Assume that a sample is used to estimate a population mean $\mu$. Find the $98 \%$ confidence interval for a sample of size 972 with a mean of 72.6 and a standard deviation of 5.3. Enter your answer as a tri-linear inequality accurate to 3 decimal places. \[ <\mu< \] Question Help: Message instructor

Step 1 ：We are given a sample size of 972, a sample mean of 72.6, a standard deviation of 5.3, and we are asked to find the 98% confidence interval for the population mean.

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

Step 3 ：For a 98% confidence level, the Z-score is approximately 2.33.

Step 4 ：Substituting the given values into the formula, we get \(72.6 \pm 2.33 \frac{5.3}{\sqrt{972}}\).

Step 5 ：Solving this, we get a margin of error of approximately 0.396.

Step 6 ：Subtracting this margin of error from the sample mean, we get a lower bound of 72.204.

Step 7 ：Adding this margin of error to the sample mean, we get an upper bound of 72.996.

Step 8 ：Thus, the 98% confidence interval for the population mean \(\mu\) is given by the inequality \(72.204 < \mu < 72.996\).

Step 9 ：\(\boxed{72.204 < \mu < 72.996}\) is the final answer.