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For our example the hypotheses are: <br />Ho: a1=a2=o3=a4=a5 <br />H8: population variances are heterogeneous <br /> N#t West N#1 T/S R.A. T/S R.A. RN2 E <br />E ueres 445.57 17.97 17.97 71.91 500.88 1054,30 <br />v idf) 5 1 L 1 4 12 <br />S 89.11 17.97 17.97 TI.91 125.22 <br />l0 1.95 1.25 1.25 1.86 2.09 <br />vl S 9.75 1.75 1.25 1.86 8.36 22.47 <br />1/r 0.2 1 1 I 0,25 3.45 <br />where: n= number of samples <br />v= the degrees of freedom (n-1) <br />S2-sample variance <br />S2p= 1054.3/12 =87.86 <br />log S2p= 1.94 <br />B= 2,3026[(1.94*12)-22.47] = 1.86 <br />C= 1+1((3)(4) * (3.45-1112) = 3.65 <br />B~= 1.86/3.647 = 0.51 <br />The critical value X2 value with 4 degrees of freedom and 95 `% confidence is 9.48. <br />Therefore, do not reject Ho: o1=a2=o3=a4=a5 .995< P < .999 <br />The sample vaziances are equal, therefore they come from the same population, and the <br />population is homogenous. <br />ANALYSIS OF VARIANCE (IIOMOGENEITY OF MEANS) <br />The sample data for this demonstration are the vegetation cover data from the Northern <br /># 1 Mine Area- West Side (N#1 West), Northern # 1 Mine- Topsoil Stockpile Area <br />(N#1 T/S), Refuse Area- Topsoil Stockpile Area (R. A. T/S), Refuse Area (R.A.), and <br />Rienau # 2 Mine Area (R#2). These data aze presented in full in the vegetation data <br />section of this submittal. <br />