二元线性回归模型案例

1、二元线性回归模型范例我国货物周转量问题研究。一、建立模型通过经济分析可知，工农业总产值、运输线路长度是影响我国货物周转量的主要因素用y表示货物周转量，x表示工农业总产值，x表示运输线路长度，可建立如下二元线性12回归理论模型y=x+Bx+p（1i011i22ii收集我国某一时间13年的货物运输量（单位：1亿吨公里）、工农业总产值（单位：亿元）、运输线路长度（单位：万公里）的统计资料数据，并同时给出离差形式数据。年份货物周转量yi国内生产总值x1i运输线路长度x2ix=x一x1i1i1x2i二x一x2i2y=y一yii12236150469.54-4328.8461-42.7446-6726.5

2、38423463223574.79-3597.8461-37.4946-5499.538434565313886.79-2694.8461-25.4946-4397.5384472974467105.47-1365.8461-6.8146-1665.5384569044536111.12-1296.8461-1.1646-2058.5384679694978117.92-854.84615.6354-993.5384768295634123.2-198.846110.9154866.46168109076379120.25546.15397.96541944.46169115177077124

3、.711244.153912.42542554.461610116167580128.431747.153916.14542653.461611124038291130.922458.153918.63543440.461612132959211131.563378.153919.27544332.46161314512107971354964.153922.71545549.46161Eny=x=1x=28962.53845832.8461112.2846二、参数估计多元线性回归模型的参数估计公式2）3）eeye2c=ciiin-k-1iin-k-1（4）（5）（6）2dxy)(exii)-

4、(Ey)(e)=2ii女iii1_2i22(Sx丿(Yx2i)-(Exx)27）1i2i由于=y-px-Px22(xXhk=2工x2iL一工xxii2iL工xxii2iLy2厶xiiL丿8）入eeye2b2=1卩nk1n一k一1Cov(B)=b2(XX)-1其中，c为（XX）-1主对角线上元素。iiye2s厂Mr()：c-in-i-r对于二元线性回归模型，由正规方程组可以得到参数估计公式如下2=(Exy)(Ex2i)-(Exy)(Exx)=iii九i1222(Ex1i)(Ex2i)-(Exx)21i2i1i2i其中L=(yx2)(yx2)-(yxx)21i2i1i2i所以291)Exc=211

5、22厶厶(Ex1i)(Ex2i)-(Exx)21i2i1i2i2zxTOC o 1-5 h zc22=1(92)2222(Zx.)(zx2.)-(Zxx)21i2i1i2i统计计算表如下：年份2x1i2x2i2yixy1iixy2iixx1i2i118738908.561827.100845246318.85299118149.52287523.1933185034.7950212944496.561405.845030244922.611978.6492.78206202.9925134899.800437262195.503649.974619338343.9811850689.21112

6、113.482568704.023441865535.56946.43882774018.1622274869.12811349.97809307.694851681809.8071.35634237580.3442669607.4962397.37381510.30706730761.854731.7577987118.552389323.4264-5598.9863-4817.3997739539.7715119.1460750755.7043-172292.59457.7750-2170.48478298284.082563.44763780930.9141061975.28515488

7、.41444350.334391547918.927154.39066525274.0563178143.36231740.207215459.1099103052546.75260.67407040858.4634636005.78342841.198928208.4986116042520.596347.278111836776.028457184.164114.378145808.68121211411923.94371.541018770223.5214635722.0583509.930365115.26771324642823.94515.989430796524.05275483

8、81.44126058.2400112762.7415Z90259265.695794.9399(TSS)182329645.2125894250.1987198.1777664173.3692TOC o 1-5 h z计算各参数：由公式(7)得到A125894250.1X5794.9399-987198.1777x664173.36920q*1=90259265.69x5794.9399-664173.36922=a987198.1777x90259265.69-125894250.1x664173.36926699372=90259265.69x5794.9399-664173.36922

9、=旷=8962.5384-0.9018x5832.8461-66.9939x112.2846=-382O0780根据上述计算结果，二元回归模型为：y=3820.078+0.9018x+66.9937xi1i2i残差计算如下年份AyiAe=y.-yiiie2i12194.971141.02891683.370823205.9038257.096266098.444234824.1536-259.153667160.600347274.088122.9119524.953457714.8267-810.8267657440.008868568.9795-599.9795359975.4052795

10、14.2870314.713099044.247289988.4966918.5034843648.4499910916.7450600.25513603065669-3.566912.72271112427.561-24.5610603.24291751213300.0930-5.093025.93841314960.8061-448.8061201426.9154E(RSS)2657950.452由公式（3）（6），得到22657950.452265795.0452x265795.0452x265795.04525794.939990259265.69x5794.

11、9399664173.3692290259265.69=0.018801990259265.69x5794.9399664173.36922=292.8496617s=0.13712004s=17.112851卩2三、拟合优度的检验tss=工（y-y）2i2y=182329661.6iRss=(yy)2ii=ze2=2657950.452iEss=z(yy)2i=Tss-Rss=182329645.2-2657950.452=179671716.248S=一-RSS=2657950.452十10=265795.0452rnk1S=丄TSS=182329645.2十12=15194137.1tn

12、10.982507S265795.0452R2=1一f=1-S15194137.1t四、显著性检验1F检验179671716.248/22657950.452/10337.98921若给定a=0.05,查F（2，10）的临界值，F0.05=4.10o显然F=337.98920F=4.10,0.05因此回归方程显著成立。2t检验0.i=s0.9018:乙e20.13712004c11106.577066.9937Xe217.112851c22103.9148对给定的a=0.05，查t(10)的临界值，t(10)=2.228o0050.0252由于1、12均大于仁25（1），因此两个参数卩1、卩2

13、显著的不为零Eviews软件计算结果DependentVariable:YMethod:LeastSquaresDate:12/08/03Time:19:05Sample:20012013Includedobservations:13VariableCoefficientStd.Errort-StatisticProb.X10.9018330.1371206.5769640.0001X266.9937017.112853.9148180.0029C-3820.0781236.779-3.0887310.0115R-squared0.985422Meandependentvar8962.538AdjustedR-squared0.982507S.D.dep