应用多元统计分析报告SAS作业

标准文案含42.351.150.79含111.961.481.04矿含42.351.150.79含111.961.481.04矿类型 序号Cu1Ag岩石化学成分的含量数据Bi 类型 序号CuAgBi12.580.900.95 82.251.981.0622.901.231.00 92.161.801.0633.551.151.00 不 102.331.741.1053.541.850.79 矿 121.941.401.0062.702.231.30133.001.301.0072.701.700.48142.781.701.48〔先验概率取为相等，并假定两类样本的协方差阵相等；试推断该标本是含矿还是不含矿？问题求解datad59;inputgroupx1-x3@@;datad59;inputgroupx1-x3@@;cards;12.580.90.9512.91.23113.551.15112.351.150.7913.541.850.7912.72.231.312.71.70.4822.251.981.0622.161.81.0622.331.741.121.961.481.0421.941.41231.3122.781.71.48；procprocprintdata=d59;run;procdiscrimdata=d59pool=yesdistancelist;classgroup;varx1-x3;run;大全总体均值相等的检验结果：D=3.19774，F=3.10891，p=0.0756<0.10，故在显著性水平=0.10时量总体的均值向量有显著差异，即认为争论这两个三元总体的判别问题是有意义的。线性判别函数为：Y31.110513.7895Cu8.2120Ag11.3311Bi,1Y28.737510.3139Cu8.9904Ag16.8578Bi.2判别结果为含矿的613对给定样本判别归类Cu，Ag，Bi2.95、2.15、1.541 2

贝叶斯判别的解D* D*, ,D* 为1 k D*t

X|Y(X)Y(X),jt,j1, ,k (t1, ,k)，t j由于Y1

44.6742Y2

46.9788，因此待判的样品判为不含矿。47,4,63〔2态总体。样品号X1X样品号X1X2X3X4类别名16.0-11.519.090.012-11.0-18.525.0-36.03390.2-17.017.03.024-4.0-15.013.054.0150.0-14.020.035.0260.5-11.519.037.037-10.0-19.021.0-42.0380.0-23.05.0-35.01920.0-22.08.0-20.0310-100.0-21.47.0-15.0111-100.0-21.515.0-40.021213.0-17.218.02.0213-5.0-18.515.018.011410.0-18.014.050.0115-8.0-14.016.056.01160.6-13.026.021.0317-40.0-20.022.0-50.031-8.0-14.016.056.0292.2-17.018.03.03-14.0-18.525.0-36.033后比较之。问题求解判别分析及判别归类datad510;inputx1-x4datad510;inputx1-x4group@@;procprint;procprint;run;procdiscrimdata=d510simplecards;6 -11.519901-11-18.525-36390.2-171732-4-15135410 -14203520.5-11.519373-10-1921-4230 -235-35120-228-203-100-21.47-151-100-21.515-40213-17.21822-5-18.51518110-1814501-8-14165610.6-1326213-40-2022-503-8-141656.92.2-17183.-14-18.5；25-36.pcovpcovwsscppsscpwcovdistancelist;classgroup;varx1-x4;run;23

=1.34

(3) 的12 0F0.63177p=0.651>0.10，故在显著性水平=0.10时量总体没有太大意义的。182，2911，2，3二次判别函数判别由第一问SAS数进展判别。此时贝叶斯判别的解D* D*, ,D* 为1 k D*t其中

(X)Z(X),jj1, ,k ,k)，t jZ(X)lnqj j

f(X)dj 012lnq2 j

lnSj

(X)jprocdiscrimdata=d510simplepool=noprocdiscrimdata=d510simplepool=nodistancelist;程序运行结果如以以下图。由此可知，171,1,3某城市的环保监测站与1982年在全市均匀地布置了14个监测点，每日三5〔监测点〕每种污染元素实测1515为该取样点的大气污染元素的含量〔3。表中最终一列给出的类号是〔123。3大气污染数据污染元素二氧化硫氮氧化物飘尘类别样品号(X)1(X)2(X)310.0450.0430.265220.0660.0390.264230.0940.0610.194240.0030.0030.102350.0480.0150.106360.2100.0660.263170.0860.0720.274280.1960.0720.211190.1870.0820.3011100.0530.0600.2092110.0200.0080.1123120.0350.0150.1703130.2050.0680.2841140.0880.0580.2152150.1010.0520.181160.0450.0050.122〔假设三个总体为多元正态总体，其协方差阵相等，先验概率取为各类样本的比例，并列出回判结果。的含量〔3属于哪一类。问题求解datad511;inputx1-x3datad511;inputx1-x3group@@;cards;0.0450.0430.26520.0660.0390.26420.0940.0610.19420.0030.0030.10230.0480.0150.10630.2100.0660.2631procprint;procprint;run;procdiscrimdata=d511simpledistancelist;classgroup;0.0860.0720.27420.1960.0720.21110.1870.0820.30110.0530.0600.20920.0200.0080.11230.0350.015