Question

1. ANOVA is a useful tool in performance comparisons. A nonprofit foundation funds a program to reduce accidents among school-aged children. The employees in the program conduct regular education workshops for students and teachers. The daily number of accidents per 1000 students is used to assess the performance of the program in reducing accidents among students. The program is operated in school A, which collected the following data on 10 randomly selected days: 3.82, 1.01, 5.96, 8.99, 8.85, 5.54, 0.14, 4.07, 7.65, and 1.39.

You are a performance analyst whose job it is to assess the performance of this program in school A. To conduct your evaluation, you collect data from two schools (School B and School C) that do not have these programs (that is, the control groups). The following data are for school B: 6.58, 0.59, 7.00, 7.17, 7.83, 8.68, 1.17, 5.50, 6.84, and 4.09. For School C the data are as follows: 9.47, 4.86, 8.65, 5.54, 6.45, 2.71, 8.63, 9.30, 7.37, and 7.09. Assume normal distributions of the data in each group and determine whether the program in school A has a lower number of accidents.

Answer 1

Answer ANOVA null hypothesis: - Thee is no significant difference between the means, ie, 4,= 42=43 Alternative hytotheis! - There is a significant difference between at least one of the means 4. & um m 2,3 . Descriptive! - School A :- n, - 10 TA = {xi - 4.74 20 SA - 3.217 School B!- n2 = 10 YB = 5:54 50 58 = 186 2.75615 Schcoe @ : - 03:10 Xe : 7.0070 SI = 2.16678 ì = XA+ AB+ XC A 15.76467

Sum of squares :- between groups : n C KA-7)? + ,678 - 70+ - nocte fj? :10[C417420-5176467) + [5.5480-57846 +(70070 -5.76467)" ] 26. 375 mean square = 26.375 K-1 K:3 -> 26.375 13187 within gensup = {cxij. FjJ² Total : & cxij" - 51764075? - 230.2160 Thus, within group · Tetap between group = 230.2160 - 20.375 • 203.841

ArvovA TABLE Source of variation Sum of Squares 2018 dif meansquare F 11947 Sig 193 131197 Betweengroups 261375 fisso within groups 2031841 a 7 29 Total 330.2160 Ag bualue C-193) > dCOos) we fail to grejed the nuse thesis. Thus there is no Sigrupicant dieperence between the meang' Post HOC : - vull hypothesis: There is a no significant difference he щ, 54а е 4, 3 ча Alternative hypothesis: There is a significant difcerence betwe en el sur le 4 furt Similarly 100 hy to thew's Thee is no significare din cenne beruru 41 9 4 3 ne l - 431 Alternative hytothesis. There is a significant dicgo.core between eleng "e ulug'

95% Confidence Interval (J) Mean Difference (I- VARO 0001 1 VARO 0001 Tukey HSD .7817 Std. Error 1.22879 1.22879 1.22879 1.22879 | 1.22879 | 1.22879 1.22879 1.22879 1.22879 1.22879 | 1.22879 1.22879 Sig. 792 .175 .792 .469 .175 469 .519 .076 .519 .244 .076 .244 - 80300 -2.26500 .80300 -1.46200 2.26500 1.46200 -80300 -2.26500 .80300 -1.46200 2.26500 1.46200 Lower Bound | Upper Bound -3.8497 2.2437 -5.3117 -2.2437 3.8497 -4.5087 1.5847 --7817 5.3117 -1.5847 4.5087 -3.3243 1.7183 -4.7863 .2563 -1.7183 3.3243 -3.9833 1.0593 - 2563 4.7863 -1.0593 3.9833 LSD 1

RESULT As pualue in both Tukey HSD and uso are greater than dc0.05) lie we fail to reject the rule hypothesis Thus, we cannot determine whether the brogeam in Schocol A has a lower number of accidents