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An Employee Benefits Research Institute survey of 1250 workers over the age of 25 collected opinions on the health care system in America and on retirement planning \((A A R P\) Bulletin, January 2007 ). a. The American health care system was rated as poor by 388 of the respondents. Construct a \(95 \%\) confidence interval for the proportion of workers over 25 who rate the American health care system as poor. b. Eighty-two percent of the respondents reported being confident of having enough money to meet basic retirement expenses. Construct a \(95 \%\) confidence interval for the proportion of workers who are confident of having enough money to meet basic retirement expenses. c. Compare the margin of error in part (a) to the margin of error in part (b). The sample size is 1250 in both cases, but the margin of error is different. Explain why.

Short Answer

Expert verified
Part (a) has a confidence interval of approximately (0.290, 0.342); Part (b) of approximately (0.799, 0.841). The margin of error differs due to varying sample proportions.

Step by step solution

01

Calculate Sample Proportion for Part (a)

First, identify the sample proportion (\(\hat{p}\)) for workers who rate the health care system as poor. Divide the number of respondents who rated it as poor by the total number of respondents: \(\hat{p} = \frac{388}{1250}\).
02

Construct 95% Confidence Interval for Part (a)

To construct the confidence interval, use the formula for a proportion: \(\hat{p} \pm Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\). With a 95% confidence level, \(Z\) is approximately 1.96. Calculate the standard error and find the interval.
03

Calculate Sample Proportion for Part (b)

The sample proportion is given as 0.82 (or 82%) for workers confident in having enough retirement money. Use this to construct the confidence interval.
04

Construct 95% Confidence Interval for Part (b)

Again, apply the confidence interval formula: \(0.82 \pm 1.96 \times \sqrt{\frac{0.82(1-0.82)}{1250}}\). Compute the standard error and determine the interval.
05

Compare Margins of Error

The margin of error is different due to the sample proportions being different. The standard error, which contributes to the margin of error, is calculated as \(\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) and is dependent on \(\hat{p}\). Differences in \(\hat{p}\) result in different standard errors.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
The sample proportion, often represented by the symbol \( \hat{p} \), is a crucial concept in statistics that represents the ratio of respondents who exhibit a particular characteristic within a sample. In the context of survey data, the sample proportion helps us estimate what might be true for the whole population based on the sample we have. To calculate the sample proportion, we simply divide the number of favorable outcomes by the total number of observations in the sample.

In our example, for part (a), we computed the sample proportion of workers who rated the American health care system as poor. We did this by dividing the 388 workers who gave a negative rating by the total 1250 workers surveyed. This calculation is essential as it forms the basis for constructing confidence intervals and determining the overall opinions of the population.

For part (b), the sample proportion was given directly as 0.82 (or 82%) for employees confident about meeting retirement expenses. This indicates a large proportion feel secure, which influences the confidence interval calculation by affecting both the standard error and the margin of error.
Margin of Error
The margin of error provides us with a range within which we can be confident that the true population parameter lies. It's a critical concept in statistics that helps us understand and convey the potential variability in our sample estimate. The margin of error is influenced by both the sample size and the sample proportion and is an integral part of the confidence interval formula.

When looking at the 95% confidence intervals in our exercise, margin of error is calculated as part of \( \hat{p} \pm Z \times \text{(standard error)} \), where \( Z \) is the z-value from the standard normal distribution for the desired confidence level. In both parts (a) and (b), the sample size remains 1250, yet the margin of error differs due to differences in the sample proportions. This is because the margin of error takes into account the variation in the sample proportion, influenced by \( \sqrt{\hat{p}(1-\hat{p})} \), affecting the overall interval width.

It is quite normal for two different sample proportions to result in different margins of error even if the sample size and confidence level are the same.
Standard Error
The standard error is a measure of the statistical accuracy of an estimate, indicating how much the sample proportion is expected to fluctuate from the actual population proportion. In terms of construction, the standard error of the sample proportion is calculated as \( \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( \hat{p} \) is the sample proportion and \( n \) is the sample size.

In our survey exercise, even though the sample size is consistent across both scenarios (1,250 respondents), the standard error varies. This is due to the differing sample proportions: 0.31 in part (a) and 0.82 in part (b). As this formula shows a direct dependency on the sample proportion, any variation directly impacts the standard error.

Understanding standard error aids in interpreting the reliability of the confidence interval. A smaller standard error implies the sample proportion is a more precise approximation of the population proportion, while a larger standard error suggests greater variability in sample estimates. This is crucial when conveying the reliability of results derived from survey data.

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Most popular questions from this chapter

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