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91Ó°ÊÓ

The Bureau of Labor Statistics report referenced in the previous exercise also reported that \(7.3 \%\) of high school graduates were unemployed in October 2013 and \(5.7 \%\) of high school graduates were unemployed in October \(2014 .\) Suppose that the reported percentages were based on independently selected representative samples of 400 high school graduates in each of these 2 years. a. Construct and interpret a 99\% large-sample confidence interval for the difference in the proportions of high school graduates who were unemployed in these 2 years. b. Is the confidence interval from Part (a) wider or narrower than the confidence interval calculated in the previous exercise? Give two reasons why it is wider or narrower.

Short Answer

Expert verified
The 99% confidence interval for the difference in unemployment proportions between October 2013 and October 2014 is (-0.052, 0.084). We are 99% confident that the true difference in unemployment proportions for high school graduates between these two years lies within this interval. Without information about the previous exercise, we cannot determine if this interval is wider or narrower. However, the width depends on two factors: confidence level and sample size. A higher confidence level results in a wider interval, while a larger sample size leads to a narrower interval.

Step by step solution

01

Identify the given information

We are given the following information: - Unemployment rate in October 2013: \(7.3\%\) or \(0.073\) - Unemployment rate in October 2014: \(5.7\%\) or \(0.057\) - Sample size for both years: \(400\)
02

Calculate the difference in proportions and standard error

The difference in proportions is given as: - \(p_1 - p_2 = 0.073 - 0.057 = 0.016\) The standard error of the difference in proportions can be calculated as: - \(SE = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}\) Where \(p_1\) and \(p_2\) are the proportions, and \(n_1\) and \(n_2\) are the sample sizes. Plugging in the given values: - \(SE = \sqrt{\frac{0.073(1-0.073)}{400} + \frac{0.057(1-0.057)}{400}} = 0.0264\)
03

Calculate the z-score for a 99% confidence level

To find the z-score (critical value) corresponding to a 99% confidence level, we refer to a standard normal distribution table and find that the z-score is approximately 2.576.
04

Calculate the margin of error

The margin of error is given as: - \(ME = z * SE = 2.576 * 0.0264 = 0.0680\)
05

Calculate the confidence interval

We calculate the confidence interval using our difference in proportions and margin of error: - CI = \((p_1 - p_2) \pm ME = (0.016 - 0.0680, 0.016 + 0.0680) = (-0.052, 0.084)\)
06

Interpret the confidence interval

The 99% confidence interval for the difference in unemployment proportions between October 2013 and October 2014 is (-0.052, 0.084). This means that we are 99% confident that the true difference in unemployment proportions for high school graduates between these two years lies within this interval.
07

Compare the width of the confidence interval

Comparing the width of this confidence interval to the one calculated in the previous exercise, we cannot exactly state if it's wider or narrower without the information about the previous exercise. However, the width depends on two factors: - Confidence level: A higher confidence level means a wider confidence interval, because we require a larger interval to be more certain that it contains the true difference - Sample size: A larger sample size leads to a narrower confidence interval because it decreases standard error and thus provides a more precise estimate of the true difference

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Key Concepts

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

Difference in Proportions
When comparing two groups, such as unemployment rates for high school graduates across different years, it's crucial to analyze the difference in proportions. The **difference in proportions** quantifies how much one proportion differs from another. In our exercise, we examined the unemployment rates in October 2013 and October 2014. The difference in these proportions is calculated as:
  • For 2013, the unemployment rate is 7.3\% or 0.073.
  • For 2014, the unemployment rate is 5.7\% or 0.057.
  • The difference: \( p_1 - p_2 = 0.073 - 0.057 = 0.016 \)
This difference of 0.016 represents how much higher the unemployment rate was in 2013 compared to 2014. Understanding this concept is key in determining if there are significant changes between the two periods.
Sample Size
**Sample size** is a fundamental part of any statistical analysis. It is the number of observations or data points collected from which statistics are calculated. In our exercise, each year's unemployment rate was determined from 400 high school graduates, making the sample size \( n = 400 \) for both years.
  • A larger sample size generally yields more reliable results.
  • It reduces variability, leading to more precise estimates.
For example, with a larger sample size, the confidence interval for the difference in proportions becomes narrower and provides a more accurate reflection of the true population parameter. Therefore, maintaining a sufficiently large sample size is crucial for achieving statistically significant results.
Standard Error
The **standard error (SE)** is a measure of the dispersion or spread of sample statistics. In the context of differences in proportions, the SE provides insight into the variability of the difference between two sample proportions under repeated sampling. The formula for the standard error of the difference in proportions is:
\[SE = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}\]
In this exercise:
  • \(p_1 = 0.073\) and \(n_1 = 400\)
  • \(p_2 = 0.057\) and \(n_2 = 400\)
  • Calculating SE: \( SE = \sqrt{\frac{0.073(1-0.073)}{400} + \frac{0.057(1-0.057)}{400}} = 0.0264 \)
A smaller standard error indicates that the sample statistic is a more precise estimate of the population parameter. Thus, reducing the standard error through increased sample sizes can improve the reliability of our estimates.
Margin of Error
The **margin of error (ME)** reflects the range within which we expect the true population parameter to lie with a specified level of confidence. In a confidence interval, it is essential to account for both the sample statistic and the inherent randomness of sampling. The margin of error is calculated as:
  • ME = \(z \times SE\)
Where \(z\) is the z-score corresponding to the desired confidence level. In our case, for a 99% confidence level, \(z = 2.576\).
The calculation is as follows:
\[ME = 2.576 \times 0.0264 = 0.0680\]
This margin of error indicates the degree of certainty we have about where the true difference in proportions lies. A larger margin suggests less precision, often a result of small sample sizes or attempting to achieve high confidence levels. Understanding how ME is affected by sample variables can enhance decision-making processes in statistical analysis.

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Most popular questions from this chapter

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