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.

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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.)

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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

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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:

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

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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.

$t_{\text{critical}} \approx \pm 2.797$

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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.

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Step 6: 99% Confidence Interval for the Mean

The confidence interval is given by:

$\bar{X} \pm t_{\text{critical}} \times SE$

$0.32 \pm 2.797 \times 0.018$

$0.32 \pm 0.0503$

$(0.2697, 0.3703)$

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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.

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Do you have any questions or need further clarifications?

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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?

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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.