Solution
For this case we cna use the following formula for the confidence interval:
![\operatorname{mean}\pm t\cdot\frac{s}{\sqrt[]{n}}]()
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:
![28.6\pm2.606\cdot\frac{6.3}{\sqrt[]{165}}](https://img.qammunity.org/2023/formulas/mathematics/college/4wgdopirgmd1ej39gmncum77wb0xdjyx5k.png)
Solving we got:

And after round to one decimal we got:
27.3 <= mu <= 29.9