Sample Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 Lincoln Omaha 42 40 39 39 38 39 40 39 42 37 39 38 38 37 37 38 40 37 41 37 38 40 38 39 41 40 40 38 40 39 40 38 40 39 39 41 42 39 40 40 42 37 38 37 44 43 39 45 42 42 40 37 39 38 41 39 37 45 40 37 38 41 45 42 38 44 39 43 40 37 37 40 44 41 41 40 44 39 44 37 40 44 38 44 45 40 42 37 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 38 40 38 40 41 40 42 42 39 39 38 42 41 37 40 38 42 38 42 42 37 39 37 39 39 42 38 41 37 39 39 40 41 40 39 39 41 40 38 37 38 40 37 41 37 42 40 40 45 38 39 37 38 37 43 39 39 45 42 44 42 43 38 39 41 45 41 38 39 41 37 37 43 43 43 43 41 41 41 40 40 38 38 39 40 37 44 37 40 42 90 91 92 93 94 95 96 97 98 99 100 42 41 39 42 40 39 39 42 42 38 41 45 38 41 40 42 37 40 45 38 40 42 Question Here are 100 samples of the public officials work for the City of Lincoln and Omaha government. This data presents their working hours per week. Please test the hypothesis that the average weekly working hours of all 450 public officials work for Omaha government and al Size Lincoln Omaha Mean 100 100 Standard Deviation 39.41 1.6086571349 40.59 2.5469133626 Answer using 8 steps on the textbook on the texbook page 202-206 1. H0: 1=2, The average weekly working hours of all public officials work for Omaha Gov. is equal to the working hours of tho H1: X1-bar X2-bar, The average weekly working hours of sample public officials work for Omaha Gov is not equal to the w 2. Set the level of risk at the .05 as instructed 3. select the appropriate test statistics: the t-test for Independent samples (samples from 2 independent groups) by using two4. Compute the test statisitic value: the week 7 lecture note ppt page 14 shows that you need to know the mean, number of pa 5. Critical Value at 5% and two-tailed test with degree of freedom of 198 from the t.2 table on the textbook page 370-372: 1.9 6. obtained t-value (-3.917) is more extreme than the critical value (1.96) 7. Therefore, we reject the null hypothesis and conclude that working hours of Lincoln's public officials work less amount of tim using '=T.TEST' function on the lecture PPT =T.Test (array 1, array 2, tails, type) ==> =T.TEST(B2:B101,C2:C101,2,2) 0.000123304 means that there is only .01% chance that the null hypothesis is true. 0.000123304 t-Test: Two-Sample Assuming Equal Variances Mean Variance Observations Pooled Variance Hypothesized Mean Difference df t Stat Variable 1 Variable 2 39.41 40.59 2.5877777778 6.4867676768 100 100 4.5372727273 0 198 -3.91714427 P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail 0.000061652 1.6525857836 0.000123304 1.9720174778 or Omaha government and all 450 public officaials work for Lincoln government is equal. (Test at the .05 level) Variance 2.5877777778 6.4867676768 Degree of Freedom nominator Denominator 1 Denominator 2 Denominator Final t-value 198 -1.18 4.537273 0.02 0.30124 -3.917144 age 202-206 al to the working hours of those work for Lincoln Gov. aha Gov is not equal to the working hours of those sample public officials work for Lincoln Gov. endent groups) by using two-tailed test (Non-directional approach) know the mean, number of participants, and variance. Then, calculate the obtained value. It is -.03917 e textbook page 370-372: 1.96 (df = infinite, 0.05 for two-tailed) fficials work less amount of time than those who work for the Omaha Gov. l) Sample Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 Before After 42 43 39 41 41 40 42 42 39 40 40 42 41 39 41 42 39 42 41 42 42 41 43 43 42 40 39 43 41 43 42 41 42 43 43 43 41 40 42 41 41 41 43 41 43 39 43 D 42 41 42 42 41 43 42 40 42 42 43 40 38 40 38 42 42 40 40 41 38 40 43 39 43 42 39 38 41 38 39 42 40 38 38 42 38 38 39 40 41 40 42 43 42 40 43 D^2 0 -2 3 1 0 3 0 -2 3 2 3 -2 -3 1 -3 0 3 -2 -1 -1 -4 -1 0 -4 1 2 0 -5 0 -5 -3 1 -2 -5 -5 -1 -3 -2 -3 -1 0 -1 -1 2 -1 1 0 0 4 9 1 0 9 0 4 9 4 9 4 9 1 9 0 9 4 1 1 16 1 0 16 1 4 0 25 0 25 9 1 4 25 25 1 9 4 9 1 0 1 1 4 1 1 0 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 43 39 41 39 42 40 41 41 42 40 40 40 41 40 41 41 39 42 39 41 39 39 39 39 42 43 42 43 41 43 43 40 42 42 39 43 41 42 41 40 41 42 39 39 42 43 40 43 40 41 39 42 39 38 41 43 40 41 42 39 40 42 38 38 41 41 40 39 39 38 38 41 40 42 41 38 38 39 38 40 39 42 40 40 43 40 41 40 41 42 43 42 43 43 39 38 -3 2 -2 3 -3 -2 0 2 -2 1 2 -1 -1 2 -3 -3 2 -1 1 -2 0 -1 -1 2 -2 -1 -1 -5 -3 -4 -5 0 -3 0 1 -3 2 -2 0 0 0 0 4 3 1 0 -1 -5 9 4 4 9 9 4 0 4 4 1 4 1 1 4 9 9 4 1 1 4 0 1 1 4 4 1 1 25 9 16 25 0 9 0 1 9 4 4 0 0 0 0 16 9 1 0 1 25 96 97 98 99 100 41 41 42 41 42 39 42 41 42 39 -2 1 -1 1 -3 4 1 1 1 9 Question Mayor received the report and surprized the average working hour of government workers are too long. Therefore, she implemented 'working less policy' that allows public officials work less. This data presents the working hour working hours of all 450 public officials work for Omaha government get decreased after the 'working less policy' is implem Size Before After 100 100 D D^2 Answer using 8 steps on the textbook 1. H0: _before=_after, The average weekly working hours of all public officials work after the policy implementation is eq H1: X_after < X_before, The average weekly working hours of sample public officials work after the policy implementatio 2. Set the level of risk at the .05 as instructed 3. select the appropriate test statistics: the t-test for dependent samples (same samples for pre-test and post-test) by using 4. Compute the test statisitic value: the week 7 lecture note ppt page 29 shows that you need to know the differences, diff 5. Critical Value at 5% and one-tailed test with degree of freedom of 99 from the t.2 table on the textbook page 370-372: 1 than the X_before, the critical value is "-1.66" 6. obtained t-value (-3.917) is more extreme than the critical value (-1.66) 7. Therefore, we reject the null hypothesis and conclude that working hours after the policy implementation is smaller than using '=T.TEST' function on the lecture PPT =T.Test (array 1, array 2, tails, type) ==> =T.TEST(C2:C101,B2:B101,1,1) The Post-test should come to the array 1. 0.000681643 means that there is only .07% chance that the null hypothesis is true. 0.000681643 t-Test: Paired Two Sample for Means Mean Variance Observations Pearson Correlation Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail Variable 1 Variable 2 40.42 41.15 2.7309090909 1.805555556 100 100 -0.083244749 0 99 -3.29573235 0.000681643 1.660391156 0.0013632859 t Critical two-tail 1.9842169516 workers are too long. ess. This data presents the working hours per week before and after the policy. Please test the hypothesis that the average weekly d after the 'working less policy' is implemented. (Test at the .05 level) -73 539 degree of feeedom nominator denominator 1 t 99 -73 22.14986 -3.295732 ork after the policy implementation is equal to the working hours of those before the implementation. cials work after the policy implementation is smaller than the working hours of those before the implementation. mples for pre-test and post-test) by using one-tailed test (directional approach) t you need to know the differences, differences squared, and number of cases. Then, calculate the obtained value. It is -3.29573. 2 table on the textbook page 370-372: 1.66 (df = 99, thus check the df of 100, 0.05 for one-tailed). As it is testing the X_after is smaller he policy implementation is smaller than the working hours before. is that the average weekly entation. ined value. It is -3.29573. is testing the X_after is smaller