A biologist examines 15 sedimentary samples for lead concentration. The mean lead concentration for the sample data is $0.511 \mathrm{cc} / \mathrm{cubic}$ meter with a standard deviation of 0.0598 . Determine the $98 \%$ confidence interval for the population mean lead concentration. Assume the population is approximately normal.

Step 2 of 2 : Construct the $98 \%$ confidence interval. Round your answer to three decimal places.

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Step 1 ：First, we need to find the z-score that corresponds to a 98% confidence interval. This value is approximately \(2.33\).

Step 2 ：The formula for a confidence interval is \(CI = \bar{x} \pm Z \cdot \left(\frac{\sigma}{\sqrt{n}}\right)\), where \(\bar{x}\) is the sample mean, \(Z\) is the z-score, \(\sigma\) is the standard deviation, and \(n\) is the sample size.

Step 3 ：In this case, \(\bar{x} = 0.511\), \(Z = 2.33\), \(\sigma = 0.0598\), and \(n = 15\).

Step 4 ：We first calculate the value of the standard deviation divided by the square root of the sample size: \(\frac{\sigma}{\sqrt{n}} = \frac{0.0598}{\sqrt{15}} = 0.0154\).

Step 5 ：Then, we multiply this value by the z-score: \(Z \cdot \left(\frac{\sigma}{\sqrt{n}}\right) = 2.33 \cdot 0.0154 = 0.0359\).

Step 6 ：Finally, we add and subtract this value from the sample mean to find the confidence interval: \(CI = 0.511 \pm 0.0359\).

Step 7 ：So, the 98% confidence interval for the population mean lead concentration is \(\boxed{(0.475, 0.547)}\), rounded to three decimal places.