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Chapter 9: Problem 44

During a national debate on changes to health care, a cable news service performs an opinion poll of 500 small-business owners. It shows that \(65 \%\) of small-business owners do not approve of the changes. Develop a \(95 \%\) confidence interval for the proportion opposing health care changes. Comment on the result.

Short Answer

Expert verified
The 95% confidence interval is (60.83%, 69.17%).

Step by step solution

01

Identify the Sample Proportion

The problem states that 65% of the small-business owners oppose the changes. Thus, the sample proportion, \( \hat{p} \), is \( 0.65 \).
02

Determine the Sample Size

The problem specifies that the opinion poll was conducted with 500 small-business owners. Therefore, the sample size \( n \) is 500.
03

Find the Z-Score for Confidence Level

For a 95% confidence interval, the Z-score, which represents the critical value for a normal distribution, is approximately 1.96.
04

Calculate the Standard Error

The standard error (SE) of the sample proportion is calculated using the formula \( \text{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \). Substituting the known values, we have \( \text{SE} = \sqrt{\frac{0.65 \times 0.35}{500}} \).
05

Compute the Standard Error

Computing the value from the above step: \( \text{SE} = \sqrt{\frac{0.2275}{500}} = \sqrt{0.000455} \approx 0.0213 \).
06

Calculate the Margin of Error

The margin of error (ME) is given by \( \text{ME} = Z \times \text{SE} \). Substituting the values, \( \text{ME} = 1.96 \times 0.0213 \approx 0.0417 \).
07

Determine the Confidence Interval

The confidence interval is calculated as \( (\hat{p} - \text{ME}, \hat{p} + \text{ME}) \). Using the values: \( (0.65 - 0.0417, 0.65 + 0.0417) \), the confidence interval is \( (0.6083, 0.6917) \).
08

Comment on the Result

The 95% confidence interval for the proportion of small-business owners opposing the health care changes is from 60.83% to 69.17%. This suggests that if we were to sample all small-business owners, we are 95% confident that the true proportion opposing is within this range.

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Key Concepts

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

Sample Proportion
In the world of statistics, a sample proportion is a way of summarizing data from a sample survey. When we survey a group, like the 500 small-business owners in our example, we calculate the proportion of the sample that exhibits a particular trait or opinion. In this case, it is those who do not approve of health care changes.

The sample proportion is denoted as \(\hat{p}\). It's simply the number of favorable responses divided by the total number of responses. For instance, if 325 out of 500 small-business owners oppose the changes, then \(\hat{p} = \frac{325}{500} = 0.65\) or 65%.
  • It's a point estimate of the true population proportion.
  • Helps in making predictions about the entire population based on a sample.
Standard Error
The standard error (SE) is vital in statistics because it tells us how much the sample proportion might vary from the true population proportion. Essentially, it measures the accuracy of the sample proportion as an estimator of the population proportion.

To calculate the standard error of a sample proportion, we use the formula:\[\text{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}.\]For our example with \(\hat{p} = 0.65\) and sample size \(n = 500\), the standard error can be calculated to be approximately 0.0213.
  • Smaller standard errors indicate more precise estimates.
  • It decreases as the sample size increases, reflecting less variability in larger samples.
Margin of Error
The margin of error (ME) provides a range that is likely to contain the true population proportion. It's calculated as the product of the Z-score and the standard error.

We use this equation to find the margin of error:\[\text{ME} = Z \times \text{SE}.\]Given a 95% confidence level (with a Z-score of 1.96) and a standard error of 0.0213 in our problem, the margin of error is approximately 0.0417. This means that we expect the true population proportion to fall within 4.17% of our sample proportion.
  • Reflects the extent of uncertainty around the sample estimate.
  • Larger margins of error suggest less precision, while smaller margins indicate more confidence in the estimate.
Z-score
The Z-score is an essential element in constructing confidence intervals. It's a statistical measurement that describes the number of standard deviations away from the mean a data point is.

In the context of confidence intervals, the Z-score translates the confidence level into a critical value. For common confidence levels:
  • 90% confidence level has a Z-score of 1.645.
  • 95% confidence level corresponds to a Z-score of 1.96.
  • 99% confidence level has a Z-score of 2.576.
When developing a 95% confidence interval, like in our sample problem, we use a Z-score of 1.96 to ensure that 95% of the interval's estimates should include the true population parameter if the process were repeated numerous times.

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Most popular questions from this chapter

Passenger comfort is influenced by the amount of pressurization in an airline cabin. Higher pressurization permits a closer-to-normal environment and a more relaxed flight. A study by an airline user group recorded the equivalent air pressure on 30 randomly chosen flights. The study revealed a mean equivalent air pressure of 8,000 feet with a standard deviation of 300 feet. a. Develop a \(99 \%\) confidence interval for the population mean equivalent air pressure. b. How large a sample is needed to find the population mean within 25 feet at \(95 \%\) confidence?

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The estimate of the population proportion should be within plus or minus. \(05,\) with a \(95 \%\) level of confidence. The best estimate of the population proportion is .15. How large a sample is required?

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We want to estimate the population mean within \(5,\) with a \(99 \%\) level of confidence. The population standard deviation is estimated to be \(15 .\) How large a sample is required?
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