Question

here are summary statistics for the weights of pepsi in randomly selected cans: ( n = 36 ), ( overline{x}=0.82409 \text{lb} ), ( s = 0.00569 \text{lb} ). use a confidence level of ( 90% ) 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}=1.69 )
(round to two decimal places as needed.)
b. find the margin of error
( e=\text{lb} )
(round to five decimal places as needed.)

Explanation:

Step1: Recall the formula for margin of error

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

Step2: Substitute the given values

We are given \(t_{\alpha/2}=1.69\), \(s = 0.00569\), and \(n = 36\)

First, calculate \(\sqrt{n}=\sqrt{36}=6\)

Then, \(\frac{s}{\sqrt{n}}=\frac{0.00569}{6}\approx0.0009483\)

Now, \(E=t_{\alpha/2}\times\frac{s}{\sqrt{n}}=1.69\times0.0009483\)

Step3: Perform the multiplication

\(E = 1.69\times0.0009483=0.001602627\approx0.00160\)

Answer:

\(E = 0.00160\)