5. A researcher wanted to know if individuals have better reaction times depending on when the test is taken -- in the morning or in the afternoon. He randomly assigned 16 participants to take the same reaction time test in either the morning or the afternoon. The following results were obtained, where lower scores indicate faster reaction time.Assume that the parametric test assumptions ARE violated. Conduct an appropriate statistical test using an alpha of .05 and the eight steps of hypothesis testing.Note: more information given in table than is needed| Individual | Morning (X1) | R1 | X1² | Individual | Afternoon (X2) | R2 | X2² | D | Ranks | D² | +- ||------------|--------------|-----|------|------------|----------------|-----|------|----|-------|-----|-----|| 1 | 79 | | 6241 | 9 | 91 | | 8281 | -12| | 144 | || 2 | 88 | | 7744 | 10 | 92 | | 8464 | -4 | | 16 | || 3 | 95 | | 9025 | 11 | 87 | | 7569 | 8 | | 64 | || 4 | 78 | | 6084 | 12 | 87 | | 7569 | -9 | | 81 | || 5 | 81 | | 6561 | 13 | 94 | | 8836 | -13| | 169 | || 6 | 83 | | 6889 | 14 | 86 | | 7396 | -3 | | 9 | || 7 | 80 | | 6400 | 15 | 82 | | 6724 | -2 | | 4 | || 8 | 84 | | 7056 | 16 | 93 | | 8649 | -9 | | 81 | || **Total

Given information: Individual Morning Individual Afternoon 1 79 9 91 2 88 10 92 3 95 11 87 4 78 12 87 5 81 13 94 6 83 14 86 7 80 15 82 8 84 16 93 Since the parametric assumptions are violated, hence we will conduct Mann-Whitney U test for two samples. Null hypothesis: H0: The reaction time for individuals is same for morning time and afternoon time. Alternate hypothesis: HA: The reaction time for individuals is different for morning time and afternoon time. The level of significance is 0.05. Use the given data to form the excel table: Individual Morning Rank for Morning Individual Afternoon Rank for Afternoon 1 79 2 9 91 12 2 88 11 10 92 13 3 95 16 11 87 9.5 4 78 1 12 87 9.5 5 81 4 13 94 15 6 83 6 14 86 8 7 80 3 15 82 5 8 84 7 16 93 14 Sum 668 50 712 86 Mean 83.5 6.25 89 10.75 Determine the test statistics using the table. R1=50R2=86 U1=n1·n2+n1n1+12-R1=8·8+8(8+1)2-50=64+36-50=50U2=n1·n2+n2n2+12-R2=8·8+8(8+1)2-86=64+36-86=14U=Minimum of U1 and U2=U1=14 μR=n1·n22 =8×82=32σU=n1n2n1+n2+112 =8·88+8+112=108812=9.521904571≈9.52z=U-μRσR=14-329.52=-1.89 Since we have a two tailed test, the P-value is the probability that the z-score is less than -1.89 or greater than 1.89. P-value = P(z < - 1.89) + P(z > 1.89) P-value = 2*P(z <- 1.89) Use excel formula, "=NORM.S.DIST(-1.89,TRUE)" to determine the P-value. P-value = 2(0.0294) Thus, the P-value = 0.0588 Interpret results. Since the P-value (0.0588) is greater than the significance level (0.05), hence we failed to reject the null hypothesis. From the above test we do not have sufficient evidence in the favor of the claim that the reaction time for individuals is different for morning time and afternoon time.