Questions: Russell is doing some research before buying his first house. He is looking at two different areas of the city, and he wants to know if there is a significant difference between the mean prices of homes in the two areas. For the 36 homes he samples in the first area, the mean home price is 161,000. Public records indicate that home prices in the first area have a population standard deviation of 20,865. For the 39 homes he samples in the second area, the mean home price is 156,000. Again, public records show that home prices in the second area have a population standard deviation of 23,310. Let Population 1 be homes in the first area and Population 2 be homes in the second area. Construct a 99% confidence interval for the true difference between the mean home prices in the two areas. Round the endpoints of the interval to the nearest whole number, if necessary. Lower Endpoint = Upper Endpoint =

Russell is doing some research before buying his first house. He is looking at two different areas of the city, and he wants to know if there is a significant difference between the mean prices of homes in the two areas. For the 36 homes he samples in the first area, the mean home price is 161,000. Public records indicate that home prices in the first area have a population standard deviation of 20,865. For the 39 homes he samples in the second area, the mean home price is 156,000. Again, public records show that home prices in the second area have a population standard deviation of 23,310. Let Population 1 be homes in the first area and Population 2 be homes in the second area. Construct a 99% confidence interval for the true difference between the mean home prices in the two areas. Round the endpoints of the interval to the nearest whole number, if necessary.

Lower Endpoint =

Upper Endpoint =
Transcript text: Russell is doing some research before buying his first house. He is looking at two different areas of the city, and he wants to know if there is a significant difference between the mean prices of homes in the two areas. For the 36 homes he samples in the first area, the mean home price is $\$ 161,000$. Public records indicate that home prices in the first area have a population standard deviation of $\$ 20,865$. For the 39 homes he samples in the second area, the mean home price is $\$ 156,000$. Again, public records show that home prices in the second area have a population standard deviation of \$23,310. Let Population 1 be homes in the first area and Population 2 be homes in the second area. Construct a $99 \%$ confidence interval for the true difference between the mean home prices in the two areas. Round the endpoints of the interval to the nearest whole number, if necessary. Answer Tables Keypad Keyboard Shortcuts Lower Endpoint = Upper Endpoint =
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Solution

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Solution Steps

Step 1: Define the Problem

Russell is interested in determining if there is a significant difference between the mean prices of homes in two different areas of the city. We have the following data:

  • Area 1:

    • Sample size (\(n_1\)) = 36
    • Mean price (\(\bar{x}_1\)) = \$161,000
    • Population standard deviation (\(\sigma_1\)) = \$20,865
  • Area 2:

    • Sample size (\(n_2\)) = 39
    • Mean price (\(\bar{x}_2\)) = \$156,000
    • Population standard deviation (\(\sigma_2\)) = \$23,310
Step 2: Calculate the Confidence Interval

To construct a 99% confidence interval for the difference between the mean home prices, we use the formula:

\[ (\bar{x}_1 - \bar{x}_2) \pm z \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \]

Where:

  • \(z\) is the critical value for a 99% confidence level, which is approximately 3.0.

Substituting the values:

\[ (161000 - 156000) \pm 3.0 \sqrt{\frac{20865^2}{36} + \frac{23310^2}{39}} \]

Calculating the difference in means:

\[ \bar{x}_1 - \bar{x}_2 = 5000 \]

Calculating the standard error:

\[ \sqrt{\frac{20865^2}{36} + \frac{23310^2}{39}} \approx 27141.0 \]

Thus, the confidence interval becomes:

\[ 5000 \pm 3.0 \cdot 27141.0 \]

Calculating the margin of error:

\[ 3.0 \cdot 27141.0 \approx 8141.0 \]

Step 3: Determine the Confidence Interval Endpoints

The confidence interval is:

\[ (5000 - 8141.0, 5000 + 8141.0) = (-3141.0, 13141.0) \]

Final Answer

The 99% confidence interval for the difference in mean home prices is:

\[ \boxed{(-8141, 18141)} \]

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