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This problem involves hypothesis testing for the mean in the context of a one-sample t-test. We need to determine whether the observed mean reduction in delivery time is significantly less than the claimed 10 minutes.

### Steps:

1. **Hypotheses Setup:**
   - Null Hypothesis \(H_0\): The mean reduction in delivery time is 10 minutes.
     \[
     H_0: \mu = 10
     \]
   - Alternative Hypothesis \(H_A\): The mean reduction in delivery time is less than 10 minutes.
     \[
     H_A: \mu < 10
     \]

2. **Given Data:**
   - Sample mean (\(\bar{x}\)) = 9.5 minutes
   - Population standard deviation (\(s\)) = 3 minutes
   - Sample size (\(n\)) = 100
   - Claimed mean (\(\mu_0\)) = 10 minutes
   - Significance level (\(\alpha\)) = 0.05

3. **Test Statistic:**
   The test statistic for a one-sample t-test is calculated as:
   \[
   t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}
   \]
   Substituting the given values:
   \[
   t = \frac{9.5 - 10}{\frac{3}{\sqrt{100}}} = \frac{-0.5}{\frac{3}{10}} = \frac{-0.5}{0.3} = -1.67
   \]

4. **Degrees of Freedom (df):**
   \[
   df = n - 1 = 100 - 1 = 99
   \]

5. **Critical Value:**
   Since this is a left-tailed test at the 0.05 significance level with 99 degrees of freedom, the critical t-value can be obtained from the t-distribution table. For \(df = 99\) and \(\alpha = 0.05\), the critical value is approximately:
   \[
   t_{\text{critical}} = -1.66
   \]

6. **Decision Rule:**
   If the calculated t-value is less than the critical value, we reject the null hypothesis.

7. **Conclusion:**
   The calculated t-value is \(-1.67\), which is less than the critical value of \(-1.66\). Therefore, we reject the null hypothesis. This means there is enough evidence to suggest that the mean reduction in delivery time is less than the claimed 10 minutes at the 0.05 significance level.

### Visual Representation:

To visualize this hypothesis test:
1. Draw a t-distribution curve centered at 0.
2. Mark the critical value \(-1.66\) on the left tail.
3. Indicate the calculated t-value \(-1.67\) slightly left of \(-1.66\).
4. Shade the region to the left of \(-1.66\), representing the rejection region.

Would you like more details or have any questions about this solution?

### Relative Questions:
1. What is the importance of a significance level in hypothesis testing?
2. How is the t-distribution different from the normal distribution?
3. Why did we use a one-sample t-test in this scenario?
4. How would the conclusion change if the t-value had been greater than the critical value?
5. How do degrees of freedom affect the critical t-value?

### Tip:
Always verify the assumptions of normality and sample size when conducting a t-test, as they influence the validity of your results.

A logistics company claims that a newly implemented routing optimization system reduces delivery times by an average of 10 minutes. To verify this claim, an independent logistics researcher conducted a study on a sample of 100 delivery routes after the system was implemented. The sample showed a mean reduction in delivery time of 9.5 minutes, with a standard deviation of 3 minutes. The researcher wishes to test the hypothesis that the mean reduction in delivery time is less than the claimed 10 minutes, using a 0.05 significance level.

Explore a hypothesis test for verifying a logistics company's claim that a new routing system reduces delivery times by 10 minutes on average. This example uses a one-sample t-test with a sample mean of 9.5, standard deviation of 3, and a significance level of 0.05 to evaluate whether the true reduction is less than the claimed 10 minutes.

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