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For our example the hypotheses are: <br />Ho: al=a2=a3=a4=a5=a6 <br />He: population variances are heterogeneous <br /> N#1 West Nql T/S R.A. T/S R.A. RN2 N#1 East E <br />E wares 445.57 17.97 17.97 71.9! soo.aa 59.29 1113.59 <br />r (dt) 5 1 I 1 4 4 16 <br />S 89.11 17.97 17.97 71.91 175.22 14.82 <br />1 S 1.95 ].25 1.25 1.86 2.09 1.17 <br />v S 9.75 1.25 1.25 L.86 8.36 4.68 27.15 <br />1/v 0.2 I 1 f 0.25 0.25 3.70 <br />where: n= number of samples <br />v= the degrees of freedom (n-1) <br />S2°sample variance <br />S2p= 1113.59/16 = b9.59 <br />log S2p= 1.84 <br />B= 2.3026[(1,84*16)-27.15] = 5.27 <br />C= l+1!(3)(5) • (3.7 - 1116) = 1.24 <br />Bc= 5.2711.24 = 4.25 <br />The critical value X2 value with 5 degrees of freedom and 95% confidence is l 1.07, <br />'t'herefore, do not reject Ha: a1=a2=ag=aq=a5=a6 .75< P < .50 <br />The sample variances are equal, therefore they come from the same population, and the <br />population is homogenous with respect to variance. <br />ANALYSIS OF VARIANCE finclutlin~ NORTIiERN # l EAST SmEI <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), and Northern # 1 Mine Area- East Side (N#1 East) <br />These data are presented in full in the vegetation data section of this submittal. <br />