Chapter 9 - Review - Review Exercises - Page 470: 1a

$t_{\frac{α}{2}}=t_{0.005}=2.898$

$n=18$, so: $d.f.=n-1=17$ $level~of~confidence=(1-α).100$% $99$% $=(1-α).100$% $0.99=1-α$ $α=0.01$ $t_{\frac{α}{2}}=t_{0.005}=2.898$ (According to Table VI, for d.f. = 17 and area in right tail = 0.005)