Question

here are summary statistics for the weights of pepsi in randomly selected cans: n = 36, x = 0.82412 lb, s = 0.00672 lb. use a confidence level of 95% to complete parts (a) through (d) below. a. identify the critical value tα/2 used for finding the margin of error. tα/2 = 2.03 (round to two decimal places as needed.) b. find the margin of error. e = □ lb (round to five decimal places as needed.)

Explanation:

Step1: Recall the formula for margin of error

The formula for margin of error \(E\) when population standard deviation \(\sigma\) is known is \(E = t_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}\).

Step2: Substitute the given values

We are given \(t_{\alpha/2}=2.03\), \(\sigma = 0.00672\) lb, and \(n = 36\).
First, calculate \(\frac{\sigma}{\sqrt{n}}\): \(\frac{0.00672}{\sqrt{36}}=\frac{0.00672}{6}=0.00112\).
Then, find \(E\): \(E=2.03\times0.00112\).

Step3: Perform the multiplication

\(E = 2.03\times0.00112=0.0022736\approx0.00227\) (rounded to five decimal places).

Answer:

\(E = 0.00227\) lb