MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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**Problem Statement** 1. A sample of 36 vacation homes built during the past two years in the coastal region had a mean construction cost of $159,000 with a standard deviation of $27,000. Construct a 99% confidence interval for the mean construction cost for all vacation homes built in this region during the past two years. What is the best estimate, and what is the margin of error? --- **Detailed Explanation for Educational Context** In this problem, we are tasked with determining a confidence interval for the mean construction cost of vacation homes. The given data includes: - A sample size (\(n\)) of 36 homes. - A mean sample cost (\(\bar{x}\)) of $159,000. - A standard deviation (\(s\)) of $27,000. **Solution Approach:** 1. **Best Estimate:** The best estimate of the population mean is the sample mean itself, which is $159,000. 2. **Margin of Error (ME):** To construct the 99% confidence interval, we will use the formula: \[ ME = z \times \frac{s}{\sqrt{n}} \] Here, \(z\) is the z-score corresponding to a 99% confidence level, and can be found from the standard normal distribution table (approximately 2.576 for 99%). \[ ME = 2.576 \times \frac{27,000}{\sqrt{36}} \] \[ ME = 2.576 \times \frac{27,000}{6} \] \[ ME = 2.576 \times 4,500 = 11,592 \] 3. **Confidence Interval (CI):** The confidence interval is calculated as: \[ CI = \bar{x} \pm ME = 159,000 \pm 11,592 \] Therefore, the 99% confidence interval for the mean construction cost is: \[ (147,408, 170,592) \] **Conclusion:** The best estimate for the mean construction cost is $159,000, and the margin of error is $11,592. This provides us with a 99% confidence interval of $147,408 to $170,592, indicating we can be 99% confident that the true mean
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Transcribed Image Text:**Problem Statement** 1. A sample of 36 vacation homes built during the past two years in the coastal region had a mean construction cost of $159,000 with a standard deviation of $27,000. Construct a 99% confidence interval for the mean construction cost for all vacation homes built in this region during the past two years. What is the best estimate, and what is the margin of error? --- **Detailed Explanation for Educational Context** In this problem, we are tasked with determining a confidence interval for the mean construction cost of vacation homes. The given data includes: - A sample size (\(n\)) of 36 homes. - A mean sample cost (\(\bar{x}\)) of $159,000. - A standard deviation (\(s\)) of $27,000. **Solution Approach:** 1. **Best Estimate:** The best estimate of the population mean is the sample mean itself, which is $159,000. 2. **Margin of Error (ME):** To construct the 99% confidence interval, we will use the formula: \[ ME = z \times \frac{s}{\sqrt{n}} \] Here, \(z\) is the z-score corresponding to a 99% confidence level, and can be found from the standard normal distribution table (approximately 2.576 for 99%). \[ ME = 2.576 \times \frac{27,000}{\sqrt{36}} \] \[ ME = 2.576 \times \frac{27,000}{6} \] \[ ME = 2.576 \times 4,500 = 11,592 \] 3. **Confidence Interval (CI):** The confidence interval is calculated as: \[ CI = \bar{x} \pm ME = 159,000 \pm 11,592 \] Therefore, the 99% confidence interval for the mean construction cost is: \[ (147,408, 170,592) \] **Conclusion:** The best estimate for the mean construction cost is $159,000, and the margin of error is $11,592. This provides us with a 99% confidence interval of $147,408 to $170,592, indicating we can be 99% confident that the true mean
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