You measure 30 backpacks' weights, and find they have a mean weight of 73 ounces. Assume the population standard deviation is 7.1 ounces. Based on this, construct a $95 \%$ confidence interval for the true population mean backpack weight. Give your answers as decimals, to two places \[ \pm \] ounces Question Help: Video 9 Message instructor Calculator Submit Question

Step 1 ：Given that the sample mean (\(\bar{x}\)) is 73 ounces, the population standard deviation (\(\sigma\)) is 7.1 ounces, and the sample size (\(n\)) is 30.

Step 2 ：The Z-score for a 95% confidence interval is approximately 1.96.

Step 3 ：Substitute these values into the formula for a confidence interval: \(\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}\)

Step 4 ：Calculate the margin of error: \(1.96 \times \frac{7.1}{\sqrt{30}} \approx 2.54\) ounces

Step 5 ：Subtract the margin of error from the sample mean to get the lower bound of the confidence interval: \(73 - 2.54 = 70.46\) ounces

Step 6 ：Add the margin of error to the sample mean to get the upper bound of the confidence interval: \(73 + 2.54 = 75.54\) ounces

Step 7 ：Final Answer: The 95% confidence interval for the true population mean backpack weight is approximately \(\boxed{70.46}\) ounces to \(\boxed{75.54}\) ounces.