To construct a 90% confidence interval for the difference between the mean home prices in the two areas, we'll use the formula for a confidence interval for the difference between two means when the population standard deviations are known.

Given Information

• Sample size in Area 1, $n_1 = 38$

• Mean price in Area 1, $\bar{x}_1 = 153,600$

• Population standard deviation in Area 1, $\sigma_1 = 39,475$

• Sample size in Area 2, $n_2 = 31$

• Mean price in Area 2, $\bar{x}_2 = 150,800$

• Population standard deviation in Area 2, $\sigma_2 = 34,695$

Step 1: Set Up the Confidence Interval Formula

The confidence interval for the difference between two means when population standard deviations are known is: $(\bar{x}_1 - \bar{x}_2) \pm z \cdot \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$ where $z$ is the z-score corresponding to the confidence level.

Step 2: Find the z-score for a 90% Confidence Level

For a 90% confidence level, the z-score $z$ that corresponds to the upper 5% (since it’s two-tailed, we split the remaining 10% equally on both sides) is: $z = 1.645$

Step 3: Calculate the Difference in Sample Means

$\bar{x}_1 - \bar{x}_2 = 153,600 - 150,800 = 2,800$

Step 4: Calculate the Standard Error

$\text{Standard Error} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$ Plugging in the values: $= \sqrt{\frac{39,475^2}{38} + \frac{34,695^2}{31}}$ Calculating each part individually:

1. $\frac{39,475^2}{38} \approx 41,003,872.37$
2. $\frac{34,695^2}{31} \approx 38,820,129.84$

So the standard error becomes: $\text{Standard Error} = \sqrt{41,003,872.37 + 38,820,129.84} \approx \sqrt{79,824,002.21} \approx 8,933.8$

Step 5: Calculate the Margin of Error

$\text{Margin of Error} = z \cdot \text{Standard Error} = 1.645 \cdot 8,933.8 \approx 14,692$

Step 6: Construct the Confidence Interval

Now we can construct the confidence interval: $(\bar{x}_1 - \bar{x}_2) \pm \text{Margin of Error} = 2,800 \pm 14,692$ So the confidence interval is: $(2,800 - 14,692, 2,800 + 14,692) = (-11,892, 17,492)$

Answer

The 90% confidence interval for the difference between the mean home prices in the two areas is approximately: $(-11,892, 17,492)$

This means we are 90% confident that the true difference in mean home prices between the two areas falls within this interval.

Would you like further details or have any questions about the steps?

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Here are some questions that can expand your understanding of this concept:

1. How would the confidence interval change if the confidence level were 95% instead of 90%?
2. Why do we use a z-score instead of a t-score in this problem?
3. What would happen to the confidence interval if the sample sizes were smaller?
4. How does the population standard deviation affect the width of the confidence interval?
5. What conclusions can be drawn if the confidence interval includes zero?

Tip: When constructing confidence intervals, larger sample sizes generally result in narrower intervals, as they provide more precise estimates of the population parameters.