Answer: (1812.967, 1907.033)
Step-by-step explanation:
The confidence interval for population mean is given by :-
[tex]\overline{x}\pm t_{df,\ \alpha} \dfrac{s}{\sqrt{n}}[/tex]
, where [tex]\overline{x}[/tex] = Sample mean
n= Sample size.
s = Sample standard deviation
[tex]t_{df,\ \alpha}[/tex] = Critical t-value for degree of freedom(n-1).
As per given , we have
n= 35
[tex]\overline{x}=1860[/tex]
s=102
Significance level : [tex]\alpha=0.01[/tex]
By t- distribution table , for degree of freedom 43 and [tex]\alpha=0.01[/tex] , we have
[tex]t_{df,\ \alpha}=t_{34,\ 0.005} =2.728[/tex]
Substitute all the values in the above formula , we will get
[tex]1860\pm (2.728)\dfrac{102}{\sqrt{35}}[/tex]
[tex]=1860\pm (2.728)(17.2411)[/tex]
[tex]=1860\pm47.034[/tex]
[tex]=(1860-47.033\ 1860+47.033)= (1812.967,\ 1907.033)[/tex]
Hence, the 99% confidence interval estimate of the mean of the population is (1812.967, 1907.033).
