一元线性回归分析

1、123456 ab、yabxyabx721(,)()niiiQ a byabx( , )min( , )Q a bQ a b用最小二乘法估计用最小二乘法估计a、b，使，使2,1, 2,(0,) ,iiiiiyabxinN各相 互 独 立8分别取分别取Q关于关于a、b的偏导数，并令其为的偏导数，并令其为0，有，有112()02()0niiiniiiiQyabxaQyabxxb 211nniiiiiaxbynxaxbx y91122211()()()nnxyiiiiiinnxxiiiiSxxyyx ynx ySxxxnx,xyxxSbayxbS1221,niiiniix ynx ybayxbxnx

2、10yabx ayxbyabx()yyb xx( , )x y11iiyabxiiyy2211nneiiiiiiQyyyabx22(2)eQn222,2eeQQEnEn1222eQn13xxxyyyniininiiiiniiiniiiniiieSbSbSxxbxxyybyyxxbyyyyyyyyQ2122112121212)(2)()()(2)()()()()(14xyxxSbS( )eyyxxQSb S152222(2)(2)eQnn可以证明可以证明从而从而2( ,/)xxbN bS222(2)(2) (2)xxxxb bb bnnSt nS2(0,1)xxbbNS162 (2)xxbTSt

3、 n22(2)xxbtStn2(2)tn172(2)ttn22( , )(2),(2)xxxxb bb t nb t nSS1821212211()()nyyiiniiiinniiiiieSyyyyyyyyyySQ回192121niineiiiSyyQyy回20222222(1),(2)1(1,2)(2)(2)eeeSQnSSFnQQnn回回回21(1,2)(2)eSFFnQn回22(2)eSQn回232425得率020406080100050100150200得率262722211111450145,67367.31010218500 10 1458250101570 10 145 67.3

4、 398539850.483825067.3 145 0.4832.735nxxiinxyiiixyxxxySxnxSx ynx ySbSayxb 282.7350.48367.30.483(145)yxyx 222122247225 10 67.31932.18250,0.483( )1932.1 0.48382507.4667.470.93428nyyiixxeyyxxeSynySbQSb SQn2920.483,8250,0.934xxbS0.483825045.3940.934xxbtS0.0250.025,28,(8)2.3062nt0.9340.934( , )0.483 2.30

5、6,0.483 2.30682508250(0.458, 0.508)b b300.483,8250 xxbS22( )0.48382501924.6xxSbS回31yabx32yabx0 abx00yabx3320000200,(0,)(,)yabxNyN abx34000yabx00yabx2020()2100()2100 , 0, 1xxxxxxnSxxnSyN abxyyN35202000()122200()1(0,1)1(2)(2)(2) (2) 1xxxxxxnSxxnSyyUNnVnUTVnyyt n36200000(2)1()()1P TtnP yxyyx 202()10()(

6、2)1xxxxnSxtn0000(),()yxyx37oxyyx1( )y x( )yx2( )yabx12( )( )( )( )( )( )y xy xxyxy xx382022()111,(2)xxxxnStnZ2200,yZyZ220.025000.00150010.95,1.962, (2 ,2 )10.997,2.973, (3 ,3 )ZZyyZZyy3912(,)yy12(,)xx121P yyy 4012 xxx21yyyo( )y xabx2( )( )( )yxy xx1( )( )( )y xy xx4112yy202()11(2)1xxxxnSyabxtn 202()12(2)1xxxxnSyab