If $\mathrm{n}=15, \bar{x}=44$, and $\mathrm{s}=12$, construct a confidence interval at a $98 \%$ confidence level. Assume the data came from a normally distributed population. Give your answers to three decimal places.
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Step 1 ：Given values are: sample size (n) = 15, sample mean (\(\bar{x}\)) = 44, and standard deviation (s) = 12. The Z-score for a 98% confidence level is 2.33.

Step 2 ：First, calculate the margin of error using the formula: \(Z \times \frac{s}{\sqrt{n}}\). Substituting the given values, we get: \(2.33 \times \frac{12}{\sqrt{15}}\) which equals approximately 7.219.

Step 3 ：Next, calculate the lower and upper bounds of the confidence interval using the formulas: \(\bar{x} - \text{margin of error}\) and \(\bar{x} + \text{margin of error}\) respectively.

Step 4 ：For the lower bound, we get: \(44 - 7.219\) which equals approximately 36.781.

Step 5 ：For the upper bound, we get: \(44 + 7.219\) which equals approximately 51.219.

Step 6 ：Final Answer: The confidence interval at a 98% confidence level is approximately \(\boxed{(36.781, 51.219)}\).