Given $\displaystyle \alpha={0.1}$,  $\displaystyle {\chi_{{{1}-\frac{\alpha}{{2}}}}^{{2}}}={\chi_{{0.95}}^{{2}}}={16.151}$, $\displaystyle {{s}_{{i}}^{{2}}}={44.89}$, and $\displaystyle {n}={28}$ compute the upper bound for a 90% confidence interval using the formula

 $\displaystyle {U}{B}={\frac{{{{s}_{{i}}^{{2}}}\cdot{\left({n}-{1}\right)}}}{{{\chi_{{{1}-\frac{\alpha}{{2}}}}^{{2}}}}}}=$ (Round the answer to 2 decimal places.)