*A tire manufacturer wishes to investigate the tread life of*

*its tires. A sample of 25 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. Y*ou give your conclusion and compute the confidence interval with the alpha is 99%.

### Problem Overview

You are given a sample of **25 tires** with the following information:  
- **Sample mean** (\(\bar{X}\)) = 0.32 inches  
- **Sample standard deviation** (\(s\)) = 0.09 inches  
- **Sample size** (\(n\)) = 25  
- **Population mean** (\(\mu\)) to test = 0.30 inches  
- **Confidence level** = 99% (\(\alpha = 0.01\))  

You are asked to determine if it would be reasonable to conclude that the population mean is **0.30 inches** and compute the **99% confidence interval**.

---

### Step 1: Hypothesis Setup  
We will perform a **one-sample t-test** since the population standard deviation is unknown, and the sample size is small.

1. **Null Hypothesis**: \(H_0: \mu = 0.30\)  
2. **Alternative Hypothesis**: \(H_1: \mu \neq 0.30\)  
(Since the test is two-tailed, we are checking if the mean differs from 0.30 in either direction.)

---

### Step 2: Test Statistic Formula  
The t-test statistic for a one-sample t-test is given by:  

\[
t = \frac{\bar{X} - \mu}{\frac{s}{\sqrt{n}}}
\]

Where:  
- \(\bar{X} = 0.32\) is the sample mean  
- \(\mu = 0.30\) is the hypothesized population mean  
- \(s = 0.09\) is the sample standard deviation  
- \(n = 25\) is the sample size  

---

### Step 3: Compute the Test Statistic  
First, calculate the standard error (SE) of the mean:  

\[
SE = \frac{0.09}{\sqrt{25}} = \frac{0.09}{5} = 0.018
\]

Now, calculate the t-statistic:  

\[
t = \frac{0.32 - 0.30}{0.018} = \frac{0.02}{0.018} \approx 1.11
\]

---

### Step 4: Determine Critical Value  
For a two-tailed test with \(n - 1 = 24\) degrees of freedom at **99% confidence level** (\(\alpha = 0.01\)):  
- Use a t-table to find the critical value for \(t\) with 24 degrees of freedom and \(\alpha/2 = 0.005\) in each tail.  

\[
t_{\text{critical}} \approx \pm 2.797
\]

---

### Step 5: Make the Decision  
- **If** \(|t| > t_{\text{critical}}\), we reject the null hypothesis.  
- **Otherwise**, we fail to reject the null hypothesis.

In this case:  

\[
|1.11| < 2.797
\]

Thus, **we fail to reject the null hypothesis**. This means there is not enough evidence to conclude that the population mean tread remaining differs from 0.30 inches.

---

### Step 6: 99% Confidence Interval for the Mean  
The confidence interval is given by:

\[
\bar{X} \pm t_{\text{critical}} \times SE
\]

\[
0.32 \pm 2.797 \times 0.018
\]

\[
0.32 \pm 0.0503
\]

\[
(0.2697, 0.3703)
\]

---

### Conclusion  
The **99% confidence interval** for the population mean tread remaining is approximately \( (0.27, 0.37) \). Since the hypothesized mean of **0.30 inches** is within this interval, it is reasonable to conclude that the population mean tread remaining could indeed be 0.30 inches after 50,000 miles. Therefore, we fail to reject the null hypothesis.

---

### Do you have any questions or need further clarifications?

---

Here are **5 related questions** to expand your understanding:
1. How does increasing the confidence level affect the width of the confidence interval?
2. What happens to the t-statistic when the sample size increases?
3. Why do we use a t-distribution instead of a z-distribution in this problem?
4. How would the result change if the sample mean were 0.28 inches instead?
5. How do we interpret a p-value in a hypothesis test like this?

---

**Tip**: When working with small sample sizes (typically \(n < 30\)), it's safer to use the t-distribution instead of the z-distribution for more accurate results.

A tire manufacturer wishes to investigate the tread life of its tires. A sample of 25 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? Compute the confidence interval with an alpha of 99%.

Explore a hypothesis test to determine if the population mean tread remaining on tires after 50,000 miles is 0.30 inches. This example involves a sample of 25 tires, a sample mean of 0.32 inches, and a standard deviation of 0.09 inches. Calculate the 99% confidence interval and perform a t-test for accurate results.

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