Question

here are summary statistics for the weights of pepsi in randomly selected cans: n = 36, \\(\bar{x}\\) = 0.82412 lb, s = 0.00567 lb. use a confidence 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 = \\(\square\\) lb\
(round to five decimal places as needed.)

Explanation:

Step1: Recall the margin of error formula for t - interval

The formula for the margin of error \( E \) when using the t - distribution is \( E=t_{\alpha/2}\times\frac{s}{\sqrt{n}} \), where \( t_{\alpha/2} \) is the critical value, \( s \) is the sample standard deviation, and \( n \) is the sample size.

Step2: Substitute the given values

We are given that \( t_{\alpha/2} = 1.69 \), \( s=0.00567 \) lb, and \( n = 36 \). First, calculate \( \sqrt{n}=\sqrt{36}=6 \). Then, substitute into the formula:
\( E = 1.69\times\frac{0.00567}{6} \)
First, calculate \( \frac{0.00567}{6}=0.000945 \). Then, multiply by \( 1.69 \): \( 1.69\times0.000945 = 0.00160705 \)

Step3: Round to five decimal places

Rounding \( 0.00160705 \) to five decimal places gives \( 0.00161 \) (since the sixth decimal place is 7, which is greater than 5, we round up the fifth decimal place).

Answer:

\( 0.00161 \)