Question 1 $0 / 1$ pt $り 2 \rightleftarrows 99$ You measure 20 textbooks' weights, and find they have a mean weight of 73 ounces. Assume the population standard deviation is 9.3 ounces. Based on this, construct a $99 \%$ confidence interval for the population mean textbook weight. Give your answers as decimals, to two places \[ <\mu< \]
Solution
Step 1 ：Given values are: sample mean (x_bar) = 73, Z-score (Z) = 2.576, population standard deviation (sigma) = 9.3, and sample size (n) = 20.
Step 2 ：Calculate the margin of error using the formula: Margin of Error = Z * (sigma / sqrt(n)).
Step 3 ：Substitute the given values into the formula: Margin of Error = 2.576 * (9.3 / sqrt(20)) = 5.36 (rounded to two decimal places).
Step 4 ：Calculate the lower limit of the confidence interval using the formula: Lower Limit = x_bar - Margin of Error.
Step 5 ：Substitute the given values into the formula: Lower Limit = 73 - 5.36 = 67.64.
Step 6 ：Calculate the upper limit of the confidence interval using the formula: Upper Limit = x_bar + Margin of Error.
Step 7 ：Substitute the given values into the formula: Upper Limit = 73 + 5.36 = 78.36.
Step 8 ：The 99% confidence interval for the population mean textbook weight is approximately \(\boxed{[67.64, 78.36]}\) ounces.
