Answer: [tex](628.48,\ 661.52)[/tex]
Step-by-step explanation:
Given : Sample size : [tex]n=16[/tex] , which is a small sample (, 30) so we use t-test.
Sample mean : [tex]\overline{x}=645 \text{ hours}[/tex]
Standard deviation : [tex]\sigma = 31 \text{ hours}[/tex]
Significance level : [tex]\alpha=1-0.95=0.05[/tex]
Critical value : [tex]t_{n-1,\alpha/2}=2.131[/tex]
The confidence interval for population mean is given by :-
[tex]\overline{x}\pm t_{n-1,\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=645\pm(2.131)\dfrac{31}{\sqrt{16}}\\\\\approx645\pm16.52\\\\=(645-16.52,\ 645+16.52)\\\\=(628.48,\ 661.52)[/tex]
Hence, a 95% confidence interval for the population mean [tex]\mu[/tex] = [tex](628.48,\ 661.52)[/tex]
