数学建模 实验报告

精选文库数学建模实验报告实验序号：实验8 实验项目名称：统计回归模型学号1210012143姓名詹建妹专业、班12信计实验地点实4-401指导教师吴春红实验时间2014.4.29一、实验目的及要求通过对具体实例的分析，学会运用统计回归方法建立模型的方法。二、实验设备（环境）及要求多媒体机房，单人单机，独立完成三、实验内容与步骤1. 表1列出了某城市18位3544岁经理的年平均收入x1（千元），风险偏好度x2和人寿保险额y（千元）的数据，其中风险偏好度是是根据每个发给经理的问卷调查表综合评估得到的，它的数值越大，就越偏爱高风险，研究人员想研究此年龄段中的经理所投保的人寿保险额与年均收入及风险偏好度之间的关系。研究者预计，经理年均收入和人寿保险之间存在着二次关系，并有把握的认为风险偏好度对人寿保险额有线性效应，但对于风险偏好度对人寿保险额是否有二次效应以及两个自变量是否对人寿保险额有交互效应，心中没底。序号yX1X2119666.290726340.9645325272.9961048445.0106512657.204461426.852574938.122484935.8406926675.7969104937.40851110554.3762129846.1867137746.1304141430.3663155639.06051624579.38011713352.76681813355.91662. 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。（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x19771120.96127.32221.41303321.96132.74421.52129.419781522.391352622.76137.13723.48141.24823.66142.819791924.1145.521024.01145.331124.54148.341224.3146.4198011325150.221425.64153.131526.36157.341626.98160.7198111727.52164.221827.78165.631929.24168.742028.78171.7四、实验结果与数据处理1. Matlab代码： X1=66.290 40.964 72.996 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 0.0002 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 -6.0579 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 F=289.1934 p0.0000 s=182.0773由于置信水平a=0.05,处理结果p=0.00,p X1=66.290 40.964 72.996 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; X2=7 5 10 6 4 5 4 6 9 5 2 7 4 3 5 1 8 6; Y=196 63 252 84 126 14 49 49 266 49 105 98 77 14 56 245 133 133; X=ones(18,1) X2 X1 (X1.2); b,bint,r,rint,stats=regress(Y,X)处理结果：b = -62.3489 5.6846 0.8396 0.0371bint = -73.5027 -51.1952 5.2604 6.1089 0.3951 1.2840 0.0330 0.0412r = -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 -1.6456rint = -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 -4.2149 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 1.9773stats = 1.0e+04 * 0.0001 1.1070 0.0000 0.0003B038.743459.7383 ,137.2251B113.52183.3538 . 30.3975R=0.2% F=2.9 p=0.0001 s=5721参数参数参考值参数置信区间-62.3489-73.5027 ,-51.1952 5.68465.2604 , 6.1089 0.83960.3951 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Matlab代码： y =20.9600 21.4000 21.9600 21.5200 22.3900 22.7600 23.4800 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,X)处理结果：b = -2.2816 0.1822bint = -3.4309 -1.1324 0.1745 0.1900r = 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 -0.2260rint = -0.3886 0.4780 -0.4486 0.4340 -0.3859 0.5072 -0.2030 0.6470 -0.3791 0.5222 -0.3956 0.5134 -0.4283 0.4918 -0.5396 0.3800 -0.5898 0.3262 -0.6384 0.2677 -0.6536 0.2496 -0.5563 0.3647 -0.5488 0.3723 -0.4376 0.4843 -0.4786 0.4346 -0.4727 0.4296 -0.5591 0.3204 -0.5510 0.3221 0.6132 0.9481 -0.6315 0.1794stats = 1.0e+03 *0.0010 2.4381 0.0000 0.00