ion 7.2
Question 8, 7.2.7-T
HW Score: $53.85 \%, 7$ of 13
Part 2 of 4
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Here are summary statistics for the weights of Pepsi in randomly selected cans: $n=36, \bar{x}=0.82409 \mathrm{lb}, s=0.00567 \mathrm{lb}$. Use a confidence level of $99 \%$ to complete parts (a) through (d) below.
a. Identify the critical value $t_{\alpha / 2}$ used for finding the margin of error.
\[
t_{\alpha / 2}=2.72
\]
(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 ：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 \(\boxed{0.00257 \, \mathrm{lb}}\).