ANOVA Testing

University of PhoenixResearch and Evaluation The Army would like to relocate me. My choices ar Georgia, azimuth and California. I result determine the most cost sparing domicile to live. Since transportation is a necessity in without delay?s world I thought that I would pop attain with the prices of gas. I would like to know if the opine gas prices argon equal in all(prenominal) three states. I will act the Analysis of Variance (ANOVA) sieve. The reason I chose this extra running game is because the data met all three requirements to perform the ANOVA. The requirements met were: ?1. the populations pass off the normal distribution; 2. the population have equal standardised deviations; 3. the populations atomic number 18 independent? (Lind, Marchal, & Wathen, 2004, P. 392). Next I induce to quiverher 25 samples of gas prices from each state. I am use .05 as the level of significance. The data collected is as follows:ArizonaGeorgiaCalifornia$1.65$1.79$1.99$ 1.89$1.82$2.03$1.89$1.83$2.04$1.91$1.84$2.05$1.92$1.85$2.09$1.93$1.86$2.12$1.94$1.87$2.15$1.95$1.87$2.49$1.95$1.87$2.85$1.97$1.87$2.39$1.97$1.81$2.03$1.98$1.85$2.03$1.98$1.85$2.03$2.02$1.76$2.04$2.02$1.76$2.79$2.02$1.77$2.71$2.03$1.79$2.49$2.03$1.86$2.05$2.05$1.87$2.55$2.06$1.87$2.49$2.07$1.88$2.03$2.07$1.89$2.13$2.09$1.82$2.15$2.09$1.94$2.14$2.11$2.07$2.37Using the above data it is determined thatH sub 0 is µA = µI = µCH sub 1 is at least atomic number 53 mean is different. The critical value is 2.03; therefore we can pause that if the F calculation is greater than 2.03 the null is rejected. The data epitome gumshoe in excel was used for the ANOVA and the results atomic number 18 at a lower place:Anova: star FactorSUMMARYGroupsCountSumAverageVarianceArizona2549.591.98360.009074Georgia2546.261.85040.003971California2556.232.24920.071949ANOVASource of VariationSSdfMSFP-valueF critBetween Groups2.06105921.03052936.374191.21E-113.123907Within Groups2.039856720.028331 fit4.100 91574From this I know that the SST is 2.06; ! the SSE is 2.03 and the SS Total is 4.10. This also shows that F = 36.37; therefore, the null is rejected. Their for I cannot conclude that all the gist are equal. Now to determine the best place to live since the center are not equal I determine which of the means were different. To figure this out I performed some other test using Mega stat, shown below. Post hoc analysisTukey simultaneous comparison t-values (d.f. = 72)GeorgiaArizonaCalifornia1.850... If you want to get a full essay, order it on our website: OrderCustomPaper.com

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