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，x1 表示工农业总产值，x2 表示运输线路长度，可建立如下二元线性回归理论模型yi 01x1i2x2i i（ 1）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x2i_x1i x1i x1x2ix2ix2_yi yi y12236150469.54-4328.8461-42.7

2、446-6726.538423463223574.79-3597.8461-37.4946-5499.538434565313886.79-2694.8461-25.4946-4397.5384472974467105.47-1365.8461-6.8146-1665.5384569044536111.12-1296.8461-1.1646-2058.5384679694978117.92-854.84615.6354-993.5384768295634123.2-198.846110.9154866.46168109076379120.25546.15397.96541944.4616911

3、5177077124.711244.153912.42542554.461610116167580128.431747.153916.14542653.461611124038291130.922458.153918.63543440.461612132959211131.563378.153919.27544332.46161314512107971354964.153922.71545549.46161ny=8962.5384x1 =5832.8461x2 =112.2846二、参数估计多元线性回归模型的参数估计公式：2）其中所以?2 ee =ei2nk1nk1Cov(?)2(XX) 1其

4、中，cii 为 (XX) 1 主对角线上元素。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eeei2cii n k 1ciin k 1si Var( i)= ciin kei 12(x1iyi)(x2i) (x2iyi)(x1ix2i )22(x1i)(x2i ) (x1i x2i)22(x(X X)k1222i yi )( x1i) (x1i yi)(x1i x2i )22(x1i)( x2i) (x1i x2i )21 x1x22 x2i Lx1i x2iLx1i x2iL2x1iL2L (x1i)(2x2i) (x1i x2i )3)4)5)( 6)8)7)2x2ic11=

5、22(x1i)(x2i) (x1i x2i)29 1)x1ic22 =22( x1i)( x2i ) (x1i x2i )2统计计算表如下：年 份2x1i2x2i2 yix1i yix2i yix1i x2i118738908.561827.100845246318.85299118149.52287523.1933185034.7950212944496.561405.845030244922.611978.6492.78206202.9925134899.800437262195.503649.974619338343.9811850689.21112113.482568704.02344

6、1865535.56946.43882774018.1622274869.12811349.97809307.694851681809.8071.35634237580.3442669607.4962397.37381510.30706730761.854731.7577987118.552389323.4264-5598.9863-4817.3997739539.7715119.1460750755.7043-172292.59457.7750-2170.48478298284.082563.44763780930.9141061975.28515488.41444350.334391547

7、918.927154.39066525274.0563178143.36231740.207215459.1099103052546.75260.67407040858.4634636005.78342841.198928208.4986116042520.596347.278111836776.028457184.164114.378145808.68121211411923.94371.541018770223.5214635722.0583509.930365115.26771324642823.94515.989430796524.0527548381.44126058.2400112

8、762.741590259265.695794.9399(TSS)182329645.2125894250.1987198.1777664173.3692计算各参数：由公式（7）得到0.9018= 66.9937125894250.1 5794.9399 987198.1777 664173.36921=2190259265.69 5794.9399 664173.36922987198.1777 90259265.69 125894250.1 664173.3692290259265.69 5794.9399 664173.369220 =8962.5384-0.9018 5832.8461

9、-66.9939 112.2846= - 3820.078根据上述计算结果，二元回归模型为：yi3820.078 0.9018x1i 66.9937x2i残差计算如下年 份yieiyi - yi2 ei12194.971141.02891683.370823205.9038257.096266098.444234824.1536-259.153667160.600347274.088122.9119524.953457714.8267-810.8267657440.008868568.9795-599.9795359975.405279514.2870314.713099044.2472899

10、88.4966918.5034843648.4499910916.7450600.25513603065669-3.566912.72271112427.561-24.5610603.24291751213300.0930-5.093025.93841314960.8061-448.8061201426.9154(RSS) 2657950.4523）（6） ，得到22 2657950.452=10265795.0452Var( 1) =5794.93992 265795.0452190259265.69 5794.9399 664173.36922=0.0188019

11、Var( 2) =265795.045290259265.6990259265.69 5794.9399 664173.36922=292.8496617s 0.13712004s 17.112851TSS= (yi y)2 2 =yi =182329661.6RSS=(yi yi)2=ei 2 =2657950.4522 ESS= (yi y)=TSS-RSS=182329645.2 - 2657950.452=179671716.2481SrRSS 2657950.452 10 =265795.0452nk1 1StTSS=182329645.2 12=15194137.1n12SR21

12、r =1St265795.0452= 0.98250715194137.11 F 检验337.98921179671716.248/22657950.452/10若给定0.05，查F( 2， 10)的临界值，F0.054.10。显然 F=337.98920F0.054.10，因此回归方程显著成立ti?i?it1t2sic11ei2ciink1?10.9018ei20.137120046.5770c2210?266.9937e2 17.1128513.914810对给定的0.05，查t0.05(10)的临界值，t0.025(10)2.228。t1 、 t2均大于 t0.025 (10)，因此 两

13、个参数1 、2显著的不为零Eviews软件计算结果Dependent Variable: YMethod: Least SquaresDate: 12/08/03 Time: 19:05Sample: 2001 2013Included observations: 13VariableCoefficientStd. Errort-Statistic Prob.X10.9018330.1371206.5769640.0001X266.9937017.112853.9148180.0029C-3820.0781236.779-3.0887310.0115R-squared0.985422Mean dependent var8962.538Adjusted R-squared0.982507S.D. dependent v