To find the 99% confidence interval for the population mean, we need to perform the following steps from statistics theory:
1. Identifying the given values:
- The sample mean [tex]\(\overline{x}\)[/tex] (denoted as Overbar)
- The sample standard deviation [tex]\(s\)[/tex] (denoted as S)
- The sample size [tex]\(n\)[/tex]
- The z\*-score for the 99% confidence level from the provided table, which is 2.58
2. Understanding the formula for the confidence interval:
The formula for a confidence interval for the population mean is given by:
[tex]\[
\overline{x} \pm z^* \frac{s}{\sqrt{n}}
\][/tex]
where [tex]\(\overline{x}\)[/tex] is the sample mean, [tex]\(z^*\)[/tex] is the z-score corresponding to the desired confidence level, [tex]\(s\)[/tex] is the sample standard deviation, and [tex]\(n\)[/tex] is the sample size.
3. Calculating the margin of error:
The margin of error (ME) can be calculated as:
[tex]\[
\text{ME} = z^* \frac{s}{\sqrt{n}}
\][/tex]
For a 99% confidence level, [tex]\(z^* = 2.58\)[/tex]. Therefore, the margin of error is:
[tex]\[
\text{ME} = 2.58 \frac{s}{\sqrt{n}}
\][/tex]
4. Finding the confidence interval:
The confidence interval is computed by subtracting and adding the margin of error from/to the sample mean. Therefore, the 99% confidence interval is:
[tex]\[
\left( \overline{x} - 2.58 \frac{s}{\sqrt{n}}, \overline{x} + 2.58 \frac{s}{\sqrt{n}} \right)
\][/tex]
Thus, the correct answer is:
[tex]\[
\overline{x} \pm 2.58 \frac{s}{\sqrt{n}}
\][/tex]
From the given options, the correct one is:
[tex]\[
\overline{x} \pm \frac{2.58 \times s}{\sqrt{n}}
\][/tex]
