An automobile manufacturer has given its jeep a 56.1 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the actual MPG for this jeep since it is believed that the jeep performs under the manufacturer’s MPG rating. After testing 200 jeeps, they found a mean MPG of 56.0. Assume the population standard deviation is known to be 2.1. Is there sufficient evidence at the 0.02 level to support the testing firm’s claim?
An automobile manufacturer has given its jeep a 56.1 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the actual MPG for this jeep since it is believed that the jeep performs under the manufacturer’s MPG rating. After testing 200 jeeps, they found a mean MPG of 56.0. Assume the population standard deviation is known to be 2.1. Is there sufficient evidence at the 0.02 level to support the testing firm’s claim?
The Correct Answer and Explanation is:
Answer:
Yes, there is sufficient evidence at the 0.02 level to support the testing firm’s claim that the actual MPG is less than the manufacturer’s rating.
Explanation
To evaluate the testing firm’s claim, we perform a one-sample z-test for the mean. The goal is to determine whether the average miles per gallon (MPG) is less than the manufacturer’s claimed value of 56.1 MPG.
Step 1: State the Hypotheses
- Null Hypothesis (H₀): μ = 56.1
- Alternative Hypothesis (H₁): μ < 56.1
This is a left-tailed test since the claim is that the actual MPG is less than the stated value.
Step 2: Gather the Data
- Sample mean (𝑥̄) = 56.0
- Population standard deviation (σ) = 2.1
- Sample size (n) = 200
- Significance level (α) = 0.02
Step 3: Compute the Test Statistic
z=xˉ−μσ/n=56.0−56.12.1/200=−0.12.1/14.142≈−0.10.1485≈−0.673z = \frac{\bar{x} – \mu}{\sigma / \sqrt{n}} = \frac{56.0 – 56.1}{2.1 / \sqrt{200}} = \frac{-0.1}{2.1 / 14.142} ≈ \frac{-0.1}{0.1485} ≈ -0.673
Step 4: Find the Critical Value and Decision Rule
At α = 0.02 for a left-tailed test, the critical z-value is: zcritical≈−2.05z_{critical} ≈ -2.05
Step 5: Make a Decision
Since: ztest=−0.673>zcritical=−2.05z_{test} = -0.673 > z_{critical} = -2.05
We fail to reject the null hypothesis.
Conclusion
There is not sufficient evidence at the 0.02 significance level to support the testing firm’s claim that the actual MPG is lower than 56.1. Although the sample mean (56.0) is slightly lower than the claimed 56.1 MPG, the difference is not statistically significant given the sample size and population variability.
This analysis shows that small differences may arise due to natural variability, and strong claims require stronger statistical evidence—especially at a low significance level like 0.02.
