Answer: (1.375, 18.625 )
Step-by-step explanation:
The confidence interval for population mean is given by :-
[tex]\overline{x}\pm t^*SE[/tex] (1)
, where [tex]\overline{x}[/tex] = sample mean
t* = critical value.
SE = standard error
As per given , we have
n= 21
Degree of freedom : df = n-1=21-1=20
[tex]\overline{x}=10[/tex]
Significance level : [tex]\alpha=1-0.90=0.10[/tex]
Standard error : SE= 5
The critical t-value for confidence interval is a two-tailed value with respect to the given degree of freedom i.e. [tex]t^*=t_{(\alpha/2,df)}[/tex] .
Using student's t-distribution table ,
[tex]t_{(\alpha/2,df)}=t_{(0.10/2, 20)}=t_{(0.05, 20)}=\pm1.725[/tex]
The 90% confidence interval will be :
[tex]10\pm (1.725) (5)[/tex] [by formula in (1)]
[tex]10\pm 8.625[/tex]
[tex]\approx(10- 8.625,\ 10+ 8.625)=(1.375,\ 18.625 )[/tex]
Hence, the limits of the confidence interval=(1.375, 18.625 )
