Answer: (145820, 170180)
Step-by-step explanation:
Formula for confidence interval for population mean ( when population standard deviation is unknown ) :
[tex]\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}[/tex]
, where n= sample size
[tex]\overline{x}[/tex] = sample mean
s= sample standard deviation
[tex]t_{\alpha/2}[/tex]= two-tailed t-value for significance level of ([tex]\alpha[/tex]).
We assume that the salary of CFA charterholders is normally distributed .
Let x represents the salary of CFA charterholders .
As per given , we have
n= 36
[tex]\overline{x}=\$158,000[/tex]
Degree of freedom : df = 35 [ df= n-1]
s= $36,000
Significance level : [tex]\alpha=1-0.95=0.05[/tex]
Using t-distribution table ,
[tex]t_{\alpha/2, df}=t_{0.025,\ 35}=2.030[/tex]
95% confidence interval for the average salary of a CFA charterholders :-
[tex]158000\pm (2.030)\dfrac{36000}{\sqrt{36}}\\\\=158000\pm 12180\\\\=(158000-12180,158000+12180) =(145820,\ 170180)[/tex]
Hence, the 95% confidence interval for the average salary of a CFA charterholder : (145820, 170180)