1、应用回归分析习题413答案4.13(1)用普通最小二乘法建立 y与x回归方程.(2)用残差图及DW检验诊断序列的自相关data dxh;in put obs x y;cards ;1127.320.962130 21.43132.721.964129.421.525135 22.396137.122.767141.123.488142.823.669145.524.110145.324.0111148.324.5412146.424.2813150.22514153.125.6415157.326.4616160.726.9817164.227.5218165.627.7819168.728.
2、2420172 28.78run ;procprint ;run ;procgplotdata =dxh;ploty*x;run ;procregdata =dxh; |modely=x/clb p r spec dw; |outputout=out r =residual; |run ;方差分析源自由度平方 和均方F值Pr F模型1110.59832110.5983211648.6 |t|95%置信限In tercept1-1.434830.24196-5.93.0001-1.94316-0.92650x10.176160.00163107.93 F模型113.1333013.1333024
3、67.41.0001误差170.090490.00532校正合计1813.22379P值 |t| 95%置信限In tercept估计值-0.300060.17268误差0.17763 -1.69 0.10940.00348 49.67 F模型11380.746041380.74604235188 |t|95%置信限0.166840.00034402 484.96 F模型12.115932.11593381.34 |t|In tercept10.032890.025851.270.2203difx10.160960.0082419.53 F模型22205552110277610.150.000
4、2误差495326177108697校正合计517531729P值 |t|95%置信限In tercep1-574.0623349.2707-1.60.1067-1275.9482127.8234t95446x11191.0984973.309172.610.012143.77821338.41878x212.045140.910692.250.02930.215043.87524回归分析方程y=-574.06239 +19105849 + 2 04514兀Durbi n-Watson D0.745观测数52第一阶自相关0.615DW值=0.745 F模型22865658143282921.5
5、5 |t|95%置信限In tercep1-178.775290.3381-1.90.0536-360.41232.86189t2982x1_t_11211.1104347.74734.42 F模型24033892201694625.04 |t|In tercept17.6981039.754210.190.8473difxl1209.8910644.143164.75.0001difx211.398980.582822.400.0203回归方程: g = 7-6M101209拠1。&4斥 *13989fiA其中如二只-Ph,唇二耳一也,-洛(4 )比较以上各方法所建回归方程的优良性在回归模型不存在序列相关时,普通最小二乘法比迭代法和一阶差分法操作起来更简便, 但是,当一个回归模型存在序列相关性时,普通最小二乘法所建立的回归方程就不适用了, 这时需要使用迭代法或一阶差分法 由于一阶差分法的应用条件是自相关系数 P=1 ,当P接近1时,一阶差分法比迭代法好,当原模型存在较高程度的一阶自相关的情况时,一般使 用一阶差分法而不用迭代法,因为一阶差分法比迭代法简单而且迭代法需要用样本估计自相 关系数P,对P的估计误差会影响迭代法的使用效率, 同时迭代法的算法时间复杂度比一阶