Answer: a. $40,800 b. 36
Step-by-step explanation:
Given : a 99% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$34,393, $47,207].
[tex]\sigma= \$14,900[/tex]
a. Since Point estimate of of the mean = Average of upper limit and lower limit of the interval.
Therefore , the point estimate of the mean salary for all college graduates in this town = [tex]\dfrac{34393+47207}{2}=\dfrac{81600}{2}[/tex]
= 40,800
hence, the point estimate of the mean salary for all college graduates in this town = $40,800
b. Since lower limit = Point estimate - margin of error, where Margin of error is the half of the difference between upper limit and lower limit.
Margin of error[tex]=\dfrac{47207-34393}{2}=6407[/tex]
Also, margin of error = [tex]z\times\dfrac{\sigma}{\sqrt{n}}[/tex], where z= critical z-value for confidence level and n is the sample size.
z-value for 99% confidence level = 2.576
So,
[tex]6407=2.576\times\dfrac{14900}{\sqrt{n}}\\\\\Rightarrow\ \sqrt{n}=2.576\times\dfrac{14900}{6407}=5.99\\\\\Rightarrow\ n=(5.99)^2=35.8801\approx 36[/tex]
The sample size used for the analysis =36
