Question

here are summary statistics for randomly selected weights of newborn girls: ( n = 36 ), ( overline{x}=3150.0 ) g, ( s = 695.5 ) g. use a confidence level of ( 99% ) 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}=2.72 )
(round to two decimal places as needed.)
b. find the margin of error.
( e = 315.3 ) g
(round to one decimal place as needed.)
c. find the confidence interval estimate of ( mu ).
( square mathrm{g}<mu<square mathrm{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 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 \(\bar{x}=3150.0\) g and \(E = 315.3\) g.

Substitute into the formula:
\(\bar{x}-E=3150.0 - 315.3\)
\(\bar{x}+E=3150.0+ 315.3\)

Step3: Calculate the values

\(3150.0 - 315.3=2834.7\)
\(3150.0+315.3 = 3465.3\)

Answer:

\(2834.7\) g \(<\mu<3465.3\) g