According to a leasing firm's reports, the mean number of miles driven annually in its leased cars is 12,440 miles with a standard deviation of 2380 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, , is less than 12,440 miles. He takes a random sample of 38 cars under the new contracts. The cars in the sample had a mean of 12,021 annual miles driven. Assume that the population is normally distributed. Is there support for the claim, at the 0.5 level of significance, that the population mean number of miles driven annually by cars under the new contracts, is less than 12,440 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. (a)State the null hypothesis H0 and the alternative hypothesis H1. (b)Determine the type of test statistic to use. (c)Find the value of the test statistic. (Round to three or more decimal places.) (d)Find the critical value. (Round to three or more decimal places.) (e)Can we support the claim that the population mean number of miles driven annually by cars under the new contracts is less than 12,440 miles? YES/NO

To perform the hypothesis test, we'll go through each step systematically.

### Given Data:
- **Population mean (\(\mu_0\))**: 12,440 miles
- **Sample mean (\(\bar{x}\))**: 12,021 miles
- **Population standard deviation (\(\sigma\))**: 2380 miles
- **Sample size (\(n\))**: 38
- **Significance level (\(\alpha\))**: 0.05

### (a) State the null hypothesis \(H_0\) and the alternative hypothesis \(H_1\):
- **Null Hypothesis (\(H_0\))**: \(\mu = 12,440\) miles (The mean number of miles driven annually under the new contracts is 12,440 miles or more.)
- **Alternative Hypothesis (\(H_1\))**: \(\mu < 12,440\) miles (The mean number of miles driven annually under the new contracts is less than 12,440 miles.)

### (b) Determine the type of test statistic to use:
Since we know the population standard deviation and the sample size is large (\(n = 38\)), we will use the **z-test** for the test statistic.

### (c) Find the value of the test statistic:
The test statistic for a z-test is calculated using the formula:

\[
z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}
\]

Substituting the given values:

\[
z = \frac{12,021 - 12,440}{\frac{2380}{\sqrt{38}}}
\]

Let's calculate the z-value.

\[
z = \frac{-419}{\frac{2380}{\sqrt{38}}}
\]

\[
z = \frac{-419}{386.164} \approx -1.085
\]

### (d) Find the critical value:
Since this is a one-tailed test with a significance level of 0.05, we will find the critical value \(z_{critical}\) for \(\alpha = 0.05\) in the lower tail.

From the standard normal distribution table, the critical value for \(z_{critical}\) at \(\alpha = 0.05\) in a one-tailed test is approximately:

\[
z_{critical} = -1.645
\]

### (e) Can we support the claim that the population mean number of miles driven annually by cars under the new contracts is less than 12,440 miles? 
To determine whether we can reject the null hypothesis, we compare the test statistic \(z\) with the critical value:

- **Decision rule**: If \(z < z_{critical}\), reject \(H_0\).

In this case:

- \(z = -1.085\)
- \(z_{critical} = -1.645\)

Since \(z = -1.085\) is **greater** than \(z_{critical} = -1.645\), we **fail to reject** the null hypothesis.

### Conclusion:
- **Answer**: **NO**, we cannot support the claim that the population mean number of miles driven annually by cars under the new contracts is less than 12,440 miles at the 0.05 level of significance.

Would you like more details or explanations on any of these steps?

### Related Questions:
1. What would the conclusion be if the significance level were 0.10 instead of 0.05?
2. How would you interpret the p-value in this context?
3. How does increasing the sample size affect the power of the test?
4. What are the assumptions required to perform a z-test?
5. How would the test change if the population standard deviation were unknown?
6. What is the potential impact of a Type II error in this context?
7. How do we determine if the normality assumption is valid for this test?
8. Could you perform a two-tailed test, and how would the hypotheses and conclusions differ?

**Tip:** Always ensure the normality of the data or a sufficiently large sample size before using the z-test, especially when the population standard deviation is known.

Perform a hypothesis test to determine if the mean number of miles driven annually under new contracts is less than 12,440 miles. Calculate the z-test statistic and compare it with the critical value at a 0.05 significance level. Data includes a population mean of 12,440 miles, sample mean of 12,021 miles, and population standard deviation of 2380 miles.

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