Question

problem 5

an analyst wants to compare the average home prices in bloomington and west lafayette indiana. random samples of recent home sales were collected from each city.

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$$\begin{tabular}{lccc} \\hline city & sample size (\\(n\\)) & mean price (\\(\\bar{x}\\)) & standard deviation (\\(s\\)) \\\\ \\hline bloomington & 40 & 350,000 & 75,000 \\\\ west lafayette & 35 & 360,000 & 90,000 \\\\ \\hline \\end{tabular}$$

assume the samples were drawn from approximately normally distributed populations, and the population variances are different.

a. state the hypotheses to test whether the average home price differs between the two cities.
b. compute the test statistic.
c. determine the p-value for the test statistic using 66 degrees of freedom.
d. state the decision and conclusion (\\(\alpha = 0.05\\)).
e. construct a 90\\% confidence interval for the difference in average home prices between the two cities.

Explanation:

⚡ Using what you learned: Hypothesis Testing · Point Estimation and Interval Estimation

Step 1: State the Hypotheses (Part a)

Let \( \mu_1 \) be the population mean home price in Bloomington and \( \mu_2 \) be the population mean home price in West Lafayette.

To test if the average home price differs between the two cities:

$$ H_0: \mu_1 - \mu_2 = 0 \quad (\text{or } \mu_1 = \mu_2) $$

$$ H_a: \mu_1 - \mu_2 eq 0 \quad (\text{or } \mu_1 eq \mu_2) $$

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Step 2: Compute the Test Statistic (Part b)

Since the population variances are assumed to be different, we use the two-sample \( t \)-test with unequal variances (Welch's \( t \)-test):

$$ t = \frac{(\bar{X}_1 - \bar{X}_2) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} $$

Given data:

• Bloomington (Group 1): \( n_1 = 40 \), \( \bar{X}_1 = 350,000 \), \( s_1 = 75,000 \)
• West Lafayette (Group 2): \( n_2 = 35 \), \( \bar{X}_2 = 360,000 \), \( s_2 = 90,000 \)

Calculate the standard error:

$$ SE = \sqrt{\frac{75,000^2}{40} + \frac{90,000^2}{35}} = \sqrt{\frac{5,625,000,000}{40} + \frac{8,100,000,000}{35}} $$

$$ SE = \sqrt{140,625,000 + 231,428,571.4} = \sqrt{372,053,571.4} \approx 19,288.69 $$

Calculate the \( t \)-statistic:

$$ t = \frac{350,000 - 360,000}{19,288.69} = \frac{-10,000}{19,288.69} \approx -0.518 $$

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Step 3: Determine the p-value (Part c)

Using the \( t \)-distribution with \( df = 66 \) for a two-tailed test:

$$ p\text{-value} = 2 \times P(T > |-0.518|) = 2 \times P(T > 0.518) $$

Using standard \( t \)-distribution tables or software:

$$ P(T > 0.518) \approx 0.303 $$

$$ p\text{-value} \approx 2 \times 0.303 = 0.606 $$

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Step 4: State the Decision and Conclusion (Part d)

Compare the \( p \)-value to the significance level \( \alpha = 0.05 \):

• Since \( p\text{-value} \approx 0.606 > 0.05 \), we fail to reject the null hypothesis \( H_0 \).
• Conclusion: There is not enough statistical evidence to conclude that the average home price differs between Bloomington and West Lafayette.

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Step 5: Construct a 90% Confidence Interval (Part e)

The formula for the confidence interval for the difference between two independent means with unequal variances is:

$$ (\bar{X}_1 - \bar{X}_2) \pm t_{\alpha/2, \, df} \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} $$

For a 90% confidence level (\( \alpha = 0.10 \)) and \( df = 66 \):

$$ t_{0.05, \, 66} \approx 1.668 $$

Using the values calculated in Step 2:

$$ (\bar{X}_1 - \bar{X}_2) = 350,000 - 360,000 = -10,000 $$

$$ SE \approx 19,288.69 $$

Calculate the margin of error (\( ME \)):

$$ ME = 1.668 \times 19,288.69 \approx 32,173.53 $$

Calculate the confidence interval limits:

$$ \text{Lower Limit} = -10,000 - 32,173.53 = -42,173.53 $$

$$ \text{Upper Limit} = -10,000 + 32,173.53 = 22,173.53 $$

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Answer:

• a. Hypotheses:

$$ H_0: \mu_1 - \mu_2 = 0 \quad \text{vs.} \quad H_a: \mu_1 - \mu_2 eq 0 $$

• b. Test Statistic:

$$ t \approx -0.518 $$

• c. p-value:

$$ p\text{-value} \approx 0.606 $$

• d. Decision & Conclusion:

Since \( p\text{-value} > 0.05 \), fail to reject \( H_0 \). There is no significant difference in the average home prices between the two cities.

• e. 90% Confidence Interval:

$$ [-42,173.53, \, 22,173.53] $$