Question

here are summary statistics for randomly selected weights of newborn girls: ( n = 36 ), ( overline{x}=3150.0 mathrm{~g} ), ( s = 695.5 mathrm{~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}= )
(round to two decimal places as needed.)

Explanation:

Step1: Calculate the significance level ($\alpha$)

The confidence level is $99\%$, so $\alpha = 1 - 0.99=0.01$.

Step2: Calculate the degrees of freedom ($df$)

Given $n = 36$, then $df=n - 1=36 - 1 = 35$.

Step3: Find $t_{\alpha/2}$

Since $\alpha=0.01$, then $\alpha/2=0.005$. Using a $t$-distribution table or a calculator with $t$-distribution function (e.g., in Excel: =T.INV.2T(0.01,35)), we find the critical value.

Answer:

$t_{\alpha/2}= 2.72$