用最小二乘法求解线性模型及对模型的分析

1、用最小二乘法求解线性模型及对模型的分析作者:邓春亮1、研究30名儿童体重为因变量与身高为自变量的关系，儿童体重与身高的记录如下表：编号体重Y (kg)身高X(cm) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 22.60 21.50 19.10 21.80 21.50 20.10 18.80 22.00 21.30 24.00 23.30 22.50 22.90 19.50 22.90 22.30 22.70 23.50 21.50 25.50 25.00 26.10 27.9

2、0 26.80 27.20 24.40 24.40 23.00 26.30 28.80 119.80 121.70 121.40 124.40 120.00 117.00 118.00 118.80 124.20 124.80 124.70 123.10 125.30 124.20 127.40 128.20 126.10 128.60 129.40 126.90 126.50 128.20 131.40 130.80 133.90 130.40 131.30 130.20 136.00 138.00试用计算机完成下面统计分析：（1） 应用最小二乘法求经验回归方程；（2） 以拟合值为横坐标，残

3、差为纵坐标，作残差图，分析Gauss-Markou假设对本例的适用性；（3） 考虑因变量的变换,再对新变量和重复（1）和（2）的统计分析;（4） 将Box-Cox变换应用到本例，计算变换参数的值，并做讨论。说明：第一题的数据和结果文件见附件1，下面第二题的数据文件和结果文件见附件2，必要时可参看。解：（1）在SPSS窗口中录入数据，首先进行异常值检测，探查对回归估计有异常大影响的数据。先利用SPSS画出体重与身高的散点图图1从图1可以看出没有明显不一致的点。也可以通过SPSS软件计算COOK统计量，看下表表1编号残差学生化残差centerCOOK统计量 1 2 3 4 5 6 7 8 9 10

4、 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 1.88378 .03312 -2.24835 -.73361 .70477 .49003 -1.20506 1.67887 -1.15460 1.30835 .64786 .48000 .01081 -2.95460 -.81887 -1.73494 -.50526 -.69298 -3.00905 1.97867 1.63670 2.06506 2.60078 1.73783 .91306 -.50413 -.85971 -1.82512 -.81662 .89321.

5、 1.27241 .02204 -1.49944 -.48247 .47518 .34183 -.82971 1.14538 -.75974 .85964 .42576 .31710 .00710 -1.94417 -.53756 -1.14067 -.33146 -.45611 -1.98610 1.29824 1.07368 1.35771 1.73550 1.15517 .62283 -.33434 -.57329 -1.20912 -.57199 .64671. .05491 .02770 .03138 .00489 .05161 .11182 .08920 .07294 .00594

6、 .00310 .00351 .01355 .00143 .00594 .00139 .00434 .00008 .00643 .01183 .00038 .00003 .00434 .03249 .02522 .07268 .02088 .03121 .01887 .11878 .17316. .07835 .00002 .07778 .00463 .01048 .00992 .04807 .07800 .01180 .01397 .00347 .00247 .00000 .07726 .00520 .02547 .00190 .00431 .09329 .02940 .01989 .036

7、08 .10611 .04150 .02300 .00320 .01134 .04026 .02935 .05442.从上面数据看残差值和中心化的杠杆率center的值没有异常大的，数据,这里(= center+1/n)， COOK统计量值也没有异常大的数据，一般来说，残差值和杠杆率越大，COOK统计量就越大，残差值和杠杆率越小，COOK统计量就越小。可见这些数据是比较一致的。接下来对这些数据求解经验回归方程。然后利用最小二乘法，在SPSS中Analyze菜单下依次选择Regress:2-Stage Least Square,选择因变量和自变量执行可输出结果如下表：表2MODEL: MOD_3

8、.Equation number: 1Dependent variable. 体重YListwise Deletion of Missing DataMultiple R .80301R Square .64483Adjusted R Square .63215Standard Error 1.55047 Analysis of Variance: DF Sum of Squares Mean SquareRegression 1 122.20765 122.20765Residuals 28 67.31102 2.40396F = 50.83587 Signif F = .0000- Var

9、iables in the Equation -Variable B SE B Beta T Sig T身高X .395087 .055412 .803014 7.130 .0000(Constant) -26.615154 7.007449 -3.798 .0007Correlation Matrix of Parameter Estimates 身高X身高X 1.0000000这里可以看出所求经验回归方程的常数项(Constant) 为-26.615154，身高X的系数为0.395087。故经验回归方程为：=-26.615154+0.395087（2）通过SPSS,可得拟合值与残差如下表表

10、3：拟合值与残差表体重 Y 身高X拟合值 残差 22.60 21.50 19.10 21.80 21.50 20.10 18.80 22.00 21.30 24.00 23.30 22.50 22.90 19.50 22.90 22.30 22.70 23.50 21.50 25.50 25.00 26.10 27.90 26.80 27.20 24.40 24.40 23.00 26.30 28.80119.80 121.70 121.40 124.40 120.00 117.00 118.00 118.80 124.20 124.80 124.70 123.10 125.30 124.20

11、 127.40 128.20 126.10 128.60 129.40 126.90 126.50 128.20 131.40 130.80 133.90 130.40 131.30 130.20 136.00 138.00 20.71622 21.46688 21.34835 22.53361 20.79523 19.60997 20.00506 20.32113 22.45460 22.69165 22.65214 22.02000 22.88919 22.45460 23.71887 24.03494 23.20526 24.19298 24.50905 23.52133 23.3633

12、0 24.03494 25.29922 25.06217 26.28694 24.90413 25.25971 24.82512 27.11662 27.90679 1.88378 .03312 -2.24835 -.73361 .70477 .49003 -1.20506 1.67887 -1.15460 1.30835 .64786 .48000 .01081 -2.95460 -.81887 -1.73494 -.50526 -.69298 -3.00905 1.97867 1.63670 2.06506 2.60078 1.73783 .91306 -.50413 -.85971 -1

13、.82512 -.81662 .89321以拟合值为横坐标，残差为纵标，得残差图图2从图中可以看出，残差图没有明显的不一致的征兆，则可以认为Gauss-Markou假设对本例基本上是合理的。（3）作变换,这时用同样的方法可求得经验回归方程为：=-0.314471+0.040641其预测值与残差如下表U拟合值残差4.754.644.374.674.644.484.344.694.624.904.834.744.794.424.794.724.764.854.645.055.005.115.285.185.224.944.944.805.135.37.4.554264.631484.619294.

14、741214.562394.440474.481114.513624.733084.757474.753404.688384.777794.733084.863134.895654.810304.911904.944414.842814.826564.895655.025705.001315.127304.985055.021634.976935.212645.29392.4.554264.631484.619294.741214.562394.440474.481114.513624.733084.757474.753404.688384.777794.733084.863134.89565

15、4.810304.911904.944414.842814.826564.895655.025705.001315.127304.985055.021634.976935.212645.29392.以拟合值为横坐标，残差为纵坐标，作残差图得图3从图3看，此时的残差图也没有明显的不一致的趋势，认为Gauss-Markou假设对本例基本上是合理的。（4）将因变量进行Box-Cox变换，变换后原来的因变量变为计算不同值对应的残差平方和，这里分别取值为-1.5，-1，-0.5，0，0.5，1，1.5，计算分别计算，然后计算对应的残差平方和，这里n=30,计算得到如表所示，这里表示。表5编号UZ1Z2Z

16、3Z4Z5Z6Z71234567891011121314151617181920212223242526272829304.754.644.374.674.644.484.344.694.624.904.834.744.794.424.794.724.764.854.645.055.005.115.285.185.224.944.944.805.135.37.1707.191705.941702.581706.301705.941704.111702.091706.531705.701708.571707.911707.081707.501703.221707.501706.861707.30

17、1708.101705.941709.851709.441710.301711.541710.811711.081708.931708.931707.611710.451712.08.513.21511.99508.85512.34511.99510.25508.40512.56511.76514.59513.92513.10513.52509.43513.52512.89513.31514.12511.99515.91515.49516.39517.72516.93517.23514.96514.96513.62516.55518.32.176.17174.98172.05175.31174

18、.98173.33171.64175.53174.76177.56176.88176.06176.48172.57176.48175.85176.27177.07174.98178.92178.48179.43180.86180.00180.32177.93177.93176.58179.59181.52.72.2571.0968.3571.4271.0969.5367.9871.6370.8873.6472.9672.1572.5668.8372.5671.9472.3573.1671.0975.0574.5975.5977.1376.2076.5474.0374.0372.6675.767

19、7.87.36.1435.0132.4535.3235.0133.5432.1235.5334.8137.5436.8436.0436.4432.8936.4435.8436.2437.0435.0138.9938.5139.5641.2340.2140.5837.9337.9336.5439.7542.04.21.6020.5018.1020.8020.5019.1017.8021.0020.3023.0022.3021.5021.9018.5021.9021.3021.7022.5020.5024.5024.0025.1026.9025.8026.2023.4023.4022.0025.3

20、027.80.14.7413.6711.4213.9613.6712.3411.1514.1513.4816.1415.4414.6415.0411.7915.0414.4514.8415.6413.6717.6917.1718.3320.2719.0819.5116.5516.5515.1418.5421.27.通过SPSS软件运行得到的方差分析表，可知道相应的残差平方和，具体数据如下表所示：表6-1.5-1-0.500.511.5RSS73.14370.51468.63867.49267.05167.31168.277通过表6的简单比较可以看出当时，残差平方和达到最小，因此我们可以近似地认

21、为0.5就是变换参数的最优选择.2、研究儿童的体重与身高和胸围之间的关系是具有一定现实意义的，因为这种关系使我们能够用简单的方法从和的值去估计一个儿童的体重，下表是一组观测数据:表编号体重身高胸围 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 3022.6021.5019.1021.8021.5020.1018.8022.0021.3024.0023.3022.5022.9019.5022.9022.3022.7023.5021.5025.5025.0026.1027.9026.802

22、7.2024.4024.4023.0026.3028.80119.80121.70121.40124.40120.00117.00118.00118.80124.20124.80124.70123.10125.30124.20127.40128.20126.10128.60129.40126.90126.50128.20131.40130.80133.90130.40131.30130.20136.00138.0060.5055.5056.5060.5057.7057.0057.1061.7058.4060.8060.0060.0065.2053.7059.5060.1057.4060.405

23、2.0061.5063.9063.0063.1061.5065.8062.6059.5062.5060.0063.70试用计算机完成下面的统计分析：（1） 先假设Y与和有如下线性关系：，做最小二乘分析，并做相应的残差图。试计算Box-Cox变换参数的值.（2） 对（1）中计算出的变换参数值，做相应的Box-Cox变换，并对变换后的因变量做对和的最小二乘回归，并做残差图。解：（1）先计算中心化杠杆率center和COOK统计量的值表2-1编号拟合值残差学生化残差centerCOOK统计量12345678910111213141516171819202122232425262728293021.5

24、170320.2753020.5475322.8902020.5637519.4149619.7496521.6526622.0707623.1181422.7988622.3212424.8592320.3703823.4239623.8798422.2761524.1077821.3076323.9982724.7471324.9290125.9204325.1624727.6435225.4410224.5881725.3451426.1720728.107691.082971.22470-1.44753-1.09020.93625.68504-.94965.34734-.77076.8

25、8186.50114.17876-1.95923-.87038-.52396-1.57984.42385-.60778.192371.50173.252871.170991.979571.63753-.44352-1.04102-.18817-2.34514.12793.69231.999761.12888-1.31902-.97103.85394.64630-.88474.33076-.68711.78544.44527.15985-1.85996-.83580-.46553-1.40437.38119-.54071.210881.33663.231521.054401.797031.471

26、37-.42282-.93760-.17089-2.10925.12316.67800.07509.07238.05156.00889.05330.11302.09125.12873.01058.00882.00418.01640.12356.14264.00413.00510.02725.00666.33436.00754.06029.02950.04463.02553.13059.02995.04540.02738.14686.17443.04052.05021.05380.01386.02305.02387.03713.00705.00723.00905.00258.00045.2145

27、9.04973.00281.02627.00312.00406.00862.02538.00185.02484.09102.04514.01168.01980.00083.09585.00111.04018从表中2-1的计算结果可以看出，第19个观测的杠杆率最高为0.33436.。因此，第19个样本点最有可能对模型拟合造成较大的影响。然后求解经验回归方程，从运行结果的方差分析表2-2（ANOVA(b)）可以看出F统计量的P-值（Sig.）为0.000，这表明模型在总体中是显著的。表2-2 ANOVA(b)表2-3从回归系数计算分析表2-3（Coefficients(a)），可知道回归方程的常数

28、项为-36.133，自变量身高和胸围相对应的未标准化的回归系数（Unstandardized Coefficients）分别为0.299、0.362，因而回归方程为且从表中可知3个系数的t统计量的P值均为0.000，这表明模型在总体中是显著的。以拟合值为横坐标，残差为纵坐标，作残差图：图2-1 残差图从图2-1可以看出，残差图从左至右逐渐散开呈漏斗状，这是误差方差不相等的征兆。考虑将因变量进行Box-Cox变换，跟第一题的（4）问同样。这里同样分别取值为-1.5，-1，-0.5，0，0.5，1，1.5，计算分别计算，然后计算对应的残差平方和，这里n=30,计算得到如表1-5所示，然后计算对应自变量和的残差平方和。得方差分析表如下从上面的方差分析表中可以得到对应的残差平方和表2-4-1.5-1-0.500.511.5RSS37.90036.24835.20634.74734.85735.53436.787从这个表中可的简单比较可以看出当时，残差平方和达到最小，而对应的残差平方和次之为34.857，且从的方差分析表可知它们对应的P值都为0.000，都具有显著性。现在再看和时,对应因变量和对应的回归系数分析表。从上面两个表可知，因变量为，