The bank manager wants to show that the new system reduces typical customer waiting times to less than 6 minutes. One way to do this is to demonstrate that the mean of the population of all customer waiting times is less than 6. Letting this mean be *µ*, in this exercise we wish to investigate whether the sample of 100 waiting times provides evidence to support the claim that *µ* is less than 6.

For the sake of argument, we will begin by assuming that *µ* equals 6, and we will then attempt to use the sample to contradict this assumption in favor of the conclusion that *µ* is less than 6. Recall that the mean of the sample of 100 waiting times is

= 5.46 and assume that σ, the standard deviation of the population of all customer waiting times, is known to be 2.47.

**(a)** Consider the population of all possible sample means obtained from random samples of 100 waiting times. What is the shape of this population of sample means? That is, what is the shape of the sampling distribution of

?

**(b)** Find the mean and standard deviation of the population of all possible sample means when we assume that *µ* equals 6. **(Round your answer to 4 decimal places.)**

**(c)** The sample mean that we have actually observed is

= 5.46. Assuming that *µ* equals 6, find the probability of observing a sample mean that is less than or equal to   = 5.46. **(Round "z-value" to 2 decimals and final answer to 4 decimal places.)**

**(d)** If *µ* equals 6, what percentage of all possible sample means are less than or equal to 5.46? What do you conclude about whether the new system has reduced the typical customer waiting time to less than 6 minutes? **(Round your answer to 2 decimal places.)**

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.

---

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.

Explore how hypothesis testing is used to analyze customer waiting times, focusing on a sample mean of 100 waiting times with a known population standard deviation. Learn about the Central Limit Theorem and calculate probabilities to determine if a new system reduces waiting times.

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