For Part (a), the correct method to use is the Z-method (or Z-interval method) because we know the population standard deviation.

To construct a confidence interval, the Z-method is appropriate in cases where:

1. We know the population standard deviation, $\sigma$.
2. The sample size is large (typically $n \geq 30$), or if the sample size is small, the population is approximately normal.

For Part (b), to construct a 95% confidence interval for the mean tuition, we use the formula: $\bar{X} \pm Z \times \frac{\sigma}{\sqrt{n}}$ where:

• $\bar{X}$ is the sample mean,
• $Z$ is the Z-score for a 95% confidence level (approximately 1.96),
• $\sigma$ is the population standard deviation,
• $n$ is the sample size.

If you provide the values for $\bar{X}$, $\sigma$, and $n$, I can help you calculate the confidence interval precisely.

Would you like detailed steps or clarification on any part of this?

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Here are five related questions to deepen your understanding:

1. How does the confidence level affect the width of the confidence interval?
2. Why is the Z-method preferred over the t-method when the population standard deviation is known?
3. What happens to the confidence interval if the sample size increases?
4. How would we proceed if only the sample standard deviation was known?
5. Why is the sample mean an unbiased estimator of the population mean?

Tip: Knowing the population standard deviation makes calculations simpler and allows us to use the Z-method directly, even for moderately-sized samples.