商务与经济统计
商务与经济统计案例题答案
商务与经济统计案例题答案1、．（年温州一模考）以下不属于会计特点的是（）[单选题] *A会计具有一整套科学实用的专门方法B会计是以货币作为主要计量尺度C会计具有连续性、系统性、综合性、全面性D会计是一种经济管理活动(正确答案)2、企业的下列固定资产中，不计提折旧的是（）。
[单选题] *A.闲置的房屋B.租入的设备C.临时出租的设备D.已提足折旧仍继续使用的设备(正确答案)3、已达到预定可使用状态但未办理竣工决算的固定资产，应根据（）作暂估价值转入固定资产，待竣工决算后再作调整。
[单选题] *A.市场价格B.计划成本C.估计价值(正确答案)D.实际成本4、企业购进货物用于集体福利时，该货物负担的增值税额应当计入（）。
[单选题] *A.应交税费——应交增值税B.应付职工薪酬(正确答案)C.营业外支出D.管理费用5、委托加工应纳消费税产品（非金银首饰）收回后，如直接对外销售，其由受托方代扣代交的消费税，应计入（）。
[单选题] *A.生产成本B.应交税费——应交消费税C.委托加工物资(正确答案)D.主营业务成本6、当法定盈余公积达到注册资本的（）时，可以不再提取。
[单选题] *A.10％B.20％C.50％(正确答案)D.30％7、计提固定资产折旧时，可以先不考虑固定资产残值的方法是（）。
[单选题] *A.年限平均法B.工作量法C.双倍余额递减法(正确答案)D.年数总和法8、企业发生的公益捐赠支出应计入（）。
[单选题] *A.销售费用B.营业外支出(正确答案)C.其他业务成本D.财务费用9、企业专设销售机构发生的办公费应计入（）科目。
[单选题] *A.管理费用B.财务费用C.制造费用D.销售费用(正确答案)10、根据准则的规定，企业收到与日常活动无关的政府补助应当计入（）科目。
[单选题] *A.投资收益B.其他业务收入C.主营业务收入D.营业外收入(正确答案)11、下列各项税金中不影响企业损益的是（）。
商务与经济统计选择题
商务与经济统计选择题选择题课堂练习第一章数据不统计学1. 样本容量 ba. 可以比总体容量大b. 总是比总体容量小c. 可以比总体容量大,也可以比总体容量小d. 总是和总体容量相同 2. 一个总体 ca. 不样本相同b. 是一个随机样本c. 特定研究中所有感兴趣的数据单位的集合d. 以上都不是3. 对数据进行相对简单的组织、概括和表述的统计方法称为 ba. 统计推断b. 描述性统计c. 抽样d. 以上都不是4. 根据样本信息对总体特征进行推测和估计的过程称为 ca. 描述性统计b. 随机样本c. 统计推断d. 抽样5. 收集数据时所依赖的对象称为 c a. 发量b. 数据集c. 数据单位d. 以上都不是6. 对数据单位感兴趣的某个特征量是 aa. 一个发量b. 一个数据单位c. 一个数据集d. 以上都不是7. 数学运算适合应用于 b a. 定性数据b. 定量数据c. 定性数据和定量数据d. 以上都不是8. 人的年龄是定量数据 aa. 是b. 否9. 在同一时刻收集以下的数据:同学的姓名,性别,年龄,成绩等数据,这些数据是aa. 横截面数据b. 时间序列数据第四章概率论介绍1. 所有样本点的集合称为 ca. 样本b. 事件c. 样本空间d. 试验2. 所有样本点概率相等的概率分配方法称为 b a. 主观方法b. 古典概率方法c. 相对频数方法d. 以上都不是3. 每个样本点，即试验结果，的概率必须是 d a. 任何大于零的数b. 小于零c. 大于1d. 在0和1之间4. 下面哪个条件成立时可以认为事件X和Y相亏独立？ Ba. P(Y|X) = P(X)b. P(Y|X) = P(Y)c. P(X|Y) = P(Y) 5. 从5个字母 (A, B, C, D, E)中取出两个字母,一共有多少种不同的取法？ Da. 20b. 7c. 5d. 106. P(A) = 0.6, P(B) = 0.5, 则 P(AuB) = Ea. 0.3b. 0.5c. 0.6d. 1.1e. 不能确定7. P(A) = 0.6, P(B) 0.5, P(AnB) = 0.3, 则 P(AuB) = c a. 0.5b. 0.6c. 0.8d. 1.1e. 不能确定8. P(A) = 0.6, P(B) = 0.5, P(AÇB) = 0.3, 则 P(A|B) = b a. 0.5b. 0.6c. 0.8d. 1.1e. 不能确定9. P(A) = 0.6, P(B) = 0.5, P(A n B) = 0.3, 则事件A和B为亏斥事件 ba. 正确b. 错误10. P(A) = 0.6, P(B) = 0.5, P(A n B) = 0.3, 则事件A和B亏为独立事件a a. 正确b. 错误11. P(A) = 0.2, P(B) = 0.5, 且事件A和B为亏斥事件,则 P(A n B) = a a.b. 0.3c. 0.7d. 112. P(A) = 0.2, P(B) = 0.5, 且事件A和B为亏斥事件,则 P(A u B) = c a.b. 0.3c. 0.7d. 113. P(A) = 0.2, P(B) = 0.5, 且事件A和B为亏斥事件,则 P(A|B) = a a.b. 0.3c. 0.7d. 114. 如果事件A和B为亏斥事件,则它们一定是相亏独立事件 b a. 正确b. 错误15. 一个学生在放假前认为,有 50% 的概率去于南度假,有30%的概率去西藏度假,有20%概率去新疆度假。
安德森-商务与经济统计公式汇总
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商务与经济统计ppt第20章
control
• First honorary member of ASQ
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 9
Quality Frameworks
Malcolm Baldrige National Quality Award
• Established in 1987 and given by the U.S. president
to organizations that apply and are judged to be outstanding in seven areas:
Slide 7
Quality Philosophies
Joseph Juran
• Helped educate the Japanese on quality
management shortly after World War II
• Proposed a simple definition of quality:
Slide 10
Quality Frameworks
ISO 9000
• A series of five standards published in 1987 by the
International Organization for Standardization in Geneva, Switzerland.
商务与经济统计第13版案例
商务与经济统计第13版案例引言商务与经济统计的重要性商务与经济统计是现代商业运营和经济决策的重要工具。
通过收集、分析和解释数据，商务与经济统计帮助企业和政府了解市场趋势、消费者行为和经济发展趋势，从而为决策者提供重要的信息支持。
商务与经济统计第13版案例概述商务与经济统计第13版案例是一本以实际案例作为基础，介绍商务与经济统计重要原理和方法的教材。
通过分析案例，读者可以了解商务与经济统计在实际运用中的一些重要问题和解决方法。
案例1：市场调研与分析为什么市场调研与分析重要？市场调研与分析是商务与经济统计的基础。
通过市场调研，企业可以了解市场需求、竞争情况和潜在机会，为产品开发和市场定位提供依据。
市场调研与分析过程1.定义研究目标和问题2.设计研究方案3.收集数据4.数据分析与解释5.提出建议和决策支持市场调研与分析案例以某电子产品公司为例，通过市场调研与分析，他们发现消费者对于更轻薄、更高分辨率的电子设备需求增加。
基于这些结果，公司调整了产品开发方向，推出了一款更轻薄、更高分辨率的电子产品，取得了巨大成功。
案例2：经济指标的解读什么是经济指标？经济指标是衡量经济健康状况和发展趋势的重要指标。
例如，国内生产总值（GDP）、通货膨胀率和失业率等都是常用的经济指标。
经济指标的解读方法1.趋势分析：通过比较不同时间段内的指标数据，观察指标是否呈现上升或下降趋势。
2.比较分析：将同一指标在不同地区或不同产业中的数据进行比较，了解不同区域或行业之间的差异。
3.引用分析：将指标数据与其他相关指标进行比较，分析它们之间的关联性和影响因素。
4.批判性分析：对指标数据的可靠性和有效性进行评估，考察数据收集和统计方法是否科学准确。
经济指标的解读案例以某国GDP增长率为例，近几年该国GDP增长率保持在5%左右，呈现稳定增长态势。
通过比较分析，我们发现农业和服务业对GDP增长的贡献最大，而工业的增长速度较缓慢。
我们可以对这些结果进行引用分析，了解农业和服务业的发展对经济增长的影响因素和潜在机会。
商务与经济统计课后答案
商务与经济统计课后答案【篇一：商务经济统计学复习题】．简答题1．简要谈谈你对统计与统计学的初步认识。
2.谈谈你对统计的三种含义的理解，并举出现实经济生活中你所了解到的运用统计学的一个例子3．试就统计数据的四种类型给出统计整理与显示的方法（统计图要求划出示意图）。
4．概述数据的离散程度的常用的测度方法(异众比率标准差离散系数)。
5．什么是个体指数? 什么是总指数？它们的作用分别是什么？6．试简要说明总量指标、平均指标和相对指标的在统计学中的作用。
7．只能用统计条形图和饼图来展示的是哪种类型的数据？画出这两种图形的示意图。
8.自己用一个实例画出统计条形图和饼图的示意图，它们通常可以用来展示哪种类型的数据？ 9．某高校毕业生就业指导中心想对2007届本校大学本科毕业生的毕业去向做一网上调查，请你为此设计一份半开放式（即：既含有封闭式问题又含有开放式问题）调查问卷。
（要求涉及学生的性别、专业、意向中的毕业去向：如出国、考研、自主创业、自主择业，以及意向中的就业领域、工薪待遇、单位性质、工作地区等等信息）。
二．填空题1.将下列指标按要求分类（只填写标号即可）（1）我国高等院校2006届本科毕业生就业率；（2）某贺岁片在国内上演第一周的票房收入；（3） 2006年第3季度一汽大众销售的某品牌小汽车台数占其全部小汽车销售量的比率；,（5）进藏铁路开通后第一周，每天乘火车前往西藏的旅客的累计人数；（6）第3季度某商场的月平均销售额。
哪些是时点时标；哪些是时期指标；哪些是平均指标；哪些是相对指标。
2．统计调查方式除了重点调查，典型调查之外，另三种主要方式是 3．加权调和平均公式为4．异众比率公式是其含义是5．一组数据中非众数组所占的比率叫做，它可测度分类数据的趋势；离散系数测度的是总体的平均离散程度，它的计算公式是v?=。
6．将下列指标分类：（1）2005年我国人均占有粮食产量（2）我国第五次人口普查总人口数（3）股价指数（4）销售量指数（5）单位产品成本（6）某商店全年销售额(7)某企业在岗职工人数和下岗职工人数的比例 (8)我国高等院校“十五”期间年平均招生人数哪些是时期指标哪些是时点指标；哪些是一般平均数, 哪些是序时平均数；哪些是相对指标 7．个体指数是反映项目或变量变动的相对数；反映多种项目或变量变动的相对数是。
商务与经济统计15版
商务与经济统计15版（原创实用版）目录1.商务与经济统计的重要性2.商务与经济统计的发展历程3.商务与经济统计的基本概念与方法4.商务与经济统计在实际应用中的案例分析5.商务与经济统计的未来发展趋势与挑战正文【商务与经济统计的重要性】商务与经济统计是一门重要的学科，它对商务和经济的发展起着至关重要的作用。
商务与经济统计能够为政府和企业提供有关经济活动的重要信息，帮助他们制定政策和策略，以促进经济的发展。
此外，商务与经济统计还能够帮助企业了解市场需求和竞争情况，为企业的运营和发展提供依据。
【商务与经济统计的发展历程】商务与经济统计的发展历程可以追溯到 19 世纪末 20 世纪初。
在那个时期，随着工业革命的完成，经济的发展进入了一个新的阶段，商务与经济统计应运而生。
随着经济的发展和科技的进步，商务与经济统计也不断发展，现在已经成为一个独立的学科。
【商务与经济统计的基本概念与方法】商务与经济统计的基本概念包括统计对象、统计指标、统计方法等。
统计对象是指统计活动的主体，统计指标是统计活动的客体，统计方法是指进行统计活动的方式和手段。
商务与经济统计的方法包括描述性统计、推断性统计、回归分析等。
【商务与经济统计在实际应用中的案例分析】商务与经济统计在实际应用中的案例分析可以帮助我们更好地理解商务与经济统计的应用。
例如，通过分析某个行业的统计数据，我们可以了解该行业的发展趋势和市场需求；通过分析某个企业的统计数据，我们可以了解该企业的运营状况和竞争力。
【商务与经济统计的未来发展趋势与挑战】随着经济的发展和科技的进步，商务与经济统计的未来发展趋势是朝着更加精细化、智能化的方向发展。
商务与经济统计13版pdf
商务与经济统计13版pdf引言概述：商务与经济统计是商务和经济领域中的重要学科，对于学习商务和经济的人来说，掌握统计知识是必不可少的。
本文将介绍商务与经济统计13版pdf的内容，并从五个大点进行详细阐述，最后进行总结。
正文内容：1. 商务与经济统计的基本概念1.1 统计的定义和作用1.2 商务与经济统计的研究对象1.3 商务与经济统计的发展历程1.4 商务与经济统计的重要性2. 商务与经济统计的数据收集与整理2.1 数据收集的方法和途径2.2 数据整理的基本原则2.3 数据整理的工具和技巧2.4 数据收集与整理的注意事项3. 商务与经济统计的数据分析与解读3.1 数据分析的基本方法3.2 数据分析的常用指标3.3 数据分析的技巧和注意事项3.4 数据解读的方法和途径4. 商务与经济统计的预测与决策4.1 预测的基本原理4.2 预测的方法和技巧4.3 预测的应用领域4.4 决策的基本原则4.5 决策的方法和技巧5. 商务与经济统计在实际应用中的案例分析5.1 商务与经济统计在市场调研中的应用5.2 商务与经济统计在投资决策中的应用5.3 商务与经济统计在商业竞争中的应用5.4 商务与经济统计在经济政策制定中的应用总结：通过对商务与经济统计13版pdf的内容进行介绍，我们了解到商务与经济统计在商务和经济领域中的重要性。
从基本概念、数据收集与整理、数据分析与解读、预测与决策以及实际应用中的案例分析等五个大点进行详细阐述，我们可以更好地掌握商务与经济统计的知识和技巧。
在实际应用中，商务与经济统计能够帮助我们进行市场调研、投资决策、商业竞争和经济政策制定等方面的工作。
因此，学习商务与经济统计对于商务和经济领域的从业人员来说至关重要。
商务与经济统计15版
商务与经济统计15版摘要：1.商务与经济统计的重要性2.商务与经济统计的定义和分类3.商务与经济统计的方法和工具4.商务与经济统计的应用和案例5.商务与经济统计的发展趋势正文：商务与经济统计是在商务和经济活动中收集、整理、分析和解释数据的过程，其目的是为了更好地理解市场、管理企业、制定政策和预测未来趋势。
在现代商业和经济环境中，商务与经济统计的重要性日益凸显，它为企业和政府提供了决策依据和战略指导。
商务与经济统计主要包括两个方面：一是商务统计，主要涉及企业经营活动、市场营销、消费者行为等方面的数据收集和分析；二是经济统计，主要涉及国民经济核算、产业结构、区域发展等方面的数据收集和分析。
商务与经济统计可以分为描述性统计、推断性统计和预测性统计三大类，分别用于描述现状、分析差异和预测趋势。
商务与经济统计的方法和工具多种多样，主要包括数据收集方法（如问卷调查、抽样调查、行政记录等）和数据分析方法（如描述性统计、推断性统计、回归分析、时间序列分析等）。
随着信息技术的发展，商务与经济统计越来越依赖于计算机和互联网，大数据、云计算和人工智能等技术为商务与经济统计提供了新的方法和工具。
商务与经济统计的应用领域非常广泛，涉及到企业经营、市场研究、政策制定、宏观经济分析等方面。
例如，企业可以通过商务与经济统计了解市场需求、优化生产过程、评估营销策略等；政府可以通过商务与经济统计了解经济运行状况、制定产业政策、调控宏观经济等。
商务与经济统计还可以为学术研究、新闻报道和社会公众提供有价值的信息。
随着全球化、信息化和数据化的发展，商务与经济统计面临着新的挑战和机遇。
未来的商务与经济统计将更加注重数据的实时性、完整性和准确性，更加注重数据的挖掘和分析，更加注重数据的可视化和交互性。
商务统计学参考书
商务统计学参考书商务统计学是统计学在商业和经济领域的应用分支，其目的是通过对数据进行分析，了解商业和经济现象的本质和规律，并作出合理的决策。
以下是推荐的商务统计学参考书：1. 《商务与经济统计（第12版）》：该书是全球应用最广的统计学教材之一，内容涵盖了描述性统计、概率论、贝叶斯统计学、回归分析、实验设计和抽样调查等商务统计学的基础知识。
此外，该书还通过丰富的实际案例和应用，帮助学生和从业人员理解和掌握统计学的原理和方法。
2. 《商务与金融统计（第10版）》：该书主要介绍了如何运用统计学方法对商务与金融数据进行有效分析。
内容涵盖了金融市场和机构的介绍、投资组合理论、资本资产定价模型、市场有效性假说、风险管理和测量等金融领域的核心主题。
此外，该书还提供了大量的实际案例和练习题，帮助学生培养解决实际问题的能力。
3. 《统计学：从入门到实践》：该书从统计学的基本概念入手，详细介绍了描述性统计、概率论、参数估计、假设检验、回归分析等内容。
同时，该书还结合了大量的实际案例和数据，通过练习和实践的方式，帮助读者更好地理解和掌握统计学的知识。
4. 《商务与经济统计：学习指导》：该书是《商务与经济统计》的配套学习指导书，提供了丰富的习题、案例分析和答案，有助于读者加深对统计学知识的理解和掌握。
5. 《统计学：从数据到结论（第3版）》：该书以数据为基础，系统介绍了统计学的基本概念、方法和应用。
通过丰富的实际案例和数据，该书引导读者逐步了解从数据收集、整理、分析和解释的全过程，并提供了各种统计方法的选择和应用。
以上书籍仅供参考，建议根据个人需求和兴趣选择合适的参考书。
同时，也可以参加相关的课程和培训，以更深入地了解商务统计学在实际工作中的应用。
商务与经济统计习题答案（第8版，中文版）SBE8-SM12
商务与经济统计习题答案（第8版，中文版）SBE8-SM12Chapter 12 Tests of Goodness of Fit and Independence Learning Objectives 1. Know how to conduct a goodness of fit test. 2. Know how to use sample data to test for independence of two variables. 3. Understand the role of the chi-square distribution in conducting tests of goodness of fit and independence. 4. Be able to conduct a goodness of fit test for cases where the population is hypothesized to have either a multinomial, a Poisson, or a normal probability distribution. 5. For a test of independence, be able to set up a contingency table, determine the observed and expected frequencies, and determine if the two variables are independent. Solutions: 1. Expected frequencies: e1 = 200 (.40) = 80, e2 = 200 (.40) = 80 e3 = 200 (.20) = 40 Actual frequencies: f1 = 60, f2 = 120, f3 = 20 = 9.__ with k - 1 = 3 - 1 = 2 degrees of freedom Since = 35 9.__ reject the null hypothesis. The population proportions are not as stated in the null hypothesis. 2. Expected frequencies: e1 = 300 (.25) = 75, e2 = 300 (.25) = 75 e3 = 300 (.25) = 75, e4 = 300 (.25)= 75 Actual frequencies: f1 = 85, f2 = 95, f3 = 50, f4 = 70 = 7.__ with k - 1 = 4 - 1 = 3 degrees of freedom Since c2 = 15.33 7.__ reject H0 We conclude that the proportions are not all equal. 3. H0 = pABC = .29, pCBS = .28, pNBC = .25, pIND = .18 Ha = The proportions are not pABC = .29, pCBS = .28, pNBC = .25, pIND = .18 Expected frequencies: 300 (.29) = 87, 300 (.28) = 84 300 (.25) = 75, 300 (.18) = 54 e1 = 87, e2 = 84, e3 = 75, e4 = 54 Actual frequencies: f1 = 95, f2 = 70, f3 = 89, f4 = 46 = 7.81 (3 degrees of freedom) Do not reject H0; there is no significant change in the viewing audience proportions. 4. Observed Expected Hypothesized Frequency Frequency Category Proportion (fi) (ei) (fi - ei)2 / ei Brown 0.30 177 151.8 4.18 Yellow 0.20 135 101.2 11.29 Red 0.20 79 101.2 4.87 Orange 0.10 41 50.6 1.82 Green 0.10 36 50.6 4.21 Blue 0.10 38 50.6 3.14 Totals: 506 29.51 = 11.07 (5 degrees of freedom) Since 29.51 11.07, we conclude that the percentage figures reported by the pany have changed. 5. Observed Expected Hypothesized Frequency Frequency Category Proportion (fi) (ei) (fi - ei)2 / ei Full Service 1/3 264 249.33 0.86 Discount 1/3 255 249.33 0.13 Both 1/3 229 249.33 1.66 Totals: 748 2.65 = 4.61 (2 degrees of freedom) Since 2.65 4.61, there is no significant difference in preference among thethree service choices. 6. Observed Expected Hypothesized Frequency Frequency Category Proportion (fi) (ei) (fi - ei)2 / ei News and Opinion 1/6 20 19.17 .04 General Editorial 1/6 15 19.17 .91 Family Oriented 1/6 30 19.17 6.12 Business/Financial 1/6 22 19.17 .42 Female Oriented 1/6 16 19.17 .52 African-American 1/6 12 19.17 2.68 Totals: 115 10.69 = 9.24 (5 degrees of freedom) Since 10.69 9.24, we conclude that there is a difference in the proportion of ads with guilt appeals among the six types of magazines. 7. Expected frequencies: ei = (1 / 3) (135) = 45 With 2 degrees of freedom, = 5.99 Do not reject H0; there is no justification for concluding a difference in preference exists.8. H0: p1 = .03, p2 = .28, p3 = .45, p4 = .24 df = 3 = 11.34 Reject H0 if c2 11.34 Rating Observed Expected (fi - ei)2 / ei Excellent 24 .03(400) = 12 12.00 Good 124 .28(400) = 112 1.29 Fair 172 .45(400) = 180 .36 Poor 80 .24(400) = 96 2.67 400 400 c2 = 16.31 Reject H0; conclude that the ratings differ. A parison of observed and expected frequencies show telephone service is slightly better with more excellent and good ratings. 9. H0 = The column variable is independent of the row variable Ha = The column variable is not independent of the row variable Expected Frequencies: A B C P 28.5 39.9 45.6 Q 21.5 30.1 34.4 = 7.__ with(2 - 1) (3 - 1)= 2 degrees of freedom Since c2 = 7.86 7.__ Reject H0 Conclude that the column variable is not independent of the row variable. 10. H0 = The column variable is independent of the row variable Ha = The column variable is not independent of the row variable Expected Frequencies: A B C P 17.5000 30.6250 21.8750 Q 28.7500 50.3125 35.9375 R 13.7500 24.0625 17.1875 = 9.__ with (3 - 1) (3 - 1)= 4 degrees of freedom Since c2 = 19.78 9.__ Reject H0 Conclude that the column variable is not independent of f the row variable. 11. H0 : Type of ticket purchased is independent of the type of flight Ha: Type of ticket purchased is not independent of the type of flight. Expected Frequencies: e11 = 35.59 e12 = 15.41 e21 = 150.73 e22 = 65.27 e31 = 455.68 e32 = 197.32 Observed Expected Frequency Frequency Ticket Flight (fi) (ei) (fi - ei)2 / ei First Domestic 29 35.59 1.22 First International 22 15.41 2.82 Business Domestic 95 150.73 20.61 Business International 121 65.27 47.59 Full Fare Domestic 518 455.68 8.52 Full Fare International 135 197.32 19.68 Totals: 920 100.43 = 5.99 with (3 - 1)(2 - 1) = 2 degrees of freedom Since 100.43 5.99, we conclude that the type of ticket purchased is not independent of the type of flight. 12. a. Observed Frequency (fij) Domestic European Asian Total Same 125 55 68 248 Different140 105 107 352 Total 265 160 175 600 Expected Frequency (eij) Domestic European Asian Total Same 109.53 66.13 72.33 248 Different 155.47 93.87 102.67 352 Total 265 160 175 600 Chi Square (fij - eij)2 / eij Domestic European Asian Total Same 2.18 1.87 0.26 4.32 Different 1.54 1.32 0.18 3.04 c2 = 7.36 Degrees of freedom = 2 = 5.99 Reject H0; conclude brand loyalty is not independent of manufacturer. b. Brand Loyalty Domestic 125/265 = .472 (47.2%) ¬ Highest European 55/160 = .344 (34.4%) Asian 68/175 = .389 (38.9%) 13. Industry Major Oil Chemical Electrical Computer Business 30 22.5 17.5 30 Engineering 30 22.5 17.5 30 Note: Values shown above are the expected frequencies. = 11.3449 (3 degrees of freedom: 1 x 3 = 3) c2 = 12.39 Reject H0; conclude that major and industry not independent. 14. Expected Frequencies: e11 = 31.0 e12 = 31.0 e21 = 29.5 e22 = 29.5 e31 =13.0 e32 = 13.0 e41 = 5.5 e42 = 5.5 e51 = 7.0 e52 = 7.0 e61 =14.0 e62 = 14.0 Observed Expected Frequency Frequency Most Difficult Gender (fi) (ei) (fi - ei)2 / ei Spouse Men 37 31.0 1.16 Spouse Women 25 31.0 1.16 Parents Men 28 29.5 0.08 Parents Women 31 29.5 0.08 Children Men 7 13.0 2.77 Children Women 19 13.0 2.77 Siblings Men 8 5.5 1.14 Siblings Women 3 5.5 1.14 In-Laws Men 4 7.0 1.29 In-Laws Women 10 7.0 1.29 OtherRelatives Men 16 14.0 0.29 Other Relatives Women 12 14.0 0.29 Totals: 200 13.43 = 11.0705 with (6 - 1) (2 - 1) = 5 degrees of freedom Since 13.43 11.0705. we conclude that gender is not independent of the most difficult person to buy for. 15. Expected Frequencies: e11 = 17.16 e12 = 12.84 e21 = 14.88 e22 = 11.12 e31 = 28.03 e32 = 20.97 e41 = 22.31 e42 = 16.69 e51 = 17.16 e52 = 12.84 e61 = 15.45 e62 = 11.55 Observed Expected Frequency Frequency Magazine Appeal (fi) (ei) (fi - ei)2 / ei News Guilt 20 17.16 0.47 News Fear 10 12.84 0.63 General Guilt 15 14.88 0.00 General Fear 11 11.12 0.00 Family Guilt 30 28.03 0.14 Family Fear 19 20.97 0.18 Business Guilt 22 22.31 0.00 Business Fear 17 16.69 0.01 Female Guilt 16 17.16 0.08 Female Fear 14 12.84 0.11 African-American Guilt 12 15.45 0.77 African-American Fear 15 11.55 1.03 Totals: 201 3.41 = 15.09 with (6 - 1) (2 - 1) = 5 degrees of freedom Since 3.41 15.09, the hypothesis of independence cannot be rejected. 34. a. Observed Frequency (fij) Pharm Consumer Computer Total Correct 207 136 151 178 672 Incorrect 3 4 9 12 28 Total 210 140 160 190 700 Expected Frequency (eij) Pharm Consumer Computer Total Correct 201.6 134.4 153.6 182.4 672 Incorrect 8.4 5.6 6.4 7.6 28 Total 210 140 160 190 700 Chi Square (fij - eij)2 / eij Pharm ConsumerComputer Total Correct .14 .02 .04 .11 .31 Incorrect 3.47 .46 1.06 2.55 7.53 c2 = 7.85 Degrees of freedom = 3 = 7.__ Do not reject H0; conclude order fulfillment is not independent of industry. b. The pharmaceutical industry is doing the best with 207 of 210 (98.6%) correctly filled orders. 17. Expected Frequencies: Part Quality Supplier Good Minor Defect Major Defect A 88.76 6.07 5.14 B 173.09 11.83 10.08 C 133.15 9.10 7.75 c2 = 7.96 = 9.__ (4 degrees of freedom: 2 x 2 = 4) Do not reject H0; conclude that the assumption of independence cannot be rejected 18. Expected Frequencies: Party Affiliation Education Level Democratic Republican Independent Did not plete high school 28 28 14 High school degree 32 32 16 College degree 40 40 20 c2 = 13.42 = 13.2767 (4 degrees of freedom: 2 x 2 = 4) Reject H0; conclude that party affiliation is not independent of education level. 19. Expected Frequencies: e11 = 11.81 e12 = 8.44 e13 = 24.75 e21 = 8.40 e22 = 6.00 e23 = 17.60 e31 = 21.79 e32 = 15.56 e33 = 45.65 Observed Expected Frequency Frequency Siskel Ebert (fi) (ei) (fi - ei)2 / ei Con Con 24 11.81 12.57 Con Mixed 8 8.44 0.02 Con Pro 13 24.75 5.58 Mixed Con 8 8.40 0.02 Mixed Mixed 13 6.00 8.17 Mixed Pro 11 17.60 2.48 Pro Con 10 21.79 6.38 Pro Mixed 9 15.56 2.77 Pro Pro 64 45.65 7.38 Totals: 16045.36 = 13.28 with (3 - 1) (3 - 1) = 4 degrees of freedom Since 45.36 13.28, we conclude that the ratings are not independent.20. First estimate m from the sample data. Sample size = 120. Therefore, we use Poisson probabilities with m = 1.3 to pute expected frequencies. x Observed Frequency Poisson Probability Expected Frequency Difference (fi - ei) 0 39 .2725 32.700 6.300 1 30 .3543 42.516 -12.516 2 30 .2303 27.636 2.364 3 18 .0998 11.976 6.024 4 or more 3 .0430 5.160 - 2.160 = 7.__ with 5 - 1 - 1 = 3 degrees of freedom Since c2 = 9.0348 7.__ Reject H0 Conclude that the data do not follow a Poisson probability distribution. 21. With n = 30 we will use six classes with 16 2/3% of the probability associated with each class. = 22.80 s = 6.2665 The z values that create 6 intervals, each with probability .1667 are -.98, -.43, 0, .43, .98 z Cut off value of x -.98 22.8 - .98 (6.2665) = 16.66 -.43 22.8 - .43 (6.2665) = 20.11 0 22.8 + 0 (6.2665) = 22.80 .43 22.8 + .43 (6.2665) = 25.49 .98 22.8 + .98 (6.2665) = 28.94 Interval Observed Frequency Expected Frequency Difference less than 16.66 3 5 -2 16.66 - 20.11 7 5 2 20.11 - 22.80 5 5 0 22.80 - 25.49 7 5 2 25.49- 28.94 3 5 -2 28.94 and up 5 5 0 = 9.__ with 6 - 2 - 1 = 3 degrees of freedom Since c2 = 3.20 £ 9.__ Do not reject H0 The claim that the data es from a normaldistribution cannot be rejected. 22. Use Poisson probabilities with m = 1. c2 = 4.30 = 5.__ (2 degrees of freedom) Do not reject H0; the assumption of a Poisson distribution cannot be rejected. 23. x Observed Poisson Probabilities Expected 0 15 .1353 13.53 1 31 .2707 27.07 2 20 .2707 27.07 3 15 .1804 18.04 4 13 .0902 9.02 5 or more 6 .0527 5.27 c2 = 4.98 = 7.__ (4 degrees of freedom) Do not reject H0; the assumption of a Poisson distribution cannot be rejected. 24. = 24.5 s = 3 n = 30 Use 6 classes Interval Observed Frequency Expected Frequency less than 21.56 5 5 21.56 - 23.21 4 5 23.21 - 24.50 3 5 24.50 - 25.79 7 5 25.79 - 27.44 7 5 27.41 up 4 5 c2 = 2.8 = 6.__ (3 degrees of freedom: 6 - 2 - 1 = 3) Do not reject H0; the assumption of a normal distribution cannot be rejected. 25. = 71 s = 17 n = 25 Use 5 classes Interval Observed Frequency Expected Frequency less than 56.7 7 5 56.7 - 66.5 7 5 66.5 - 74.6 1 5 74.6 - 84.5 1 5 84.5 up 9 5 c2 = 11.2 = 9.__ (2 degrees of freedom) Reject H0; conclude the distribution is not a normal distribution. 26. Observed 60 45 59 36 Expected 50 50 50 50 c2 = 8.04 = 7.__ (3 degrees of freedom) Reject H0; conclude that the order potentials are not the same in each sales territory. 27. Observed 48 323 79 16 63 Expected 37.03 306.82 126.96 21.16 37.03 Since 41.69 13.2767, reject H0. Mutual fundinvestors＇attitudes toward corporate bonds differ from their attitudes toward corporate stock. 28. Observed 20 20 40 60 Expected 35 35 35 35 Since 31.43 7.__, reject H0. The park manager should not plan on the same number attending each day. Plan on a larger staff for Sundays and holidays. 29. Observed 13 16 28 17 16 Expected 18 18 18 18 18 c2 = 7.44 = 9.__ Do not reject H0; the assumption that the number of riders is uniformly distributed cannot be rejected. 30. Observed Expected Hypothesized Frequency Frequency Category Proportion (fi) (ei) (fi - ei)2 / ei Very Satisfied 0.28 105 140 8.75 Somewhat Satisfied 0.46 235 230 0.11 Neither 0.12 55 60 0.42 Somewhat Dissatisfied 0.10 90 50 32.00 Very Dissatisfied 0.04 15 20 1.25 Totals: 500 42.53 = 9.49 (4 degrees of freedom) Since 42.53 9.49, we conclude that the job satisfaction for puter programmers is different than the job satisfaction for IS managers. 31. Expected Frequencies: Quality Shift Good Defective 1st 368.44 31.56 2nd 276.33 23.67 3rd 184.22 15.78 c2 = 8.11 = 5.__ (2 degrees of freedom) Reject H0; conclude that shift and quality are not independent. 32. Expected Frequencies: e11 = 1046.19 e12 = 632.81 e21 = 28.66 e22 = 17.34 e31 = 258.59 e32 = 156.41 e41 = 516.55 e42 = 312.45 Observed Expected Frequency FrequencyEmployment Region (fi) (ei) (fi - ei)2 / ei Full-Time Eastern 1105 1046.19 3.31 Full-time Western 574 632.81 5.46 Part-Time Eastern 31 28.66 0.19 Part-Time Western 15 17.34 0.32 Self-Employed Eastern 229 258.59 3.39 Self-Employed Western 186 156.41 5.60 Not Employed Eastern 485 516.55 1.93 Not Employed Western 344 312.45 3.19 Totals: 2969 23.37 = 7.81 with (4 - 1) (2 - 1) = 3 degrees of freedom Since 23.37 7.81, we conclude that employment status is not independent of region. 33. Expected frequencies: Loan Approval Decision Loan Offices Approved Rejected Miller 24.86 15.14 McMahon 18.64 11.36 Games 31.07 18.93 Runk 12.43 7.57 c2 = 2.21 = 7.__ (3 degrees of freedom) Do not reject H0; the loan decision does not appear to be dependent on the officer. 34. a. Observed Frequency (fij) Never Married Married Divorced Total Men 234 106 10 350 Women 216 168 16 400 Total 450 274 26 750 Expected Frequency (eij) Never Married Married Divorced Total Men 210 127.87 12.13 350 Women 240 146.13 13.87 400 Total 450 274 26 750 Chi Square (fij - eij)2 / eij Never Married Married Divorced Total Men 2.74 3.74 .38 6.86 Women 2.40 3.27 .33 6.00 c2 = 12.86 Degrees of freedom = 2 = 9.21 Reject H0; conclude martial status is not independent of gender. b. Martial Status Never Married Married Divorced Men66.9% 30.3% 2.9% Women 54.0% 42.0% 4.0% Men 100 - 66.9 = 33.1% have been married Women 100 - 54.0 = 46.0% have been married 35. Expected Frequencies: = 9.__ (4 degrees of freedom) Since 9.76 9.__, reject H0. Banking tends to have lower P/E ratios. We can conclude that industry type and P/E ratio are related. 36. Expected Frequencies: Days of the Week County Sun Mon Tues Wed Thur Fri Sat Total Urban 56.7 47.6 55.1 56.7 60.1 72.6 44.2 393 Rural 11.3 9.4 10.9 11.3 11.9 14.4 8.8 78 Total 68 57 66 68 72 87 53 471 c2 = 6.20 = 12.5916 (6 degrees of freedom) Do not reject H0; the assumption of independence cannot be rejected.37. = 76.83 s = 12.43 Interval Observed Frequency Expected Frequency less than 62.54 5 5 62.54 - 68.50 3 5 68.50 - 72.85 6 5 72.85 - 76.83 5 5 76.83 - 80.81 5 5 80.81 - 85.16 7 5 85.16 - 91.12 4 5 91.12 up 5 5 c2 = 2 = 11.0705 (5 degrees of freedom) Do not reject H0; the assumption of a normal distribution cannot be rejected. 38. Expected Frequencies: Los Angeles San Diego San Francisco San Jose Total Occupied 165.7 124.3 186.4 165.7 642 Vacant 34.3 25.7 38.6 34.3 133 Total 200.0 150.0 225.0 200.0 775 = 7.__ with 3 degrees of freedom Since c2 = 7.78 £ 7.__ Do not reject H0. We cannot conclude that office vacancies are dependent on metropolitan area, but it is close: the p-value isslightly larger than .05. 39. a. x Observed Frequencies Binomial Prob. n = 4, p = .30 Expected Frequencies 0 30 .2401 24.01 1 32 .4116 41.16 2 25 .2646 26.46 3 10 .0756 7.56 4 3 .0081 .81 100 100.00 The expected frequency of x = 4 is .81. Combine x = 3 and x = 4 into one category so that all expected frequencies are 5 or more. x Observed Frequencies Expected Frequencies 0 30 24.01 1 32 41.16 2 25 26.46 3 or 4 13 8.37 100 100.00 b. With 3 degrees of freedom, = 7.__. Reject H0 if c2 7.__. Do not reject H0; conclude that the assumption of a binomial distribution cannot be rejected.。
商务与经济统计
Unit 2 第二单元 Estimation:Population Mean, Proportion and Variance总体均值，总体比例和总体方差的
参数估计----学时4
8.1 Interval Estimationof a Population Mean:Large-Sample Case总体均值的区间估计:大样本--2学时练习8.1 8.2 Interval Estimationof a Population Mean:Small-Sample Case总体均值的区间估计:小样本练习8.2 8.3 Determining the Sample Size样本数量的确定---2学时练习8.3 8.4 Interval Estimationof a Population Proportion总体比例 的区间估计练习8.4
总复习 Ex4.1-4.3 Ex5.2-5.4 Ex6.2 Ex7.5-7.6 Ex3.1-3.5 Ex8.1-8.4 Ex9.1-9.6 Ex10.3-10.5、11.1 Ex12.4-12.9 附录1:SAS的数据输出 附录2用SAS作直方图的过程
模拟试卷 参考书1:电子版概率论与数理统计 参考书2:电子版SAS操作入门
一个数据分析的例
下表是快餐店在3年中三类业务的销量百分数据(数据名Q203)
本例中我们采用统计中图表法来对数据中的某些特征进行对比分 析。 譬如借助Excel采用作带状图进行分类对比 以上数据做图如下
下面带状图是业务类别为组,比较每个业务类别内三个 年度销量百分数
下面带状图是以年份为组比较同一年三个业务类 别的销量百分数
Unit 5 第五单元 imple Linear Regression Method 简单线性回归模型---2学时 12.2 Least Squares Method最小二乘法 12.3 Coefficient of Determination系数的确定 12.4 Model Assumption模型假设---2学时 12.5 Testing for Significance显著性检验 练习12.4-12.5
商务与经济统计习题答案(第8版,中文版)SBE8
商务与经济统计习题答案（第8版，中文版）SBE8Chapter 14 Simple Linear Regression Learning Objectives 1. Understand how regression analysis can be used to develop an equation that estimates mathematically how two variables are related.2. Understand the differences between the regression model, the regression equation, and the estimated regression equation.3. Know how to fit an estimated regression equation to a set of sample data based upon the least-squares method.4. Be able to determine how good a fit is provided by the estimated regression equation and compute the sample correlation coefficient from the regression analysis output.5. Understand the assumptions necessary for statistical inference and be able to test for a significant relationship.6. Learn how to use a residual plot to make a judgement as to the validity of the regression assumptions, recognize outliers, and identify influential observations.7. Know how to develop confidence interval estimates of y given a specific value of x in both the case of a mean value of y and an individual value of y.8. Be able to compute the sample correlation coefficient from the regression analysis output.9. Know the definition of the following terms: independent and dependent variable simple linear regression regression model regression equation and estimated regression equation scatter diagram coefficient of determination standard error of the estimate confidence interval prediction interval residual plot standardized residual plot outlier influential observation leverage Solutions: 1 a. b. There appears to be a linear relationship between x and y. c. Many different straight lines can be drawn to provide a linear approximation of the relationship between x and y; in part d we will determine theequation of a straight line that “best” represents the relationship according to the least squares criterion. d. Summations needed to compute the slope and y-intercept are: e. 2. a. b. There appears to be a linear relationship between x and y. c. Many different straight lines can be drawn to provide a linear approximation of the relationship between x and y; in part d we will determine the equation of a straight line that “best” represents the relationship according to the least squares criterion. d. Summations needed to compute the slope and y-intercept are: e. 3. a. b. Summations needed to compute the slope and y-intercept are: c. 4. a. b. There appears to be a linear relationship between x and y. c. Many different straight lines can be drawn to provide a linear approximation of the relationship between x and y; in part d we will determine the equation of a straight line that “best” represents the relationship according to the least squares criterion. d. Summations needed to compute the slope and y-intercept are: e. pounds 5. a. b. There appears to be a linear relationship between x and y. c. Many different straight lines can be drawn to provide a linear approximation of the relationship between x and y; in part d we will determine the equation of a straight line that “best” represents the relationship according to the least squares criterion. Summations needed to compute the slope and y-intercept are: d. A one million dollar increase in media expenditures will increase case sales by approximately 14.42 million. e. 6. a. b. There appears to be a linear relationship between x and y. c. Summations needed to compute the slope and y-intercept are: d. A one percent increase in the percentage of flights arriving on time will decrease the number of complaints per 100,000 passengers by 0.07. e 7. a. b. Let x = DJIA and y = SP. Summations needed to compute the slope and y-intercept are: c. or approximately 1500 8. a. Summations needed to compute the slopeand y-intercept are: b. Increasing the number of times an ad is aired by one will increase the number of household exposures by approximately 3.07 million. c. 9. a. b. Summations needed to compute the slope and y-intercept are: c. 10. a. b. Let x = performance score and y = overall rating. Summations needed to compute the slope and y-intercept are: c. or approximately 84 11. a. b. There appears to be a linear relationship between the variables. c. The summations needed to compute the slope and the y-intercept are: d. 12. a. b. There appears to be a positive linear relationship between the number of employees and the revenue. c. Let x = number of employees and y = revenue. Summations needed to compute the slope and y-intercept are: d. 13. a. b. The summations needed to compute the slope and the y-intercept are: c. or approximately $13,080. The agent＇s request for an audit appears to be justified. 14. a. b. The summations needed to compute the slope and the y-intercept are: c. 15. a. The estimated regression equation and the mean for the dependent variable are: The sum of squares due to error and the total sum of squares are Thus, SSR = SST - SSE = 80 - 12.4 = 67.6 b. r2 = SSR/SST = 67.6/80 = .845 The least squares line provided a very good fit; 84.5% of the variability in y has been explained by the least squares line. c. 16. a. The estimated regression equation and the mean for the dependent variable are: The sum of squares due to error and the total sum of squares are Thus, SSR = SST - SSE = 114.80 - 6.33 = 108.47 b. r2 = SSR/SST = 108.47/114.80 = .945 The least squares line provided an excellent fit; 94.5% of the variability in y has been explained by the estimated regression equation. c. Note: the sign for r is negative because the slope of the estimated regression equation is negative. (b1 = -1.88) 17. The estimated regression equation and the mean for the dependent variable are: The sum of squares due to error and thetotal sum of squares are Thus, SSR = SST - SSE = 11.2 - 5.3 = 5.9 r2 = SSR/SST = 5.9/11.2 = .527 We see that 52.7% of the variability in y has been explained by the least squares line. 18. a. The estimated regression equation and the mean for the dependent variable are: The sum of squares due to error and the total sum of squares are Thus, SSR = SST - SSE = 335,000 - 85,135.14 = 249,864.86 b. r2 = SSR/SST = 249,864.86/335,000 = .746 We see that 74.6% of the variability in y has been explained by the least squares line. c. 19. a. The estimated regression equation and the mean for the dependent variable are: The sum of squares due to error and the total sum of squares are Thus, SSR = SST - SSE = 47,582.10 - 7547.14 = 40,034.96 b. r2 = SSR/SST = 40,034.96/47,582.10 = .84 We see that 84% of the variability in y has been explained by the least squares line. c. 20. a. Let x = income and y = home price. Summations needed to compute the slope and y-intercept are: b. The sum of squares due to error and the total sum of squares are Thus, SSR = SST - SSE = 11,373.09 – 2017.37 = 9355.72 r2 = SSR/SST = 9355.72/11,373.09 = .82 We see that 82% of the variability in y has been explained by the least squares line. c. or approximately $173,500 21. a. The summations needed in this problem are: b. $7.60 c. The sum of squares due to error and the total sum of squares are: Thus, SSR = SST - SSE = 5,648,333.33 - 233,333.33 = 5,415,000 r2 = SSR/SST = 5,415,000/5,648,333.33 = .9587 We see that 95.87% of the variability in y has been explained by the estimated regression equation. d. 22. a. The summations needed in this problem are: b. The sum of squares due to error and the total sum of squares are: Thus, SSR = SST - SSE = 1998 - 1272.4495 = 725.5505 r2 = SSR/SST = 725.5505/1998 = 0.3631 Approximately 37% of the variability in change in executive compensation is explained by the two-year change in the return on equity. c. It reflects a linearrelationship that is between weak and strong. 23. a. s2 = MSE = SSE / (n - 2) = 12.4 / 3 = 4.133 b. c. d. t.025 = 3.182 (3 degrees of freedom) Since t = 4.04 t.05 = 3.182 we reject H0: b1 = 0 e. MSR = SSR / 1 = 67.6 F = MSR / MSE = 67.6 / 4.133 = 16.36 F.05 = 10.13 (1 degree of freedom numerator and 3 denominator) Since F = 16.36 F.05 = 10.13 we reject H0: b1 = 0 Source of Variation Sum of Squares Degrees of Freedom Mean Square F Regression 67.6 1 67.6 16.36 Error 12.4 3 4.133 Total 80.0 4 24. a. s2 = MSE = SSE / (n - 2) = 6.33 / 3 = 2.11 b. c. d. t.025 = 3.182 (3 degrees of freedom) Since t = -7.18 -t.025 = -3.182 we reject H0: b1 = 0 e. MSR = SSR / 1 = 8.47 F = MSR / MSE = 108.47 / 2.11 = 51.41 F.05 = 10.13 (1 degree of freedom numerator and 3 denominator) Since F = 51.41 F.05 = 10.13 we reject H0: b1 = 0 Source of Variation Sum of Squares Degrees of Freedom Mean Square F Regression 108.47 1 108.47 51.41 Error 6.333 2.11 Total 114.804 25. a. s2 = MSE = SSE / (n - 2) = 5.30 / 3 = 1.77b. t.025 = 3.182 (3 degrees of freedom) Since t = 1.82 t.025 = 3.182 we cannot reject H0: b1 = 0; x and y do not appear to be related.c. MSR = SSR/1 = 5.90 /1 = 5.90 F = MSR/MSE = 5.90/1.77 = 3.33 F.05 = 10.13 (1 degree of freedom numerator and 3 denominator) Since F = 3.33 F.05 = 10.13 we cannot reject H0: b1 = 0; x and y do not appear to be related. 26. a. s2 = MSE = SSE / (n - 2) = 85,135.14 / 4 = 21,283.79 t.025 = 2.776 (4 degrees of freedom) Since t = 3.43 t.025 = 2.776 we reject H0: b1 = 0 b. MSR = SSR / 1 = 249,864.86 / 1 = 249.864.86 F = MSR / MSE = 249,864.86 / 21,283.79 = 11.74 F.05 = 7.71 (1 degree of freedom numerator and 4 denominator) Since F = 11.74 F.05 = 7.71 we reject H0: b1 = 0 c. Source of Variation Sum of Squares Degrees of Freedom Mean Square F Regression *****.86 1 *****.86 11.74 Error *****.14 4 *****.79 Total ***** 5 27. The sum of squares due to error and the total sum of squares are: SSE = SST =2442 Thus, SSR = SST - SSE = 2442 - 170 = 2272 MSR = SSR / 1 = 2272 SSE = SST - SSR = 2442 - 2272 = 170 MSE = SSE / (n - 2) = 170 / 8 = 21.25 F = MSR / MSE = 2272 / 21.25 = 106.92 F.05 = 5.32 (1 degree of freedom numerator and 8 denominator) Since F = 106.92 F.05 = 5.32 we reject H0: b1 = 0. Years of experience and sales are related. 28. SST = 411.73 SSE = 161.37 SSR = 250.36 MSR = SSR / 1 = 250.36 MSE = SSE / (n - 2) = 161.37 / 13 = 12.413 F = MSR / MSE = 250.36 / 12.413= 20.17 F.05 = 4.67 (1 degree of freedom numerator and 13 denominator) Since F = 20.17 F.05 = 4.67 we reject H0: b1 = 0. 29. SSE = 233,333.33 SST = 5,648,333.33 SSR = 5,415,000 MSE = SSE/(n - 2) = 233,333.33/(6 - 2) = 58,333.33 MSR = SSR/1 = 5,415,000 F = MSR / MSE = 5,415,000 / 58,333.25 = 92.83 Source of Variation Sum of Squares Degrees of Freedom Mean Square F Regression 5,415,000.00 1 5,415,000 92.83 Error 233,333.33 4 58,333.33 Total 5,648,333.33 5 F.05 = 7.71 (1 degree of freedom numerator and 4 denominator) Since F = 92.83 7.71 we reject H0: b1 = 0. Production volume and total cost are related. 30. Using the computations from Exercise 22, SSE = 1272.4495 SST = 1998 SSR = 725.5505 = 45,833.9286 t.025 = 2.571 Since t = 1.69 2.571, we cannot reject H0: b1 = 0 There is no evidence of a significant relationship between x and y. 31. SST = 11,373.09 SSE = 2017.37 SSR = 9355.72 MSR = SSR / 1 = 9355.72 MSE = SSE / (n - 2) = 2017.37/ 16 = 126.0856 F = MSR / MSE = 9355.72/ 126.0856 = 74.20 F.01 = 8.53 (1 degree of freedom numerator and 16 denominator) Since F = 74.20 F.01 = 8.53 we reject H0: b1 = 0. 32. a. s = 2.033 b. 10.6 ± 3.182 (1.11) = 10.6 ± 3.53 or 7.07 to 14.13 c. d. 10.6 ± 3.182 (2.32) = 10.6 ± 7.38 or 3.22 to 17.9833. a. s = 1.453 b. 24.69 ± 3.182 (.68) = 24.69 ± 2.16 or 22.53 to 26.85c. d. 24.69 ± 3.182 (1.61) = 24.69 ± 5.12 or 19.57 to 29.81 34. s = 1.332.28 ±3.182 (.85) = 2.28 ± 2.70 or -.40 to4.98 2.28 ± 3.182 (1.58) =2.28 ± 5.03 or -2.27 to 7.31 35. a. s = 145.89 2,033.78 ± 2.776 (68.54) = 2,033.78 ± 190.27 or $1,843.51 to $2,224.05 b. 2,033.78 ± 2.776 (161.19) = 2,033.78 ± 447.46 or $1,586.32 to $2,481.24 36. a. b. s = 3.5232 80.859 ± 2.160 (1.055) = 80.859 ± 2.279 or 78.58 to 83.14 c.80.859 ± 2.160 (3.678) = 80.859 ± 7.944 or 72.92 to 88.80 37. a. s2 = 1.88 s = 1.37 13.08 ± 2.571 (.52) = 13.08 ± 1.34 or 11.74 to 14.42 or $11,740 to $14,420 b. sind = 1.47 13.08 ± 2.571 (1.47) = 13.08 ± 3.78 or 9.30 to 16.86 or $9,300 to $16,860 c. Yes, $20,400 is much larger than anticipated. d. Any deductions exceeding the $16,860 upper limit could suggest an audit. 38. a. b. s2 = MSE = 58,333.33 s = 241.52 5046.67 ± 4.604 (267.50) = 5046.67 ± 1231.57 or $3815.10 to $6278.24 c. Based on one month, $6000 is not out of line since $3815.10 to $6278.24 is the prediction interval. However, a sequence of five to seven months with consistently high costs should cause concern. 39. a. Summations needed to compute the slope and y-intercept are: b. SST = 39,065.14 SSE = 4145.141 SSR = 34,920.000 r2 = SSR/SST = 34,920.000/39,065.141 = 0.894 The estimated regression equation explained 89.4% of the variability in y; a very good fit. c. s2 = MSE = 4145.141/8 = 518.143 270.63 ± 2.262 (8.86) = 270.63 ± 20.04 or 250.59 to 290.67 d. 270.63 ± 2.262 (24.42) = 270.63 ± 55.24 or 215.39 to 325.87 40. a. 9 b. = 20.0 + 7.21x c. 1.3626 d. SSE = SST - SSR = 51,984.1 - 41,587.3 = 10,396.8 MSE = 10,396.8/7 = 1,485.3 F = MSR / MSE = 41,587.3 /1,485.3 = 28.00 F.05 = 5.59 (1 degree of freedom numerator and 7 denominator) Since F = 28 F.05 = 5.59 we reject H0: B1 = 0. e. = 20.0 + 7.21(50) = 380.5 or $380,500 41. a. = 6.1092 + .8951x b. t.025 = 2.306 (1 degree of freedom numerator and 8 denominator) Since t = 6.01 t.025 = 2.306 we reject H0: B1 = 0 c. = 6.1092 + .8951(25) = 28.49 or $28.49 per month 42 a. = 80.0 + 50.0x b. 30 c. F = MSR / MSE = 6828.6/82.1 = 83.17 F.05 = 4.20 (1 degreeof freedom numerator and 28 denominator) Since F = 83.17 F.05 = 4.20 we reject H0: B1 = 0. Branch office sales are related to the salespersons. d. = 80 + 50 (12) = 680 or $680,000 43. a. The Minitab output is shown below: The regression equation is Price = - 11.8 + 2.18 Income Predictor Coef SE Coef T P Constant -11.80 12.84 -0.92 0.380 Income 2.1843 0.2780 7.86 0.000 S = 6.634 R-Sq = 86.1% R-Sq(adj) = 84.7% Analysis of Variance Source DF SS MS F P Regression 1 2717.9 2717.9 61.75 0.000 Residual Error 10 440.1 44.0 Total 11 3158.0 Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 75.79 2.47 ( 70.29, 81.28) ( 60.02, 91.56) b. r2 = .861. The least squares line provided a very good fit. c. The 95% confidence interval is 70.29 to 81.28 or $70,290 to $81,280. d. The 95% prediction interval is 60.02 to 91.56 or $60,020 to $91,560. 44. a/b. The scatter diagram shows a linear relationship between the two variables. c. The Minitab output is shown below: The regression equation is Rental$ = 37.1 - 0.779 Vacancy% Predictor Coef SE Coef T P Constant 37.066 3.530 10.50 0.000 Vacancy% -0.7791 0.2226 -3.50 0.003 S = 4.889 R-Sq = 43.4% R-Sq(adj) = 39.8% Analysis of Variance Source DF SS MS F P Regression 1 292.89 292.89 12.26 0.003 Residual Error 16 382.37 23.90 Total 17 675.26 Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 17.59 2.51 ( 12.27, 22.90) ( 5.94, 29.23) 2 28.26 1.42 ( 25.26, 31.26) ( 17.47, 39.05) Values of Predictors for New Observations New Obs Vacancy% 1 25.0 2 11.3 d. Since the p-value = 0.003 is less than a = .05, the relationship is significant. e. r2 = .434. The least squares line does not provide a very good fit. f. The 95% confidence interval is 12.27 to 22.90 or $12.27 to $22.90. g. The 95% prediction interval is 17.47 to 39.05 or $17.47 to $39.05. 45. a. b. The residuals are 3.48, -2.47, -4.83, -1.6, and 5.22 c. With only 5 observations it is difficult to determine if the assumptions are satisfied.However, the plot does suggest curvature in the residuals that would indicate that the error term assumptions are not satisfied. The scatter diagram for these data also indicates that the underlying relationship between x and y may be curvilinear. d. The standardized residuals are 1.32, -.59, -1.11, -.40, 1.49. e. The standardized residual plot has the same shape as the original residual plot. The curvature observed indicates that the assumptions regarding the error term may not be satisfied. 46. a. b. The assumption that the variance is the same for all values of x is questionable. The variance appears to increase for larger values of x. 47. a. Let x = advertising expenditures and y = revenue b. SST = 1002 SSE = 310.28 SSR = 691.72 MSR = SSR / 1 = 691.72 MSE = SSE / (n - 2) = 310.28/ 5 = 62.0554 F = MSR / MSE = 691.72/ 62.0554= 11.15 F.05 = 6.61 (1 degree of freedom numerator and 5 denominator) Since F = 11.15 F.05 = 6.61 we conclude that the two variables are related. c. d. The residual plot leads us to question the assumption of a linear relationship between x and y. Even though the relationship is significant at the .05 level of significance, it would be extremely dangerous to extrapolate beyond the range of the data. 48.a. b. The assumptions concerning the error term appear reasonable.49. a. Let x = return on investment (ROE) and y = price/earnings (P/E) ratio. b. c. There is an unusual trend in the residuals. The assumptions concerning the error term appear questionable. 50. a. The ***** output is shown below: The regression equation is Y = 66.1 + 0.402 X Predictor Coef Stdev t-ratio p Constant 66.10 32.06 2.06 0.094 X 0.4023 0.2276 1.77 0.137 s = 12.62 R-sq = 38.5% R-sq(adj) = 26.1% Analysis of Variance SOURCE DF SS MS F p Regression 1 497.2 497.2 3.12 0.137 Error 5 795.7 159.1 Total 6 1292.9 Unusual Observations Obs. X Y Fit Stdev.Fit Residual St.Resid 1 135 145.00 120.42 4.87 24.58 2.11R R denotes an obs. with a large st. resid. The standardizedresiduals are: 2.11, -1.08, .14, -.38, -.78, -.04, -.41 The first observation appears to be an outlier since it has a large standardized residual. b.2.4+ - * *****D- - - 1.2+ - - - - * 0.0+ * - - * * - * - -1.2+ * - --+---------+---------+---------+---------+---------+----YHAT 110.0 115.0 120.0 125.0 130.0 135.0 The standardized residual plot indicates that the observation x = 135,y = 145 may be an outlier; note that this observation has a standardized residual of 2.11. c. The scatter diagram is shown below - Y - * - - 135+ - - * * - - 120+ * * - - - * - 105+ - - * ----+---------+---------+---------+---------+---------+--X 105 120 135 150 165 180 The scatter diagram also indicates that the observation x = 135,y = 145 may be an outlier; the implication is that for simple linear regression an outlier can be identified by looking at the scatter diagram. 51. a. The Minitab output is shown below: The regression equation is Y = 13.0 + 0.425 X Predictor Coef Stdev t-ratio p Constant 13.002 2.396 5.43 0.002 X 0.4248 0.2116 2.01 0.091 s = 3.181 R-sq = 40.2% R-sq(adj) = 30.2% Analysis of Variance SOURCE DF SS MS F p Regression 1 40.78 40.784.03 0.091 Error 6 60.72 10.12 Total 7 101.50 Unusual Observations Obs. X Y Fit Stdev.Fit Residual St.Resid 7 12.0 24.00 18.10 1.205.90 2.00R 8 22.0 19.00 22.35 2.78 -3.35 -2.16RX R denotes an obs. with a large st. resid. X denotes an obs. whose X value gives it large influence. The standardized residuals are: -1.00, -.41, .01, -.48, .25, .65, -2.00, -2.16 The last two observations in the data set appear to be outliers since the standardized residuals for these observations are 2.00 and -2.16, respectively. b. Using *****, we obtained the following leverage values: .28, .24, .16, .14, .13, .14, .14, .76 ***** identifies an observation as having high leverage if hi 6/n; for these data, 6/n = 6/8 = .75. Since the leverage for the observation x = 22, y = 19 is .76, ***** would identify observation 8 as a high leverage point. Thus, we conclude thatobservation 8 is an influential observation. c. 24.0+ * - Y - - - 20.0+ * - * - * - - 16.0+ * - * - - * - 12.0+ * - +---------+---------+---------+---------+---------+------X 0.0 5.0 10.0 15.0 20.0 25.0 The scatter diagram indicates that the observation x = 22, y = 19 is an influential observation. 52. a. The Minitab output is shown below: The regression equation is Amount = 4.09 + 0.196 MediaExp Predictor Coef SE Coef T P Constant 4.089 2.168 1.89 0.096 MediaExp 0.***** 0.03635 5.38 0.001 S = 5.044 R-Sq = 78.3% R-Sq(adj) = 75.6% Analysis of Variance Source DF SS MS F P Regression 1 735.84 735.84 28.93 0.001 Residual Error 8 203.51 25.44 Total 9 939.35 Unusual Observations Obs MediaExp Amount Fit SE Fit Residual St Resid 1 120 36.30 27.55 3.30 8.75 2.30R R denotes an observation with a large standardized residual b. Minitab identifies observation 1 as having a large standardized residual; thus, we would consider observation 1 to be an outlier. 53. a. The Minitab output is shown below: The regression equation is Exposure = - 8.6 + 7.71 Aired Predictor Coef SE Coef T P Constant -8.55 21.65 -0.39 0.703 Aired 7.7149 0.5119 15.07 0.000 S = 34.88 R-Sq = 96.6% R-Sq(adj) = 96.2% Analysis of Variance Source DF SS MS F P Regression 1 ***** ***** 227.17 0.000 Residual Error 8 9735 1217 Total 9 ***** Unusual Observations Obs Aired Exposure Fit SE Fit Residual St Resid 1 95.0 758.8 724.4 32.0 34.4 2.46RX R denotes an observation with a large standardized residual X denotes an observation whose X value gives it large influence. b. Minitab identifies observation 1 as having a large standardized residual; thus, we would consider observation 1 to be an outlier. Minitab also identifies observation 1 as an influential observation. 54. a. The Minitab output is shown below: The regression equation is Salary = 707 + 0.00482 MktCap Predictor Coef SE Coef T P Constant 707.0 118.0 5.99 0.000 MktCap 0.***-***** 0.***-***** 5.96 0.000 S = 379.8 R-Sq = 66.4% R-Sq(adj) = 64.5% Analysis of Variance Source DF SS MS F P Regression 1 ***-***** ***-***** 35.55 0.000 Residual Error 18 ***-***** ***** Total 19 ***-***** Unusual Observations Obs MktCap Salary Fit SE Fit Residual St Resid 6 ***** 3325.0 3149.5 338.6 175.5 1.02 X 17 ***** 116.2 1289.5 86.4 -1173.3 -3.17R R denotes an observation with a large standardized residual X denotes an observation whose X value gives it large influence. b. Minitab identifies observation 6 as having a large standardized residual and observation 17 as an observation whose x value gives it large influence. A standardized residual plot against the predicted values is shown below: 55. No. Regression or correlation analysis can never prove that two variables are casually related. 56. The estimate of a mean value is an estimate of the average of all y values associated with the same x. The estimate of an individual y value is an estimate of only one of the y values associated with a particular x. 57. To determine whether or not there is a significant relationship between x and y. However, if we reject B1 = 0, it does not imply a good fit. 58. a. The Minitab output is shown below: The regression equation is Price = 9.26 + 0.711 Shares Predictor Coef SE Coef T P Constant 9.265 1.099 8.43 0.000 Shares 0.7105 0.1474 4.82 0.001 S = 1.419 R-Sq = 74.4% R-Sq(adj) = 71.2% Analysis of Variance Source DF SS MS F P Regression 1 46.784 46.784 23.22 0.001 Residual Error 8 16.116 2.015 Total 9 62.900 b. Since the p-value corresponding to F = 23.22 = .001 a = .05, the relationship is significant. c. = .744;a good fit. The least squares line explained 74.4% of the variability in Price. d. 59. a. The Minitab output is shown below: The regression equation is Options = - 3.83 + 0.296 Common Predictor Coef SE Coef T P Constant -3.834 5.903 -0.65 0.529 Common 0.***** 0.02648 11.17 0.000 S = 11.04 R-Sq = 91.9% R-Sq(adj) = 91.2% Analysis of Variance Source DF SS MS F P Regression 1 ***** ***** 124.72 0.000 ResidualError 11 1341 122 Total 12 ***** b. ; approximately 40.6 million shares of options grants outstanding. c. = .919; a very good fit. The least squares line explained 91.9% of the variability in Options. 60. a. The Minitab output is shown below: The regression equation is IBM = 0.275 + 0.950 SP 500 Predictor Coef StDev T P Constant 0.2747 0.9004 0.31 0.768 SP 500 0.9498 0.3569 2.66 0.029 S = 2.664 R-Sq = 47.0% R-Sq(adj) = 40.3% Analysis of Variance Source DF SS MS F P Regression 1 50.255 50.255 7.08 0.029 Error 8 56.781 7.098 Total 9 107.036 b. Since the p-value = 0.029 is less than a = .05, the relationship is significant. c. r2 = .470. The least squares line does not provide a very good fit. d. Woolworth has higher risk with a market beta of 1.25. 61. a. b. It appears that there is a positive linear relationship between the two variables. c. The Minitab output is shown below: The regression equation is High = 23.9 + 0.898 Low Predictor Coef SE Coef T P Constant 23.899 6.481 3.69 0.002 Low 0.8980 0.1121 8.01 0.000 S = 5.285 R-Sq = 78.1% R-Sq(adj) = 76.9% Analysis of Variance Source DF SS MS F P Regression 1 1792.3 1792.3 64.18 0.000 Residual Error 18 502.7 27.9 Total 19 2294.9 d. Since the p-value corresponding to F = 64.18 = .000 a = .05, the relationship is significant. e. = .781; a good fit. The least squares line explained 78.1% of the variability in high temperature. f. 62. The ***** output is shown below: The regression equation is Y = 10.5 + 0.953 X Predictor Coef Stdev t-ratio p Constant 10.528 3.745 2.81 0.023 X 0.9534 0.1382 6.90 0.000 s = 4.250 R-sq = 85.6% R-sq(adj) = 83.8% Analysis of Variance SOURCE DF SS MS F p Regression 1 860.05 860.05 47.62 0.000 Error 8 144.47 18.06 Total 9 1004.53 Fit Stdev.Fit 95% C.I. 95% P.I. 39.13 1.49 ( 35.69, 42.57) ( 28.74, 49.52) a. = 10.5 + .953 x b. Since the p-value corresponding to F = 47.62 = .000 a = .05, we reject H0: b1 = 0. c. The 95% prediction interval is 28.74 to 49.52 or $2874 to $4952 d. Yes,since the expected expense is $3913. 63. a. The Minitab output is shown below: The regression equation is Defects = 22.2 - 0.148 Speed Predictor Coef SE Coef T P Constant 22.174 1.653 13.42 0.000 Speed -0.***** 0.04391 -3.37 0.028 S = 1.489 R-Sq = 73.9% R-Sq(adj) = 67.4% Analysis of Variance Source DF SS MS F P Regression 1 25.130 25.130 11.33 0.028 Residual Error 4 8.870 2.217 Total 5 34.000 Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 14.783 0.896 ( 12.294, 17.271) ( 9.957, 19.608) b. Since the p-value corresponding to F = 11.33 = .028 a = .05, the relationship is significant. c. = .739; a good fit. The least squares line explained 73.9% of the variability in the number of defects. d. Using the Minitab output in part (a), the 95% confidence interval is 12.294 to 17.271. 64. a. There appears to be a negative linear relationship between distance to work and number of days absent. b. The ***** output is shown below: The regression equation is Y = 8.10 - 0.344 X Predictor Coef Stdev t-ratio p Constant 8.0978 0.8088 10.01 0.000 X -0.***** 0.07761 -4.43 0.002 s = 1.289 R-sq = 71.1% R-sq(adj) = 67.5% Analysis of Variance SOURCE DF SS MS F p Regression 1 32.699 32.699 19.67 0.002 Error 8 13.301 1.663 Total 9 46.000 Fit Stdev.Fit 95% C.I. 95% P.I. 6.377 0.512 ( 5.195, 7.559) ( 3.176, 9.577) c. Since the p-value corresponding to F = 419.67 is .002 a = .05. We reject H0 : b1 = 0. d. r2 = .711. The estimated regression equation explained 71.1% of the variability in y; this is a reasonably good fit. e. The 95% confidence interval is 5.195 to 7.559 or approximately 5.2 to 7.6 days. 65. a. Let X = the age of a bus and Y = the annual maintenance cost. The ***** output is shown below: The regression equation is Y = 220 + 132 X Predictor Coef Stdev t-ratio p Constant 220.00 58.48 3.76 0.006 X 131.67 17.80 7.40 0.000 s = 75.50 R-sq = 87.3% R-sq(adj) = 85.7% Analysis of Variance SOURCE DF SS MS F p Regression 1 ***** ***** 54.75 0.000 Error 8 ***** 5700 Total 9***** Fit Stdev.Fit 95% C.I. 95% P.I. 746.7 29.8 ( 678.0, 815.4) ( 559.5, 933.9) b. Since the p-value corresponding to F = 54.75 is .000 a = .05, we reject H0: b1 = 0. c. r2 = .873. The least squares line provided a very good fit. d. The 95% prediction interval is 559.5 to 933.9 or $559.50 to $933.90 66. a. Let X = hours spent studying and Y = total points earned The ***** output is shown below: The regression equation is Y = 5.85 + 0.830 X Predictor Coef Stdev t-ratio p Constant 5.847 7.972 0.73 0.484 X 0.8295 0.1095 7.58 0.000 s = 7.523 R-sq = 87.8% R-sq(adj) = 86.2% Analysis of Variance SOURCE DF SS MS F p Regression 1 3249.7 3249.7 57.42 0.000 Error 8 452.8 56.6 Total 9 3702.5 Fit Stdev.Fit 95% C.I. 95% P.I. 84.65 3.67 ( 76.19, 93.11) ( 65.35, 103.96) b. Since the p-value corresponding to F = 57.42 is .000 a = .05, we reject H0: b1 = 0. c. 84.65 points d. The 95% prediction interval is 65.35 to 103.96 67. a. The Minitab output is shown below: The regression equation is Audit% = - 0.471 +0.000039 Income Predictor Coef SE Coef T P Constant -0.4710 0.5842 -0.81 0.431 Income 0.***-***** 0.***-***** 2.23 0.038 S = 0.2088 R-Sq = 21.7% R-Sq(adj) = 17.4% Analysis of Variance Source DF SS MS F P Regression 1 0.***** 0.***** 4.99 0.038 Residual Error 18 0.***** 0.04358 Total 19 1.00200 Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 0.8828 0.0523 ( 0.7729, 0.9927) ( 0.4306, 1.3349) b. Since the p-value = 0.038 is less than a = .05, the relationship is significant. c. r2 = .217. The least squares line does not provide a very good fit. d. The 95% confidence interval is .7729 to .9927.。
商务与经济统计ppt第13章A
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Data obtained from observational or experimental studies can be used for the analysis.
We want to use the sample results to test the following hypotheses:
Slide 8
Analysis of Variance: A Conceptual Overview Sampling Distribution of x Given H0 is True
Sample means are close together because there is only
In an observational study, no attempt is made to control the factors.
Cause-and-effect relationships are easier to establish in experimental studies than in observational studies.
In an experimental study, one or more factors are controlled so that data can be obtained about how the factors influence the variables of interest.
商务与经济统计课件 (6)
生活中的统计：经常需要对两个班级同一学科考试平均成绩进行比 较而不计较成绩的绝对高低；又如：对男女两组人群进行肺活量大 小的比较以鉴别二者是否存在显著差异但也不计较每组人群肺活量 的绝对高低等等问题都属于均值的比较问题
n n
难点：有关公式的理解，特别是两总体联合方差的表达形式 重点：均值比较的区间估计法
n n
重点：假设检验的基本思路、工作原理、方法特点； 难点：检验时所用的繁杂公式以及公式之间的细微差别
n
9.1 零假设和择假设
n n n
对研究性假设的检验 对陈述正确性的检验 对决策情况下的检验
n
零假设和备择假设类型 设 0 表示在零假设和备择假设中考虑的某一特定数值。一般来说，对 总体均值的假设检验采取下面的三种形式之一：
sx1 x2
2 (n1 1) s12 (n2 1) s2 n1 n2 2
n
当 2 2 1 2
2
时
s2( 1 1 ) n1 n2
x x
1
2
的点估计为 s x
1
x2
则估计出的区间为：
x1 x 2 t 2 s x1 x2
n n
式中，t的值基于自由度为 n1 n 2 2 的分布，为置信度 μ1-μ2的区间估计:小样本情形且σ1,σ2 的值已知
n
样本容量较小时(n<30)，用样本标准差s来估计总体标准差。入股总 体具有整台概率分布也是合理的，那么就可以用t分布来推断总体的 均值。在这种情况下，检验统计量是：
n
p值和T统计量 n 双尾检验 9.6 单个总体比例的检验
n
n
t
x 0 s n
有如下三种形式
商务与经济统计15版
商务与经济统计15版摘要：I.引言A.统计学的定义B.商务与经济统计的重要性C.本书的目标读者II.数据收集与整理A.数据来源B.数据类型C.数据整理III.描述性统计分析A.频数与频率分布B.图表法C.统计量度IV.概率论基础A.随机实验与样本空间B.事件与概率C.概率公理体系V.抽样与抽样分布A.抽样方法B.抽样分布C.参数估计VI.假设检验A.假设检验的基本思想B.常见的检验方法C.检验的误差VII.相关与回归分析A.相关分析B.一元线性回归C.多元线性回归VIII.时间序列分析与预测A.时间序列的基本概念B.平稳性检验C.时间序列模型与预测IX.统计软件的应用A.常见统计软件介绍B.软件在统计分析中的应用C.注意事项X.总结与展望A.统计学在经济和管理中的作用B.统计学的发展趋势C.学习建议正文：商务与经济统计（第15 版）从数据收集与整理、描述性统计分析、概率论基础、抽样与抽样分布、假设检验、相关与回归分析、时间序列分析与预测以及统计软件的应用等方面，系统地介绍了商务与经济统计的基本概念、原理和方法。
统计学是一门关于数据的科学，旨在通过对数据的收集、整理、分析、解释和展示，揭示数据背后的规律和趋势。
商务与经济统计作为统计学的一个重要分支，旨在为经济和管理决策提供依据。
数据收集与整理是统计分析的基础。
数据来源包括企业内部数据、政府公开数据和市场调查数据等。
数据类型主要包括定量数据和定性数据。
数据整理包括数据清洗、数据转换和数据汇总等步骤。
描述性统计分析主要通过频数与频率分布、图表法和统计量度等方法，对数据进行概括和描述。
概率论基础是统计学的理论基石，涉及随机实验与样本空间、事件与概率以及概率公理体系等内容。
抽样与抽样分布、假设检验、相关与回归分析以及时间序列分析与预测等方法，主要应用于对数据进行深入分析和预测。
这些方法在商务和经济领域具有广泛的应用价值。
统计软件的应用使得统计分析变得更加简便和高效。
商务与经济统计 第13版案例分析答案
商务与经济统计第13版案例分析答案1. 引言本文将对《商务与经济统计第13版》的案例分析题进行答案解析。
通过对案例进行深入剖析，旨在帮助读者更好地理解和应用商务与经济统计知识，提高解决实际问题的能力。
2. 案例1: 公司市场调研分析2.1 案例描述案例中，一家制造业公司希望进一步扩大市场份额。
为了做出决策，他们雇佣了一家市场调研公司进行研究。
市场调研公司对潜在客户进行了问卷调查，并收集了一系列数据，包括客户的年龄、性别、收入、购买习惯等。
2.2 答案分析针对这个案例，可以使用商务与经济统计中的一些基本方法和工具进行分析，例如：•描述性统计分析：对收集到的数据进行整理、汇总和描述，包括计算平均值、中位数、众数等，并绘制相关的图表，如频率分布表、饼图等。
•推论统计分析：使用概率分布、假设检验等方法，对样本数据进行推断，从而得出对总体的推论结论。
•回归分析：建立回归模型来研究不同变量之间的关系，并通过模型拟合和预测来支持决策。
3. 案例2: 零售店销售数据分析3.1 案例描述这个案例中，一个零售店希望利用已有的销售数据进行分析，以提高经营效益。
他们收集了一段时间内的销售记录，包括销售额、时间、产品种类、地理位置等信息。
3.2 答案分析针对这个案例，可以使用商务与经济统计的方法对销售数据进行分析，如下所示：•时间序列分析：通过对销售数据的时间序列进行建模和分析，可以了解销售趋势、季节性变动等，并预测未来的销售情况。
•空间分析：通过对销售数据在地理位置上的分布进行分析，可以了解销售的地域特点，为制定营销策略提供依据。
•预测分析：基于历史销售数据，可以利用回归分析、指数平滑法等方法，进行销售额的预测，为经营决策提供支持。
4. 案例3: 制造业质量控制分析4.1 案例描述在这个案例中，一家制造业公司关注产品的质量控制问题。
他们收集了一批产品的样本，并针对产品质量的关键指标进行了测试，如尺寸、硬度、重量等。
4.2 答案分析针对这个案例，可以运用商务与经济统计的方法对产品质量进行分析如下：•抽样统计分析：利用样本数据进行参数估计、假设检验等，研究样本数据与总体质量的关系。
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1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
2005
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2006
Composite Index Number
• Made up of two or more commodities • A simple index using the total price or total quantity of all the series (commodities) • Disadvantage: Quantity of each commodity purchased is not considered
Now compute the simple composite index by dividing each total by the January 2005 total. For example, December 2006:
12 / 06price 100 1/ 05price 99.64 100 95.49 © 2011 Pearson Education, Inc 104.3
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Statistics for Business and Economics
Chapter 13 Time Series: Descriptive Analyses, Models, & Forecasting
© 201 5 M -0 5 M -0 6 M -0 6 S05 S06 J05 J05 J06 N J06 N -0 6
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Weighted Composite Price Index
A weighted composite price index weights the prices by quantities purchased prior to calculating totals for each time period. The weighted totals are then used to compute the index in the same way that the unweighted totals are used for simple composite indexes.
– e.g. Number of cell phones produced annually
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Steps for Calculating a Simple Index Number
1. Obtain the prices or quantities for the commodity over the time period of interest. 2. Select a base period. 3. Calculate the index number for each period according to the formula Index number at time t
• •
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Time Series
• • • • Data generated by processes over time Describe and predict output of processes Descriptive analysis
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Content
13.7 Forecasting Trends: Simple Linear Regression 13.8 Seasonal Regression Models 13.9 Autocorrelation and the Durbin-Watson Test
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Composite Index Number Example
The table on the next slide shows the closing stock prices on the last day of the month for Daimler–Chrysler, Ford, and GM between 2005 and 2006. Construct the simple composite index using January 2005 as the base period. (Source: )
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Simple Index Numbers 1990–2006
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Simple Index Numbers 1990–2006
Gasoline Price Simple Index 250.0 200.0 150.0 100.0 50.0 0.0
Simple Composite Index Solution
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Simple Composite Index Solution
Simple Composite Index Numbers 2005 – 2006
120.0 100.0 80.0 60.0 40.0 20.0 0.0
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Simple Composite Index Solution
First compute the total for the three stocks for each date.
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Simple Composite Index Solution
Symbolically,
Yt I t 100 Y0
where It is the index number at time t, Yt is the time series value at time t, and Y0 is the time series value at the base period.
Simple Index Number Solution
2006 Index Number
2006price 2.572 100 100 198 1.299 1990price
Indicates price had risen by 98% (100 – 198) between 1990 and 2006.
• Example: Consumer Price Index (CPI)
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Steps for Calculating a Laspeyres Index
Index Number
• • • Measures change over time relative to a base period Price Index measures changes in price
– e.g. Consumer Price Index (CPI)
Quantity Index measures changes in quantity
Time series value at time t 100 Time series value at base period
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Steps for Calculating a Simple Index Number
© 2011 Pearson Education, Inc
Learning Objectives
• Focus on methods for analyzing data generated by a process over time (i.e., time series data). Present descriptive methods for characterizing time series data. Present inferential methods for forecasting future values of time series data.
Year 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 $ 1.299 1.098 1.087 1.067 1.075 1.111 1.224 1.199 1.03 1.136 1.484 1.42 1.345 1.561 1.852 2.27 2.572
Content
13.1 Descriptive Analysis: Index Numbers 13.2 Descriptive Analysis: Exponential Smoothing 13.3 Time Series Components 13.4 Forecasting: Exponential Smoothing 13.5 Forecasting Trends: Holt’s Method 13.6 Measuring Forecast Accuracy: MAD and RMSE
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Simple Index Number Example