First, find the degrees of freedom.
[tex]df=n-1=53-1=52[/tex]
The t-value of 52 degrees of freedom is 2.01.
Then, we use the following formula.
[tex]\bar{x}\pm t\cdot\frac{s}{\sqrt[]{n}}=71.6\pm2.01\cdot\frac{16.6}{\sqrt[]{53}}=71.6\pm4.6[/tex]
Therefore, the confidence interval is
[tex]67\leq\mu\leq76.2[/tex]
