Problem 44
Question
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.
Step-by-Step Solution
Verified Answer
The 95% confidence interval is (60.83%, 69.17%).
1Step 1: Identify the Sample Proportion
The problem states that 65% of the small-business owners oppose the changes. Thus, the sample proportion, \( \hat{p} \), is \( 0.65 \).
2Step 2: 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.
3Step 3: 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.
4Step 4: 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 \( \text{SE} = \sqrt{\frac{0.65 \times 0.35}{500}} \).
5Step 5: Compute the Standard Error
Computing the value from the above step: \( \text{SE} = \sqrt{\frac{0.2275}{500}} = \sqrt{0.000455} \approx 0.0213 \).
6Step 6: Calculate the Margin of Error
The margin of error (ME) is given by \( \text{ME} = Z \times \text{SE} \). Substituting the values, \( \text{ME} = 1.96 \times 0.0213 \approx 0.0417 \).
7Step 7: 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 \( (0.6083, 0.6917) \).
8Step 8: 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.
Key Concepts
Sample ProportionStandard ErrorMargin of ErrorZ-score
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%.
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.
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.
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:
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.
Other exercises in this chapter
Problem 42
Pharmaceutical companies promote their prescription drugs using television advertising. In a survey of 80 randomly sampled television viewers, 10 indicated that
View solution Problem 43
HighTech Inc. randomly tests its employees about company policies. Last year, in the 400 random tests conducted, 14 employees failed the test. Develop a \(99 \%
View solution Problem 45
There are 20,000 eligible voters in York County, South Carolina. A random sample of 500 York County voters revealed 350 plan to vote to return Louella Miller to
View solution Problem 46
In a poll to estimate presidential popularity, each person in a random sample of 1,000 voters was asked to agree with one of the following statements: 1\. The p
View solution