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

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

where:

• $Z$ is the z-score corresponding to the desired confidence level (in this case, 99.5% confidence).
• $\sigma$ is the population standard deviation.
• $E$ is the margin of error (how close we want to be to the true mean).

Given Values

1. Population standard deviation $\sigma = 45.3$
2. Confidence level = 99.5%
   • The z-score for 99.5% confidence is approximately 2.807 (from standard z-tables).
3. Margin of error $E = 2.5$

Calculation

1. Substitute the values into the formula: $n = \left( \frac{2.807 \times 45.3}{2.5} \right)^2$

2. Calculate inside the parentheses first: $n = \left( \frac{127.3071}{2.5} \right)^2$

3. Divide: $n = (50.92284)^2$

4. Square the result: $n \approx 2593.24$

Since sample size must be a whole number, we round up: $n = 2594$

Final Answer

The required sample size $n$ is 2594.

Would you like more detailed steps or have any questions?

Related Questions:

1. What is the z-score for different confidence levels, like 90% or 95%?
2. How would changing the margin of error affect the required sample size?
3. Why do we round up in sample size calculations instead of rounding to the nearest integer?
4. What if the population standard deviation is unknown—how does that impact the sample size formula?
5. How does the confidence level influence the z-score and, consequently, the sample size?

Tip: Always round up the sample size to ensure the estimate meets the desired confidence level and margin of error.