A random sample of 9 observations from a normally distribute

A random sample of 9 observations from a normally distributed population produced a meant x = 12.4 and standard deviation s = 2.8. Find a 99% confidence interval for the population mean mu (round off to two decimal places). (a) (11.36, 13.44) (b) (9.77, 15.03), (c) (9.70, 15.10) (d) (9.27, 15.53) (e) (9.37, 15.43)

Solution

The degree of freedom =n-1=9-1=8

Given a=1-0.99=0.01, t(0.005, df=8) =3.355 (from student t table)

So the lower bound is

xbar -t*s/vn= 12.4 -3.355*2.8/sqrt(9 ) =9.27

So the upper bound is

xbar +t*s/vn=12.4 +3.355*2.8/sqrt(9 ) =15.53

Answer: (d) (9.27, 15.53)