FORCE HW 8 # 1 1.Description of data. Obtained 10 cadaveric femers from 3 different epoch groups and big businessman interrogationed them. Group Y force MA E 193.6 125.4 59 137.5 126.5 87.2 122 115.9 84.4 145.4 98.8 78.1 117 94.3 51.9 105.4 99.9 57.1 99.9 83.3 54.7 74 72.8 78.6 74.4 83.5 53.7 112.8 96 1182 900.4 700.7 center 118.2 100.0444444 70.07 Average 1245.926667 365.6402778 270.3912 Variance Xj Total Average 95.96896552 118.2 100.044 70.07 n*(Xi-Xbar)^2 95.968 4942.618 95.968 149.524 95.968 6707.064 11799.21 SSA SSW=SST-SSA SSW 16572.02 193.6 137.5 122 145.4 117 105.4 99.9 74 74.4 112.8 125.4 126.5 115.9 98.8 94.3 99.9 83.3 72.8 83.5 59 87.2 84.4 78.1 51.9 57.1 54.7 78.6 53.7 96 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 95.968 (Xi-Xbar)^2 9532.007 1724.907 677.665 2443.523 442.345 88. 96262 15.46062 482.593 465.1786 283.3162 866.2426 932.203 397.2846 8.020224 2.782224 15.46062 160.4782 536.7562 155.451 1366.633 76.87782 133.8186 319.2654 1941.989 1510.721 1703.048 301.6474 1786.584 0.001024 28371.22 SST Page 1 FORCE 2.Assumptions. Normal distribution breakaway random samples stand for for each group 3.Hypotheses.
H0: u1 = u2 = u3 HA: ui non equal to un 4.Test statistic. VR = MSA/MSW 5.Distribution of test statistic. df = 2 on top and 26 on diffuse F dis tribution 6.Decision rule. alpha = 0.05 F evaluate = 3.37 7.Calculation of test statistic. analysis of variance Source SS with sample SSA indoors sample SS W SST df 2 (k-1) 26 (N-k) 28 (N-1) SS ! MS 11799.21 5899.603 16572.02 637.3852 28371.22 VR 9.255945866 P-Value 0.000921435 8.Statistical decision. 9.25>3.37 there fore unprofitable hypotesis can be rejected. 9.Conclusion. solely population means argon not equal, and there is residuum in femer strength among age groups 10.Determination of p value. Page 2 FORCE 0.00092If you want to get a full essay, coiffe it on our website: BestEssayCheap.com
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