Answer:
standard deviation of the sample = s = 3.04
Step-by-step explanation:
given,
μ, is given as (18.985, 21.015)
sample mean = [tex]\dfrac{(18.958+21.015)}{2}[/tex]
=19.9865
margin of error = 19.9865 - 18.958 =1.0285
The degree of freedom = n -1 = 36 - 1 = 35
a = 0.05, t(0.025, df = 35) = 2.03 (from student t table)
margin error = [tex]\dfrac{t\times s}{\sqrt{n}}= 1.0285[/tex]
= [tex]\dfrac{2.03 \times s}{\sqrt{36}}= 1.0285[/tex]
= [tex]\dfrac{2.03 \times s}{6}= 1.0285[/tex]
= [tex]s = \dfrac{1.0285\times 6}{2.03}[/tex]
= s = 3.04
standard deviation of the sample = s = 3.04
