Answer: [tex](18.17,\ 20.03)[/tex]
Step-by-step explanation:
Given : Sample size : [tex]n=9[/tex] , which is a small sample (<30), so we use t-test.
Sample mean : [tex]\overline{x}=19.1\text{ years}[/tex]
Standard deviation : [tex]\sigma =1.5\text{ years}[/tex]
Significance level : [tex]\alpha=1-0.99=0.01[/tex]
Critical value : [tex]t_{n-1, \alpha/2}=t_{8, 0.005}=1.86[/tex]
The formula to find the confidence interval for population mean :-
[tex]\overline{x}\pm t_{n-1,\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=19.1\pm(1.86)\dfrac{1.5}{\sqrt{9}}\\\\\approx19.1\pm0.93\\\\=(19.1-0.93,\ 19.1+0.93)\\\\=(18.17,\ 20.03)[/tex]
Hence,the 99% confidence interval for the population mean = [tex](18.17,\ 20.03)[/tex]
