Question

here are summary statistics for the weights of pepsi in randomly selected cans: n = 36, x = 0.823
use a confidence level of 95% to complete parts (a) through (d) below.
( t _ { alpha / 2 } = 2.03 )
(round to two decimal places as needed.)
b. find the margin of error.
( e = 0.00194 ) lb
(round to five decimal places as needed.)
c. find the confidence interval estimate of ( mu )
( square mathrm { lb } < mu < square mathrm { lb } )
(round to five decimal places as needed.)

Explanation:

Step1: Recall the formula for confidence interval

The formula for the confidence interval when the population standard deviation \(\sigma\) is unknown (using \(t -\)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 given values

We are given \(\bar{x}=0.823\) (assuming this value from the context as it was cut - off in the problem statement) and \(E = 0.00194\).

Substitute into the formula: \((0.823-0.00194)<\mu<(0.823 + 0.00194)\)

Step3: Calculate the values

\(0.823-0.00194=0.82106\)

\(0.823 + 0.00194=0.82494\)

Answer:

\(0.82106\) lb\(<\mu<0.82494\) lb