To determine the confidence level represented by the given confidence interval for the population mean \(\mu\), follow these steps:
1. Understand the given data:
- Sample size (\(n\)): 12
- Sample mean (\(\bar{x}\)): 185 milligrams
- Sample standard deviation (\(s\)): 17.6 milligrams
- Confidence interval: \(173.8 \text{ mg} < \mu < 196.2 \text{ mg}\)
2. Calculate the margin of error (E):
[tex]\[
E = \text{sample mean} - \text{lower bound of the interval}
\][/tex]
Substituting the values:
[tex]\[
E = 185 - 173.8 = 11.2 \text{ mg}
\][/tex]
3. Calculate the t-score corresponding to the margin of error:
The formula for the t-score is:
[tex]\[
t = \frac{E}{s / \sqrt{n}}
\][/tex]
Substituting the values:
[tex]\[
t = \frac{11.2}{17.6 / \sqrt{12}} \approx 2.204
\][/tex]
4. Determine the degrees of freedom (df):
[tex]\[
\text{df} = n - 1 = 12 - 1 = 11
\][/tex]
5. Find the p-value associated with the t-score for the degrees of freedom:
Using the t-distribution table, or a statistical software or calculator, find the p-value corresponding to \(t = 2.204\) with 11 degrees of freedom. For a one-tailed test, the p-value for \(t = 2.204\) and 11 degrees of freedom is approximately 0.025.
6. Convert the p-value to the confidence level:
Since the confidence interval is two-tailed, double the p-value obtained:
[tex]\[
\text{Confidence Level} = 1 - (2 \times 0.025) = 0.95 = 95\%
\][/tex]
Thus, the confidence interval \(173.8 \text{ mg} < \mu < 196.2 \text{ mg}\) represents a 95\% confidence level. Therefore, the correct answer is:
[tex]\[ \boxed{95\%} \][/tex]
