To determine the minimum sample size required to estimate an unknown population mean [tex]\(\mu\)[/tex], we start with the given information:
- Margin of error (E): [tex]$126
- Confidence level: 99%
- Population standard deviation (\(\sigma\)): $[/tex]512
The formula to calculate the minimum sample size ([tex]\(n\)[/tex]) is derived from the formula for the margin of error in a normal distribution:
[tex]\[ E = z \cdot \frac{\sigma}{\sqrt{n}} \][/tex]
To solve for [tex]\(n\)[/tex], we rearrange this formula as follows:
[tex]\[ n = \left( \frac{z \cdot \sigma}{E} \right)^2 \][/tex]
Step-by-Step Solution:
1. Determine the z-score:
For a 99% confidence level, we need to find the critical value [tex]\(z\)[/tex] (z-score). The z-score corresponds to the point such that the area under the standard normal curve in both tails is [tex]\(1 - 0.99 = 0.01\)[/tex] (0.005 in each tail). Consulting z-tables or using statistical tools, the z-score for a 99% confidence level is approximately 2.576.
2. Substitute the values into the formula:
[tex]\[
n = \left( \frac{2.576 \cdot 512}{126} \right)^2
\][/tex]
3. Calculate the right-hand side:
- Compute [tex]\( \frac{2.576 \cdot 512}{126} \)[/tex]:
[tex]\[
\frac{2.576 \cdot 512}{126} \approx 10.465
\][/tex]
- Square this value:
[tex]\[
10.465^2 \approx 109.56
\][/tex]
4. Round up to the next whole number:
Since the sample size must be a whole number and we always round up to ensure the margin of error is not exceeded, we round 109.56 up to the next whole number:
[tex]\[
n = 110
\][/tex]
Therefore, the minimum sample size required to estimate the population mean [tex]\(\mu\)[/tex] with a margin of error of \[tex]$126 at a 99% confidence level, given a population standard deviation of \$[/tex]512, is [tex]\( \boxed{110} \)[/tex].
