Solution
For this case we cna use the following formula for the confidence interval:
The degrees of freedom are: df = 165-1= 164
For this case the significance level is 1-0.99 = 0.01 and the value of alpha/2 = 0.005 then the critical value is:
t =2.606
Replacing we have:
Solving we got:
And after round to one decimal we got:
27.3 <= mu <= 29.9