## Statistics (12th Edition)

df = n - 1 = 10 - 1 = 9 $t_{α/2} = 0.05$ From student's t distribution table, we have: $t_{0.05} = 1.833$ $E = t. \frac{s_{d}}{n} = 1.833 . \frac{2.2136}{\sqrt 10}$ $= 1.833 . 0.7 = 1.283$1$Confidence interval will be:$\bar d - E = -3.7 - 1.2831 = -4.9831\bar d - E = -3.7 + 1.2831 = -2.4169\$