Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an advance indication of illiness such as gout, leukemia, or lymphoma.t Over a period of months, an adult male pati has taken twelve blood tests for uric acid. The mean concentration was $\bar{x}=5.35 \mathrm{mg} / \mathrm{dl}$. The distribution of uric acid in healthy adult males can be assumed to be normal, with $\sigma=1.81$ mg/dl. (a) Find a $95 \%$ confidence interval for the population mean concentration of uric acid in this patient's blood. What is the margin of error? (Round your answers to two decimal places.) lower limit upper limit margin of error

Step 1 ：Given that the mean concentration of uric acid in the patient's blood is \(\bar{x}=5.35 \mathrm{mg/dl}\), the standard deviation is \(\sigma=1.81 \mathrm{mg/dl}\), and the number of blood tests taken is \(n=12\).

Step 2 ：We are asked to find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. The formula for the confidence interval is \(\bar{x} \pm z \frac{\sigma}{\sqrt{n}}\), where \(z\) is the z-score corresponding to the desired confidence level. For a 95% confidence level, \(z\) is approximately 1.96.

Step 3 ：Let's calculate the margin of error, which is \(z \frac{\sigma}{\sqrt{n}}\).

Step 4 ：Substituting the given values into the formula, we get the margin of error to be approximately 1.02 mg/dl.

Step 5 ：Next, we calculate the lower limit of the confidence interval, which is \(\bar{x} - z \frac{\sigma}{\sqrt{n}}\). Substituting the given values into the formula, we get the lower limit to be approximately 4.33 mg/dl.

Step 6 ：Similarly, we calculate the upper limit of the confidence interval, which is \(\bar{x} + z \frac{\sigma}{\sqrt{n}}\). Substituting the given values into the formula, we get the upper limit to be approximately 6.37 mg/dl.

Step 7 ：Final Answer: The 95% confidence interval for the population mean concentration of uric acid in this patient's blood is approximately from \(\boxed{4.33 \mathrm{mg/dl}}\) to \(\boxed{6.37 \mathrm{mg/dl}}\). The margin of error is approximately \(\boxed{1.02 \mathrm{mg/dl}}\).