To solve the problem and calculate the margin of error (ME) for a 95% confidence interval for $\mu$, we use the formula:

$ME = z^* \cdot \frac{\sigma}{\sqrt{n}}$

Information provided:

• $n = 92$ (sample size)
• $\sigma = 20$ (population standard deviation)
• Confidence level = 95%

The $z^*$ value for a 95% confidence interval (from standard Z-tables) is approximately 1.96.

Step-by-step calculation:

1. Compute the standard error (SE): $SE = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{92}}$ First, find $\sqrt{92} \approx 9.5916$: $SE = \frac{20}{9.5916} \approx 2.085$

2. Compute the margin of error (ME): $ME = z^* \cdot SE = 1.96 \cdot 2.085 \approx 4.085$

Final Answer:

The margin of error is approximately 4.085.

Would you like additional explanation or steps for clarification?

Related Questions:

1. What is the role of $z^*$ in determining the confidence interval?
2. How does increasing the sample size affect the margin of error?
3. What is the difference between a 95% and 99% confidence interval in terms of $z^*$?
4. How would the margin of error change if $\sigma$ were smaller?
5. Can the formula for margin of error be used for non-normal distributions?

Tip: Always ensure that the sample size $n$ is large enough for the Central Limit Theorem to apply when working with confidence intervals.