To address the problem, we need to conduct a hypothesis test for the mean customer waiting time. We will use the sample data and standard deviation to determine if the mean waiting time has significantly decreased.

(a) Shape of the Sampling Distribution

The shape of the population of all possible sample means obtained from random samples of 100 waiting times is approximately normal. This is due to the Central Limit Theorem, which states that the distribution of the sample mean will be approximately normal if the sample size is sufficiently large (typically $n \geq 30$).

(b) Mean and Standard Deviation of the Sampling Distribution

When we assume that $\mu = 6$:

• The mean of the sampling distribution of the sample mean $\bar{x}$ is equal to the population mean $\mu$: $\mu_{\bar{x}} = \mu = 6$

• The standard deviation of the sampling distribution of the sample mean (also known as the standard error) is given by: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$ where $\sigma$ is the population standard deviation and $n$ is the sample size. Here, $\sigma = 2.47$ and $n = 100$: $\sigma_{\bar{x}} = \frac{2.47}{\sqrt{100}} = \frac{2.47}{10} = 0.2470$

(c) Probability of Observing a Sample Mean Less Than or Equal to 5.46

First, we need to calculate the z-value corresponding to $\bar{x} = 5.46$:

$z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{5.46 - 6}{0.2470} = \frac{-0.54}{0.2470} \approx -2.19$

Next, we use the standard normal distribution to find the probability $P(Z \leq -2.19)$. Using standard normal distribution tables or a calculator, we find:

$P(Z \leq -2.19) \approx 0.0143$

(d) Percentage of All Possible Sample Means Less Than or Equal to 5.46 and Conclusion

The probability of observing a sample mean less than or equal to 5.46 is 0.0143, which is equivalent to 1.43%.

Since this percentage (1.43%) is quite small, we can conclude that it is unlikely to observe such a sample mean if the true population mean were 6 minutes. Therefore, there is significant evidence to suggest that the new system has indeed reduced the typical customer waiting time to less than 6 minutes.

Summary of Results

• Shape of Sampling Distribution: Approximately normal
• Mean of Sampling Distribution: $6$
• Standard Deviation of Sampling Distribution: $0.2470$
• Probability of Sample Mean $\leq 5.46$: $0.0143$
• Percentage of Sample Means $\leq 5.46$: $1.43\%$
• Conclusion: Significant evidence suggests the new system reduces the waiting time to less than 6 minutes.

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Do you want more details or have any questions?

1. What is the Central Limit Theorem and why is it important here?
2. How do you interpret the z-value in hypothesis testing?
3. What assumptions are made in this hypothesis test?
4. How would the conclusions change if the sample size were smaller?
5. What are Type I and Type II errors in the context of this test?
6. How can we determine if a result is statistically significant?
7. What other methods could be used to verify the effectiveness of the new system?
8. What if the standard deviation was unknown? How would that affect the test?

Tip: Always ensure that the assumptions of your statistical tests are met before drawing conclusions.