To find out how large a sample size is required to estimate the population mean with a specified margin of error and confidence level, we'll follow these steps:
1. Identify the given information:
- Population standard deviation, [tex]\(\sigma = 69.5\)[/tex]
- Desired confidence level, [tex]\(99.5\% = 0.995\)[/tex]
- Margin of error, [tex]\(E = 2.5\)[/tex]
2. Determine the z-score corresponding to the specified confidence level:
The z-score is a critical value from the standard normal distribution that corresponds to the desired confidence level. For a [tex]\(99.5\%\)[/tex] confidence interval, the z-score can be found using standard z-tables or statistical software. We lookup the z-value such that the area under the standard normal curve from the center to the z-value covers [tex]\(99.5\%\)[/tex]:
[tex]\[
z = 2.807
\][/tex]
3. Calculate the required sample size using the formula for the margin of error:
The formula for the margin of error in terms of the sample size [tex]\(n\)[/tex] is given by:
[tex]\[
E = z \frac{\sigma}{\sqrt{n}}
\][/tex]
Rearranging this formula to solve for the sample size [tex]\(n\)[/tex], we get:
[tex]\[
n = \left( \frac{z \cdot \sigma}{E} \right)^2
\][/tex]
4. Substitute the values into the formula:
[tex]\[
n = \left( \frac{2.807 \cdot 69.5}{2.5} \right)^2
\][/tex]
5. Perform the arithmetic calculation:
[tex]\[
n = \left( \frac{194.4865}{2.5} \right)^2 = \left( 77.7946 \right)^2 = 6056.5411
\][/tex]
6. Round up to the next whole number:
Since sample size must be an integer, we round up to ensure the margin of error criteria is met:
[tex]\[
n \approx 6090
\][/tex]
Thus, the required sample size is [tex]\(n = 6090\)[/tex].
