of error is .03 at \(98 \%\) confidence. a. How large a sample should be selected if it is anticipated that roughly \(70 \%\) of the firm's card holders carry a nonzero balance at the end of the month? b. How large a sample should be selected if no planning value for the proportion could be specified?

A Phoenix Wealth Management/Harris Interactive survey of 1500 individuals with net worth of \(\$ 1\) million or more provided a variety of statistics on wealthy people (Business Week, September 22,2003 ). The previous three-year period had been bad for the stock market, which motivated some of the questions asked. a. The survey reported that \(53 \%\) of the respondents lost \(25 \%\) or more of their portfolio value over the past three years. Develop a \(95 \%\) confidence interval for the proportion of wealthy people who lost \(25 \%\) or more of their portfolio value over the past three years. b. The survey reported that \(31 \%\) of the respondents feel they have to save more for retirement to make up for what they lost. Develop a \(95 \%\) confidence interval for the population proportion. c. Five percent of the respondents gave \(\$ 25,000\) or more to charity over the previous year. Develop a \(95 \%\) confidence interval for the proportion who gave \(\$ 25,000\) or more to charity. d. Compare the margin of error for the interval estimates in parts (a), (b), and (c). How is the margin of error related to \(\bar{p} ?\) When the same sample is being used to estimate a variety of proportions, which of the proportions should be used to choose the planning value \(p^{*} ?\) Why do you think \(p^{*}=.50\) is often used in these cases?

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