To perform the hypothesis test, we need to calculate the value of the test statistic. This involves the following steps:

1. Set up the hypotheses:

• Null Hypothesis $H_0$: The mean delivery time is 40 minutes, $\mu = 40$.
• Alternative Hypothesis $H_a$: The mean delivery time is greater than 40 minutes, $\mu > 40$.

2. Determine the test statistic:

Given that the sample size is small ($n = 8$), we'll use the $t$-test. The test statistic $t$ is calculated using the formula:

$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}$

Where:

• $\bar{x}$ is the sample mean.
• $\mu_0$ is the hypothesized population mean (40 minutes).
• $s$ is the sample standard deviation.
• $n$ is the sample size.

3. Plug in the values:

$\bar{x} = 47.6 \text{ minutes}$ $\mu_0 = 40 \text{ minutes}$ $s = 11.5 \text{ minutes}$ $n = 8$

Now, substitute these into the formula:

$t = \frac{47.6 - 40}{\frac{11.5}{\sqrt{8}}}$

Let's calculate this step by step.

$t = \frac{7.6}{\frac{11.5}{\sqrt{8}}}$

$t = \frac{7.6}{\frac{11.5}{2.828}}$

$t = \frac{7.6}{4.066}$

$t \approx 1.869$

4. Conclusion:

The value of the test statistic $t$ is approximately 1.869 (rounded to three decimal places).

Would you like to proceed with the next step, or do you have any questions?

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Related Questions:

1. How do you determine the critical value for this test?
2. What is the significance of the 0.005 level of significance in this context?
3. How does the sample size affect the test statistic and the test's power?
4. What assumptions must be met for the $t$-test to be valid?
5. How would the test result change if the standard deviation was smaller?

Tip: Remember that the critical value depends on the chosen significance level and the degrees of freedom (sample size - 1).