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 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 = 0.00161 ) lb
(round to five decimal places as needed.)

c. find the confidence interval estimate of ( mu ).
( square ) lb ( < mu < square ) lb
(round to five decimal places as needed.)

Explanation:

Step1: Recall the formula for confidence interval

The formula for the confidence interval of the population mean \(\mu\) when using the t - distribution is \(\bar{x}-E<\mu<\bar{x} + E\), where \(\bar{x}\) is the sample mean and \(E\) is the margin of error.

Step2: Substitute the values of \(\bar{x}\) and \(E\)

We know that \(\bar{x}=0.82412\) lb and \(E = 0.00161\) lb.
First, calculate the lower bound: \(\bar{x}-E=0.82412 - 0.00161=0.82251\)
Then, calculate the upper bound: \(\bar{x}+E=0.82412+ 0.00161 = 0.82573\)

Answer:

\(0.82251\) lb\(<\mu<0.82573\) lb