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

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

where:

• $Z_{\alpha/2}$ is the critical value from the standard normal distribution for a given confidence level,
• $\sigma$ is the population standard deviation,
• $E$ is the desired margin of error.

Given values:

• Confidence level = 99%, so $Z_{\alpha/2} \approx 2.576$,
• Population standard deviation $\sigma = 1.75$,
• Desired margin of error $E = 0.4$.

Solution

1. Substitute the values into the formula:

   $n = \left( \frac{2.576 \cdot 1.75}{0.4} \right)^2$

2. Calculate inside the parentheses:

   $\frac{2.576 \cdot 1.75}{0.4} = 11.27$

3. Square the result:

   $n = (11.27)^2 \approx 127.04$

4. Round up to ensure the margin of error condition is met, since sample size $n$ must be a whole number:

   $n \approx 128$

Conclusion

The required sample size is 128.

Would you like further details on confidence intervals or margin of error?

Here are some related questions to consider:

1. How would the sample size change with a different confidence level, say 95%?
2. What happens to the sample size if the margin of error requirement is tightened to 0.2?
3. How is this calculation affected if the population standard deviation is unknown?
4. How would increasing the sample size impact the margin of error for the same data?
5. What are typical applications for calculating a sample size for population means?

Tip: In general, a higher confidence level or smaller margin of error requires a larger sample size.