Question

here are summary statistics for the weights of pepsi in randomly selected cans: n = 36, x̄ = 0.82409 lb, s = 0.00569 lb. use a confidence level of 90% to complete parts (a) through (d) below.
a. identify the critical value t_{α/2} used for finding the margin of error.
t_{α/2}=1.69
(round to two decimal places as needed.)
b. find the margin of error.
e = 0.00160 lb
(round to five decimal places as needed.)
c. find the confidence interval estimate of μ.
□ lb < μ < □ lb
(round to five decimal places as needed.)

Explanation:

Step1: Recall the formula for confidence interval

The formula for the confidence interval for the population mean \(\mu\) when the population standard deviation \(\sigma\) is unknown 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 \(\bar{x}=0.82409\) lb and \(E = 0.00160\) lb.
For the lower bound: \(\bar{x}-E=0.82409 - 0.00160\)

$$0.82409-0.00160=0.82249$$

For the upper bound: \(\bar{x}+E=0.82409+ 0.00160\)

$$0.82409 + 0.00160=0.82569$$

Answer:

\(0.82249\) lb\(<\mu<0.82569\) lb