Question 1192209: Records of 40 used passenger cars and 40 used pickup trucks (none used commercially) were
randomly selected to investigate whether there was any difference in the mean time in years that
they were kept by the original owner before being sold. For cars the mean was 5.3 years with
standard deviation 2.2 years. For pickup trucks the mean was 7.1 years with standard deviation
3.0 years.
a. Construct the 95% confidence interval for the difference in the means based on these data.
b. Test the hypothesis that there is a difference in the means against the null hypothesis that
there is no difference. Use the 1% level of significance.
c. Compute the observed significance of the test in part (b).
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to address this two-sample t-test problem:
**a. Construct the 95% Confidence Interval:**
Since the sample sizes are both greater than 30, we can use a z-interval to approximate the t-interval. We'll use the following formula:
(x̄₁ - x̄₂) ± z * √((s₁²/n₁) + (s₂²/n₂))
Where:
* x̄₁ and x̄₂ are the sample means
* s₁ and s₂ are the sample standard deviations
* n₁ and n₂ are the sample sizes
* z is the z-score for the desired confidence level (1.96 for 95%)
Plugging in the values:
(5.3 - 7.1) ± 1.96 * √((2.2²/40) + (3.0²/40))
-1.8 ± 1.96 * √(0.121 + 0.225)
-1.8 ± 1.96 * √0.346
-1.8 ± 1.96 * 0.588
-1.8 ± 1.153
The 95% confidence interval is (-2.953, -0.647).
**b. Hypothesis Test:**
* **Null Hypothesis (H₀):** There is no difference in the means (μ₁ - μ₂ = 0)
* **Alternative Hypothesis (H₁):** There is a difference in the means (μ₁ - μ₂ ≠ 0) - This is a two-tailed test.
* **Significance Level (α):** 0.01
We'll use a z-test for the difference of two means, as the sample sizes are large. The test statistic is calculated the same way as in part (a), but we subtract zero from the difference in means:
z = (x̄₁ - x̄₂ - 0) / √((s₁²/n₁) + (s₂²/n₂))
z = (5.3 - 7.1) / √((2.2²/40) + (3.0²/40))
z ≈ -3.06
* **Critical Values:** For a two-tailed test at α = 0.01, the critical values are ±2.576.
* **Decision:** Since the absolute value of the test statistic (|z| = 3.06) is greater than the critical value (2.576), we reject the null hypothesis.
* **Conclusion:** There is sufficient evidence at the 1% level of significance to conclude that there is a difference in the mean time that cars and pickup trucks are kept by their original owners.
**c. Observed Significance (p-value):**
The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the one calculated, assuming the null hypothesis is true. Since this is a two-tailed test, we need to find the area in both tails of the standard normal distribution that is beyond our z-score of -3.06.
Using a z-table or calculator:
P(z < -3.06) ≈ 0.0011
Since it's a two-tailed test, we double this value:
p-value ≈ 2 * 0.0011 = 0.0022
The observed significance is 0.0022. This is less than our significance level of 0.01, which is why we rejected the null hypothesis in part (b).
|
|
|
