SOLUTION: An automobile tyre manufacturer claims that the average life of a certain grade of tyre is greater than 25000 km. when used under normal driving conditions on a car of certain weig

Question 1185842: An automobile tyre manufacturer claims that the average life of a certain grade of tyre is greater than 25000 km. when used under normal driving conditions on a car of certain weight. A random sample of 15 tyres was tested, and a mean and standard deviation of 27000 and 5000 km. respectively were computed. Can we conclude that the manufacturer’s product is as good as claimed?
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to conduct a hypothesis test to determine if the manufacturer's claim is valid:
**1. Hypotheses:**
* **Null Hypothesis (H0):** The average tire life is less than or equal to 25,000 km (μ ≤ 25000).
* **Alternative Hypothesis (H1):** The average tire life is greater than 25,000 km (μ > 25000). This is a right-tailed test.
**2. Significance Level (alpha):** The problem doesn't explicitly state a significance level. A common choice is α = 0.05 (5%), but you could use a different value depending on the desired level of confidence. I'll proceed with α = 0.05.
**3. Test Statistic:** Since the sample size is small (n = 15) and the population standard deviation is unknown, we use a t-test. The test statistic is:
```
t = (x̄ - μ) / (s / √n)
```
Where:
* x̄ is the sample mean (27,000 km)
* μ is the population mean under H0 (25,000 km)
* s is the sample standard deviation (5,000 km)
* n is the sample size (15)
**4. Degrees of Freedom:** df = n - 1 = 15 - 1 = 14
**5. Critical Value:** For a one-tailed (right-tailed) t-test with α = 0.05 and df = 14, we consult a t-table or calculator. The critical value is approximately t = 1.761.
**6. Decision Rule:** Reject H0 if the calculated t-statistic is greater than the critical value (1.761).
**7. Calculation:**
```
t = (27000 - 25000) / (5000 / √15)
t = 2000 / (5000 / 3.873)
t = 2000 / 1290.99
t ≈ 1.55
```
**8. Conclusion:**
The calculated t-statistic (1.55) is *less than* the critical value (1.761). Therefore, we *fail to reject* the null hypothesis.
**Interpretation:**
At a 5% significance level, there is not enough evidence from the sample to conclude that the manufacturer's claim is true. While the sample mean tire life (27,000 km) is greater than 25,000 km, the variability in the sample (standard deviation of 5,000 km) and the small sample size mean that this difference could reasonably be due to chance. We do not have sufficient statistical evidence to say the manufacturer's tires are better than 25000km on average.
**Important Note:** If a different significance level (e.g., α = 0.10) were used, the critical value would be lower, and the conclusion might be different. It's crucial to state the significance level being used.

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