According to a leasing firm's reports, the mean number of miles driven annually in its leased cars is 12,500 miles with a standard deviation of 2340 miles. The company recently starting using new contracts which require customers to have the cars serviced at their own expense. The company's owner believes the mean number of miles driven annually under the new contracts, M, is less than 12,500 miles. He takes a random sample of 17 cars under the new contracts. The cars in the sample had a mean of 11,004 annual miles driven. Assume that the population is normally distributed. Is there support for the claim, at the 0.01 level of significance, that the population mean number of miles driven annually by cars under the new contracts, is less than 12,500 miles? Assume that the population standard deviation of miles driven annually was not affected by the change to the contracts.

Perform a one-tailed test. Then complete the parts below.

Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.)

To determine if there is support for the claim that the mean number of miles driven annually by cars under the new contracts is less than 12,500 miles, we can perform a one-tailed hypothesis test. We'll use the following steps to carry out the test:

### 1. **State the Hypotheses:**

- **Null Hypothesis (\( H_0 \))**: \( \mu = 12,500 \) miles  
  (The mean number of miles driven annually is 12,500 miles.)
  
- **Alternative Hypothesis (\( H_1 \))**: \( \mu < 12,500 \) miles  
  (The mean number of miles driven annually is less than 12,500 miles.)

### 2. **Set the Significance Level:**
  
- The significance level \( \alpha \) is given as 0.01.

### 3. **Calculate the Test Statistic:**

Since the population standard deviation is known, we'll use the **Z-test** for the mean. The test statistic \( Z \) is calculated as follows:

\[
Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}}
\]

Where:
- \( \bar{X} = 11,004 \) miles (sample mean)
- \( \mu = 12,500 \) miles (population mean under the null hypothesis)
- \( \sigma = 2340 \) miles (population standard deviation)
- \( n = 17 \) (sample size)

\[
Z = \frac{11,004 - 12,500}{\frac{2340}{\sqrt{17}}}
\]

Now, let's calculate this:

\[
Z = \frac{-1,496}{\frac{2340}{\sqrt{17}}} \approx \frac{-1,496}{567.571} \approx -2.636
\]

### 4. **Determine the Critical Value:**

Since this is a one-tailed test (left-tailed), we look up the critical value for \( Z \) at \( \alpha = 0.01 \).

For a significance level of 0.01, the critical value \( Z_{critical} \) for a left-tailed test is approximately \( -2.33 \).

### 5. **Make the Decision:**

- **If \( Z \) < \( Z_{critical} \)**, we reject the null hypothesis.
- **If \( Z \) > \( Z_{critical} \)**, we fail to reject the null hypothesis.

In this case:

\[
Z = -2.636 \quad \text{and} \quad Z_{critical} = -2.33
\]

Since \( Z \) is less than \( Z_{critical} \), we reject the null hypothesis.

### 6. **Conclusion:**

At the 0.01 level of significance, there is sufficient evidence to support the claim that the mean number of miles driven annually by cars under the new contracts is less than 12,500 miles.

---

Would you like further details on any of these steps, or do you have any questions?

### Related Questions:
1. What assumptions are necessary for performing this Z-test?
2. How would the conclusion change if the significance level were \( \alpha = 0.05 \)?
3. What is the p-value associated with this test statistic?
4. How would the test differ if the population standard deviation were unknown?
5. Can you describe a situation where a two-tailed test would be more appropriate?

**Tip:** Always ensure the sample size is sufficiently large when using the Z-test, especially if the population distribution is not perfectly normal.

Learn how to perform a one-tailed Z-test to determine if there is sufficient evidence to support the claim that the mean number of miles driven annually by cars under new contracts is less than 12,500 miles. Includes step-by-step calculations and interpretation using a significance level of 0.01.

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