You measure 21 backpacks' weights, and find they have a mean weight of 45 ounces. Assume the population standard deviation is 8.3 ounces. Based on this, construct a $99 \%$ confidence interval for the true population mean backpack weight.
Give your answers as decimals, to two places
$\pm$ ounces

Step 1 ：We are given that the sample mean (\(\bar{x}\)) is 45 ounces, the population standard deviation (\(\sigma\)) is 8.3 ounces, and the sample size (\(n\)) is 21.

Step 2 ：We are asked to construct a 99% confidence interval for the true population mean backpack weight. The Z-score for a 99% confidence level is approximately 2.576.

Step 3 ：We use the formula for a confidence interval, which is \(\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}\).

Step 4 ：Substituting the given values into the formula, we get \(45 \pm 2.576 \frac{8.3}{\sqrt{21}}\).

Step 5 ：Solving the above expression, we get the margin of error to be approximately 4.67 ounces.

Step 6 ：Subtracting and adding this margin of error from the sample mean, we get the confidence interval to be approximately 40.33 ounces to 49.67 ounces.

Step 7 ：\(\boxed{\text{The 99% confidence interval for the true population mean backpack weight is approximately 40.33 ounces to 49.67 ounces.}}\)