31回归分析的基本思想及初步应用(2)

3.1回归分析的基本思想及初步应用(2),e=y-(bx+a)称为随机误差,温故知新,一.用心温故,R2越大模型越好,残差平方和越小精确度越高,3.相关指数R2,引例:从某大学中随机选出8名女大学生，其身高和体重数据如下表：,(1)求每个点(xi,yi)的残差,(2)画出残差的散点图,(3)求出相关指数R2,说明身高在多大程度上解释了体重的变化.,二.探求新知,-6.373,2.627,2.419,-4.618,1.137,6.627,-2.883,0.382,-8,-6,-4,-2,2,4,6,8,O,2,1,3,4,6,5,7,8,9,10,编号,残差,.,.,.,.,.,.,.,.,.R2=0.64,表明女大学生的身高解释了64%的体重变化。,残差点比较均匀地落在(以x轴为中心）水平带状区域内.模型较合适,带状区域的宽度越窄,模型拟合精度越高,回归方程的预报精度越高,.,4,3,2,1,0,-1,-2,-3,-4,01002003004005006007008009001000,45,40,35,30,25,20,15,10,5,0,-5,0102030405060708090100,2500,2000,1500,1000,500,0,-500,-1000,0102030405060708090100,200,150,100,50,0,-50,-100,-150,0102030405060708090100,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,（）分析下列残差图,所选用的回归模型效果最好的是（）,牛刀小试,（2）有下列说法：在残差图中，残差点比较均匀地落在水平的带状区域内，说明选用的模型比较合适。相关指数R2来刻画回归的效果,R2值越大，说明模型的拟合效果越好。比较两个模型的拟和效果，可以比较残差平方的大小，残差平方和越小的模型，拟合效果越好。正确的是（）,被害棉花,红铃虫喜高温高湿，适宜各虫态发育的温度为25一32，相对湿度为80一100，低于20和高于35卵不能孵化，相对湿度60以下成虫不产卵。冬季月平均气温低于一48时，红铃虫就不能越冬而被冻死。,创设情景,1953年，18省发生红铃虫大灾害，受灾面积300万公顷，损失皮棉约二十万吨。,因材施教,例2现收集了一只红铃虫的产卵数y和温度xoC之间的7组观测数据列于下表：,（1）试建立产卵数y与温度x之间的回归方程；并预测温度为28oC时产卵数目。（2）你所建立的模型中温度在多大程度上解释了产卵数的变化？,问题呈现：,画散点图,假设线性回归方程为：,选模型,合作探究,方案1,当x=28时，y=19.8728-463.7393,残差,编号,1,2,3,4,5,6,7,10,20,30,40,50,60,70,80,-10,-20,-30,-40,-50,-60,90,100,线性模型,53.46,17.72,-12.02,-48.76,-46.5,-57.11,93.28,19818.9,相关指数R20.7464,所以，一次函数模型中温度解释了74.64%的产卵数变化。,方案2,问题3,产卵数,气温,合作探究,t=x2,方案2解答,平方变换：令t=x2，产卵数y和温度x之间二次函数模型y=bx2+a就转化为产卵数y和温度的平方t之间线性回归模型y=bt+a,作散点图，并由计算器得：,将t=x2代入线性回归方程得：y=0.367x2-202.54,y和t之间的线性回归方程为y=0.367t-202.54，,当x=28时，y=0.367282-202.5485,残差,编号,1,2,3,4,5,6,7,10,20,30,40,50,60,70,80,-10,-20,-30,-40,-50,-60,90,100,二次函数模型,47.696,19.400,-5.832,-41.000,-40.104,-58.265,77.968,15448.4,相关指数R2=0.802所以二次函数模型中温度解释了80.2%的产卵数变化。,产卵数,气温,指数函数模型,方案3,合作探究,对数,方案3解答,当x=28oC时，y44,残差,编号,1,2,3,4,5,6,7,10,20,30,40,50,60,70,80,-10,-20,-30,-40,-50,-60,90,100,指数函数模型,-0.1944,1.7248,-9.1894,8.8521,-14.1219,33.2573,1471.5,指数回归模型中温度解释了98.5%的产卵数的变化,0.4987,最好的模型是哪个?,线性模型,二次函数模型,指数函数模型,比一比,最好的模型是哪个?,结论：无论从图形上直观观察，还是从数据上分析，指数函数模型是更好的模型。,数学思想：,数学方法：,数形结合的思想，化归思想及整体思想,数形结合法，转化法，换元法,数学知识：,建立回归模型及残差图分析的基本步骤,不同模型拟合效果的比较方法：相关指数和残差的分析,非线性模型向线性模型的转换方法,课堂总结,1.在画两个变量的散点图时，下面叙述正确的事（）(A)预报变量在x轴上，解释变量在y轴上（B）解释变量在x轴上，预报变量在y轴上（C）可以选择两个变量中任意一个变量在x轴上（D）可以选择两个变量中任意一个变量在y轴上2.一位母亲记录了她儿子3到9岁的身高，数据如下表。,由此她建立了身高与年龄的回归模型，她用这个模型预测儿子10岁时的身高，则下面的叙述正确的是（）（A）身高一定是145.83cm（B）身高在145.83CM以上（C）身高在145.83cm左右（D）身高在145.83cm以下,学以致用,3.在建立两个变量x与y的回归模型中，分别选择了4个不同模型，它们的相关指数如下，其中拟和得最好的模型是（）,(A)模型1的相关指数,为0.98,为0.80,为0.50,4.如果发现散点图中所有的样本点都在一条直线上，请回答下列问题：,（1）解释变量和预报变量的关系是，残差平方和是_（2）解释变量和预报变量之间的相关系数是