Here are summary statistics for the weights of Pepsi in randomly selected cans: \( \mathrm{n}=36, \overline{\mathrm{x}}=0.82414 \mathrm{lb}, \mathrm{s}=0.00572 \mathrm{lb} \). Use a confidence level of \( 95 \% \) to complete part a. Identify the critical value \( \mathrm{t}_{\alpha / 2} \) used for finding the margin of error. \( \mathrm{t}_{\alpha / 2}=2.03 \) (Round to two decimal places as needed.) b. Find the margin of error. \( \mathrm{E}=\square \mathrm{lb} \) (Round to five decimal places as needed.)

To find the margin of error, you can use the formula: \[ E = t_{\alpha/2} \cdot \left( \frac{s}{\sqrt{n}} \right) \] Here, \( t_{\alpha/2} = 2.03 \), \( s = 0.00572 \, \text{lb} \), and \( n = 36 \). First, calculate the standard error: \[ \frac{s}{\sqrt{n}} = \frac{0.00572}{\sqrt{36}} = \frac{0.00572}{6} = 0.0009533333 \, \text{lb} \] Now, multiply this by the critical value: \[ E = 2.03 \cdot 0.0009533333 \approx 0.0019377 \, \text{lb} \] Rounding this to five decimal places, the margin of error is: \( E \approx 0.00194 \, \text{lb} \)