Answer
$(114.99784,434.56216)$
Work Step by Step
The degree of freedom is the sample size minus $2$: $df=n-2=12-2=10$.
The critical value is given in the row $df=10$ and in the column with $c=0.9$ of the table. Thus $t=1.812$.
The confidence interval is then: $b\pm tSE_b=274.78\pm1.812\cdot88.18$, thus the confidence interval is $(114.99784,434.56216)$
