Answer: [tex](16.926\, 19.074)[/tex]
Step-by-step explanation:
Given : Level of significance : [tex]1-\alpha:0.95[/tex]
Then , significance level : [tex]\alpha: 1-0.95=0.05[/tex]
Since , sample size : [tex]n=23[/tex] , which is small sample (n<30) so the test applied here is a t-test.
Sample mean : [tex]\overline{x}=18[/tex]
Standard deviation: [tex]\sigma: 3[/tex]
Degree of freedom= [tex]n-1=23-1=22[/tex]
Using t-distribution table , Critical value : [tex]\text{t-score}=t_{n-1, \alpha/2}=1.717144[/tex]
The confidence interval for population mean is given by :-
[tex]\overline{x}\pm t_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
[tex]18\pm(1.717144)\dfrac{3}{\sqrt{23}}\approx18\pm 1.074=(16.926\, 19.074)[/tex]
Hence, a 90% confidence interval for the population mean=[tex](16.926\, 19.074)[/tex]
