To determine the minimum sample size required to estimate the population mean with a specified level of confidence and margin of error, we can use the following formula:

$n = \left( \frac{Z \times \sigma}{E} \right)^2$

Where:

• $n$ = minimum sample size
• $Z$ = Z-value corresponding to the desired confidence level
• $\sigma$ = population standard deviation
• $E$ = margin of error

Given the problem:

• Confidence level = 95% (which corresponds to $Z = 1.96$)
• $\sigma = 3.8$
• $E = 1$

Plugging in the values:

$n = \left( \frac{1.96 \times 3.8}{1} \right)^2$

Let's calculate that.

$n = \left( 7.448 \right)^2 = 55.49$

Since the sample size must be a whole number, we round up to the next whole number, which gives:

$n = 56$

So, the correct answer is 56.

Would you like further details or have any questions?

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Here are 8 related questions for further exploration:

1. How does increasing the confidence level affect the sample size required?
2. What happens to the sample size if the population standard deviation is unknown?
3. How is the margin of error calculated in different contexts?
4. What are the implications of using a smaller margin of error?
5. Why is it important to round up when calculating sample size?
6. How would the sample size change if the margin of error was increased to 2 units?
7. Can this formula be used for non-normally distributed populations? Why or why not?
8. What factors could lead to underestimating the required sample size?

Tip: Always ensure to round up the calculated sample size to ensure the margin of error is not exceeded.