20) You measure 20 randomly selected textbooks' weights, and find they have a mean weight of 48 ounces. Assume the population standard deviation is 4.8 ounces. Based on this, construct a $99 \%$ confidence interval for the true population mean textbook weight. Give your answers as decimals, round to two decimal places The confidence interval is Select an answer $v$ Interpretation: With $99 \%$ confidence, you can say that the population mean textbook weight is between and ounces. Use the following online calculator to solve this problem. Confidence Interval Calculator omn " CALCULATOR
Solution
Step 1 ：We are given that the sample mean (mean) is 48 ounces, the population standard deviation (std_dev) is 4.8 ounces, the sample size (n) is 20, and the confidence level is 99%.
Step 2 ：The Z-score for a 99% confidence level is approximately 2.576. This value can be found in Z-score tables or calculated using a statistical function.
Step 3 ：We use the formula for a confidence interval for a population mean, which is: \(mean ± Z * (std_dev / \sqrt{n})\).
Step 4 ：Substituting the given values into the formula, we get: \(48 ± 2.576 * (4.8 / \sqrt{20})\).
Step 5 ：Calculating the margin of error, we get approximately 2.7648533328189395.
Step 6 ：Subtracting this margin of error from the mean, we get the lower bound of the confidence interval: \(48 - 2.7648533328189395 = 45.24\) ounces.
Step 7 ：Adding the margin of error to the mean, we get the upper bound of the confidence interval: \(48 + 2.7648533328189395 = 50.76\) ounces.
Step 8 ：Final Answer: The $99 \%$ confidence interval for the true population mean textbook weight is between \(\boxed{45.24}\) ounces and \(\boxed{50.76}\) ounces. With $99 \%$ confidence, we can say that the population mean textbook weight is between 45.24 and 50.76 ounces.
