Assume that a sample is used to estimate a population mean $\mu$. Find the $95 \%$ confidence interval for a sample of size 65 with a mean of 70.2 and a standard deviation of 7.6. Enter your answer as an openinterval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place). \[ \text { 95\% C.I. }= \] Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places. Question Help: Message instructor Submit Question Jump to Answer

Step 1 ：Given that the sample mean (x_bar) is 70.2, the standard deviation (sigma) is 7.6, the sample size (n) is 65, and the z-score for a 95% confidence level (z) is 1.96.

Step 2 ：Calculate the margin of error using the formula: \( z \times \frac{\sigma}{\sqrt{n}} \). Substituting the given values, we get a margin of error of approximately 1.85.

Step 3 ：Calculate the lower limit of the confidence interval by subtracting the margin of error from the sample mean. This gives us a lower limit of approximately 68.35.

Step 4 ：Calculate the upper limit of the confidence interval by adding the margin of error to the sample mean. This gives us an upper limit of approximately 72.05.

Step 5 ：Round the lower and upper limits to one decimal place to match the precision of the given sample statistics. This gives us a final 95% confidence interval for the population mean of \(\boxed{(68.4, 72.0)}\).