求回归直线方程

求回归直线方程 “求直线的回归方程”的教学设计 一(教学内容分析 本节课的主要内容为用最小二乘法求线性回归方程。所以，在内容重点的侧重上，本节课与上节课有较大的区别:上节课侧重于估算方法设计，在不同的数据处理过程中，体会回归直线作为变量相关关系代表这一概念特征;本节课侧重于估算方法评价与实际应用，在评价中使学生体会核心思想，理解核心概念。 考虑到本节课的教学侧重点与新课程的要求，对线性回归方程系数的计算公式，可直接给出。由于公式的复杂性，一方面，既要通过教学设计合理体现知识发生过程，不搞“割裂”;另一方面，要充分利用计算机或计算器，简化繁琐的求解系数过程，简化过于形式化的证明说理过程。 基于上述内容分析，确定本节课的教学重点为知道最小二乘法思想，并能根据给出的线性回归方程的系数公式建立线性回归方程。 二(教学目标分析 本节课要求学生了解最小二乘法思想，掌握根据给出的线性回归方程系数公式建立线性回归方程，理解线性回归方程概念和回归思想，在以上过程中体会随机思想: 1(能用数学符号刻画出“从整体上看，各点与此直线的点的偏差”的表达方式; 2(知道最小二乘法的思想，了解其公式的推导过程; 3(能结合具体案例，根据回归方程系数公式建立回归方程; 4(利用回归方程预测，体现用“确定关系研究相关关系”的回归思想; 三(重点，难点分析 在经历用不同估算方法描述两个变量线性相关的过程后，在学生现有知识能力范围内，如何选择一个最优方法，成为知识发展的逻辑必然。知识发展的要求与学生能力和经验的欠缺成为本节课将会遇到的最大矛盾。在教学中，要防止两种倾向:一是直接套用回归系数公式求解回归方程而回避说理过程;二是过多纠缠于数学刻画过程，甚至在课堂内花大量时间对回归系数公式进行证明说理。这两种倾向，都脱离了实际情况，前者忽略了“最小二乘法思想”，迷失了本节课的教学目标;后者人为拔高教材要求，脱离了本节课教学要求。 所以，本节课的教学难点是:如何通过数学方法刻画“从整体上看，各点与此直线的距离最小”并在此过程中了解最小二乘法思想。通过“降次举特例说明，进行合情推理”是学生突破此难点的一个方法。 四(教学过程设计 1(课题引入 问题1:(投影上节课探究结果)如何评价这些“直线”的优劣,理由呢, 问题2:能否从几何直观角度用文字语言叙述你的理由, 问题3:“从整体上看，各点与此直线的距离最小”中，距离等于偏差吗,作为of the works. Casement sash installation: required fixed hinge on the door and window frames, Windows Embedded in the box after a temporary fix, adjust the appropriate door after ... 12, scaffolding the construction phase of the project this project proposes to adopt the following three types of scaffolds for construction: part a, Foundation construction with double rows of single pipe scaffolding, to provide construction jobs and support. B, up to four-layer cantilever scaffold, six-layer cantilever, suspension height of 18.6 metres, does not exceed the standard 20 m c, exterior eave construction by Beijing feitian red construction machinery manufacturing ZLD500-630 electric hanging blue (see next figure). (A) the double rows of steel tubular scaffold with couplers 1, material preparation: erection of scaffold fittings should be prepared in advance, classified according to specifications, neatly stacked, steady, pile site does not have water. Material quality: tube selection φ 48x3.5 standard pipe, wall thickness and no cracks, no warpage, rounding and other defects. Pipe brush paint the pipe connected fastener using a right angle fastener, joint fasteners, rotate the fasteners, meet quality standards, no cracks, threads oil, no accusations, torsional flexibility. Mesh has closed key factory certificate, required construction safety supervision and management departments permit card shall not be used, safety nets and scaffolding with a nylon rope around the tie firmly, nylon connectors should 判断优劣的标准，可以等同吗, 设计意图:在上节课“计算预测值与实际值偏差”的经验基础上，通过学生对“从整体上看，各点与此直线的点的距离的最小”这一新标准与旧经验的冲突和联系，对“优劣问题”展开反思:从旧经验“单个点”到新标准“所有点”，突出“整体”二字;从旧经验“偏差计算”到新标准“点线距离”，对比几何描述直观性和代数表达便捷性，揭示出两者是同一标准的不同表述。 师生活动:在上节课铺垫的基础上，学生不难回想到上节课比较不同“回归直线”优劣的方法——通过计算点与直线对应点纵坐标差比较偏差。在此铺垫基础上，教师可结合图形，用代数符号y、标记，为下一步代数表达做好准备。第二问更具ii 有几何直观性，学生也易于接受此标准，达成“几何”与“代数”的转化、“距离”与“偏差”的转化。若学生对“距离”与“偏差”有疑问，教师可提出问题3，通过观察课本92页图2.3,6，简单介绍偏差处理法的优越性和等价性即可。 2(知识发展 设回归直线方程为，(x，y)表示第i个样本点， ii 问题1:你能用代数式来刻画“从整体上看，各点与此直线的偏差最小”吗, 问题2:偏差有正有负，我们可以怎么规避,比较绝对值处理和平方处理，我们选择哪种合适, 设计意图:几何问题代数化，为下一步探究作好准备，经历“几何直观”转化为“代数表达”过程，体验“最小二乘法思想”。 师生活动:在引入的设问中，已经解决了转化的问题，由于上节课学生有“用具体数据来计算偏差”的经验，学生易于抽象出各点偏差表示式y, =y,(bx+a)iiii(i=1，2，„，n)，进而不难得出:Q=(y,bx,a)+(y,bx,a)+„+(y,bx1122nn,a)。问题2可在投影屏上举极端例子说明，学生会发现此处理方法的局限性，学生可能会提出多种方法，教师肯定其观点，说明去绝对值对后续研究不便，可类比“方差”处理方式，采用平方处理方法，教师投影: 问题3:从代数上说，偏差最小既哪个量最小,当样本点的坐标(x，y)确定时，ii上述表达式可否化为关于a、b的二次式呢, 设计意图:体会最小二乘法思想，不经历公式化简无法真正理解，而直接从n个点的公式化简，教学要求、教学时间、学生能力都没达到这个高度。而由具体到抽象，由特殊到一般，是学生顺利完成认知过程的一般性原则。通过此问，让学生了解化简的结果，在此过程中，既熟悉了新符号，又通过观察展开式，能唤起学生已有认知结构中关于处理带参数的二次多项式最小值问题的数学处理方法，揭示n个点的代数式本质也是关于a、b的二次多项式，从而了解最小二乘法思想，突破教学难点。 of the works. Casement sash installation: required fixed hinge on the door and window frames, Windows Embedded in the box after a temporary fix, adjust the appropriate door after ... 12, scaffolding the construction phase of the project this project proposes to adopt the following three types of scaffolds for construction: part a, Foundation construction with double rows of single pipe scaffolding, to provide construction jobs and support. B, up to four-layer cantilever scaffold, six-layer cantilever, suspension height of 18.6 metres, does not exceed the standard 20 m c, exterior eave construction by Beijing feitian red construction machinery manufacturing ZLD500-630 electric hanging blue (see next figure). (A) the double rows of steel tubular scaffold with couplers 1, material preparation: erection of scaffold fittings should be prepared in advance, classified according to specifications, neatly stacked, steady, pile site does not have water. Material quality: tube selection φ 48x3.5 standard pipe, wall thickness and no cracks, no warpage, rounding and other defects. Pipe brush paint the pipe connected fastener using a right angle fastener, joint fasteners, rotate the fasteners, meet quality standards, no cracks, threads oil, no accusations, torsional flexibility. Mesh has closed key factory certificate, required construction safety supervision and management departments permit card shall not be used, safety nets and scaffolding with a nylon rope around the tie firmly, nylon connectors should 2 师生活动:教师指出:可采用n个偏差的平方和Q=(y,bx,a)+(y,bx,112222a)+„+(y,bx,a)表示n个点与相应直线在整体上的接近程度:记nn Q=(向学生说明的意义)。 在此基础上，给出可求出使Q为最小值时的a、b的值的线性回归方程系数公式: 问题4:这个公式不要求记忆，但要会运用这个公式进行运算，那么，要求a,b的值，你会按怎样的顺序求呢, 设计意图:公式不要求推导，又不要求记忆，学生对这个公式缺少感性的认识，通过这个问题，使学生从感性的层次上对公式有所了解。 师生活动:由于这个公式比较复杂，因此在运用这个公式求a,b时，必须要有条理，先求什么，再求什么，比如，我们可以按照 顺序来求，再代入公式。 问题5:回归直线通过样本点中心，观察此公式，比照平均数与样本数据之间的关系，你能发现回归直 线方程如何体现这点, 设计意图:在不确定问题探讨中出现的确定性性质，比较有戏剧性，能再次激发学生的探究欲望，而此问题的探究，使得学生在“回归直线是两个变量具有相关关系的代表”的理解上，上升到“回归直线是双变量样本点的中心”这一高度，深化对回归直线和回归思想的理解，完成学生认知结构的再次建构。 3(知识深化: 问题1:观察公式，根据表一数据，需要计算哪些新数据，才能求出线性回归方程系数,计算量大不大,我们用计算器来代替这重复的劳动，请大家一起跟我来操作 人体的脂肪百分比和年龄 表一: 年龄 23 39 45 50 54 57 60 脂肪 9(5 21(2 27(5 28(2 30(2 30(8 35(2 设计意图:公式形式化程度高、表达复杂，通过分解，可加深对公式结构的理解。同时，通过例题，反映数据处理的繁杂性，体现计算器处理的优越性。 of the works. Casement sash installation: required fixed hinge on the door and window frames, Windows Embedded in the box after a temporary fix, adjust the appropriate door after ... 12, scaffolding the construction phase of the project this project proposes to adopt the following three types of scaffolds for construction: part a, Foundation construction with double rows of single pipe scaffolding, to provide construction jobs and support. B, up to four-layer cantilever scaffold, six-layer cantilever, suspension height of 18.6 metres, does not exceed the standard 20 m c, exterior eave construction by Beijing feitian red construction machinery manufacturing ZLD500-630 electric hanging blue (see next figure). (A) the double rows of steel tubular scaffold with couplers 1, material preparation: erection of scaffold fittings should be prepared in advance, classified according to specifications, neatly stacked, steady, pile site does not have water. Material quality: tube selection φ 48x3.5 standard pipe, wall thickness and no cracks, no warpage, rounding and other defects. Pipe brush paint the pipe connected fastener using a right angle fastener, joint fasteners, rotate the fasteners, meet quality standards, no cracks, threads oil, no accusations, torsional flexibility. Mesh has closed key factory certificate, required construction safety supervision and management departments permit card shall not be used, safety nets and scaffolding with a nylon rope around the tie firmly, nylon connectors should 师生活动:可让学生观察公式，充分讨论，得出要计算:n、、、、 五个新数据。而后教师可偕同学生，对计算器操作方式提供示范，师生共同完成。 问题2:利用计算器，根据表二，请同学们独立解决求出表中两变量的回归方程: 表二: 年龄 27 41 49 53 56 58 61 脂肪 17(8 25(9 26(3 29(6 31(4 33(5 34(6 师生活动:教师利用Excle软件，示范操作，并适时给出回归直线答案，检测正确与否。 年龄 23 39 45 50 54 57 60 脂肪 9(5 21(2 27(5 28(2 30(2 30(8 35(2 回归直线为:y=0.6541x-4.5659 年龄 27 41 49 53 56 58 61 脂肪 17(8 25(9 26(3 29(6 31(4 33(5 34(6 回归直线为:y=0.4767x+4.9476 师生活动:教师利用Excle软件，合并表中数据，求出此时的回归直线，比较回归直线异同。 问题3:同样问题背景，为什么回归直线不止一条,回归方程求出后，变量间的相关关系是否就转变成确定关系, 问题4:若给出的样本数据相关程度较弱，按照公式能否求出系数a、b,此时的直线方程是回归直线吗, 设计意图:明确样本点的选择影响回归直线方程，体现统计的随机思想。同时，明确其揭示的是相关关系而非函数的确定关系，而且最小二乘法只是某一标准下的一种数据处理方法，使学生更全面的理解回归直线这一核心概念。在重复求解回归直线的过程中，使学生掌握利用计算器求回归直线的操作方法，了解计算机处理方法。 五(习题检测设计 1、下表是某小卖部6天卖出热茶的杯数与当天气温的对比表(用计算器直接求回归直线): of the works. Casement sash installation: required fixed hinge on the door and window frames, Windows Embedded in the box after a temporary fix, adjust the appropriate door after ... 12, scaffolding the construction phase of the project this project proposes to adopt the following three types of scaffolds for construction: part a, Foundation construction with double rows of single pipe scaffolding, to provide construction jobs and support. B, up to four-layer cantilever scaffold, six-layer cantilever, suspension height of 18.6 metres, does not exceed the standard 20 m c, exterior eave construction by Beijing feitian red construction machinery manufacturing ZLD500-630 electric hanging blue (see next figure). (A) the double rows of steel tubular scaffold with couplers 1, material preparation: erection of scaffold fittings should be prepared in advance, classified according to specifications, neatly stacked, steady, pile site does not have water. Material quality: tube selection φ 48x3.5 standard pipe, wall thickness and no cracks, no warpage, rounding and other defects. Pipe brush paint the pipe connected fastener using a right angle fastener, joint fasteners, rotate the fasteners, meet quality standards, no cracks, threads oil, no accusations, torsional flexibility. Mesh has closed key factory certificate, required construction safety supervision and management departments permit card shall not be used, safety nets and scaffolding with a nylon rope around the tie firmly, nylon connectors should 气温/? ,1 26 18 13 10 4 杯数 20 24 34 38 50 64 (1)画散点图;(2)从散点图中发现温度与热饮销售杯数之间关系的一般规律; (3)求回归方程; (4)按照回归方程，计算温度为10度时销售杯数。为什么与表中不同,如果某天的气温是,5?时，预测这天小卖部卖出热茶的杯数; 设计意图:通过此题，让学生完整经历求回归直线过程。其中第4问，让学生体会到即使是相比下“最优”的所获得的回归直线，也存在着一定的误差，从中体会无论方法的优劣，统计学中随机性无法避免。而在预测值的计算中，体现了回归直线的应用价值。 六、课堂小结设计 请你对本节课的内容进行小结，并谈谈你的收获以及课后的学习安排。 设计意图:让学生进行小结，谈谈体会，帮助他们回归反思、归纳概括。 of the works. Casement sash installation: required fixed hinge on the door and window frames, Windows Embedded in the box after a temporary fix, adjust the appropriate door after ... 12, scaffolding the construction phase of the project this project proposes to adopt the following three types of scaffolds for construction: part a, Foundation construction with double rows of single pipe scaffolding, to provide construction jobs and support. B, up to four-layer cantilever scaffold, six-layer cantilever, suspension height of 18.6 metres, does not exceed the standard 20 m c, exterior eave construction by Beijing feitian red construction machinery manufacturing ZLD500-630 electric hanging blue (see next figure). (A) the double rows of steel tubular scaffold with couplers 1, material preparation: erection of scaffold fittings should be prepared in advance, classified according to specifications, neatly stacked, steady, pile site does not have water. Material quality: tube selection φ 48x3.5 standard pipe, wall thickness and no cracks, no warpage, rounding and other defects. Pipe brush paint the pipe connected fastener using a right angle fastener, joint fasteners, rotate the fasteners, meet quality standards, no cracks, threads oil, no accusations, torsional flexibility. Mesh has closed key factory certificate, required construction safety supervision and management departments permit card shall not be used, safety nets and scaffolding with a nylon rope around the tie firmly, nylon connectors should