一元线性回归分析.doc

实习二 线性回归分析一、 一元线性回归分析1、目的：通过对一元线性回归分析程序设计及完成算例，掌握一元线性回归分析的基本原理和方法。2、算例：算例一（1）、原始数据样品号1234567891011（千卡）8.307.87.77.26.86.25.65.55.004.704.30.030.050.080.100.150.200.270.300.340.400.45（2）按程序格式排列数据如下（3）处理以后的数据如下图算例二（1）、原始数据样品号1234567距离x23457810含量y106.42108.20109.58109.50110.00109.93110.49样品号8910111213距离x111415161819含量y110.59110.60110.90110.76111.00111.20（2）按程序格式排列数据如下（3）处理以后的数据如下图算例三收集到广西晚二迭世含煤建造30个工程点煤系厚度（H）和煤层厚度（M）的资料，如表三（提示：用抛物线方程）作二元一次线性回归。样点编号12345678910煤厚（M）1.801.050.560.02.300.922.502.933.262.28煤系厚（H）121.257.363.138.0154.6138.8134.6126.9196.1328.7样点编号11121314151617181920煤厚（M）0.401.103.490.460.02.802.540.924.332.84煤系厚（H）528.5480.0420.0418.520.099.1378.5596.3155.8357.7样点编号21222324252627282930煤厚（M）3.352.061.633.701.181.84.312.222.704.00煤系厚（H）240.9154.8116.7183.7127.0165.4192.4179.1230.0226.0试建立煤层厚度（M）对含煤建造厚度（H）的回归方程，并作显著性检验。（1）、原始数据30,11,1.80,121.22,1.05,57.33,0.56,63.14,0.0,38.0,5,2.30,154.66,0.92,138.27,2.50,134.68,2.93,126.99,3.26,196.110,2.28,328.711,0.40,528.512,1.10,480.013,3.49,420.014,0.46,418.615,0.0,20.016,2.80,99.117,2.54,378.518,0.92,596.319,4.33,155.820,2.84,357.721,3.35,240.922,2.06,154.823,1.63,116.724,3.7,183.725,1.18,127.026,1.8,165.427,4.31,192.428.2.22,179.129,2.70,230.030,4.00,226.0（3）处理以后的数据如下图详细运算结果如下：线性回归或逐步回归计算: 回归方程 : Y= 64.72754+-2.402171X( 1) 复相关系数.315750006668924 标 准 差12.1478208698231 检 验 值=3.100677 拟合表: 序号实测值回归值偏差值1121.20060.40460.796257.30062.205-4.905363.10063.382-0.282438.00064.728-26.72852.30052.717-50.41760.92050.315-49.39572.50047.912-45.41282.93045.510-42.58093.26043.108-39.848102.28040.706-38.426110.40038.304-37.904121.10035.901-34.801133.49033.499-30.009140.46031.097-30.637150.00028.695-28.695162.80026.293-23.493172.54023.891-21.351180.92021.488-20.568194.33019.086-14.756202.84016.684-13.844213.35014.282-10.932222.06011.880-9.820231.6309.478-7.848243.7007.075-3.375251.1804.673-3.493261.8