数学建模实验报告文档格式.docx

1、726340.9645325272.9961048445.010612657.2041426.8524938.122835.840926675.79637.4081110554.376129846.186137746.13030.366155639.0601624579.3801713352.7661855.9162. 某公司想用全行业的销售额作为自变量来预测公司的销售额，下表给出了1977-1981年公司销售额和行业销售额的分季度数据（单位：百万元）。（1）画出数据的散点图，观察用线性回归模型拟合是否合适。（2）建立公司销售额对全行业销售额的回归模型，并用DW检验诊断随机误差项的自相关性。（

2、3）建立消除了随机误差项自相关性后的回归模型。年季t公司销售额y行业销售额x197720.96127.321.413021.96132.721.52129.4197822.3913522.76137.123.48141.223.66142.8197924.1145.524.01145.324.54148.324.3146.4198025150.225.64153.126.36157.326.98160.7198127.52164.227.78165.61929.24168.72028.78171.7四、实验结果与数据处理1. Matlab代码： X1=66.290 40.964 72.996

3、45.010 57.204 26.852 38.122 35.840 75.796 37.408 54.376 46.186 46.130 30.366 39.060 79.380 52.766 55.916; Y=196 63 252 84 126 14 49 49 266 49 105 98 77 14 56 245 133 133; X=ones（18,1） X1 （X1.2）; b,bint,r,rint,stats=regress（Y,X）处理结果：b = -60.5239 1.7886 0.0302bint = -143.4598 22.4121 -1.4742 5.0513 0.

4、0002 0.0603r = 5.0447 -0.4989 20.7987 2.7433 -14.7658 4.6881 -2.6174 6.5692 17.1895 0.2908 -21.1635 11.3961 -9.3474 -7.6785 0.5151 -27.0424 14.9336 -1.0552rint = -22.6123 32.7016 -29.0151 28.0174 -3.0151 44.6125 -25.5842 31.0708 -41.2961 11.7646 -17.4529 26.8291 -30.9763 25.7415 -21.2462 34.3845 -6.

5、0579 40.4368 -28.0301 28.6116 -46.2827 3.9558 -16.1444 38.9366 -37.1409 18.4462 -33.0744 17.7174 -27.9507 28.9809 -42.7681 -11.3167 -11.6494 41.5167 -28.8865 26.7760stats =0.9747 289.1934 0.0000 182.0773参数参数参考值参数置信区间B0-60.5239-143.4598 ,22.4121B11.7886 -1.4742 ,5.0513B20.03020.0002 ,0.0603R= 0.9747

6、F=289.1934 p0.0000 s=182.0773由于置信水平a=0.05,处理结果p=0.00,p0.05=0.9747，指因变量Y的97.47%可由模型确定，Y与X1存在二次关系。,所以得到回归模型：Y=0.5239+1.7886*X1+0.0302*X12;结果表明年均收入和人寿保险额之间存在二次关系。接下来处理两个自变量X1,X2对Y是否有交互效应。因为Y与X1之间存在二次关系，所以我们设 X2=7 5 10 6 4 5 4 6 9 5 2 7 4 3 5 1 8 6; X=ones（18,1） X2 X1 -62.3489 5.6846 0.8396 0.0371 -73.5

7、027 -51.1952 5.2604 6.1089 0.3951 1.2840 0.0330 0.0412 -0.0512 0.3076 -1.3718 -0.6730 -3.7605 -1.3560 2.7129 -0.4817 0.5130 -0.3725 0.6842 2.6781 -1.0293 -0.3930 0.5561 1.3578 2.3248 -1.6456 -3.7791 3.6766 -3.5324 4.1475 -4.4124 1.6688 -4.4677 3.1217 -6.6500 -0.8710 -4.2144 1.5023 -0.7344 6.1602 -4.

8、2149 3.2516 -2.6183 3.6443 -4.1840 3.4390 -2.6447 4.0132 -0.7217 6.0779 -4.7396 2.6810 -3.8132 3.0272 -3.2676 4.3798 -0.4637 3.1793 -1.0358 5.6855 -5.2685 1.9773 1.0e+04 * 0.0001 1.1070 0.0000 0.000338.743459.7383 ,137.225113.52183.3538 . 30.3975=0.2% F=2.9 p=0.0001 s=5721-62.3489-73.5027 ,-51.19525

9、.2604 , 6.10890.3951 1.28400.03710.0330 0.0412 1.00 1107.0 0.00 0.00031.00指因变量Y可由X1与X2100%确定，F远远小于F的检验的临界值，p远小于a, 的系数均在置信区间内。可知Y与X1 ,X2有交互效应Y=-62.3489+ 5.6846X2+0.8396X1+0.0371X122.（1）散点图由散点图可看出x与y存在线性相关，可用线性回归模型拟合。（2）由散点图可看出，x与y存在正相关，所以使用一次回归模型 y =20.9600 21.4000 21.9600 21.5200 22.3900 22.7600 23.

10、4800 23.6600 24.1000 24.0100 24.5400 24.3000 25.0000 25.6400 26.3600 26.9800 27.5200 27.7800 29.2400 28.7800; x=127.3 130 132.7 129.4 135 137.1 141.2 142.8 145.5 145.3 148.3 146.4 150.2 153.1 157.3 160.7 164.2 165.6 168.7 171.7; b,bint,r,rint,stats=regress（y -2.2816 0.1822 -3.4309 -1.1324 0.1745 0.1900 0.0447 -0.0073 0.0607 0.2220 0.0716 0.0589 0.0318 -0.0798 -0.1318 -0.1853 -0.2020 -0.0958 -0.0882 0.0233 -0.0220 -0.0216 -0.1193 -0.1145 0.7807 -