Question 1179741: Suppose a real estate broker is interested in comparing the asking prices of flats in the British cities of Exeter and Cardiff. The broker conducts a small telephone survey in the two cities, asking the prices of flats. A random sample of 21 listings in Exeter resulted in a sample average price of £116,900, with a standard deviation of £2,300. A random sample of 26 listings in Cardiff resulted in a sample average price of £114,000, with a standard deviation of £1,750. The broker assumes that the prices of flats are normally distributed and that the variance in prices in the two cities is about the same. What would he obtain for a 90% confidence interval for the difference in mean prices of mid-range homes between Exeter and Cardiff? Test wheather there is a difference in the mean prices of mid-range homes of two cities for α= 0.10
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem, including calculating the confidence interval and conducting the hypothesis test:
**1. Given Information:**
* **Exeter:**
* n1 = 21
* x̄1 = £116,900
* s1 = £2,300
* **Cardiff:**
* n2 = 26
* x̄2 = £114,000
* s2 = £1,750
* Confidence level = 90%
* Significance level (α) = 0.10
* Assume normal distribution and equal variances.
**2. Calculate the Pooled Standard Deviation (sp):**
Since we assume equal variances, we use the pooled standard deviation:
sp = √[((n1 - 1) * s1²) + ((n2 - 1) * s2²)) / (n1 + n2 - 2)]
sp = √[((20 * 2300²) + (25 * 1750²)) / (21 + 26 - 2)]
sp = √[(105800000 + 76562500) / 45]
sp = √[182362500 / 45]
sp = √4052500
sp ≈ £2,013.08
**3. Calculate the Standard Error of the Difference (SE):**
SE = sp * √(1/n1 + 1/n2)
SE = 2013.08 * √(1/21 + 1/26)
SE = 2013.08 * √(0.0476 + 0.0385)
SE = 2013.08 * √0.0861
SE = 2013.08 * 0.2934
SE ≈ £590.69
**4. Find the Critical t-Value:**
* Degrees of freedom (df) = n1 + n2 - 2 = 21 + 26 - 2 = 45
* For a 90% confidence interval (α = 0.10, two-tailed), the critical t-value (t*) for df = 45 is approximately 1.68.
**5. Calculate the Margin of Error (E):**
E = t* * SE
E = 1.68 * 590.69
E ≈ £992.36
**6. Construct the Confidence Interval:**
* Difference in sample means (x̄1 - x̄2) = 116,900 - 114,000 = £2,900
* Lower Bound = (x̄1 - x̄2) - E = 2,900 - 992.36 = £1,907.64
* Upper Bound = (x̄1 - x̄2) + E = 2,900 + 992.36 = £3,892.36
**90% Confidence Interval: (£1,907.64, £3,892.36)**
**7. Hypothesis Test:**
* **Null Hypothesis (H0):** μ1 - μ2 = 0 (There is no difference in mean prices)
* **Alternative Hypothesis (H1):** μ1 - μ2 ≠ 0 (There is a difference in mean prices)
* Significance level (α) = 0.10
* Test statistic (t): t = (x̄1 - x̄2) / SE
t = 2900 / 590.69
t ≈ 4.91
**8. Find the Critical t-Values:**
* df = 45
* For α = 0.10 (two-tailed), the critical t-values are approximately ±1.68.
**9. Make a Decision:**
* Calculated t-value (4.91) > critical t-value (1.68).
* Therefore, we reject the null hypothesis.
**10. Conclusion:**
There is sufficient evidence at the α = 0.10 level to conclude that there is a difference in the mean prices of mid-range homes between Exeter and Cardiff.
**Answers:**
* 90% Confidence Interval: (£1,907.64, £3,892.36)
* Hypothesis Test: Reject the null hypothesis. There is a statistically significant difference.
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