环境风险评价作业讲解.doc

根据数据homework，利用SPSS软件，建立因变量y2与自变量 x1，x2，x3，x4和y1之间的回归方程，并对分析结果进行简要解释。 第一种方法： 非线性回归. MODEL PROGRAM a=1 b=1 c=1 d=1 e=1 f=1. 假设方程为y1=a * x1+b / x2+c / x3+d * EXP(e * x4) + f * y2. 迭代历史记录b迭代数a残差平方和参数abcdef1.0339901005035634100000.0001.0001.0001.0001.0001.0001.0001.140430.963147.06714271.473-2.936E-0071.000-4.9222.040430.963147.06714271.473-2.936E-0071.000-4.9222.14456870006.78746.825-2047.5067880.7235.549E-0071.080-3.5432.23424231.69046.486-1685.2607394.5206.286E-0081.008-3.4473.03424231.69046.486-1685.2607394.5206.286E-0081.008-3.4473.11754668.93746.467-1664.7547365.0972.852E-008.992-3.4424.01754668.93746.467-1664.7547365.0972.852E-008.992-3.4424.11641544.35646.458-1655.5067356.5592.792E-008.958-3.4385.01641544.35646.458-1655.5067356.5592.792E-008.958-3.4385.11628701.16346.454-1652.6967360.9135.631E-008.890-3.4356.01628701.16346.454-1652.6967360.9135.631E-008.890-3.4356.11633311.56246.446-1648.1357372.3972.409E-007.754-3.4296.21628531.81046.436-1637.1557357.9828.296E-008.873-3.4267.01628531.81046.436-1637.1557357.9828.296E-008.873-3.4267.11628499.19246.436-1637.9507363.5901.559E-007.840-3.4258.01628499.19246.436-1637.9507363.5901.559E-007.840-3.4258.11627991.80546.431-1634.6157368.6792.596E-007.825-3.4229.01627991.80546.431-1634.6157368.6792.596E-007.825-3.4229.11627874.12646.431-1635.8767374.7594.784E-007.795-3.42110.01627874.12646.431-1635.8767374.7594.784E-007.795-3.42110.11627564.56446.428-1634.5567382.1819.333E-007.766-3.41811.01627564.56446.428-1634.5567382.1819.333E-007.766-3.41811.11627215.59146.425-1634.0887390.5591.872E-006.737-3.41612.01627215.59146.425-1634.0887390.5591.872E-006.737-3.41612.11626820.23446.422-1633.8847399.9513.833E-006.707-3.41413.01626820.23446.422-1633.8847399.9513.833E