Question

here are summary statistics for randomly selected weights of newborn girls: ( n = 36 ), ( \bar{x} = 3197.2 ) g, ( s = 692.6 ) g. use a confidence level of 90% to complete parts (a) through (d) below

( dots )

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 = 195.1 ) g
(round to one decimal place as needed.)
c. find the confidence interval estimate of ( mu ).
( square ) g ( < mu < square ) g
(round to one decimal place as needed.)

Explanation:

Step1: Recall the formula for confidence interval

The confidence interval for the population mean \(\mu\) when the population standard deviation \(\sigma\) is unknown (we use the sample standard deviation \(s\)) is given by \(\bar{x}-E<\mu<\bar{x} + E\), where \(\bar{x}\) is the sample mean and \(E\) is the margin of error.

Step2: Substitute the known values

We know that \(\bar{x}=3197.2\) g and \(E = 195.1\) g.
First, calculate the lower bound: \(\bar{x}-E=3197.2 - 195.1=3002.1\)
Then, calculate the upper bound: \(\bar{x}+E=3197.2+195.1 = 3392.3\)

Answer:

\(3002.1\) g\(<\mu<\)\(3392.3\) g