计算机线性回归实验报告,线性回归分析实验报告doc
线性回归分析实验报告.doc
1数据分析课程实验报告实验名称线性回归分析一、实验目的1、通过实验掌握线性回归模型拟合及参数估计2、获得处理统计推断与预测的能力3、学会残差分析、掌握Box-Cox变换的方法4、学会最优回归方程的选取5、进一步熟悉SAS的应用二、用文字或图表记录实验过程和结果2.4解sas程序为datatwo_4;yx1x2;CARDS;省略了数据;RUN;PROCREGDATAtwo_4;modelyx1x2/I;OUTPUTOUTaPPREDICTEDRRESIDUALHHSTUDENTSTUDENT;RUN;PROCCAPABILITYDATAaGRAPHICS;QQPLOT;RUN;PROCGPLOTDATAa;PLOTRESIDUAL*PREDICTEDRESIDUAL*x1RESIDUAL*x2;SYMBOLVALUEDOTINONE;RUN;PROCIML;N15;USEtwo_4;READALLVARyx1x2INTOM;XM,2M,3;X2M,3;YM,1;PYXX2;CREATERESOLVEVARYXX2;APPENDFROMP;QUIT;2PROCREGDATARESOLVE;MODELYXX2;RUN;PROCPRINT;RUN;(1)参数估计的sas输出结果为ParameterEstimatesParameterStandardVariableDFEstimateErrortValuePr|t|Intercept13.452612.430651.420.1809x110.496000.0060581.92插入方差分析表AnalysisofVarianceSumofMeanSourceDFSquaresSquareFValuePrFModel253845269225679.47F0)中参数估计值可求得12t的置信度为95的置信区间为分别为012、、3.45261-2.43065*0.128即3.1414868，3.763733200.49600-0.00605*0.128即(0.37405，0.4967744)10.00920-0.00096811*0.128即(0.00907608192，0.00932391808)2(4)参数估计的sas输出结果为ParameterEstimates3ParameterStandardVariableDFEstimateErrortValuePr|t|Intercept13.452612.430651.420.1809x110.496000.0060581.92FModel25229426147195.15Obsyx1x2PREDICTEDRESIDUALSTUDENTH11622742450161.8960.104280.051940.1497421201803254122.667-2.66732-1.319810.1383732233753802224.429-1.42938-0.727730.1861341312052838131.241-0.24062-0.114830.0737456786234767.699-0.69928-0.357820.1943261692653782169.685-0.68486-0.346740.1770178198300879.7321.268060.666410.2361781923302450189.6722.328001.228330.2422491161952137119.832-3.83202-1.924820.16388105553256053.2911.709480.917330.26740112524304020253.715-1.71506-0.929660.28203122323724427228.6913.309211.891000.35396131442362660144.979-0.97934-0.469600.08250141031572088100.5332.466931.242990.16906152123702605210.9381.061940.576190.28343分析正态性的频率检验通过表中显示的数据，可知学生化残差STUDENT列中有落入(-1,1)区间的有9/150.6即60；落入(-1.5,1.5)区间的有13/150.867即86.7；落入(-2.2)区间内的有15/151即100。所以，学生化残差落在上述各区间内的频率与标准正态分布的相应概率相差不大，因此，模型误差项符合服从正态分布的假定。正态QQ图检验通过SAS系统中capacity过程可以直接做出正态QQ图，对于中求得的学生化残差，其正态QQ图如图所示5StudentizedResidual-2-1012NormalQuantiles-2.0-1.5-1.0-0.50.00.51.01.52.0由图可知，图中的点大致在一条直线上，因此说明题中线性回归模型中误差项正态分布的假定是合理的。残差图分析通过SAS系统的precgplot过程分别输出了残差与Y拟合值、残差与自变量X1,残差与自变量X1的残差图，如下图Residual-4-3-2-101234PredictedValueofy0100200300Residual-4-3-2-101234x101002003004005006Residual-4-3-2-101234x22000300040005000由图可知(1)在以因变量y为横坐标的残差图中，图中个点大致在一个水平的带状区域内，而且，没有呈现出明显的趋势，说明了因变量拟合值向量与残差向量不相关，也就是说Y与相互独立，此时，认为假设是合理的；(2)在以自变量X1、X1量为横坐标的残差图中，残差没有随自变量变化而变化的趋势，说明了假设是合理的。26解SAS源程序为DATAtwo_6;x1x2y;CARDSPROCREGDATAtwo_6;MODELyx1x2;OUTPUTOUTbPPREDICTEDRRESIDUALHHSTUDENTSTUDENT;PROCCAPABILITYDATAbGRAPHICS;QQPLOT;RUN;PROCGPLOTDATAb;PLOTRESIDUAL*PREDICTEDRESIDUAL*x1RESIDUAL*x2;SYMBOLVDOTINONE;RUN;PROCIML;N31;PI1;USEtwo_6;READALLVARx1x2yINTOM;YM,3;XJN,1,1M,12;AX*INVX*X*X;DOI1toN;PIPIMI,3;7END;TEMPPI1/N;DOLAMDA-0.5to0.5by0.01;ZYLAMDA-JN,1,1/LAMDATEMPLAMDA-1;SSEZ*IN-A*Z;LASLASLAMDASSE;END;Z0TEMPLOGY;SSE0Z0*IN-A*Z0;KLAS中求得的学生化残差，其正态QQ图如图所示9StudentizedResidual-2-10123NormalQuantiles-3-2-10123由学生化残差的正态QQ图可知，图中的线性关系并不是非常的明显，不过图中大多数点大致在一条直线上，但是还有一些点偏离了直线，由此说明题中线性回归模型中误差项正态分布的假定是基本上是合理的，但还可以做出完善。残差图分析通过SAS系统的precgplot过程分别输出了残差与Y拟合值、残差与自变量X1,残差与自变量X1的残差图，如下图Residual-7-6-5-4-3-2-1012345678PredictedValueofy010203040506070Residual-7-6-5-4-3-2-1012345678x18910111213141516171819202110Residual-7-6-5-4-3-2-1012345678x260708090由图可知(1)在以因变量y为横坐标的残差图中，图形大致形成了一个U型，说明回归函数可能是非线性的，可能需要引进某个或者某些自变量的二次项或者交叉乘积；(2)在以自变量X1、X2量为横坐标的残差图中，根据图形的显示，说明了回归函数关于基本X2基本上呈现线性，但是图像中显示的残差与X1的关系并不是很好，说明回归函数关于X1可能不是线性的，有可能要引进X1的平方项或者交叉项。得到结论需要做Box-Cox变换。2做Box-Cox变换，得到()矩阵为,zLAS-0.5477.54855-0.49468.1904-0.48458.99203-0.47449.95164-0.46441.06743-0.45432.33767-0.44423.76066-0.43415.3347-0.42407.05816-0.41398.92943-0.4390.94692-0.39383.10909-0.38375.41442-0.37367.86142-0.36360.44864-0.35353.17465-0.34346.03807-0.33339.03751-0.32332.17166-0.31325.43919-0.3318.83884-0.29312.3693611-0.28306.02953-0.27299.81815-0.26293.73406-0.25287.77612-0.24281.94324-0.23276.23432-0.22270.64831-0.21265.18419-0.2259.84096-0.19254.61763-0.18249.51328-0.17244.52696-0.16239.6578-0.15234.90491-0.14230.26746-0.13225.74462-0.12221.33561-0.11217.03965-0.1212.85601-0.09208.78396-0.08204.82282-0.07200.97191-0.06197.23059-0.05193.59824-0.04190.07426-0.03186.65809-0.02183.34918-0.01180.147013.088E-16880.284760.01174.06089LAS0.02171.176030.03168.396050.04165.720550.05163.149160.06160.681510.07158.317270.08156.056140.09153.897830.1151.842080.11149.888650.12148.03732120.13146.28790.14144.640230.15143.094170.16141.649580.17140.306380.18139.064490.19137.923870.2136.884470.21135.946310.22135.10940.23134.37380.24133.739550.25133.206770.26132.775570.27132.446090.28132.218490.29132.092960.3132.069720.31132.149010.32132.331080.33132.616230.34133.004770.35133.497040.36134.09340.37134.794240.38135.599970.39136.511030.4137.527880.41138.651020.42139.880970.43141.218270.44142.663480.45144.217210.46145.880070.47147.652730.48149.535850.49151.530150.5153.63636同时得到在不为零时，的最小值为MINSSE132.06972此时为MINLAMDA0.3，而z当0时，177.05107，所以，最终的最小值为MINSSE132.06972，为zzMINLAMDA0.3。通过SAS输出有关残差的结果如下13ObsX1X2ZPREDICTEDRESIDUALSTUDENTH18.3703.376783.41103-0.03425-0.166270.1158328.6653.376783.340760.036020.178070.1472138.8633.357173.345680.011490.057820.17686410.5724.381604.39699-0.01540-0.072470.05919510.7814.704264.82901-0.12475-0.607250.12066610.8834.817814.94795-0.13014-0.646520.15575711.0664.266714.37047-0.10376-0.503410.11480811.0754.626434.71991-0.09348-0.438130.05148911.1805.160654.955330.205320.983580.092011011.2754.842554.802480.040070.187450.047971111.3795.336754.999070.337681.601680.073831211.4764.975594.923880.051700.241910.048091311.4765.022754.923880.098870.462580.048091411.7695.011024.775960.235061.114310.072761512.0754.742535.13278-0.39025-1.815910.037651612.9745.115265.46553-0.35027-1.628190.035671712.9856.250825.892620.358201.754320.131311813.3865.665876.09660-0.43073-2.124480.143461913.7715.494595.67935-0.18475-0.872960.066662013.8645.411245.44885-0.03761-0.193300.211242114.0786.309946.074990.234951.092210.035812214.2806.068156.23522-0.16707-0.780550.045422314.5746.458206.126120.332081.555210.049952416.0726.617026.66777-0.05075-0.245770.111432516.3776.939776.98576-0.04599-0.217630.069312617.3817.782247.553940.228301.091490.088422717.5827.800267.675340.124920.599760.096032817.9807.953697.762830.190850.921620.106422918.0807.953697.804120.149570.723630.109833018.0807.541467.80412-0.26266-1.270810.109833120.6878.936149.14936-0.21322-1.107080.22706分析正态性的频率检验通过表中显示的数据，可知学生化残差STUDENT列中有落入(-1,1)区间的有20/310.645即64.5；落入(-1.5,1.5)区间的有25/310.806即80.6；落入(-2.2)区间内的有30/310.96896.8。所以，变换后学生化残差落入(-1，1)之间的更多了，而学生化残差落入(-1.5,1.5)及(-2.2)区间的个数没有变化，这样使得与标准正态分布的相应概率更加相近，因此，从正态性频率检验可以知道模型误差项符合服从正态分布的假定。正态QQ图检验通过SAS系统中capacity过程可以直接做出正态QQ图，对于中求得的学生化残差，其正态QQ图如图所示14StudentizedResidual-2-10123NormalQuantiles-3-2-10123由学生化残差的正态QQ图可知，图中各点基本位于同一条直线上，线性关系非常的明显，由此说明题中线性回归模型中误差项正态分布的假定是合理的。残差图分析通过SAS系统的precgplot过程分别输出了残差与Z拟合值、残差与自变量X1,残差与自变量X1的残差图，如下图Residual-0.5-0.4-0.3-0.2-0.10.00.10.20.30.4PredictedValueofZ34567891015Residual-0.5-0.4-0.3-0.2-0.10.00.10.20.30.4X189101112131415161718192021Residual-0.5-0.4-0.3-0.2-0.10.00.10.20.30.4X260708090由图可知(1)在以因变量Z为横坐标的残差图中，图中个点大致在一个水平的带状区域内，而且，没有呈现出明显的趋势，说明了因变量拟合值向量与残差向量不相关，也就是说经过Box-Cox变换后，Z与相互独立，此时，认为假设是合理的；(2)在以自变量X1、X1量为横坐标的残差图中，残差没有随自变量变化而变化的趋势，说明了假设是合理的。结论由此说明，Box-Cox变换的效果非常好2.7解SAS源程序为16DATAtwo_6;x1x2y;CARDS;省略了数据;PROCIML;N31;USEtwo_6;READALLVARx1x2yINTOM;X1M,1M,1;X2M,2;YM,3;PX1X2Y;CREATERESOLVEVARX1X2Y;APPENDFROMP;QUIT;PROCREGDATARESOLVE;MODELYX1X2;OUTPUTOUTbPPREDICTEDRRESIDUALHHSTUDENTSTUDENT;RUN;PROCCAPABILITYDATAbGRAPHICS;QQPLOT;RUN;PROCGPLOTDATAb;PLOTRESIDUAL*PREDICTEDRESIDUAL*X1RESIDUAL*X2;SYMBOLVDOTINONE;RUN;PROCPRINT;RUN;通过SAS输出有关残差的结果如下ObsX1X2YPREDICTEDRESIDUALSTUDENTH168.897010.38.26342.036560.783340.09266273.966510.37.43982.860171.128590.13783377.446310.27.36172.838341.143100.172364110.257216.416.10130.298700.112740.057675114.498118.819.8968-1.09683-0.427600.116766116.648319.720.9493-1.24929-0.497000.151817121.006615.615.9204-0.32043-0.124770.114638121.007518.218.9822-0.78222-0.294280.051559123.218022.621.06571.534350.589860.0917110125.447519.919.75060.149430.056130.0487811127.697924.221.50072.699271.028200.0748412129.967621.020.87300.127030.047740.0495313129.967621.420.87300.527030.198060.049531714136.896921.319.69081.609160.612180.0725015144.007519.122.9624-3.86243-1.444050.0396416166.417422.226.5003-4.30035-1.604450.0356617166.418533.830.24253.557471.403430.1374618176.898627.432.3963-4.99632-1.985480.1499519187.697125.729.1623-3.46232-1.309130.0610420190.446424.927.2568-2.35683-0.963680.1970921196.007834.532.98181.518220.566430.0356022201.648031.734.6382-2.93819-1.102260.0461823210.257436.334.08702.213010.829420.0443724256.007238.341.3238-3.02376-1.168790.1015425265.697742.644.7016-2.10163-0.795840.0638726299.298155.451.87703.523001.351700.0881127306.258255.753.42162.278350.878460.0970328320.418058.355.19173.108321.207550.1105629324.008058.355.81292.487060.968620.1150030324.008051.555.8129-4.31294-1.679740.1150031424.368777.075.56191.438080.620830.27971分析正态性的频率检验1、通过表中显示的数据，可知学生化残差STUDENT列中有落入(-1,1)区间的有18/310.581即58.1；落入(-1.5,1.5)区间的有28/310.903即90.3；落入(-2.2)区间内的有31/311100。所以，学生化残差落在上述各区间内的频率与标准正态分布的相应概率相近，因此，从正态性频率检验可以知道模型误差项符合服从正态分布的假定。2、与上一题未作变换钱相比较落入(-1,1)区间的数据个数没有改变，但落入(-1.5,1.5)区间以及(-2.2)区间内的数据个数有了明显的改善，由此可知，此模型拟合的更好。正态QQ图检验通过SAS系统中capacity过程可以直接做出正态QQ图，对于中求得的学生化残差，其正态QQ图如图所示StudentizedResidual-2.0-1.5-1.0-0.50.00.51.01.5NormalQuantiles-3-2-1012318由学生化残差的正态QQ图可知，图中的线性关系并不是很好，有一些点偏离了直线，由此说明题中线性回归模型中误差项正态分布的假定是基本上是合理的。残差图分析通过SAS系统的precgplot过程分别输出了残差与Y拟合值、残差与自变量X1,残差与自变量X1的残差图，如下图Residual-5-4-3-2-101234PredictedValueofY01020304050607080Residual-5-4-3-2-101234X1010020030040050019Residual-5-4-3-2-101234X260708090由图可知1、(1)在以因变量y为横坐标的残差图中，图形大致形成了一个U型，但并不是很明显，说明回归函数是非线性的，可能需要引进某个或者某些自变量的二次项或者交叉乘积；(2)在以自变量X1、X2量为横坐标的残差图中，根据图形的显示，说明了回归函数关于基本X2基本上呈现线性，但是图像中显示的残差与X1可能不是线性的，有可能要引进X1的平方项或者交叉项，由此说明了此模型拟合的效果是不错的。2、与上题对比可知在本题中拟合的模型的效果要由于上题变换前的模型效果。2.9解SAS源程序为DATAtwo_9;X1X2X3Y;CARDSPROCREGDATAtwo_9;MODELYX1X2X3;OUTPUTOUTbPPREDICTEDRRESIDUALHHSTUDENTSTUDENT;PROCCAPABILITYDATAbGRAPHICS;QQPLOT;RUN;PROCGPLOTDATAb;PLOTRESIDUAL*PREDICTEDRESIDUAL*X1RESIDUAL*X2RESIDUAL*X3;SYMBOLVDOTINONE;RUN;PROCREGDATAtwo_9;MODELYX1X2X3/SELECTIONADJRSQ;20RUN;PROCREGDATAtwo_9;MODELYX1X2X3/SELECTIONCP;RUN;PROCREGDATAtwo_9;MODELYX1;OUTPUTOUTa1PRESSPRESS;RUN;PROCMEANSUSSDATAa1;VARPRESS;RUN;PROCREGDATAa1;MODELYX1X2;RUN;PROCREGDATAtwo_9;MODELYX2;OUTPUTOUTa2PRESSPRESS;RUN;PROCMEANSUSSDATAa2;VARPRESS;RUN;PROCREGDATAtwo_9;MODELYX3;OUTPUTOUTa3PRESSPRESS;RUN;PROCMEANSUSSDATAa3;VARPRESS;RUN;PROCREGDATAtwo_9;MODELYX1X2;OUTPUTOUTa4PRESSPRESSPPREDICTEDRRESIDUALHHSTUDENTSTUDENT;PROCCAPABILITYDATAa4GRAPHICS;QQPLOT;RUN;PROCGPLOTDATAa4;PLOTRESIDUAL*PREDICTEDRESIDUAL*X1RESIDUAL*X2;SYMBOLVDOTINONE;RUN;RUN;PROCMEANSUSSDATAa4;VARPRESS;RUN;PROCREGDATAtwo_9;MODELYX1X3;21OUTPUTOUTa5PRESSPRESS;RUN;PROCMEANSUSSDATAa5;VARPRESS;RUN;PROCREGDATAtwo_9;MODELYX2X3;OUTPUTOUTa6PRESSPRESS;RUN;PROCMEANSUSSDATAa6;VARPRESS;RUN;PROCREGDATAtwo_9;MODELYX1X2X3;OUTPUTOUTa7PRESSPRESS;RUN;PROCMEANSUSSDATAa7;VARPRESS;RUN;PROCREGDATAtwo_9;MODELYX1X2X3/SELECTIONSTEPWISESLENTRY0.10SLSTAY0.10;RUN;PROCPRINT;RUN;通过SAS输出有关残差的结果如下ObsX1X2X3YPREDICTEDRESIDUALSTUDENTH150512.34848.0132-0.0132-0.001390.12786236462.35771.1062-14.1062-1.460880.09653340482.26663.88572.11430.215580.06798441441.87066.98263.01740.330890.19418528431.88984.45784.54220.500160.20085649542.93646.0057-10.0057-1.039300.10188742502.24659.1862-13.1862-1.333570.05262845482.45457.5835-3.5835-0.374220.11147952622.92633.5098-7.5098-0.908700.338181029502.17775.57221.42780.150220.124511129482.48977.750811.24921.177040.114931243532.46754.657812.34221.250750.056441338552.24758.7814-11.7814-1.227770.107761434512.35168.1805-17.1805-1.754520.070871553542.25740.963916.03611.719450.157171636492.06667.8383-1.8383-0.186060.054121733562.57963.994415.00561.660600.208781829461.98879.92948.07060.851200.12891221933492.16071.6196-11.6196-1.186870.071252055512.44941.71097.28910.818090.230762129522.37773.39353.60650.386810.157672244582.95247.95074.04930.436700.166872343502.36057.92572.07430.21