Here are summary statistics for the weights of Pepsi in randomly selected cans: $n=36, \bar{x}=0.82409 \mathrm{lb}, s=0.00573 \mathrm{lb}$. Use a confidence level of $95 \%$ to complete parts (a) through (d) below.
a. Identify the critical value $t_{\alpha / 2}$ used for finding the margin of error.
$t_{\alpha / 2}=2.03$
(Round to two decimal places as needed.)
b. Find the margin of error.
$E=\square \mathrm{lb}$
(Round to five decimal places as needed.)

Step 1 ：The given values are: sample size (n) = 36, sample mean (\(\bar{x}\)) = 0.82409 lb, standard deviation (s) = 0.00573 lb, and the critical value (\(t_{\alpha / 2}\)) = 2.03.

Step 2 ：We are asked to find the margin of error (E) for a confidence level of 95%.

Step 3 ：The formula for the margin of error is \(E = t_{\alpha / 2} \cdot \frac{s}{\sqrt{n}}\).

Step 4 ：Substituting the given values into the formula, we get \(E = 2.03 \cdot \frac{0.00573}{\sqrt{36}}\).

Step 5 ：Solving the above expression, we find that the margin of error E is approximately 0.00194 lb.

Step 6 ：\(\boxed{E \approx 0.00194 \text{ lb}}\)