SOLUTION: 1. A tire manufacturer wishes to investigate the tread life of its tires. A sample of 10 tires driven 50, 000 miles revealed a sample mean of 0.32 inch of tread remaining with a st

Question 1168292: 1. A tire manufacturer wishes to investigate the tread life of its tires. A sample of 10 tires driven 50, 000 miles revealed a sample mean of 0.32 inch of tread remaining with a standard deviation of 0.09 inch. Would it be reasonable for the manufacturer to conclude that after 50, 000 miles the population mean amount of tread remaining is 0.30 inches? Use α = 0.05

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's the breakdown of the hypothesis test:
**1. State the Null and Alternative Hypotheses:**
* **Null Hypothesis (H₀):** The population mean amount of tread remaining after 50,000 miles is 0.30 inches.
* $H_0: \mu = 0.30$
* **Alternative Hypothesis (Ha):** The population mean amount of tread remaining after 50,000 miles is different from 0.30 inches.
* $H_a: \mu \neq 0.30$ (This is a two-tailed test because we are interested in whether the mean is *different* from 0.30).
**2. Determine the Significance Level (α):**
* The significance level is given as $\alpha = 0.05$.
**3. Calculate the Degrees of Freedom (df):**
* The sample size ($n$) is 10.
* The degrees of freedom for a t-test (since the population standard deviation is unknown) are:
* $df = n - 1 = 10 - 1 = 9$
**4. Calculate the Test Statistic (t-value):**
* The formula for the t-statistic is:
* $t = \frac{\bar{X} - \mu_0}{s / \sqrt{n}}$
Where:
* $\bar{X}$ = sample mean = 0.32 inch
* $\mu_0$ = hypothesized population mean = 0.30 inch
* $s$ = sample standard deviation = 0.09 inch
* $n$ = sample size = 10
* $t = \frac{0.32 - 0.30}{0.09 / \sqrt{10}}$
* $t = \frac{0.02}{0.09 / 3.1623}$
* $t = \frac{0.02}{0.02846}$
* $t \approx 0.703$
**5. Determine the P-value:**
* The p-value is the probability of observing a sample mean as extreme as (or more extreme than) 0.32 inches if the true population mean were 0.30 inches. Since this is a two-tailed test, we need to consider both tails of the t-distribution.
* We look up the p-value associated with a t-statistic of approximately 0.703 with 9 degrees of freedom in a t-distribution table or using statistical software.
* For a two-tailed test with $t \approx 0.703$ and $df = 9$, the p-value is greater than 0.10 (typically around 0.50).
**6. Make a Decision:**
* We compare the p-value to the significance level ($\alpha = 0.05$).
* **Decision Rule:**
* If p-value $\leq \alpha$, reject the null hypothesis ($H_0$).
* If p-value $> \alpha$, fail to reject the null hypothesis ($H_0$).
* In this case, the p-value ($> 0.10$) is greater than $\alpha$ (0.05). Therefore, we fail to reject the null hypothesis.
**7. State the Conclusion:**
* At a significance level of 0.05, there is not enough statistical evidence from the sample data to conclude that the population mean amount of tread remaining after 50,000 miles is different from 0.30 inches.
**Therefore, it would be reasonable for the manufacturer to conclude that after 50,000 miles the population mean amount of tread remaining is 0.30 inches.**
**Summary of your required values:**
* **H₀ (Null Hypothesis):** $\mu = 0.30$
* **Ha (Alternative Hypothesis):** $\mu \neq 0.30$
* **df (Degrees of Freedom):** 9
* **α (Significance Level):** 0.05
* **p-value:** $> 0.10$ (approximately 0.50)
* **Decision:** Fail to reject H₀
* **Conclusion:** It is reasonable for the manufacturer to conclude that the population mean amount of tread remaining is 0.30 inches.
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