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Question 1193448: Randomly selected 130 student cars have ages with a mean of 7.9 years and a standard deviation of 3.4 years, while randomly selected 95 faculty cars have ages with a mean of 5.4 years and a standard deviation of 3.3 years.
1. Use a 0.05 significance level to test the claim that student cars are older than faculty cars.
- The test statistic is ______. (5 sig. figs)
- The critical value is ______. (5 sig. figs)
2. Construct a 95% confidence interval estimate of the difference μ1−μ2, where μ1 is the mean age of student cars and μ2 is the mean age of faculty cars.
- ________ <(μ1−μ2)< _________ (5 sig. figs)
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! **1. Hypothesis Testing**
* **Hypotheses:**
* H0: μ1 - μ2 ≤ 0 (Null Hypothesis: Student cars are not older than faculty cars)
* H1: μ1 - μ2 > 0 (Alternative Hypothesis: Student cars are older than faculty cars)
* **Sample Data:**
* Sample 1 (Student Cars): n1 = 130, x̄1 = 7.9 years, s1 = 3.4 years
* Sample 2 (Faculty Cars): n2 = 95, x̄2 = 5.4 years, s2 = 3.3 years
* **Calculate the Pooled Variance (assuming equal variances):**
* s_p² = [((n1 - 1) * s1²) + ((n2 - 1) * s2²)] / (n1 + n2 - 2)
* s_p² = [((130 - 1) * 3.4²) + ((95 - 1) * 3.3²)] / (130 + 95 - 2)
* s_p² ≈ 11.1225
* **Calculate the Standard Error of the Difference:**
* SE = s_p * √(1/n1 + 1/n2)
* SE = √11.1225 * √(1/130 + 1/95)
* SE ≈ 0.5069
* **Calculate the t-statistic:**
* t = (x̄1 - x̄2) / SE
* t = (7.9 - 5.4) / 0.5069
* t ≈ 4.9177
* **Determine Degrees of Freedom:**
* df = n1 + n2 - 2 = 130 + 95 - 2 = 223
* **Find the Critical Value (t-critical) using a t-distribution table or statistical software:**
* For a one-tailed test with α = 0.05 and df = 223, t-critical ≈ 1.6525
**Therefore:**
* **Test Statistic:** t ≈ 4.9177
* **Critical Value:** t-critical ≈ 1.6525
**2. Construct a 95% Confidence Interval**
* **Calculate the Margin of Error:**
* Margin of Error = t-critical * SE
* Margin of Error = 1.96 * 0.5069
* Margin of Error ≈ 0.9935
* **Calculate the Confidence Interval:**
* (x̄1 - x̄2) - Margin of Error < (μ1 - μ2) < (x̄1 - x̄2) + Margin of Error
* (7.9 - 5.4) - 0.9935 < (μ1 - μ2) < (7.9 - 5.4) + 0.9935
* 1.5065 < (μ1 - μ2) < 3.4935
**Therefore, the 95% confidence interval estimate of the difference (μ1 - μ2) is 1.5065 < (μ1 - μ2) < 3.4935.**
**Conclusion:**
* The calculated t-statistic (4.9177) is greater than the critical value (1.6525), so we **reject the null hypothesis**.
* There is sufficient evidence to support the claim that student cars are older than faculty cars at the 0.05 significance level.
* The 95% confidence interval for the difference in mean ages of student and faculty cars does not include zero, further supporting the conclusion that student cars are older.
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