## Statistics (12th Edition)

First we need to calculate difference of population 1 and population 2: 19 - 24 = -5 25 - 27 = -2 31 - 36 = -5 52 - 53 = -1 49 - 55 = -6 34 - 34 = 0 59 - 66 = -7 47 - 51 = -4 17 - 20 = -3 51 - 55 = -4 $\bar x_{d} = \frac{-5 + .... + (-4)}{10} = -3.7$ $s^{2}_{d} = \frac{(-5 - (-3.7))^2 + .... + (-4 - (-3.7))^{2}}{10 - 1} = 2.2136$ $t = \frac{-3.7}{2.2136/\sqrt 10} = \frac{-3.7}{0.7} = 0.7 = -5.2857$ df = n - 1 = 10 - 1 = 9 $t_{α} = 0.10$ From student's t distribution table, we have: $t_{0.10} = 1.383$ Since -5.2857 < 1.383, we reject Null hypothesis