Construct a 95 confidence interval for the population mean

Construct a 95% confidence interval for the population mean, ?. Assume the population has a normal distribution. A sample of 15 college students had mean annual earnings of $3120 with a standard deviation of $677. (Round the answer to the whole dollar.)
What is the lower limit of the interval?
$


What is the upper limit of the interval?
$

Solution

The degree of freedom =n-1=15-1=14

Given a=1-0.95=0.05, t(0.025, df=14) =2.14 (from student t table)

So the lower limit of the interval is

xbar- t*s/vn =3120 -2.14*677/sqrt(15) =2746

So the upper limit of the interval is

xbar+ t*s/vn =3120 +2.14*677/sqrt(15) =3494