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