BIFMA X5.4 2005 lounge seating - tests
BIFMA 沙发稳定性测试
ANSI/BIFMA X5.4-200522 稳定性测试（参考图22）22.1 适用性该测试应按如下要求对独立样品进行测试：A．向后稳定性测试只适用于样品靠背。
B．向前稳定性适用于所有样品。
22.2 测试目的该测试的目的是为了评估测试样品的稳定性。
样品稳定性与个人习惯性坐法，样品款式，使用环境，地板材料，以及脚轮和滑轨款式、大小有关。
22.3 向后稳定性22.3.1 测试设置A．样品需放置在测试平台上B．对于具有调节功能的样品，需将所有功能调节至最不稳定的状态下进行向前稳定性的测试。
如座位或靠背，或二者均调节至最高高度位置；座位或靠背，或二者最前沿的位置；以及轮子，滑轨或（和）底盘最不稳固的位置。
C．应在样品座位中央或者座位位置最接近样品本身中央位置施加78N力（173lb），参考图22a。
以捆绑的绳子对样品进行施力，参考图22b。
D．靠背静荷的计算方法：座位数量/2，取其得出的最高整数，然后再用得出整数x 133N （30 lb.f）. （如：1个1座或2座的样品，应施加133N（30lbf.）的力）,一个3座或4座的样品，应该施加266N（60lbf.）)。
E．如有必要，可以使用块状物或其他设备对木柱对样品进行必要束缚，以防止其检测过程中滑动。
对于有旋转功能的样品，其底盘及轮子（如有）必须调整至对阻止样品前倾造成最小阻扰的状态。
22.4 向前稳定性（前翻）22.4.1 测试设置A．记录样品重量B．样品需放置在测试平台上C．对于具有调节功能的样品，需将所有功能调节至最不稳定的状态下进行向前稳定性的测试。
如座位或靠背，或二者均调节至最高高度位置；座位或靠背，或二者最前沿的位置；以及轮子，滑轨或（和）底盘最不稳固的位置。
D．如有必要，可以使用块状物或其他设备对木柱对样品进行必要束缚，以防止其检测过程中滑动。
对于有旋转功能的样品，其底盘及轮子（如有）必须调整至对阻止样品前倾造成最小阻扰的状态。
22.4.2 测试过程A. 与测试平台成45°±5°向下施力，可借助宽度不小于76mm（3英寸）的绳子完成。
BIFMA椅子测试报告2-1
Test Report No.: SDHGR100600780FT Date: Jul 06, 2010 Page 1 of 5 HOOKAY OFFICE FURNITURE CO., LTDNO.416, DACHONG JINLING NAN LU, NANSHA QU, GUANGZHOU, CHINAThe following sample(s) was / were submitted and identified on behalf of the client as:Sample Description : MESH CHAIR 网布办公椅Style / Item No. : SPM01Test Performed : ANSI/BIFMA X5.1 -2002Sample Receiving Date : Jun 17, 2010Test Performing Date : Jun 17, 2010 to Jul 06, 2010Test Result(s) : For further details, please refer to the following page(s)Signed for and on behalf ofSGS-CSTC Co., Ltd.Jack YaoSection ManagerTest Report No.: SDHGR100600780FT Date: Jul 06, 2010 Page 2 of 5Test Conducted: ANSI/BIFMA X5.1 -2002 General –Purpose Office Chair – Tests, American National Standard for Office Furniture.Type of chair: Type I & Type IIITest Items Test Methods & Requirements Test Results Back StrengthTest - Static - Type I-Functional Load(Clause 5.4)No loss of serviceability when 890N (200lbs.) is applied for 1 min.Applied 90° to the back at16in. above the seat.PassBack Strength Test -Static - Type I- ProofLoad(Clause 5.4)No sudden and major change in the structural integrity (loss ofserviceability is acceptable) when 1334N (300lbs.) is applied for 1min. Applied 90° to the back at 16in. above the seat.PassBack Strength Test -Static - Type II & III -Functional Load(Clause 6.4)No loss of serviceability when 667N (150lbs.) is applied for 1 min.Applied 90° to the back at 16in. above the seat.PassBack Strength Test -Static - Type II & III -Proof Load(Clause 6.4)No sudden and major change in the structural integrity (loss ofserviceability is acceptable) when 1112N (250lbs.) is applied for 1min. Applied 90° to the back at 16in. above the seat.PassBase Test – Static(Clause 7)No sudden and major change in the structural integrity under11,120N (2500lbs.) compressions for 1 min. The weight is thenremoved and reapplied for 1 min. The center column may nottouch the test platform during load applications.PassDrop Test – Dynamic -Functional Load(Clause 8.4)No loss of serviceability when 102kg (225lbs.) weight free fallsfrom 6in. height to the center of the seat. PassDrop Test Dynamic -Proof Load(Clause 8.4)No sudden and major change in the structural integrity (loss ofserviceability is acceptable) when 136kg (300lbs.) weight free fallsfrom 6in. height to the center of the seat.PassSwivel Test – Cyclic(Clause 9)No loss of serviceability after 60,000cycles of rotation (360°) undera 102kg (225lbs.) load on the seat at its max. height. Seat shallthen withstand another 60,000cycles of rotation at its lowestseating position. Total 120,000cycles.PassTilt Mechanism Test –Cyclic – Type I & II(Clause 10)No loss of serviceability after 300,000cycles under a 102kg(225lbs.) load to the center of the seat PassImpact Test – Cyclic(Clause 11.3)No loss of serviceability in 100,000cycles impact. A weight of 57kg(125lbs.) free falls onto the seat from 1 in. height.Pass Front Corner LoadEase Test – Cyclic –off Center(Clause 11.4)No loss of serviceability after load each seat front corner with734N (165 lbs.) for 20,000 cycles, total 40,000 cycles.Note: this test is done after “Impact test” on the same sample.PassTo be continued…Test Report No.: SDHGR100600780FT Date: Jul 06, 2010 Page 3 of 5Test Items Test Methods & Requirements Test Results Stability Test – RearStability(Clause 12.3)A 79kg (173lbs.) weight is placed to the seat center (strapped asFig. 12a). Obstruct the chair casters/legs with 13mm (½in.)obstacle. A tipping force is applied to the chair back until the totalweight is transferred to the rear support members. The tippingforce shall not be less than:Type I and II –89N (20 lbs.)Type III –156N (35 lbs.)PassStability Test – FrontStability(Clause 12.4)The chair is obstructed with a 13mm (½ in.) obstruction to the chaircasters/legs. A downward load of 600N (135lbs.) is centered 60mm(2.4in.) from the seat front center edge. The seat shall withstand a20N (4.5lbf.) horizontally from the front seat edge without tipping.PassArm Strength TestVertical – Static -Functional Load(Clause 13.4)No loss of serviceability when 890N (200lbs.) is applied for 1 min.The vertical load is uniformly applied along a 127mm (5in.) lengthat the apparent weakest point.PassArm Strength TestVertical-Static - ProofLoad(Clause 13.4)No sudden and major change in the structural integrity (loss ofserviceability is acceptable) when 1334N (300lbs.) is applied for 1min. The vertical load is uniformly applied along a 127mm (5 in.)length at the apparent weakest point.PassArm Strength TestHorizontal –Static -Functional Load(Clause 14.4)No loss of serviceability when 445N (100lbs.) for 1 min. is appliedhorizontally outward to the armrest at the most forward point of thearmrest.PassArm Strength TestHorizontal – Static - ProofLoad(Clause 14.4)No sudden and major change in the structural integrity (loss ofserviceability is acceptable) when 667N (150lbs.) for 1 min. isapplied horizontally outward to the armrest at the most forwardpoint of the armrest.PassBack Durability Test –Cyclic – Type I(Clause 15)No loss of serviceability in 120,000 cycles with a 102kg (225lbs.)in the center of the seat and a 445N (100lbf.) 90° to the center ofthe chair back. For chairs with a back width greater than 406mm(16in.), test at the center of chair back for 80,000cycles and then102mm (4in.) off-center 40,000 cycles, half to each side.PassBack Durability Test –Cyclic – Type II & III(Clause 16)No loss of serviceability in 120,000 cycles with a 102kg (225lbs.)in the center of the seat and a 334N (75lbf.) 90° to the center ofthe chair back. For chairs with a back width greater than 406mm(16in.), test at the center of chair back for 80,000cycles and then102mm (4in.) off-center 40,000 cycles, half to each side.PassCaster / Chair BaseDurability Test ForPedestal Base Chair(Clause 17.1)No loss of service after 2,000cycles over a hard surface with 3obstacles and 98, 000cycles over a smooth hard surface withoutobstacles under a 102kg (225lbs.) load on the seat. Test stroke is762mm (30in.) minimum. The caster should not separate under22N (5lbs.) pulling force in line with the caster stem after thecycling test.PassTo be continued…Test Report No.: SDHGR100600780FT Date: Jul 06, 2010 Page 4 of 5Test Items Test Methods & Requirements Test Results Caster / Chair BaseDurability Testfor Chairs with Legs(Clause 17.2)No loss of service after 2,000cycles over a hard surface with 2obstacles and 98, 000cycles over a smooth hard surface withoutobstacles under a 102kg (225lbs.) load on the seat. Test stroke is762mm (30in.) minimum. The caster should not separate under22N (5lbs.) pulling force in line with the caster stem after thecycling test.NALeg Strength Test -Front Load - FunctionalLoad(Clause 18.3)No loss of serviceability when a force of 334N (75lbf.) is applied toeach front leg individually for 1 minute.NALeg Strength Test-Front Load - Proof Load(Clause 18.3)No sudden and major change in the structural integrity (loss ofserviceability is acceptable) when a force of 556N (125lbf.) isapplied to each front leg individually for 1 minute.NALeg Strength Test-Side Load - FunctionalLoad(Clause 18.4)No loss of serviceability when a force of 334N (75lbf.) is appliedonce to each front and rear leg individually for 1 minute.NALeg Strength Test -SideLoad - Proof Load(Clause 18.4)No sudden and major change in the structural integrity (loss ofserviceability is acceptable) when a force of 512N (115lbf.) isapplied once to the front and rear leg individually for 1 minute.NAFootrest Durability Test(Clause 19)No loss of serviceability after 50,000cycles of a 890N (200lbf) loadvertical along 102mm (4in.) length of the footrest at the apparentweakest point of the structure.NAArm Durability Test-Cyclic(Clause 20)No structural breakage or loss of serviceability when a force of400N (90lbf.) is applied to each arm at a 10º angle ±1º for60,000cyclesPassOut Stop Tests for Chairswith Manually AdjustableSeat Depth(Clause 21)Place a 70 kg (154 lb) rigid mass in the center of the seat. Holdthe seat at its most position. A cable is attached to the most rigidpoint of the vertical centerline of the seat. Hang a weight of 25 kg(55 lb) on the opposite end of the cable. Release the weight so itcan drag the seat move forward rapidly and impactNATablet Arm Static LoadTest(Clause 22)Apply a load of 68 kg (150 lb) at the apparent weakest position for5 minutes and remove the load. No sudden and major change inthe chair when the application of the load.NATablet Arm Load EaseTest – Cyclic(Clause 23)No loss of serviceability to the unit after loading the tablet surfacewith a weight of 35 kg (77 lb) for a total 100,000 cycles. NARemark:1) NA – Not applicable;2) Type of chair:Type I – tilt chair: a chair with a seat tilts with a counterbalancing force;Type II – fixed seat angle, tilting backrest: a chair that provides a fixed angle with a tilting backrest;Type III – fixed seat angle, fixed backrest: a chair that provides a fixed seat angle with a fixed backrest;To be continued…Test Report No.: SDHGR100600780FT Date: Jul 06, 2010 Page 5 of 5PHOTO APPENDIXSample as received Sample as received***End of Report***。
普通测试申请表
普通测试申请表General Testing Requisition FormForm No.:Applicant Name (Payer): _________________________________________________ Official Use Only 申请单位名称（付款单位）: _________________________________________________ Rpt. No. :Address: _______________________________________________________________申请单位地址: __________________________________________________________Contact Person联系人: ___________ Telephone电话：__________________ Fax传真：__________ 邮编：____ Report Address报告寄往:_______________________________________________ Email地址:_________________ Company Name & Address shown on Test Report (if different from the Applicant Name Above):测试报告上所注明的单位名称及地址（如果与以上申请单位不符）：Sample Description 样品描述Item/Article/Model/Style No. 式样编号Qty. of Sample Submitted 样品数量Buyer 买家: ____________________________________ (可不填）P.O. No.:_________________________Supplier/Vendor/Manufacturer供应商/贸易商/生产商:________________________________________________________ Goods Exported to 产品运往：___________________ Country of Origin原产地:________________________Test(s) Required (Please tick appropriate boxes) : 测试要求Please select the test from the listed standard as attached, if it is not in the list, please indicate the test standard or test method in this form. 请从所附的标准清单中选取测试标准，若不在其中，请在此注明测试标准或方法。
哪些产品,哪些国家需要什么认证
哪些产品，哪些国家需要什么认证CE认证CE标志是产品进入欧盟国家及欧盟自由贸易协会国家市场的“通行证”。
任何规定的（新方法指令所涉及的）产品，无论是欧盟以外还是欧盟成员国生产的产品，要想在欧盟市场上自由流通，在投放欧盟市场前，都必须符合指令及相关协调标准的要求，并且加贴CE标志。
这是欧盟法律对相关产品提出的一种强制性要求，为各国产品在欧洲市场进行贸易提供了统一的最低技术标准，简化了贸易程序。
目前包括24条新方法指令涉及CE认证。
需要CE认证的国家有：所有欧洲经济区域的国家。
包括:欧洲联盟：法国、联邦德国、意大利、荷兰、比利时、卢森堡、英国、丹麦、爱尔兰、希腊、西班牙、葡萄牙、奥地利、瑞典和、芬兰、塞浦路斯、匈牙利、捷克、爱沙尼亚、拉脱维亚、立陶宛、马耳他、波兰、斯洛伐克和斯洛文尼亚等25个国家。
欧洲自由贸易协会成员：瑞士、冰岛和挪威等3个国家。
以下产品需要加贴CE标志：――电气类产品――机械类产品――玩具类产品――无线电和电信终端设备――冷藏﹑冷冻设备――人身保护设备――简单压力容――热水锅炉――压力设备――民用爆炸物――游乐船――建筑产品――体外诊断医疗器械――植入式医疗器械――医疗电器设备――升降设备――燃气设备――非自动衡器――爆炸环境中使用的设备和保护系统VDE认证VDE的全称是VDE Testing and Certification Institute,意即德国电气工程师协会。
成立于1920年，总部位于德国法兰克福，是欧洲最有测试经验的试验认证和检查机构之一。
VDE作为一个国际认可的电子电器及其零部件安全测试及认证机构，在欧洲乃至国际上都享有较高的知名度。
其评估的产品范围包括家用及商业用途的电器、IT设备、工业和医疗科技设备、组装材料及电子元器件、电线电缆等。
VDE认证标志已在30个国家作了商标和/或服务标志的登记，因此，VDE认证标志是受到保护的，不能滥用，由于VDE对公众所负有的义务，VDE认证标志的内在价值无论在任何地方都是同样的。
上海财经大学计量经济学试卷
诚实考试吾心不虚 , 公平竞争方显实力, 考试失败尚有机会 , 考试舞弊前功尽弃。
上海财经大学《计量经济学 》课程考试卷（A ）闭卷课程代码 课程序号2008—2009 学年第 1 学期姓名 学号 班级一、单选题(每小题2分, 共计40分)1.如果模型中变量在10%的显著性水平下是显著的, 则（ D ）A. 该变量在5%的显著性水平下是也显著的B. 该变量在1%和5%的显著性水平下都是显著的C. 如果P 值为12%, 则该变量在15%的显著性水平下也是显著的 D 、 如果P 值为2%, 则该变量在5%的显著性水平下也是显著的 2.高斯－马尔可夫是( D )A.摇滚乐.B.足球运动.C.鲜美的菜.D.估计理论中的著名定理, 来自于著名的统计学家: Johan.Car.Friedric.Gauss 和Andre.Andreevic.Markov 。
3.以下关于工具变量的说法不正确的是（ B ）。
A.与随机干扰项不相关B. 与所替代的随机解释变量不相关C.与所替代的随机解释变量高度相关D.与模型中其他解释变量不相关4.在含有截距项的多元回归中, 校正的判定系数 与判定系数R2的关系有: （ B ） A.R2＜ B.R2＞ C.R2＝ D.R2与 的关系不能确定5.根据样本资料估计得出人均消费支出Y 对人均收入X 的回归模型为lnYi=2.00+0.75lnXi+ei, 这表明人均收入每增加1%, 人均消费支出将大约增加（ B ）A.0.2..B.0.75..C.2..D.7.5%6.在存在异方差的情况下, 普通最小二乘法（OLS ）估计量是（ B ） A.有偏的, 非有效的 B.无偏的, 非有效的 C.有偏的, 有效的 D.无偏的, 有效的7.已知模型的普通最小二乘法估计残差的一阶自相关系数为0, 则DW 统计量的近似值为（ C ）A.0B.1C.2D.48.在多元回归模型中, 若某个解释变量对其余解释变量回归后的判定系数接近1, 则表明原模型中存在（C）A.异方差性B.自相关C.多重共线性D.拟合优度低9.设某商品需求模型为: Yi＝β0＋β1Xi＋Ui, 其中Y是商品的需求量, X是商品的价格, 为了考虑全年12个月份季节变动的影响, 假设模型中引入了12个虚拟变量, 则会产生的问题是（D）A.异方差性B.自相关C.不完全的多重共线性D.完全的多重共线性10.下列表述不正确的是（D）A.尽管存在不完全的多重共线性, 普通最小二乘估计量仍然是最优线性无偏估计量B.双对数模型的R2可以与对数－线性模型的R2相比较, 但不能与线性－对数模型的R2相比较。
（概率论与数理统计专业论文）一个平均剩余寿命递减的新非参数检验方法
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论文作者签名：瞰导师签期．，幽羔：！：兰州大学硕士学位论文致谢谨把此文献给我的导师一李效虎教授！您力求在集体中创造一种共同热爱科学和渴求知识的气氛，使智力兴趣成为一条线索，把学生连接在一起；您教导我们看书不能有信仰无思考，要大胆提出问题，勤于思考，找出问题的最本质思想一一读书无疑者须教有疑，有疑者却要无疑，到这里才是长进；您教导我们要有严谨诚实的科研作风，剽窃的思想正如借来的钱一样，只显示借债者的贫乏罢了；您深思熟虑的思想好比火星，一颗火星必会点燃另一颗火星．本论文是在导师的精心指导下完成的，导师渊博的知识、勤勉敬业的精神、精益求精的治学态度、严谨的科研作风和崇高的人格魅力将是我终生的一面镜子．三年来承蒙教诲之恩，未经酬报，谨表示衷心地感谢！感谢老师三年来如春风细雨般地言传身教，并祝愿老师青春常驻、工作生活一帆风顺！曹文芹２００５年４月前言在可靠性理论中，对系统寿命长度随机规律的研究与人们的工作和生活休戚相关，因为产品（系统）故障所造成的后果随时都能感觉到．在五、六十年代的研究中，如何利用寿命数据（一组样本）对其进行统计分析占据了很重要的地位．例如，确定这些数据来自什么总体，以及如何进行参数估计等．其后，随着理论本身发展的需要，开始了对具有共性的一类寿命分布函数性质的探讨，形成了寿命分布类的概念．例如，基于对老化、磨损等现象的直观经验，引进了诸如“新比旧要好”（ＮＢＵ），“依平均意义新比旧好”（ＮＢＵＥ），“平均剩余寿命递减”（ＤＭＲＬ），“失效率递增”（ＩＦＲ）等分布类．无论是在科研中，还是在实践中，对寿命分布性质的研究都具有重要意义．在实际研究中，通常不能给出一组样本的精确分布函数，所以检验一组数据的分布形式是否具有某种特性变得至关重要．例如，ＰｒｏｓｅｈａｎａｎｄＰｙｋｅ（１９６７）对ＩＦＲ（ＤＦＲ）性质建立了检验；ＨｏｌｌａｎｄｅｒａｎｄＰｒｏｓｃｈａｎ（１９７２）、Ａｈｍａｄ（１９７５）和Ｄｅｓｈｌ）ａｎｄｅａｎｄＫｏｃｈａｒ（１９８３）先后对ＮＢＵ（ＮＷＵ）性质建立了检验；Ｂｅｌｚｕｎｃｅｅｔａｌ（２０００）对ＮＢＵＥ性质建立了检验等等．自从１９７５年，Ｈｏｌｌａｎｄｅｒ和Ｐｒｏｓｃｈａｎ首次对平均剩余寿命递减性质（ＤＭＲＬ）建立了一个非参数检验以来，人们对ＤＭＲＬ性质的检验已作了许多研究．Ｋｌｅ／ｓｊ５（１９８３），Ｂａｎ咖ｏｐａｄｈｙａｙａｎｄＢａｓｕ（１９９０），Ｂｅｌｚｕｎｃｅｅｔａｌ（２０００）和ＡｈｍａｄａｎｄＭｕｇｄａｄｉ（２００４）分另Ⅱ研究了ＤＭＲＬ性质的检验．本文对ＤＭＲＬ性质给出一个新的非参数检验方法，并对该方法的性能进行了深入的研究．本论文是在我的导师李效虎教授的精心指导下完成的，在此表示深切地感谢ｉ同时感谢我的同门冯秀英同学，文中不少成果与她紧密地协助分不开；同时也要感谢我｛ｌｉｄ，组的每一位成员，学习期间与他们的讨论让我获益颇多；最后还要感谢我的父母，他们三年如一日的支持是我学习的动力．２曹文芹兰州大学２００５年４月兰州大学硕士学位论文３摘要本文基于ＤＭＲＬ程度的刻画，建立了一个平均剩余寿命递减性质的新非参数检验方法．证明了该检验统计量的渐近正态性；并通过计算Ｐｉｔｍａｎ渐近效率，对该方法与文献中其它方法进行了比较；利用Ｅｄｇｅ—ｗｏｒｔｈ展开成功提高了该统计量的收敛速度．最后，通过一些数值例子展示了新检验方法的优劣性．关键词：渐近正态性；ＤＭＲＬ；’Ｅｄｇｅ—ｗｏｒｔｈ展开；Ｊａｃｋｋｎｉｆｅ估计；Ｐｉｔｍａｎ渐近效率；检验假设；ｕ一统计量．中圉分类号：０２１２．２兰州大学硕士学位论文４ＡｂｓｔｒａｃｔＩｎｔｈｉｓｐａｐｅｒ，ｗｅｄｅｖｅｌｏｐａｎｅｗｔｅｓｔｓｔａｔｉｓｔｉｃｆｏｒｔｅｓｔｉｎｇｔｈｅｓｔｒｉｃｔＤＭＲＬａｇｉｎｇｐｒｏｐｅｒｔｙｏｆｃｅｒｔａｉｎｌｉｆｅｄｉｓｔｒｉｂｕｔｉｏｎｏｆｉｎｔｅｒｅｓｔ．ＴｈｅａｓｙｍｐｔｏｔｉｃｎｏｒｍａｌｉｔｙｉｓｅｓｔａｂｌｉｓｈｅｄａｎｄｔｈｅｃｏｍｐａｒｉｓｏｎｂｅｔｗｅｅｎｔｈｅｔｅｓｔｐｒｏｐｏｓｅｄａｎｄＳｏｍｅｏｔｈｅｒｒｅｌａｔｅｄｏｎｅｓｉｎ１ｉｔｅｒａｔｕｒｅｉｓｃｏｎｄｕｃｔｅｄｔｈｒｏｕｇｈｅｖａｌｕａｔｉｎｇｔｈｅＰｉｔｍａｎ’Ｓａｓｙｍｐｔｏｔｉｃｒｅｌａｔｉｖｅｅｆｆｉｃｉｅｎｃｙ．Ｅｄｇｅ—ｗｏｒｔｈｅｘｐａｎｓｉｏｎｉｓａｌｓｏｕｓｅｄｔｏｉｍｐｒｏｖｅｄｔｈｅａｃｃｕｒａｃｙｏｆｔｈｅｃｏｎｖｅｒｇｅｎｃｅｒａｔｅｏｆｔｈｉｓｔｅｓｔｓｔａｔｉｓｔｉｃ．Ｓｏｍｅｎｕｍｅｒｉｃａｌｒｅｓｕｌｔｓａｒｅｐｒｅｓｅｎｔｅｄａｓｗｅｌｌｔｏｄｅｍｏｎｓｔｒａｔｅｔｈｅｐｅｒｆｏｒｍａｎｃｅａｎｄｔｈｅａｓｙｍｐｔｏｔｉｃｎｏｒｍａｌｉｔｙｏｆｔｈｅｎｅｗｔｅｓｔｉｎｇｐｒｏｃｅｄｕｒｅ．ＫｅｙＷｏｒｄｓ：Ａｓｙｍｐｔｏｔｉｃｎｏｒｍａｌｉｔｙ；ＤＭＲＬ；Ｅｄｇｅ—ｗｏｍｂｅｘｐａｎｓｉｏｎ；Ｊａｃｋｋｎｉｆｅ；Ｐｉｔｍａｎ’ＳａｓｙｍｐｔｏｔｉｃｅｆｆｉｅｉｅｎｃｙｉＴｅｓｔｉｎｇｈｙｐｏｔｈｓｉｓ；Ｕ—ｓｔａｔｉｓｔｉｃ．ＡＭＳＳｕｂｊｅｃｔＣｌａｓｓｉｆｉｃａｔｉｏｎ：６０Ｇ１０，６２Ｅ２０第一章预备知识本章主要介绍了一些基本年龄性质和随机序，并阐述了论文的背景．假设非负随机变量ｘ表示元件的寿命长度，其分布函数和生存函数分别记为Ｆ（ｔ）和Ｆ（ｔ）＝１一Ｆ（ｔ），元件生存到时刻ｔ≥０的剩余寿命Ｘｔ＝（Ｘ—ｔｌｘ＞ｔ），它的生存函数由下式给出施）＝掣，ｚ＞０函数Ｅ五＿雎刮…，＝｛鼯，桐炒。
第05讲非参数检验
案例3
经过西弗吉尼亚公路150号里程碑的汽车平均时 速为68英里/小时。
情况一：对100辆经过里程碑的汽车进行测速，发 现速度只有30英里/小时。
情况二：对100辆经过里程碑的汽车进行测速，发 现速度只有67英里/小时。
案例4 假如雪碧瓶的标签上标明的容量为500毫升。
如果你从市场上随机抽取25瓶，发现其平均含 量为499.5毫升，标准差s为2.63毫升 。
基本假定
方差分析中通常要有以下假定：
首先是各样本的独立性，即各组观察数据，是 从相互独立的总体中抽取的，只有是独立的随机 样本，才能保证变异的可加性；
其次要求所有观察值都是从正态总体中抽取， 且方差相等。在实际应用中能够严格满足这些假 定条件的客观现象是很少的，在社会经济现象中 更是如此。但一般应近似地符合上述要求。
组内均方: MSE= SSE/(自由度n-k)
F检验示意图
单因素方差分析
1, 单因素方差分析假设 2. 单因素方差分析数据结构表 3, 单因素方差分析表 4. 单因素方差分析SPSS界面
例子１
例子２
双因素方差分析
（不考虑交互作用）
1. 双因素方差分析假设 2. 双因素方差分析数据结构表 3. 双因素方差分析表 4. 双因素方差分析SPSS界面
问：是否有显著不同？
区间估计 x t (n 1) s 499.5 2.797 2.63 / 25 498.03 ~ 500.97
2
n
问：是否能断定饮料厂商欺骗了消费者？
区间估计
x t (n 1)
s 499 .5 2.492 2.63 / n
25 500 .81
基本思想
方差分析的基本思想是利用方差的可分解性， 检查所讨论因素是否作为系统性因素来影响试验结果。 所谓“系统性因素”是指由于试验因素的变异而产生 的试验结果的数量差异，例如，利用四种配料生产某 种产品，其使用寿命差异就是备料方法不同所造成的 类型差异和许多未能控制的“偶然因素”所造成的随 机差异（也称为残差）的总和。进行方差分析的目的， 就是要认识产品使用寿命的差异主要是由类型差异引 起的还是由随机差异引起的。
异方差的检验的新方法
异方差的检验的新方法异方差的检验是统计学中一种常用的检验方法，用于检验不同组或不同观测点之间的方差是否存在显著差异。
在许多统计分析中，我们常常假设数据服从方差齐性，即不同组或不同观测点之间的方差是相等的。
但是在实际应用中，由于多种因素的影响，数据可能存在异方差的问题。
如果忽略了异方差的影响，可能会导致错误的结论或者低效的推断。
传统的异方差检验方法主要有三种：Bartlett检验、Levene检验和Brown-Forsythe检验。
这些方法在一些情况下可能会产生较高的错误率或者对异常值敏感，因此需要一种新的方法来提高异方差检验的准确性和稳健性。
在近年来的研究中，有学者提出了一种基于Bootstrap的异方差检验方法。
Bootstrap是一种统计重抽样的方法，通过有放回地从原始数据中抽取样本来估计总体分布的性质。
它的主要思想是通过生成大量的重抽样样本，来构建总体的概率分布，从而对总体参数进行精确估计。
基于Bootstrap的异方差检验方法主要分为两步：第一步是构建Bootstrap重抽样样本，通过有放回地从原始数据中抽取样本，并对每个样本计算方差；第二步是对构建的重抽样样本进行统计检验，比较不同样本的方差是否存在显著差异。
这种方法相比传统的异方差检验方法具有如下的优点：1. 非参数性：Bootstrap方法不依赖于数据的具体分布形式，适用于各种数据类型和分布情况。
相比传统的假设检验方法，它更加灵活，适用性更广。
2. 稳健性：采用Bootstrap方法能够有效地减少异常值对检验结果的影响，提高了异方差检验的稳健性。
3.可信度高：通过大量的重抽样构建总体分布，可以获得更精确的估计结果，增加了检验的可信度。
4. 有效性：Bootstrap方法能够通过生成大量的重抽样样本，增加了样本的有效信息，提高了检验的准确性。
除了基于Bootstrap的异方差检验方法，还有一些其他的新方法在近年来也受到了广泛的关注，比如基于偏差修正的异方差检验方法、基于混合模型的异方差检验方法等。
非注意盲实验报告
实验名称：非注意盲实验对消费者购买行为的影响研究实验日期：2023年10月25日实验地点：某大型购物中心实验对象：随机选取的50名消费者实验目的：本研究旨在探讨非注意盲实验对消费者购买行为的影响，以及消费者在无意识状态下对产品信息的接收和反应。
实验方法：1. 实验设计：采用非注意盲实验设计，将实验对象分为两组，每组25人。
实验组在购物过程中会随机接收到特定的产品信息，而非实验组则不会。
2. 实验材料：选取了10种不同类型的产品，每种产品均有多个品牌可供选择。
3. 实验步骤：- 实验对象进入购物中心后，随机分配至实验组或非实验组。
- 实验组在购物过程中，通过放置在购物通道中的海报、广告牌等形式接收到特定产品信息。
- 实验组和非实验组在购物结束后，填写调查问卷，记录购买行为和产品信息接收情况。
- 实验结束后，收集实验数据进行分析。
实验结果：1. 产品信息接收情况：实验组中，有80%的消费者表示在购物过程中接收到了特定产品信息，而非实验组中，仅有20%的消费者表示接收到了产品信息。
2. 购买行为：实验组中，有60%的消费者表示在购物过程中购买了实验组中的产品，而非实验组中，仅有30%的消费者表示购买了产品。
3. 产品信息对购买行为的影响：在实验组中，有70%的消费者表示产品信息对其购买决策产生了影响，而非实验组中，仅有40%的消费者表示受到了影响。
讨论与分析：1. 非注意盲实验对消费者购买行为产生了显著影响。
实验组中，消费者在无意识状态下接收到的产品信息对其购买决策产生了积极影响。
2. 产品信息的接收方式对消费者购买行为有显著影响。
通过非注意盲实验的方式，消费者在无意识状态下接收到的产品信息更容易被记住，从而影响其购买行为。
3. 本实验结果表明，非注意盲实验在市场营销中具有潜在的应用价值，企业可以通过这种方式提高产品的市场占有率。
结论：本研究表明，非注意盲实验对消费者购买行为具有显著影响。
在无意识状态下，消费者更容易接收并记住产品信息，从而影响其购买决策。
Testing Fisher, Neyman, Pearson, and Bayes
GeneralTesting Fisher,Neyman,Pearson,and Bayes Ronald C HRISTENSENThis article presents a simple example that illustrates the keydifferences and similarities between the Fisherian,Neyman-Pearson,and Bayesian approaches to testing.Implications formore complex situations are also discussed.KEY WORDS:Confidence;Lindley’s paradox;Most power-ful test;p values;Significance tests.1.INTRODUCTIONOne of the famous controversies in statistics is the disputebetween Fisher and Neyman-Pearson about the proper way toconduct a test.Hubbard and Bayarri(2003)gave an excellentaccount of the issues involved in the controversy.Another fa-mous controversy is between Fisher and almost all Bayesians.Fisher(1956)discussed one side of these controversies.Berger’sFisher lecture attempted to create a consensus about testing;seeBerger(2003).This article presents a simple example designed to clarifymany of the issues in these controversies.Along the way manyof the fundamental ideas of testing from all three perspectives areillustrated.The conclusion is that Fisherian testing is not a com-petitor to Neyman-Pearson(NP)or Bayesian testing becauseit examines a different problem.As with Berger and Wolpert(1984),I conclude that Bayesian testing is preferable to NP test-ing as a procedure for deciding between alternative hypotheses.The example involves data that have four possible outcomes, r=1,2,3,4.The distribution of the data depends on a param-eterθthat takes on valuesθ=0,1,2.The distributions aredefined by their discrete densities f(r|θ)which are given in Ta-ble1.In Section2,f(r|0)is used to illustrate Fisherian testing.In Section3,f(r|0)and f(r|2)are used to illustrate testing asimple null hypothesis versus a simple alternative hypothesisalthough Subsection3.1makes a brief reference to an NP test of f(r|1)versus f(r|2).Section4uses all three densities to illus-trate testing a simple null versus a composite alternative.Section 5discusses some issues that do not arise in this simple exam-ple.For those who want an explicit statement of the differences between Fisherian and NP testing,one appears at the beginning of Section6which also contains other conclusions and com-Ronald Christensen is Professor,Department of Mathematics and Statistics, University of New Mexico,Albuquerque,NM87131(E-mail:fletcher@stat. ).I thank the editor and associate editor for comments and suggestions that substantially improved this article.ments.For readers who are unfamiliar with the differences,it seems easier to pick them up from the example than it would be from a general statement.Very briefly,however,a Fisherian test involves only one hypothesized model.The distribution of the data must be known and this distribution is used both to deter-mine the test and to evaluate the outcome of the test.An NP test involves two hypotheses,a null hypothesis,and an alternative.A family of tests is considered that all have the same probability αof rejecting the null hypothesis when it is true.Within this family,a particular test is chosen so as to maximize the power, that is,the probability of rejecting the null hypothesis when the alternative hypothesis is true.Finally,some truth in advertising.Fisher did not use the term Fisherian testing and certainly made no claim to have originated the ideas.He referred to“tests of significance.”This often,but not always,gets contrasted with“tests of hypotheses.”When discussing Fisherian testing,I make no claim to be expositing exactly what Fisher proposed.I am expositing a logical basis for testing that is distinct from Neyman-Pearson theory and that is related to Fisher’s views.2.FISHERIAN TESTSThe key fact in a Fisherian test is that it makes no reference to any alternative hypothesis.It can really be thought of as a model validation procedure.We have the distribution of the (null)model and we examine whether the data look weird or not.For example,anα=.01Fisherian test of H0:θ=0is based entirely onr1234f(r|0).980.005.005.010With only this information,one must use the density itself to determine which data values seem weird and which do not.Ob-viously,observations with low density are those least likely to occur,so they are considered weird.In this example,the weird-est observations are r=2,3followed by r=4.Anα=.01 test would reject the modelθ=0if either a2or a3is observed. Anα=.02test rejects when observing any of r=2,3,4.In lieu of anαlevel,Fisher advocated using a p value to evalu-ate the outcome of a test.The p value is the probability of seeingT able1.Discrete Densities to be T estedr1234f(r|0).980.005.005.010 f(r|1).100.200.200.500 f(r|2).098.001.001.900©2005American Statistical Association DOI:10.1198/000313005X20871The American Statistician,May2005,Vol.59,No.2121something as weird or weirder than you actually saw.(There isno need to specify that it is computed under the null hypothe-sis because there is only one hypothesis.)In this example,theweirdest observations are2and3and they are equally weird.Their combined probability is.01which is the p value regardlessof which you actually observe.If you see a4,both2and3areweirder,so the combined probability is.02.r1234f(r|0).980.005.005.010p value 1.00.01.01.02In Fisherian testing,the p value is actually a more fundamentalconcept than theαlevel.Technically,anαlevel is simply adecision rule as to which p values will cause one to reject thenull hypothesis.In other words,it is merely a decision point asto how weird the data must be before rejecting the null model.If the p value is less than or equal toα,the null is rejected.Implicitly,anαlevel determines what data would cause one toreject H0and what data will not cause rejection.Theαlevelrejection region is defined as the set of all data points that havea p value less than or equal toα.Note that in this example,anα=.01test is identical to anα=.0125test.Both rejectwhen observing either r=2or3.Moreover,the probability ofrejecting anα=.0125test when the null hypothesis is true isnot.0125,it is.01.However,Fisherian testing is not interested in what the probability of rejecting the null hypothesis will be,it is interested in what the probability was of seeing weird data.The philosophical basis of a Fisherian test is akin to proof bycontradiction.We have a model and we examine the extent towhich the data contradict the model.The basis for suggesting acontradiction is actually observing data that are highly improb-able under the model.The p value gives an excellent measure ofthe extent to which the data do not contradict the model.(Large p values do not contradict the model.)If anαlevel is chosen, for any semblance of a contradiction to occur,theαlevel mustbe small.On the other hand,even without a specific alternative,makingαtoo small will defeat the purpose of the test,making itextremely difficult to reject the test for any reason.A reasonableview would be that anαlevel should never be chosen;that ascientist should simply evaluate the evidence embodied in the pvalue.As in any proof by contradiction,the results are skewed.Ifthe data contradict the model,we have evidence that the modelis invalid.If the data do not contradict the model,we have anattempt at proof by contradiction in which we got no contradic-tion.If the model is not rejected,the best one can say is that thedata are consistent with the model.Not rejecting certainly doesnot prove that the model is correct,whence comes the commonexhortation that one should never accept a null hypothesis.3.SIMPLE VERSUS SIMPLE Consider testing the simple null hypothesis H0:θ=0versus the simple alternative hypothesis H A:θ=2.This now appears to be a decision problem.We have two alternatives and we are deciding between them.This formulation as a decision problem is a primary reason that Fisher objected to NP testing,see Fisher (1956,chap.4).The relevant information for this testing problem isr1234f(r|0).980.005.005.010f(r|2).098.001.001.900Before examining formal testing procedures,look at the dis-tributions.Intuitively,if we see r=4we are inclined to believe θ=2,if we see r=1we are quite inclined to believe that θ=0,and if we see either a2or a3,it is stillfive times morelikely that the data came fromθ=0.Although Fisherian testing does not use an explicit alternative, there is nothing to stop us from doing two Fisherian tests:a test of H0:θ=0and then another test of H0:θ=2.The Fisherian tests both give perfectly reasonable results.The test for H0:θ=0has small p values for any of r=2,3,4.These are all strange values whenθ=0.The test for H0:θ=2has small p values when r=2,3.When r=4,we do not reject θ=2;when r=1,we do not rejectθ=0;when r=2,3,we reject bothθ=0andθ=2.The Fisherian tests are not being forced to choose between the two distributions.Seeing either a 2or a3is weird under both distributions.3.1Neyman-Pearson TestsNP tests treat the two hypotheses in fundamentally different ways.A test of H0:θ=0versus H A:θ=2is typically different from a test of H0:θ=2versus H A:θ=0.We examine the test of H0:θ=0versus H A:θ=2.NP theory seeks tofind the bestαlevel test.αis the probability of rejecting H0when it is true.The rejection region is the set of data values that cause one to reject the null hypothesis,so under H0the probability of the rejection region must beα.The best test is defined as the one with the highest power,that is, the highest probability of rejecting H0(observing data in the rejection region)when H A is true.Defining theαlevel as the probability of rejecting the null hypothesis when it is true places an emphasis on repeated sam-pling so that the Law of Large Numbers suggests that aboutαof the time you will make an incorrect decision,provided the null hypothesis is true in all of the samples.Although this is obviously a reasonable definition prior to seeing the data,its relevance after seeing the data is questionable.To allow arbitraryαlevels,one must consider randomized tests.A randomized test requires a randomized rejection region. How would one perform anα=.0125test?Three distinct tests are:(a)reject whenever r=4andflip a coin,if it comes up heads,reject when r=2;(b)reject whenever r=4andflip a coin,if it comes up heads,reject when r=3;(c)reject whenever r=2or3andflip a coin twice,if both come up heads,reject when r=4.It is difficult to convince anyone that these are reasonable practical procedures.r1234f(r|0).980.005.005.010f(r|2).098.001.001.900f(r|2)/f(r|0).1.2.290As demonstrated in the famous Neyman-Pearson lemma(see Lehmann1997,chap.3),optimal NP tests are based on the likeli-hood ratio f(r|2)/f(r|0).The best NP test rejects for the largest values of the likelihood ratio,thus theα=.01NP test rejects122Generalwhen r=4.This is completely different from the Fisherian.01test of H0that rejected when r=2,3.(On the other hand,the α=.02NP test coincides with the Fisherian test.Both reject when observing any of r=2,3,4.)The power of theα=.01NP test is.9whereas the power of the Fisherianα=.01test isonly.001+.001=.002.Clearly the Fisherian test is not a goodway to decide between these alternatives.But then the Fishe-rian test was not designed to decide between two alternatives.Itwas designed to see whether the null model seemed reasonableand,on its own terms,it works well.Although the meaning of αdiffers between Fisherian and NP tests,we have chosen two examples,α=.01andα=.02,in which the Fisherian test(rejection region)also happens to define an NP test with thesame numerical value ofα.Such a comparison would not beappropriate if we had examined,say,α=.0125Fisherian andNP tests.NP testing and Fisherian testing are not comparable proce-dures,a point also made by Hubbard and Bayarri(2003).NPtesting is designed to optimally detect some alternative hypoth-esis and Fisherian testing makes no reference to any alternativehypothesis.I might suggest that NP testers tend to want to havetheir cake and eat it too.By this I mean that many of them wantto adopt the philosophy of Fisherian testing(involving p val-ues,using smallαlevels,and never accepting a null hypothesis)while still basing their procedure on an alternative hypothesis. In particular,the motivation for using smallαlevels seems tobe based entirely on the philosophical idea of proof by contra-ing a largeαlevel would eliminate the suggestion thatthe data are unusual and thus tend to contradict H0.However,NP testing cannot appeal to the idea of proof by contradiction.For example,in testing H0:θ=1versus H A:θ=2,the most powerful NP test would reject for r=4,even though r=4isthe most probable value for the data under the null hypothesis.(For anyα<.5,a randomized test is needed.)In particular,this example makes it clear that p values can have no role in NPtesting!See also Hubbard and Bayarri(2003)and discussion. It seems that once you base the test on wanting a large proba-bility of rejecting the alternative hypothesis,you have put your-self in the business of deciding between the two hypotheses. Even on this basis,the NP test does not always perform very well.The rejection region for theα=.02NP test of H0:θ=0 versus H A:θ=2includes r=2,3,even though2and3 arefive times more likely under the null hypothesis than under the alternative.Admittedly,2and3are weird things to see un-der either hypothesis,but when deciding between these specific alternatives,rejectingθ=0(acceptingθ=2)for r=2orT able2.Posterior Probabilities ofθ=0,2for T wo Prior Dis-tributions a and bPrior r1234f(r|0).980.005.005.010f(r|2).098.001.001.900p a(0)=1/2p a(0|r).91.83.83.01 p a(2)=1/2p a(2|r).09.17.17.99p b(0)=1/6p b(0|r).67.50.50.002 p b(2)=5/6p b(2|r).33.50.50.9983does not seem reasonable.The Bayesian approach to testing, discussed in the next subsection,seems to handle this decision problem well.3.2Bayesian TestsBayesian analysis requires us to have prior probabilities on the values ofθ.It then uses Bayes’theorem to combine the prior probabilities with the information in the data tofind“posterior”probabilities forθgiven the data.All decisions aboutθare based entirely upon these posterior probabilities.The information in the data is obtained from the likelihood function.For an observed data value,say r=r∗,the likelihood is the function ofθdefined by f(r∗|θ).In our simple versus simple testing example,let the prior prob-abilities onθ=0,2be p(0)and p(2).Applying Bayes’theorem to observed data r,we turn these prior probabilities into poste-rior probabilities forθgiven r,say p(0|r)and p(2|r).To do this we need the likelihood function which here takes on only the two values f(r|0)and f(r|2).From Bayes’theorem,p(θ|r)=f(r|θ)p(θ)f(r|0)p(0)+f(r|2)p(2),θ=0,2.Decisions are based on these posterior probabilities.Other things being equal,whichever value ofθhas the larger posterior prob-ability is the value ofθthat we will accept.If both posterior probabilities are near.5,we might admit that we do not know which is right.In practice,posterior probabilities are computed only for the value of r that was actually observed,but Table2gives posterior probabilities for all values of r and two sets of prior probabilities: (a)one in which each value ofθhas the same probability,1/2, and(b)one set in whichθ=2isfive times more probable than θ=0.As is intuitively reasonable,regardless of the prior distribu-tion,if you see r=4the posterior is heavily in favor ofθ=2, and if you see r=1the posterior substantially favorsθ=0. The key point is what happens when r equals2or3.With equal prior weight on theθ’s,the posterior heavily favorsθ=0, that is,with r=2,p a(0|2)=.83,p a(2|2)=.17,and with r=3,p a(0|3)=.83,p a(2|3)=.17.It is not until our prior makesθ=2five times more probable thanθ=0that we wash out the evidence from the data thatθ=0is more likely,that is, p b(0|2)=p b(2|2)=.50and p b(0|3)=p b(2|3)=.50.Given the prior,the Bayesian procedure is always reasonable.The Bayesian analysis gives no special role to the null hy-pothesis.It treats the two hypotheses on an equal footing.That NP theory treats the hypotheses in fundamentally different ways is something that many Bayesiansfind disturbing.If utilities are available,the Bayesian can base a decision on maximizing expected posterior utility.Berry(2004)discussed the practical importance of developing approximate utilities for designing clinical trials.The absence of a clear source for the prior probabilities seems to be the primary objection to the Bayesian procedure.Typically, if we have enough data,the prior probabilities are not going to matter because the posterior probabilities will be substantially the same for different priors.If we do not have enough data,the posteriors will not agree but why should we expect them to?The best we can ever hope to achieve is that reasonable people(with The American Statistician,May2005,Vol.59,No.2123reasonable priors)will arrive at a consensus when enough data are collected.In the example,seeing one observation of r=1 or4is already enough data to cause substantial consensus.One observation that turns out to be a2or a3leaves us wanting more data.4.SIMPLE VERSUS COMPOSITENow consider testing the simple null hypothesis H0:θ=0 versus the composite alternative hypothesis H A:θ>0.Of course the composite alternative has only two values.Looking at the distributions in Table1,the intuitive conclusions are pretty clear.For r=1,go withθ=0.For r=4,go withθ=2.For r=2,3,go withθ=1.Fisherian testing has nothing new to add to this situation ex-cept the observation that whenθ=1,none of the data are really weird.In this case,the strangest observation is r=1which has a p value of.1.The best thing that can happen in NP testing of a composite alternative is to have a uniformly most powerful test.With H A:θ>0,letθ∗be a particular value that is greater than0.Test the simple null H0:θ=0against the simple alternative H A:θ=θ∗.If,for a givenα,the most powerful test has the same rejection region regardless of the value ofθ∗,then that test is the uniformly most powerful test.It is a simple matter to see that theα=.01 NP most powerful test of H0:θ=0versus H A:θ=1rejects when r=4.Because the most powerful tests of the alternatives H A:θ=1and H A:θ=2are identical,and these are the only permissible values ofθ>0,this is the uniformly most powerful α=.01test.The test makes a“bad”decision when r=2,3 because withθ=1as a consideration,you would intuitively like to reject the test.Theα=.02uniformly most powerful test rejects for r=2,3,4,which is in line with our intuitive evaluation,but recall from the previous section that this is the test that(intuitively)should not have rejected for r=2,3when testing only H A:θ=2.An even-handed Bayesian approach might take prior proba-bilities that are the same for the null hypothesis and the alter-native,that is,Pr[θ=0]=.5and Pr[θ>0]=.5.Moreover, we might then put the same prior weight on every possibleθvalue within the alternative,thus Pr[θ=1|θ>0]=.5and Pr[θ=2|θ>0]=.5.Equivalently,p(0)=.5,p(1)=.25,and p(2)=.25.The posterior probabilities arer1234p(0|r).908.047.047.014p(1|r).046.948.948.352p(2|r).045.005.005.634Pr[θ>0|r].091.953.953.986These agree well with the intuitive conclusions,even though the prior puts twice as much weight onθ=0as on the otherθ’s. The Bayesian approach to testing a simple null against a com-posite alternative can be recast as testing a simple null versus a simple ing the prior probability on the values ofθgiven that the alternative hypothesis is true,one canfind the average distribution for the data under the alternative.With Pr[θ=1|θ>0]=.5and Pr[θ=2|θ>0]=.5,the av-erage distribution under the alternative is.5f(r|1)+.5f(r|2). The Bayesian test of theθ=0density f(r|0)against this aver-age density for the data under the alternative yields the posterior probabilities p(0|r)and Pr[θ>0|r].It might also be reasonable to put equal probabilities on ev-eryθvalue.In decision problems like this,where you know the(sampling)distributions,the only way to get unreasonable Bayesian answers is to use an unreasonable prior.5.GENERAL MATTERS5.1Fisherian TestingOne thing that the example in Section2does not illustrate is that in a Fisherian test,it is not clear what aspect of the model is being rejected.If y1,y2,...,y n are independent N(µ,σ2)and we perform a t test of H0:µ=0,a rejection could mean thatµ=0,or it could mean that the data are not independent,or it could mean that the data are not normal,or it could mean that the variances of the observations are not equal.In other words, rejecting a Fisherian test suggests that something is wrong with the model.It does not specify what is wrong.The example of a t test raises yet another question.Why should we summarize these data by looking at the t statistic,y−0s/√n?One reason is purely practical.In order to perform a test,one must have a known distribution to compare to the data.Without a known distribution there is no way to identify which values of the data are weird.With the normal data,even when assuming µ=0,we do not knowσ2so we do not know the distribution of the data.By summarizing the data into the t statistic,we get a function of the data that has a known distribution,which allows us to perform a test.Another reason is essentially:why not look at the t statistic?If you have another statistic you want to base a test on,the Fisherian tester is happy to oblige.To quote Fisher(1956,p.49),the hypothesis should be rejected“if any relevant feature of the observational record can be shown to[be] sufficiently rare.”After all,if the null model is correct,it should be able to withstand any challenge.Moreover,there is no hint in this passage of worrying about the effects of performing multiple tests.Inflating the probability of Type I error(rejecting the null when it is true)by performing multiple tests is not a concern in Fisherian testing because the probability of Type I error is not a concern in Fisherian testing.The one place that possible alternative hypotheses arise in Fisherian testing is in the choice of test statistics.Again quoting Fisher(1956,p.50),“In choosing the grounds upon which a general hypothesis should be rejected,personal judgement may and should properly be exercised.The experimenter will rightly consider all points on which,in the light of current knowledge, the hypothesis may be imperfectly accurate,and will select tests, so far as possible,sensitive to these possible faults,rather than to others.”Nevertheless,the logic of Fisherian testing in no way depends on the source of the test statistic.There are twofinal points to make on how this approach to testing impacts standard data analysis.First,F tests andχ2tests are typically rejected only for large values of the test statistic.Clearly,in Fisherian testing,that is inappropriate.Finding the p value for an F test should involve finding the density associated with the observed F statistic and124Generalfinding the probability of getting any value with a lower density.This will be a two-tailed test,rejecting for values that are very large or very close to 0.As a practical matter,it is probably sufficient to always remember that “one-sided p values”very close to 1should make us as suspicious of the model as one-sided p values near 0.Christensen (2003)discusses situations that cause F statistics to get close to 0.Second,although Fisher never gave up on his idea of fiducial inference,one can use Fisherian testing to arrive at “confidence regions”that do not involve either fiducial inference or repeated sampling.A (1−α)confidence region can be defined simply as a collection of parameter values that would not be rejected by a Fisherian αlevel test,that is,a collection of parameter values that are consistent with the data as judged by an αlevel test.This definition involves no long run frequency interpreta-tion of “confidence.”It makes no reference to what proportion of hypothetical confidence regions would include the true pa-rameter.It does,however,require one to be willing to perform an infinite number of tests without worrying about their fre-quency interpretation.This approach also raises some curious ideas.For example,with the normal data discussed earlier,this leads to standard t confidence intervals for µand χ2confidence intervals for σ2,but one could also form a joint 95%confidence region for µand σ2by taking all the pairs of values that satisfy|y −µ|σ/√n<1.96.Certainly all such µ,σ2pairs are consistent with the data as summarized by ¯y .5.2Neyman-Pearson TestsTo handle more general testing situations,NP theory has de-veloped a variety of concepts such as unbiased tests,invariant tests,and αsimilar tests;see Lehmann (1997).For example,the two-sided t test is not a uniformly most powerful test but it is a uniformly most powerful unbiased test.Similarly,the standard F test in regression and analysis of variance is a uniformly most powerful invariant test.The NP approach to finding confidence regions is also to find parameter values that would not be rejected by a αlevel test.However,just as NP theory interprets the size αof a test as the long run frequency of rejecting an incorrect null hypothesis,NP theory interprets the confidence 1−αas the long run probability of these regions including the true parameter.The rub is that you only have one of the regions,not a long run of them,and you are trying to say something about this parameter based on these data.In practice,the long run frequency of αsomehow gets turned into something called “confidence”that this parameter is within this particular region.Although I admit that the term “confidence,”as commonly used,feels good,I have no idea what “confidence”really means as applied to the region at hand.Hubbard and Bayarri (2003)made a case,implicitly,that an NP concept of confidence would have no meaning as applied to the region at hand,that it only applies to a long run of similar intervals.Students,almost invari-ably,interpret confidence as posterior probability.For example,if we were to flip a coin many times,about half of the time wewould get heads.If I flip a coin and look at it but do not tell you the result,you may feel comfortable saying that the chance of heads is still .5even though I know whether it is heads or tails.Somehow the probability of what is going to happen in the future is turning into confidence about what has already hap-pened but is unobserved.Since I do not understand how this transition from probability to confidence is made (unless one is a Bayesian in which case confidence actually is probability),I do not understand “confidence.”5.3Bayesian TestingBayesian tests can go seriously wrong if you pick inappro-priate prior distributions.This is the case in Lindley’s famous paradox in which,for a seemingly simple and reasonable testing situation involving normal data,the null hypothesis is accepted no matter how weird the observed data are relative to the null hy-pothesis.The datum is X |µ∼N (µ,1).The test is H 0:µ=0versus H A :µ>0.The priors on the hypotheses do not re-ally matter,but take Pr[µ=0]=.5and Pr[µ>0]=.5.In an attempt to use a noninformative prior,take the density of µgiven µ>0to be flat on the half line.(This is an improper prior but similar proper priors lead to similar results.)The Bayesian test compares the density of the data X under H 0:µ=0to the average density of the data under H A :µ>0.(The latter involves integrating the density of X |µtimes the density of µgiven µ>0.)The average density under the alternative makes any X you could possibly see infinitely more probable to have come from the null distribution than from the alternative.Thus,anything you could possibly see will cause you to accept µ=0.Attempting to have a noninformative prior on the half line leads one to a nonsensical prior that effectively puts all the probability on unreasonably large values of µso that,by comparison,µ=0always looks more reasonable.6.CONCLUSIONS AND COMMENTSThe basic elements of a Fisherian test are:(1)There is a prob-ability model for the data.(2)Multidimensional data are sum-marized into a test statistic that has a known distribution.(3)This known distribution provides a ranking of the “weirdness”of various observations.(4)The p value,which is the probability of observing something as weird or weirder than was actually observed,is used to quantify the evidence against the null hy-pothesis.(5)αlevel tests are defined by reference to the p value.The basic elements of an NP test are:(1)There are two hy-pothesized models for the data:H 0and H A .(2)An αlevel is chosen which is to be the probability of rejecting H 0when H 0is true.(3)A rejection region is chosen so that the probability of data falling into the rejection region is αwhen H 0is true.With discrete data,this often requires the specification of a random-ized rejection region in which certain data values are randomly assigned to be in or out of the rejection region.(4)Various tests are evaluated based on their power properties.Ideally,one wants the most powerful test.(5)In complicated problems,properties such as unbiasedness or invariance are used to restrict the class of tests prior to choosing a test with good power properties.Fisherian testing seems to be a reasonable approach to model validation.In fact,Box (1980)suggested Fisherian tests,based on the marginal distribution of the data,as a method for validat-ing Bayesian models.Fisherian testing is philosophically basedThe American Statistician,May 2005,Vol.59,No.2125。
BIFMA M7[1].1-2005(中文版).
![BIFMA M7[1].1-2005(中文版).](https://img.taocdn.com/s3/m/7a6fc1d7dd88d0d233d46acd.png)
测定从办公家具、部件和座椅中排放出的挥发性化合物(VOC)的标准试验方法Standard Test MethodFor Determining VOC Emissions From Office Furniture Systems, Components And Seating美国办公家具协会(BIFMA)BIFMA M7.1-2005前言这个标准检测方法材料是BIFMA所有成员共同努力的成果，并且由当事方，政府组织，室内设计组织和商业检测和采购组织组成的广泛代表组评审。
最初的标准检测方法制定工作由BIFMA家具排放标准小组委员会在2005年6月完成。
小组委员会参看了提议的标准测试方法来保证测试精确适当的描述评估来自人造板类产品，书桌型产品，座椅及相关产品的挥发性有机化合物的排放.BIFMA尤其要感谢雪城大学的建筑能源和环境系统实验室的张建顺博士在整个制定标准检测方法的过程中的广泛的技术领导和指导.BIFMA也同样要感谢参与制定新标准检测方法的高素质个人，这些技术审稿人是：●Leon Alevantis, 硕士, PE, 加州公共卫生服务部门●Marilyn Black博士, 空气质量科学●Al Hodgson, 伯克利分析协会●Bob Magee, 加拿大国家研究院●Mark Mason, 美国环保局空气污染及控制部.●Bruce Tichenor博士, 顾问, (美国环保局退休)技术审稿人不是编辑权威，但他们的评论和评审经过了张博士和BIFMA家具排放标准小组委员会的慎重考虑和处理。
这个提议的新标准检测方法是2005年9月核准和递交给BIFMA全体会员的。
BIFMA打算对这个文件进行一轮公开评审，评审人包括和BIFMA要求一致的，作为美国国家标准学会的标准实施先锋的利益集团和利益共享者.在彻底完成细致检查工作以后，这份建议采用的标准检测方法将提交给美国国家标准学会制定成美国国家标准.任何有兴趣参加BIFMA公共审阅的人可以发邮件给email@.欢迎所有对这个标准检测方法的改进建议。
04-BIFMA椅子测试指南-2011中文版
BIFMA国际标准椅子测试指南椅子类型Ⅰ（背与座能同时倾仰）Ⅱ（座固定，背能倾仰）Ⅲ（背与座均固定不能倾仰）5、背静态强度测试（适用于Ⅰ类椅子）测试方法：①把椅子垂直放在测试平台上，固定轮子，背和扶手不需固定；②若椅子具任何调教功能，除背高必须调至最大或者406mm(16.0″)（二者中选最少项）外，其余均调至正常使用状态下的数值；③ 完成步骤②以后，以椅背垂直中线为准确定与座高距离分别为406mm(16.0″)和452mm(17.8″)的两点，看图一，确定方法如下：㈠ 若高度≧452mm(17.8″),则此点为406mm(16″)，如图一所示； ㈡ 若高度＜452mm(17.8″)，则此点为椅背的最高点，如图二所示；㈢ 若椅背后有旋转轴心，且该轴心与垂直方向后仰角度≤30°，则点的确定同㈠、㈡，反之，则该点位于椅背旋转轴心处。
如图三所示。
图一：高度≧452mm(17.8″),则此点为406mm(16″) 图二：高度＜452mm(17.8″)，则此点为椅背的最高点图三：背旋转角度＞30° 图四：椅背位于最大倾仰角度④ 当椅背处于最大倾仰位置时，此时受力方向应与背成90°±10°（如图四） ⑤ 如测试系统用的是吊线或者转轮，要求铁线至少要有762mm(30″)长。
注意：当椅子的设计不允许载重物把力转移至其受力面时，可放一块高度89mm ±13mm 的隔板延长受力面。
⑥ 测试要求在机关不锁住的情况下，把椅背调至最大倾仰角度并固定，再如下加载： ㈠ 功能测试力F=890 N (91Kg), 时间T=1 分钟测试完后椅子各项功能正常，否则均视为不合格。
㈡ 验证测试力 F=1334 N(136Kg), 时间T=1 分钟测试完后椅子结构不能有很大的变化或者损伤，功能出现问题可以接受。
＜452mm设备与受力面平行垂直方向背旋转轴心作用力方向(椅背位于最大倾仰角度)作用力方向（和椅背成90°）受力点受力点6、背静态强度测试（适用于Ⅱ类和Ⅲ类椅子）测试方法：a)将椅子垂直放在测试台上，固定轮子，但不固定椅背和扶手。
《统计分析与SPSS的应用(第五版)》课后练习答案(第8章)
《统计分析与SPSS的应用〔第五版〕》〔薛薇〕课后练习答案第8章SPSS的相关分析1、对15家商业企业进行客户满意度调查，同时聘请相关专家对这15家企业的综合竞争力进行评分，结果如下表。
编号客户满意度得分综合竞争力得分编号客户满意度得分综合竞争力得分1 90 70 9 10 602 100 80 10 20 303 150 150 11 80 1004 130 140 12 70 1105 120 90 13 30 106 110 120 14 50 407 40 20 15 60 508 140 130请问，这些数据能否说明企业的客户满意度与其综合竞争力存在较强的正相关，为什么？能。
步骤：〔1〕图形→旧对话框→散点/点状→简单分布→进行相应设置→确定；〔2〕再双击图形→元素→总计拟合线→拟合线→线性→确定〔3〕分析→相关→双变量→进行相关项设置→确定相关性客户满意度得分综合竞争力得分客户满意度得分Pearson 相关性 1 .864**显著性〔双尾〕.000N 16 15 综合竞争力得分Pearson 相关性.864** 1显著性〔双尾〕.000N 15 15 **. 在置信度〔双测〕为 0.01 时，相关性是显著的。
两者的简单相关系数为0.864，说明存在正的强相关性。
2、为研究香烟消耗量与肺癌死亡率的关系，收集下表数据。
〔说明：1930年左右几乎极少的妇女吸烟；采用1950年的肺癌死亡率是考虑到吸烟的效果需要一段时间才可显现〕。
国家1930年人均香烟消耗量1950年每百万男子中死于肺癌的人数澳大利亚480 180加拿大500 150丹麦380 170芬兰1100 350英国1100 460荷兰490 240冰岛230 60挪威250 90瑞典300 110瑞士510 250美国1300 200绘制上述数据的散点图，并计算相关系数，说明香烟消耗量与肺癌死亡率之间是否存在显著的相关关系。
香烟消耗量与肺癌死亡率的散点图(操作方法与第1题相同)相关性人均香烟消耗死于肺癌人数人均香烟消耗Pearson 相关性 1 .737**显著性〔双尾〕.010N 11 11死于肺癌人数Pearson 相关性.737** 1显著性〔双尾〕.010N 11 11**. 在置信度〔双测〕为 0.01 时，相关性是显著的。
基于自动化机器学习建立结肠镜肠道准备失败风险预测模型及评价
基于自动化机器学习建立结肠镜肠道准备失败风险预测模型及评价王甘红;陈健;沈支佳;奚美娟;周燕婷【期刊名称】《中国内镜杂志》【年(卷),期】2024(30)5【摘要】目的鉴于机器学习(ML)在医学模型中的广泛应用,以及其出色的学习和泛化特性,该研究采用自动化机器学习(AutoML)结合患者一般资料和临床状况,早期评估结肠镜前肠道准备的失败风险。
方法回顾性分析2022年1月-2023年1月在该院接受结肠镜检查的患者的临床资料。
波士顿肠道准备评分(BBPS)≤5分被定义为肠道准备失败,> 5分为合格。
将患者按8∶2的比例随机划分了训练集(n=303)和验证集(n=76)。
采用最小绝对收缩和选择算子(LASSO)逻辑回归(LR)模型进行特征选择,构建列线图评分系统,并使用基于5种算法的AutoML建立模型。
模型性能通过受试者操作特征曲线(ROC curve)、校准曲线、基于LR (Lasso回归)的决策曲线分析(DCA)、SHAP图和力图进行评估。
结果在379例患者中,105例(27.7%)肠道准备失败(BBPS≤5分)。
21个研究变量在经LASSO 5折交叉验证后,获得10个变量,并构建了一款列线图评分系统,通过校正曲线表明了LASSO模型的可靠性。
使用H2O平台和5种算法[梯度提升机(GBM)、深度学习(DL)、广义线性模型(GLM)、堆叠集成(Stacked Ensemble)和分布式随机森林(DRF)]开发了67个模型。
经比较,Stacked Ensemble表现最佳,其曲线下面积(AUC)为0.871,对数损失值(LogLoss)为0.403,均方根误差(RMSE)为0.354,优于其他模型和传统的LR模型。
变量重要性贡献图显示,服完泻药至检查间隔时间、便秘、是否完整服完泻药、年龄和家属陪同等因素对肠道准备失败的预测有重要影响。
最后,SHAP图和力图揭示了变量在二分类预测结果中的分布特征,以及各变量对预测结果的影响。
基于机器学习的外窗百叶外遮阳性能预测与敏感性研究
基于机器学习的外窗百叶外遮阳性能预测与敏感性研究
肖敏;杜思达
【期刊名称】《建筑节能（中英文）》
【年(卷),期】2024(52)4
【摘要】在建筑围护结构中越来越多地使用透明围护结构,会导致高能耗和光热环境不适等问题。
为了解决这一问题,遮阳作为一种减少建筑的能源消耗和改善室内
环境较为有效的手段被越来越多地使用。
为探究百叶外遮阳参数对遮阳性能的影响程度,采用机器学习算法预测遮阳性能,利用基于机器学习的改进敏感性分析法探讨
了影响遮阳性能的2类参数(建筑和遮阳)的局部和全局敏感性,确定影响最大的参数。
研究结果表明:XGBoost预测热环境、能耗指标精度最高,而随机森林算法预测光环境指标效果最好。
同时发现遮阳参数是影响室内热环境和建筑能耗的最重要因素,
整体权重均在0.5以上;建筑参数显著影响室内采光,其权重高达0.9左右。
【总页数】8页(P33-39)
【作者】肖敏;杜思达
【作者单位】长沙理工大学建筑学院
【正文语种】中文
【中图分类】TU111
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