(iii)State the correct conclusion for this hypothesis test ("There is sufficient evidence at the a = 0.05 significance level to support..." or "There is NOT sufficient evidence at the a = 0.05 significance level to support..."). Make sure you state the conclusion in terms of Hi. (Points: 5) . Reject HO, F statistic does not fall into perimeters of F critical. (b) If the null hypothesis is rejected in part (a), use Tukey's test to determine which pairwise means differ using a familywise error rate of a = 0.05. (1) Write down the sample means in ascending order. (Points: 5) . 11.908 = X1 . 12.908 = X3 . 18.248 = X2 (2) Compute the pairwise differences for each pair of sample means. (Points: 5) . X2 - X1 = 18.248 - 11.908 = 6.34 X2 - X3 = 18.248 - 12.908 = 5.34 * X3 - X1 = 12.908 - 11.908 = 1 (3) Compute the Tukey's test statistic for each pairwise difference for each pair of sample means. (Points: 5) . X2 - X1 = 6.34 q= x2 - x1 / VMSE/2 (1; + 1j) q=6.34 / V6.7983 /2 ( 1/5 + 1/5) = 6.34 / 1.8436 (0.4) =6.34 /0.7374 = 8.5977 * X2 - X3 = 5.34 q= X2- X3/ VMSE/2 (1/mi + 1j) q=5.34 / V 6.7983 /2 ( 1/5 + 1/5) =5.34 / 0.7374 =7.2416 * X3 - X1 = 1 q= x3 - x1 / VMSE/2 (1i + 1j) q=1/ V 6.7983 /2 ( 1/5 + 1/5) =1/0.7374 = 1.3561 (4) Determine the critical value. (Points: 5) qa,v.k = q0.03,12,3 = 1.800 v =15 - 3 = 12 Critical = 3.773 Page 3 of 5(5) For each pair of sample means, state whether you reject or fail to reject the null hypothesis. Make sure you clearly state for which pair you reject or fail to reject the null hypothesis. (Points: 5) q= 8.5977 > 3.773, Reject HO 4=7.2416 > 3.773, Reject HO q= 1.3561