ANOVA Testing

Essay by password30 April 2009

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University of PhoenixResearch and Evaluation The Army would like to relocate me. My choices are Georgia, Arizona and California. I will determine the most cost efficient place to live. Since transportation is a necessity in today’s world I thought that I would start with the prices of gas. I would like to know if the mean gas prices are equal in all three states.

I will perform the Analysis of Variance (ANOVA) test. The reason I chose this particular test is because the data met all three requirements to perform the ANOVA. The requirements met were: “1. the populations follow the normal distribution; 2. the population have equal standard deviations; 3. the populations are independent” (Lind, Marchal, & Wathen, 2004, P. 392).

Next I gathered 25 samples of gas prices from each state. I am using .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 one mean is different.

The critical value is 2.03; therefore we can conclude that if the F calculation is greater than 2.03 the null is rejected.

The data analysis tool in excel was used for the ANOVA and the results are below:Anova:Single FactorSUMMARYGroupsCountSumAverageVarianceArizona2549.591.98360.009074Georgia2546.261.85040.003971California2556.232.24920.071949ANOVASource of VariationSSdfMSFP-valueF critBetween Groups2.06105921.03052936.374191.21E-113.123907Within Groups2.039856720.028331Total4.10091574From 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 means are equal.

Now to determine the best place to live since the means are not equal I determine which of the means were different. To figure this out I performed another test using Mega stat, shown below.

Post hoc analysisTukey simultaneous comparison t-values (d.f. = 72)GeorgiaArizonaCalifornia1.850...