Step 1 :Given the summary statistics for the weights of Pepsi in randomly selected cans: \(n=36\), \(\bar{x}=0.82409 \, \mathrm{lb}\), \(s=0.00567 \, \mathrm{lb}\). We are asked to use a confidence level of \(99\%\) to complete parts (a) through (d).
Step 2 :For part (a), we are asked to identify the critical value \(t_{\alpha / 2}\) used for finding the margin of error. The critical value \(t_{\alpha / 2}\) is given as \(2.72\).
Step 3 :For part (b), we are asked to find the margin of error. The margin of error \(E\) can be calculated using the formula \(E = t_{\alpha / 2} \times \frac{s}{\sqrt{n}}\).
Step 4 :Substituting the given values into the formula, we get \(E = 2.72 \times \frac{0.00567}{\sqrt{36}}\).
Step 5 :Solving the above expression, we get \(E = 0.00257 \, \mathrm{lb}\).
Step 6 :Final Answer: The margin of error is \(\boxed{0.00257 \, \mathrm{lb}}\).