Answer: (188.1384, 191.8616 )
Step-by-step explanation:
When population standard deviation is not known, then the confidence interval for population mean is given by :-
[tex]\overline{x}\pm t^* \dfrac{s}{\sqrt{n}}[/tex] (1)
, where n= sample size.
[tex]\overline{x}[/tex] = sample mean
t* = critical value.
s = sample standard deviation.
As per given , we have
n= 1000
Degree of freedom : df = n-1=999
[tex]\overline{x}=190[/tex]
s= 30
Significance level : [tex]\alpha=1-0.95=0.05[/tex]
Using t-distribution table , the critical value = [tex]t^*=t_{(\alpha/2,df)}=t_{(0.025, 999)}=1.9623[/tex]
The 95% confidence interval will be :
[tex]190\pm (1.9623) \dfrac{30}{\sqrt{1000}}[/tex]
[tex]190\pm (1.9623) (0.948683298051)[/tex]
[tex]190\pm 1.86160123577[/tex]
[tex]\approx(190- 1.8616,\ 190+ 1.8616)=(188.1384,\ 191.8616 )[/tex]
Hence, the required confidence interval = (188.1384, 191.8616 )
