The “Account Balances.xls” data set has information on the account balances of customers at a bank’s four locations. We want to test the null hypothesis that the mean account balances are equal in the four towns. Part (a) Using an α of 0.05, carry out a one-way ANOVA using Excel. You must submit your actual Excel file with the output as part of the assignment. Part (b) Do you reject the null hypothesis or not? Indicate on which part of the Excel output you base your decision. Part (c) Assuming that an acquaintance of yours has never heard of ANOVA, explain to her what the outcome in part (b) means in non-statistical terms.

(a) I do not have a current copy of Excel. The following are instructions that should get the Anova output:
(1) Enter the data in columns A,B,C, and D.(2) On the toolbar select Data -> Data analysis(3) Select Anova -> single factor(4) Enter the range A4:D23 (as it appears in your spreadsheet)(5) Select grouped by columns(6) Select or enter 0.05 for the alpha level(7) Check output range and type in E2
You should get the following information:
F=2.352657462p=0.0792776228Factor:df=3SS=2551767.07MS=850589.022Errordf=72SS=26031162.9MS=361543.929SXP=601.285231
(b) Do not reject the null hypothesis. p > alpha . (.07>.05; the probability that you could have a group such as this by chance is greater than your level of confidence.)
Compare p to the alpha level. When p <= alpha we reject the null hypothesis. 
(c) Our basic assumption is that the average values for each city (means) are the same. We assume that the spread of the data is the same for each city (variances.) The Anova test returns the probability that we could get samples from a group of cities that looks like the samples we got if the averages really were the same.
If that probability (p) is low  (lower than some expectation that we had before running the test) then we conclude the averages of at least two of the cities must be different (reject the null hypothesis.)
http://mathworld.wolfram.com/ANOVA.html