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1、异方差性的检验和消除15.表6列出了 2000年中国部分省市城镇居民每个家庭平均全年可支配收入x与消费性 支出 y 的统计数据。 利用OLS法建立人均消费支出与可支配收入的线性模型和对数线性模型； 检验模型是否存在异方差性； 如果存在异方差性，试采用适当的方法加以消除。表 6 中国城镇居民人均可支配收入与消费性支出（单位：元）地区可支配收入x消费性支出y地区可支配收入x消费性支出y内蒙古黑龙江10349.698140.505661.164724.115129.055357.794810.004912.8811718.016800.238493.496121.044348.473941.8739
2、27.754356.064020.873824.448868.195323.189279.166489.974766.265524.546218.739761.575124.244916.255169.965644.867020.225022.003830.714644.505218.798016.914276.674126.474185.734422.93输入数据：data x y绘制图形，确定模型：Plot x y （相关图）（1）线性模型： lsx yview-graph-scatter (散点图)M Equation: UyiULSD Torklile: 异方差 口回区匚objet P
3、rint(Estim ate pdre 匚乜口 esids Dependent Variable: YMethod: Least SquaresDate: 04/1 0/1 4 Time: 1 2:34Sample: 1 20Included observations: 20VariatleCoefficientStd. Errort-StatisticProb.c272.36351 59.67731 .7057130.1 053X0.7551250.02331 632.3B6900.0000R-squar&d0.983129Mean deperdentvar5199.616Adjusted
4、R-squared0.982193S.D. dependentvar1 625.275S.E. of regression21 6.8900Akaike info criterion13.69130Sum squared resid846743.0Schwarz criterion1 3.79037Log likelihood-1 34.9130Hannan-Quinr criter.1 3.71 073F- statistic1 043.912Durbin-Watson stat1.301 684Prob(F-statisti c)0.000000V = 272.3635 + 0. 7551
5、25xtts = (159.6773)(0.023316)t = (1. 705713)(32. 38690)R 2 二 0. 983129 F = 1048.912 dw = 1.301684 S.E=216.89OO对数线性模型：GENR lnx=LOG(x)GENR lny=LOG(y)Is lny c lnx Equation: UHTITLED lorkfile: 异方差巧对- J辽)区讷訥ProuObjeut PrintNarmeFre 駅&| E址imHe|For讯盘股曰佔|隔出DependentVariable: LNYMethod: Least SquaresDate: 04
6、/101 4 Time: 13:16Sample: 1 20Included obseivatiors: 20VariableCoefficientStd. Errort-StatisticProb.C0.2494340.2634950.9466350.3564LNX0.94591 70.03013231.39272o.aaooF?-squared0.982063Mean dependent var8.516948Adjusted R-suared0.981 066S.D. dependent var0.275333S.E. of regression0.038023Akaike info c
7、riterion-3.606594Sum squared resid0.026024Schwarz criterion-3.507021Log likelihood38.06594Hannan-Quinn criter.-3.587157F-statistic985.5030Durbin-Watson stat1.512696Prob(F-stati Stic)0.000000ln V = 0. 249434 + 0. 945917 ln xttS=(O.263495)(O.O3O132)t=(O.946635)(31.39272)R2 二 0. 982063R2 二 0. 981066 F=
8、985.5030 S.E=0.038023 DW=1.512696(2) 检验是否存在异方差性White 检验：Is y c x 在方程窗口中 views-residual test-heteroscedasticity-white 此时可以选择要不要包含交叉乘积项。一元的自由度为 2 二元的自由度为 5Heteroskedasticity Test: WhiteF-statistic14.B3595Prob.F(2,in0.0002Otis*R-squared12.55213Prob. Chi-Squarep)0.0018Scaled explained SS5.568079Prob. Ch
9、i-Square(2)0 0618y 2（2） = 5 99取a =0.05, n为样本数量，nr 2=12.65213 “ 0.05，即对应的p值小于0.05，表面模型存在异方差性。Goldfeld-Quandt 检验（戈德菲尔德-匡特检验）：将观察值按解释变量的大小顺序排列，将排列在中间的约1/4的数据删掉，记为c,也可不删。由样本x数据排序，n=20,c=20/4=5,取c=4，从中间去掉4个数据，确定子样1（18）Sortx将样本数据关于x排序Smpl18确定子样本 1Ls ycx求出 RSS1Smpl1320确定子样本 2ls ycx求出 RSS2计算出 F=RSS2/RSS1-k-
10、1的 F 分布表，大于查的结果则存在异n - c取a为0.05,查第一自由度和第二自由度为2 方差性。子样本1DeperidentVa riable: Y Method: Least Squares Date: 04/10/14 Time: 13:53 Sample: 1 SIncluded observations: 8VariableCoefficientStd. Errort-StatisticProb.C1277.1611540.6040.6290000.438GX0.5541260.3114321.7792870.1 255R-squared0.345397Mean dependen
11、tvar4016.814Adjusted R-squared0.236296S.D. dependent var166.17128.E. of regression145.2172Aka ike info criterion13.00666Sum squared resid126528.3Schwarz criterion13.02652Log likelihood-50.02663Hannan-Quinn criter.12.S7271F-statistic3.165861Durbin-Watson stat3.004532Prob(F-statistic)0.125501求出 RSS1=1
12、26528.6。子样本2Dependent Variable: YMethod: Least SquaresDate: 04/10/14 Time: 13:56Sample: 13 20Included obseivations: 8VariableCoefficientSid. Errort-StatisticProb.C212.2118530.88920.3997290.7032X07618930.06034812.625050.0000R-squared0.963723Mean dependertvar6760.478Adjusted R-squared0.9576768.D. depe
13、ndentvar1 556.S14S.E. of regression320.2790Akaike info criterion14.58858Sum squared resid615472.0Schwarz criterion14.60044Log likelihoocl-56.35432Hannan-Quinn criter.14.45463F-statistic159.3919Durbin-Watson stat1.722960ProbfF-statistic)0.000015确定子样本2 (1320),求出RSS2=615472，计算出F= RSS2/ RSS】=4.86,给定显著性水
14、平为 0.05，查F(6,6)二4.28, F F，所以模型存在异方差性。0.05 0.05Gleiser 检验：iews-residual test-heteroscedasticity-gleiserHeteroskedasticity Test: GlejserF-statistic22 86315Prob. F(1,1&)0.0001DbsR-squared11.19113Prob. Chi-Square(1 J0.0008Scaled explained SS-7.14S029Prob. Chi-Square(10.0075nr2 对应的 p 值小于 0.05，存在异方差性。Park 检验：iews-residual test-heteroscedasticity-harvey输入 log(x)Heteroskedasti
