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level,11/7/2009,#,统计测验,S,tatistical,T,est,单个样本平均数的测验,测验某样本所,来自的总体平均数(,)是否与某已知总体的平,均数(,0,)有显著差异的统计测验。,测验的步骤:,1.设假设H,0,:=vsH,0,A,:,0,2.计算统计量:,t,3.此统计量服从自由度为,n1,的,t,分布,可以在,t,x,0,s/n,分布表上查得出现此值的概率。当,tt,0.05,时,判,定与,0,的差异显著;若,tt,0.01,则差异极显著。,第一节,T,平方测验,T-S,quare,T,est,现在依然是讨论第一个常见例子:,1.,产品检验,:某产品某个技术指标值为,0,现从,一批该产品中抽取大小为,n,的样本,测得样本平,均数为,x,,标准差为,s,,试测验该批产品的该技,术指标平均数,是否与已知的,0,间有显著差异。,只是现在不再是单个变量而是多个变量了,即:,1.,产品检验,:某产品,m,个技术指标值为,m,维向量,0,现从一批该产品中抽取大小为,n,的样本,测,得样本平均数向量为,x,,方差协方差向量为,,试测验该批产品的这些技术指标平均数向量,是,否与已知的,0,间有显著差异。,第一节,T,平方测验,T-S,quare,T,est,如果考察的有,m,项指标,统计量用矩阵形式表,示为:,T,2,n(x,0,)S,1,(x,0,),其中(x,0,)为m1维,其转置阵为1m维,,S,1,是方差协方差矩阵的逆阵,,mm,维。,T,2,是一个数字。,因为,SQ/(n1),S,1,(n1)Q,1,此统计量又可以用平方和乘积和矩阵,Q,表示为:,T,2,n(n1)(x,0,)Q,1,(x,0,),此统计量服从Hotelling,2,分布。算出,2,后,,与,2,分布表值比较,便可以判断差异是否显著。,极少统计书提供HotellingT,2,分布表。,第一节,T,平方测验,T-S,quare,T,est,如果考察的有,m,项指标,统计量用矩阵形式表,示为:,T,2,n(x,0,)S,1,(x,0,),其中(x,0,)为m1维,其转置阵为1m维,,S,1,是方差协方差矩阵的逆阵,,mm,维。,T,2,是一个数字。,因为,SQ/(n1),S,1,(n1)Q,1,此统计量又可以用平方和乘积和矩阵,Q,表示为:,T,2,n(n1)(x,0,)Q,1,(x,0,),统计学家发现,2,分布与分布有如下关系:,nm,2,n(nm),1,FT(x,0,)S(x,0,),m(n1)m(n1),第一节,T,平方测验,T-S,quare,T,est,或:,nm,2,n(nm),1,FT(x,0,)S(x,0,),m(n1)m(n1),n(nm),1,(x,0,)Q(x,0,),m,此,F,统计量服从,df,1,m,df,2,nm,的,F,分布。,此统计量又可以用平方和乘积和矩阵,Q,表示为:,T,2,n(n1)(x,0,)Q,1,(x,0,),统计学家发现,2,分布与分布有如下关系:,nm,2,n(nm),1,FT(x,0,)S(x,0,),m(n1)m(n1),第一节,T,平方测验,T-S,quare,T,est,或:,nm,2,n(nm),1,FT(x,0,)S(x,0,),m(n1)m(n1),n(nm),1,(x,0,)Q(x,0,),m,此,F,统计量服从,df,1,m,df,2,nm,的,F,分布。,特别地,当,m1,时,此式与前面的,t,2,n(x,0,),2,(s,2,),1,完全一致。,现在可以将,2,测验的步骤归纳以下了。,第一节,T,平方测验,T-S,quare,T,est,2,测验的步骤:,1.设假设H,0,:=vsH,0,A,:,0,2.计算统计量:,n(nm),1,F(x,0,)S(x,0,),m(n1),n(nm),1,(x,0,)Q(x,0,),m,3.此统计量服从,df,1,m,df,2,nm,的,F,分布,可以,在,F,分布表上查得出现此值的概率。当,FF,时,,判定与,0,的差异显著;否判断差异不显著。,多元统计可取较低的显著水平,例如,0.1,。,第一节,T,平方测验,T-S,quare,T,est,2,测验的,例子,:,某产品的四个关键性指标的验收标准为:,0,(22.75,32.75,51.50,61.50),现从某批产品中抽取,n21,的样本,数据如下表。,试用95%的置信度测验这批产品是否符合验收标准。,样号123,2021,x,1,x,3,x,4,22.8822.7422.60,32.8132.5632.74,51.5151.4951.50,61.5161.3961.22,23.1623.13,32.7832.95,51.4851.58,61.4161.58,x,2,x,3,x,4,(表中数据参阅课本第49页表2.1),第一节,T,平方测验,T-S,quare,T,est,利用数据可以算出:,22.8219,22.75,0.0719,32.7867,51.4476,61.3767,32.75,51.50,61.50,0.0367,0.0524,0.1233,x,51.4476,0,51.50,x,0,0.0524,0.70170.54140.18360.2531,3.51102.60110.17340.3447,0.54140.71190.22810.2579,0.18360.22810.39240.3459,0.25310.25790.34590.8065,1,2.60113.67280.96940.0579,Q,Q,0.17340.96944.51211.6799,0.34470.05791.67992.0503,于是得到:,T,2,n(n1)(x,0,)Q,1,(x,0,)16.4826,第一节,T,平方测验,T-S,quare,T,est,或者利用方差协方差矩阵计算:,22.8219,22.75,0.0719,32.7867,51.4476,61.3767,32.75,51.50,61.50,0.0367,0.0524,0.1233,x,51.4476,0,51.50,x,0,0.0524,0.03510.02710.09180.0127,70.219752.02203.46846.8943,0.02710.03560.01140.0129,0.09180.01140.01960.0173,0.01270.01290.01730.0403,1,52.022073.456019.38701.1572,S,S,3.468419.387090.241133.5986,6.89431.157233.598641.0056,同样得到:,T,2,n(x,0,)S,1,(x,0,)16.4826,第一节,T,平方测验,T-S,quare,T,est,换算为,nm,2,21(214),FT16.48263.5026,m(n1)4(211),当,df,1,4,df,2,nm21417,时,,F,0.05,2.96,现在FF,0.05,,判断x所来自的总体的与,0,之间,有显著差异。,如果对四个指标分别进行单变量的平均数,t,测验,,他们的,t,值分别为t,1,1.76,t,2,0.89,t,3,1.71,t,4,2.81,当,df21120,时,,t,0.05,2.09,只有第4个变量与,验收指标有显著差异。,若删除第4个指标重作,2,测验,会发现,T,2,9.28,F2.78F,0.05,3,18,3.16,与,0,之间无显著差异。,第二节多元方差分析,M,ultiple,ANOVA,基本原理,单向分类的多元方差分析,两向分类(不带互作)的多元方差分析,两向分类(带互作)的多元方差分析,任意试验设计资料的多元方差分析,第二节多元方差分析,M,ultiple,ANOVA,基本原理,:,多元方差分析,的基本原理与单变量的方差分析的,基本原理基本一致,只是将一个变量推广到多个,变量。,让我们回顾一下关于单变量的方差分析的基本,思路。,在方差分析中数据的变异用方差来衡量。,方差分析,A,nalysisof,V,ariance,(ANOVA),方差分析的基本思路,:,将试验数据的总变异分解为已知的若干可控,因子引起的变异,扣除这些可控因子引起的变异,后,把剩余的变异当作为由误差引起的,再将要,考察的因子引起的变异与误差引起的变异比较,,如果待考察的因子引起的变异显著地大于误差引,起的变异,便判定该因子对试验指标有显著的效,应,拒绝H,0,,,接受H,A,;否则,判定该因子对试验,指标没有显著的效应,接受H,0,,拒绝H,A,。,第二节多元方差分析,M,ultiple,ANOVA,在多元方差分析中,要计算的统计量是:,Q,e,Q,h,Q,e,其中,,Q,e,为误差平方和乘积和矩阵;,Q,h,为欲测验,此,统计量服从,(m,n,1,n,2,),的,分布。其中,m,为,变量数,,n,1,为,Q,e,矩阵的自由度,,n,2,为,Q,h,矩阵的自由,度,,为显著水准。,可控因素的平方和乘积和矩阵。,注意,:这里的,中,误差项放在分子,而在一元方,差分析中的,F,统计量中,误差放在分母。所以,,F,越大越显著,而,则越小越显著。,第二节多元方差分析,M,ultiple,ANOVA,很少统计书提供,分布表。不断有人研究,分布与,其它分布的关系。国外大多数统计软件上采用Rao,提出的近似公式。该公式把,转换为,F,:,1,1/t,ut2,F,1/t,mn,2,它近似服从,F,(mn,2,ut2),分布。,其中,un,1,n,2,(mn,2,1)/2;,当,m,2,n,22,50,时,,t(m,2,n,22,4)(m,2,n,22,5),否则,t1,(mn,2,2)/4,当,ut2,不为整数时,用最接近的整数查,F,表。,注意,:当转换为,F,后,又变成,F,越大越显著了。,单向分类的多元方差分析,O,ne-way,MANOVA,举例子说明此法的步骤:,有一个完全随机设计的作物,品种,x,1,x,2,x,3,x,4,260754067,1310853059,品种试验,五个品种(A,1,A,2,A,3,A,4,A,5,),三次重复,对每,个品种记录了四项指标(产,量指标,x,抗性指标,x,质量,指标,x,3,经济指标,x,4,)。数据,如右表,现欲综合考虑这四,个指标,看这五个品种之间,是否有显著的差异。,320643941,2001223417,2310993518,2601123711,1,2,2701103924,5170653716,270653221,单向分类的多元方差分析,O,ne-way,MANOVA,一,、分别对四个变量进行单向分类平方和分解:,变异,df,平方和,来源,x,1,x,2,x,3,x,4,品种间48266.67,37916.67,总变异14,3092.40,4158.00,7250.40,19.0667,308.667,327.733,3551.73,494.67,4046.40,误差10,总变异14,46183.33,二,、分别对两两变量间进行乘积和分解:,变异,乘积和,来源,df,x,x,xx,x,x,x,x,xx,x,x,1,2,1,3,1,4,2,3,2,4,3,4,品种间4-540.00,5935.00,总变异14,-40.00,-686.67,-726.67,4653.33,426.67,5080.00,18.40-1164.73,462.33,84.267,-44.667,误差10,总变异14,314.00,332.40,5395.00,-702.40,39.600,三,、用上述品种间平方和及品种间乘积和构成,Q,h,:,单向分类的多元方差分析,O,ne-way,MANOVA,变异,df,平方和,来源,x,1,x,2,x,3,x,4,品种间48266.67,37916.67,总变异14,3092.40,4158.00,7250.40,19.0667,308.667,327.733,3551.73,494.67,4046.40,误差10,总变异14,46183.33,变异,乘积和,来源,df,x,x,xx,x,x,x,x,xx,x,x,1,2,1,3,1,4,2,3,2,4,3,4,品种间4-540.00,5935.00,总变异14,-40.00,-686.67,-726.67,4653.33,426.67,5080.00,18.40-1164.73,462.33,84.267,-44.667,误差10,总变异14,314.00,332.40,5395.00,-702.40,39.600,8266.67-540.0040.004653.33,-540.003092.4018.40-1164.73,40.0018.4019.066784.267,4653.33-1164.7384.2673551.73,Q,h,40.0018.4019.066784.267,单向分类的多元方差分析,O,ne-way,MANOVA,变异,df,平方和,来源,x,1,x,2,x,3,x,4,品种间48266.67,37916.67,总变异14,3092.40,4158.00,7250.40,19.0667,308.667,327.733,3551.73,494.67,4046.40,误差10,总变异14,46183.33,变异,乘积和,来源,df,x,x,xx,x,x,x,x,xx,x,x,1,2,1,3,1,4,2,3,2,4,3,4,品种间4-540.00,5935.00,总变异14,-40.00,-686.67,-726.67,4653.33,426.67,5080.00,18.40-1164.73,462.33,84.267,-44.667,误差10,总变异14,314.00,332.40,5395.00,-702.40,39.600,四,、用上述误差平方和及误差乘积和构成:,Q,e,单向分类的多元方差分析,O,ne-way,MANOVA,变异,df,平方和,来源,x,1,x,2,x,3,x,4,品种间48266.67,37916.67,总变异14,3092.40,4158.00,7250.40,19.0667,308.667,327.733,3551.73,494.67,4046.40,误差10,总变异14,46183.33,变异,乘积和,来源,df,x,x,xx,x,x,x,x,xx,x,x,1,2,1,3,1,4,2,3,2,4,3,4,品种间4-540.00,5935.00,总变异14,-40.00,-686.67,-726.67,4653.33,426.67,5080.00,18.40-1164.73,462.33,84.267,-44.667,误差10,总变异14,314.00,332.40,5395.00,-702.40,39.600,37916.675935.00-686.67426.67,5935.004158.00314.00462.33,-686.67314.00308.667-44.667,426.67462.33-44.667494.67,Q,e,-686.67314.00308.667-44.667,单向分类的多元方差分析,O,ne-way,MANOVA,分析步骤:,一,、分别对四个变量进行一元的单向分类方差分析:,二,、分别对两两变量间进行单向分类协方差分析:,三,、用上述品种间平方和及品种间乘积和构成,Q,h,:,四,、用上述误差平方和及误差乘积和构成:,Q,e,五,、将,Q,h,与,Q,e,相加得和矩阵,求和矩阵的行列式值,六,、求出,Q,e,的行列式值;,Q,e,1207564250,9888.67,本例中,,Q,h,Q,e,274359286664740.5,。,用它除以和矩阵的行列式值得到:,Q,e,1275642509888.67,0.0440139,Q,h,Q,e,274359286664740.5,单向分类的多元方差分析,O,ne-way,MANOVA,分析步骤:,一,、分别对四个变量进行一元的单向分类方差分析:,二,、分别对两两变量间进行单向分类协方差分析:,三,、用上述品种间平方和及品种间乘积和构成,Q,h,:,四,、用上述误差平方和及误差乘积和构成:,Q,e,五,、将,Q,h,与,Q,e,相加得和矩阵,求和矩阵的行列式值,六,、求出,Q,e,的行列式值;,Q,e,1207564250,9888.67,七,、将,换算为,F,,并算出,F,的自由度,查出,F,进,本例中,,Q,h,Q,e,274359286664740.5,。,用它除以和矩阵的行列式值得到:,0.0440139,行统计推断。,单向分类的多元方差分析,O,ne-way,MANOVA,Rao的换算公式:,本例中,,m4,n,1,10,n,2,4,1,1/t,ut2,F,1/t,mn,2,un,1,n,2,(mn,2,1)/2104(441)/29.5,t(m,2,n,22,4)(m,2,n,22,5)252/273.06,(mn,2,2)/4(441)/43.5,10.3597599.53.0623.5,F2.45,df,1,mn,2,4416,0.35975644,df,2,ut29.53.0623.522,单向分类的多元方差分析,O,ne-way,MANOVA,统计结论:,查表df,1,=16,df,2,=22F,0.05,=2.13,因为F=2.45F,0.05,=2.13,判断这五个品种之间在四项指标上具有显著差异。,对四个指标进行一元方差分析,发现只有经济效益指标X,4,的F检验是显著的。,作经济效益指标X,4,品种间的多重比较,发现品种,1的均值极显著的高于其它的品种。,两向分类的多元方差分析,T,wo-way,MANOVA,举例子说明此法的步骤:,有一个随机区组设计的作物,品,区,x,1,x,2,x,3,x,4,种,组,260754067,品种试验,五个品种(A,1,A,2,A,A,A),三次重复,对每,个品种记录了四项指标(产,量指标,x,1,抗性指标,x,2,质量,指标,x,3,经济指标,x,4,)。数据,如右表,现欲综合考虑这四,个指标,看这五个,品种,之间,是否有显著的差异。,1310853059,320643941,3,4,5,2001223417,2310993518,2601123711,2701103924,5170653716,270653221,两向分类的多元方差分析,T,wo-way,MANOVA,一,、分别对四个变量进行一元的两向分类方差分析:,变异,df,平方和,来源,x,1,x,2,x,3,x,4,品种间48266.67,区组间215163.33,3092.40,1170.00,2988.00,7250.40,19.067,124.133,184.533,327.733,3551.73,103.60,误差822753.33,46183.33,391.07,4046.40,总变异14,二,、分别对两两变量间进行两向分类协方差分析:,变异,df,乘积和,来源,x,1,x,2,x,1,x,3,x,1,x,4,x,2,x,3,x,2,x,4,x,3,x,4,品种间4-540.00,区组间2315.00,-40.00,-575.66,-111.00,-726.67,4653.33,-1121.00,18.40,333.00,-19.00,332.40,-1164.73,132.00,84.267,88.600,误差85620.00,5395.00,1547.67,5080.00,330.33,-702.40,-133.267,39.600,总变异14,三,、用上述,品种,间平方和及,品种,间乘积和构成,Q,h,:,两向分类的多元方差分析,T,wo-way,MANOVA,变异,df,平方和,来源,x,1,x,2,x,3,x,4,品种间48266.67,区组间215163.33,3092.40,19.067,3551.73,1170.00,2988.00,7250.40,124.133,184.533,327.733,103.60,391.07,误差822753.33,总变异14,46183.33,4046.40,变异,df,乘积和,来源,x,1,x,2,x,1,x,3,x,1,x,4,x,2,x,3,x,2,x,4,x,3,x,4,品种间4-540.00,区组间2315.00,-40.00,-575.66,-111.00,-726.67,4653.33,-1121.00,18.40,333.00,-19.00,332.40,-1164.73,132.00,84.267,88.600,误差85620.00,1547.67,5080.00,330.33,-702.40,-133.267,39.600,总变异14,5395.00,8266.67-540.0040.004653.33,-540.003092.4018.40-1164.73,40.0018.4019.066784.267,4653.33-1164.7384.2673551.73,Q,h,40.0018.4019.066784.267,两向分类的多元方差分析,T,wo-way,MANOVA,变异,df,平方和,来源,x,1,x,2,x,3,x,4,品种间48266.67,区组间215163.33,3092.40,1170.00,2988.00,7250.40,19.067,124.133,184.533,327.733,3551.73,103.60,误差822753.33,46183.33,391.07,4046.40,总变异14,变异,df,乘积和,来源,x,1,x,2,x,1,x,3,x,1,x,4,x,2,x,3,x,2,x,4,x,3,x,4,品种间4-540.00,区组间2315.00,-40.00,-575.66,-111.00,-726.67,4653.33,-1121.00,18.40,333.00,-19.00,332.40,-1164.73,132.00,84.267,88.600,误差85620.00,5395.00,1547.67,5080.00,330.33,-702.40,-133.267,39.600,总变异14,四,、用上述误差平方和及误差乘积和构成:,Q,e,两向分类的多元方差分析,T,wo-way,MANOVA,变异,df,平方和,来源,x,1,x,2,x,3,x,4,品种间48266.67,区组间215163.33,3092.40,1170.00,2988.00,7250.40,19.067,124.133,184.533,327.733,3551.73,103.60,误差822753.33,46183.33,391.07,4046.40,总变异14,变异,df,乘积和,来源,x,1,x,2,x,1,x,3,x,1,x,4,x,2,x,3,x,2,x,4,x,3,x,4,品种间4-540.00,区组间2315.00,-40.00,-575.66,-111.00,-726.67,4653.33,-1121.00,18.40,333.00,-19.00,332.40,-1164.73,132.00,84.267,88.600,误差85620.00,5395.00,1547.67,5080.00,330.33,-702.40,-133.267,39.600,总变异14,22753.335620.00-111.001547.67,5620.002988.00-19.00330.33,-111.00-19.00184.533-133.267,1547.67330.33-133.267391.07,Q,e,-111.00-19.00184.533-133.267,两向分类的多元方差分析,T,wo-way,MANOVA,对,品种,间差异显著性测验的分析步骤:,一,、分别对四个变量进行一元的两向分类方差分析:,二,、分别对两两变量间进行两向分类协方差分析:,三,、用上述,品种,间平方和及,品种,间乘积和构成,Q,h,:,四,、用上述误差平方和及误差乘积和构成:,Q,e,五,、将,Q,h,与,Q,e,相加得和矩阵,求和矩阵的行列式值,本例中,Q,h,Q,e,67691593519311.9,。,六,、求出,Q,e,的行列式值;,Q,e,1329611567,226.45,用它除以和矩阵的行列式值得到:,Q,e,1329611567226.45,0.0196422,Q,h,Q,e,67691593519311.9,两向分类的多元方差分析,T,wo-way,MANOVA,对,品种,间差异显著性测验的分析步骤:,一,、分别对四个变量进行一元的两向分类方差分析:,二,、分别对两两变量间进行两向分类协方差分析:,三,、用上述,品种,间平方和及,品种,间乘积和构成,Q,h,:,四,、用上述误差平方和及误差乘积和构成:,Q,e,五,、将,Q,h,与,Q,e,相加得和矩阵,求和矩阵的行列式值,本例中,,Q,h,Q,e,67691593519311.9,。,六,、求出,Q,e,的行列式值;,Q,e,1329611567,226.45,用它除以和矩阵的行列式值得到,0.0196422,。,七,、将,换算为,F,,与,F,表查得的值比较,作出统计,推断。本例中,F2.61F,0.05,16,16,2.33,差异显著。,两向分类的多元方差分析,T,wo-way,MANOVA,对,区组,间差异显著性测验的分析步骤:,一,、分别对四个变量进行一元的两向分类方差分析:,二,、分别对两两变量间进行两向分类协方差分析:,三,、用上述,区组,间平方和及,区组,间乘积和构成,Q,h,:,四,、用上述误差平方和及误差乘积和构成:,Q,e,五,、将,Q,h,与,Q,e,相加得和矩阵,求和矩阵的行列式值,本例中,,Q,h,Q,e,12075642509888.67,。,六,、求出,Q,e,的行列式值;,Q,e,1329611567,226.45,用它除以和矩阵的行列式值得到,0.1101069004,。,七,、将,换算为,F,,与,F,表查得的值比较,作出统计,推断。本例中F2.52F,0.05,8,10,3.07,差异不显著。,带互作的两向分类的多元方差分析,T,wo-way,MANOVA,with,Interaction,在两向分类资料中,当每种处理组合(即两个分,类因素的不同水平的搭配方式)有多于一个观察值,时,可以考察因子间的交互作用。,当试验中只考虑一种性状时,只收集了一个变量,的数据。单变量的方差分析方法已经在第一章讨论,过,这里讨论多变量的情况。,用一个例子说明分析的步骤。,带互作的两向分类的多元方差分析,T,wo-way,MANOVA,with,Interaction,如果这15个处理组合在田间是完全随机排列的。,它的方差分析应该是:先按单向分类的方法分解为,处理和误差两项。再将处理分解为品种间、密度间,和交互作用三项。即:,变异来源自由度,变异来源自由度,变异来源自由度,处理间15-1=14,处理间15-1=14,品种间5-1=4,密度间3-1=2,品种间,密度间,5-1=4,3-1=2,互作42=8,互作,42=8,误差29-14=15,误差29-14=15,总变异30-1=29,总变异30-1=29,带互作的两向分类的多元方差分析,T,wo-way,MANOVA,with,Interaction,这时可以对品种间的差异进行显著性测验;也可,以对密度间的差异进行显著性测验;还可以对品种,与密度间的交互作用进行显著性测验。,测验品种间差异的,0.0605798,相应的,变异来源自由度,F3.505(df,1,16,df,2,37),;,处理间15-1=14,测验密度间差异的,品种间,密度间,5-1=4,3-1=2,0.0728166,相应的,F8.117(df,1,8,df,2,24),;,0.089094,相应的,互作,42=8,测验交互作用的,误差29-14=15,;,总变异30-1=29,F1.327(df,1,32,df,2,46),任意试验设计资料的多元方差分析,MANOVAf,or,O,ther,D,esigns,对于任意试验设计的多元资料,只要能对各变量,进行方差分析,对两两变量之间进行协方差分析,,就可以将其中一个待测验因素的平方和和乘积和构,成,Q,h,矩阵,将欲作为误差的项的平方和和乘积和构,成,Q,e,矩阵,用,Q,e,Q,h,Q,e,求出,,并转换成,F,,,与某个显著水准的,下的,F,表值比较,做出统计,推断。,DT软件为我们提供了通用的方法。,
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