Question

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.

Please show all steps of your analysis. Use excel to get your results. However, use the data and tables from excel and paste them in this word file.

Answer 1

Null Hypothesis:The mean accident rate is same for all the three schools

Alternative hypothesis:Atleast one mean differs from the others

Calculation

G= 172.94

CF = G²/n=29908.24/30=996.94

TSS= = 1227.16-996.84=230.22

SST = = 1/10(10233.16 )- 996.94=26.38

SSE=TSS-SST=230.22-26.38=203.84

ANOVA Table

Source	df	SS	MS	F
Treatment	2	26.38	13.19	1.75
Error	27	203.84	7.55
Total	29	230.22

The critical value is F_2,27,.05=3.35

Since the calculated value is lesser than the critical value ,we accept the null hypothesis and conclude that mean accident rate is same in all 3 schools which implies that the program doesn't have effect