Consider 50 homes selling in an area. The average price of the 50 homes is $350,000 with a standard deviation $25,000. Find a 90% confidence interval for the average home price in the area.
Consider 50 homes selling in an area. The average price of the 50 homes is $350,000 with a standard deviation $25,000. Find a 90% confidence interval for the average home price in the area.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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![**Title: Finding a 90% Confidence Interval for the Average Home Price**
**Problem Statement:**
Consider 50 homes selling in an area. The average price of the 50 homes is $350,000 with a standard deviation of $25,000. Find a 90% confidence interval for the average home price in the area.
**Options:**
1. ($344,184.56, $355,815.44)
2. ($343,070.48, $356,929.52)
3. ($344,072.50, $355,927.50) [Correct Answer]
4. Cannot compute a confidence interval since we do not know if home prices are normally distributed.
**Explanation:**
To calculate the 90% confidence interval for the average home price, we use the formula for the confidence interval of the mean:
\[ \text{CI} = \bar{x} \pm Z \left( \frac{\sigma}{\sqrt{n}} \right) \]
Where:
- \( \bar{x} \) is the sample mean (average home price), which is $350,000.
- \( Z \) is the Z-value that corresponds to the desired confidence level (for 90%, \( Z \) ≈ 1.645).
- \( \sigma \) is the standard deviation, which is $25,000.
- \( n \) is the sample size, which is 50.
- First, calculate the standard error (SE):
\[ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{25,000}{\sqrt{50}} \approx 3,536.25 \]
- Next, calculate the margin of error (ME):
\[ \text{ME} = Z \times \text{SE} = 1.645 \times 3,536.25 \approx 5,927.50 \]
- Finally, calculate the confidence interval:
\[ \text{CI} = 350,000 \pm 5,927.50 \]
Thus, the 90% confidence interval is:
\[ (344,072.50, 355,927.50) \]
This means we can be 90% confident that the true average home price in the area falls within this range. The correct choice is the third option: **($344,072.50, $355,927.50)**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F86d6b9a7-7dfd-4e6a-9264-375a7e5a9ea5%2F8351c59d-a0ce-4ad2-8b0a-5ea46737fd59%2Fltdkubj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Finding a 90% Confidence Interval for the Average Home Price**
**Problem Statement:**
Consider 50 homes selling in an area. The average price of the 50 homes is $350,000 with a standard deviation of $25,000. Find a 90% confidence interval for the average home price in the area.
**Options:**
1. ($344,184.56, $355,815.44)
2. ($343,070.48, $356,929.52)
3. ($344,072.50, $355,927.50) [Correct Answer]
4. Cannot compute a confidence interval since we do not know if home prices are normally distributed.
**Explanation:**
To calculate the 90% confidence interval for the average home price, we use the formula for the confidence interval of the mean:
\[ \text{CI} = \bar{x} \pm Z \left( \frac{\sigma}{\sqrt{n}} \right) \]
Where:
- \( \bar{x} \) is the sample mean (average home price), which is $350,000.
- \( Z \) is the Z-value that corresponds to the desired confidence level (for 90%, \( Z \) ≈ 1.645).
- \( \sigma \) is the standard deviation, which is $25,000.
- \( n \) is the sample size, which is 50.
- First, calculate the standard error (SE):
\[ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{25,000}{\sqrt{50}} \approx 3,536.25 \]
- Next, calculate the margin of error (ME):
\[ \text{ME} = Z \times \text{SE} = 1.645 \times 3,536.25 \approx 5,927.50 \]
- Finally, calculate the confidence interval:
\[ \text{CI} = 350,000 \pm 5,927.50 \]
Thus, the 90% confidence interval is:
\[ (344,072.50, 355,927.50) \]
This means we can be 90% confident that the true average home price in the area falls within this range. The correct choice is the third option: **($344,072.50, $355,927.50)**.
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