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For our example the hypotheses are: <br />:' Ho: µ1-µ2=I~3=µ4=l+5=F~6 <br />He: the mean cover values are not equal <br />The summary of the calculations for the ANOV are presented in the following table. <br />Source of Variation Sum of ueres of Freedom Mean uare <br />Total 2006.00 21 <br />Groups 892.67 4 223.17 <br />Error 1113.33 17 65.49 <br />The test for equality of means in ANOV is the variance ration test F: <br />F= group Mean Square/error Mean Square <br />F= 223.17!65.49 = 3.41 <br />The critical value of the F statistic for 9596 confidence is: <br />Fo.osll>,a,17= 2.96 <br />Therefore, reject Ho: µ1=µ2=µ3=µ4=µ5=µ6 <br />The sample means are not equal, therefore they do not come fmm the same population, <br />and the population is not homogenous with the addition of the Northern M 1 East Area <br />samples. <br />CONCLUSIONS <br />Based on the statistical evaluations using Bartlett's Test for homogeneity of variances <br />and Analysis of Variance (homogeneity of means), the Northern !/ I Mine Area- West <br />Side, Northern 1i L Mine- Topsoil Stockpile Area, Refuse Area- Topsoil Stockpile <br />Area, Refuse Area, Rienau A' 2 Mine Area samples were found to be homogenous and <br />from the same population. The combination of the cover data from these areas for the <br />purposes of evaluation of vegetation cover success is appropriate. The Northern /t I <br />East Area was determined to have equivalent sample variance to the other areas, but a <br />significantly different sample mean. It was appropriate to exclude this area from <br />consideration for Phase II bond release. <br />