工程水文学答案.doc

'已知某河甲、乙两站的年径流模数M（见下表），甲、乙两站的年径流量在成因上具有联系。试用相关分析法推求回归方程及相关系数，并由甲站资料展延乙站资料。某河甲、乙两站的年径流模数M年份甲站年径流模数[10-3m3/(s·km2)]乙站年径流模数[10-3m3/(s·km2)]年份甲站年径流模数[10-3m3/(s·km2)]乙站年径流模数[10-3m3/(s·km2)]19753.55.419822.84.219764.66.519833.04.619773.35.019844.06.119782.94.019853.919793.14.919862.619803.85.619874.819813.04.519885.0解：甲、乙两站的年径流模数相关计算表如下年份甲站年径流模数x[10-3m3/(s·km2)]乙站年径流模数y[10-3m3/(s·km2)]KxKyKx-1Ky-1(Kx-1)2(Ky-1)2(Kx-1)2(Ky-1)2（1）（2）（3）（4）（5）（6）（7）（8）（9）（10）19753.55.41.0294121.0629920.0294120.0629920.0008650.0039680.00000319764.66.51.3529411.2795280.3529410.2795280.1245670.0781360.00973319773.35.00.9705880.984252-0.029412-0.0157480.0008650.0002480.00000019782.94.00.8529410.787402-0.147059-0.2125980.0216260.0451980.00097719793.14.90.9117650.964567-0.088235-0.0354330.0077850.0012560.00001019803.85.61.1176471.1023620.1176470.1023620.0138410.0104780.00014519813.04.50.8823530.885827-0.117647-0.1141730.0138410.0130360.00018019822.84.20.8235290.826772-0.176471-0.1732280.0311420.0300080.00093519833.04.60.8823530.905512-0.117647-0.0944880.0138410.0089280.00012419844.06.11.1764711.2007870.1764710.2007870.0311420.0403160.001256合计3450.81010000.2595160.2315700.013363平均3.45.08假设甲、乙两站的年径流模数分别为x、y。（1）由==3.4==5.08Kx=Ky=依次求出Kx和Ky（2）均值==3.4==5.08 （1）均方差：=3.4=0.57735035.08=0.814862（2）相关系数r===0.970680（3）回归系数Ry/x=r=0.970680*=1.37（4）y倚x的回归方程y=5.08+1.37(x-3.4)=0.422+1.37x(6)回归直线的均方误Sy==0.195874Sy占的=3.8558%（小于15%）（7）相关系数的误差0.018271996占r的1.8823921%(小于5%)把1985到1988的甲站年径流模数带入回归方程，算出对应的乙站年径流模数如表年份1985198619871988甲站年径流模数x[10-3m3/(s·km2)]3.92.64.85.0乙站年径流模数y[10-3m3/(s·km2)]5.7653.9846.9987.272'