Computation of test statistic and p-value. Use the formula t = (X1 - X2)/VI(s,2,) + (s222)]. t = (X1 - x2)/VI(s,21) + ($222)] t = (7.52 - 8.92)/V[(1.68/25) + (0.99/25)] t = -1.4/V0.1068 t = -4.28 The p-value is twice the area to the left of t = -4.28 since it is a two-tailed test. According to the t-table, the area below or to the left of t = -4.28 with df = 45 is 0.00005. Multiply 0.00004 by 2 to find the p-value: 2 * 0.00005 = 0.0001. Hence, p-value = 0.0001. can someone please elaborate how do they denived with this p- value? I tried using + table but giving me a diggerent ausue Decision If p-value > a => fail to reject the null hypothesis If p-value reject the null hypothesis Since the p-value is 0.0001 and a is 0.05, then p-value