统计学原理与matlab编程第5章方差分析模型

参考书目《线性模型引论》菊编著科学2004年第一节P202甲乙丙丁x=[121814171319172124panovatab[[[[[statsgnames:{4x1n:[233source:means:[1514.666719df:s:P20701x=[0p=anovatab[[[[[statsgnames:{3x1n:[24source:means:[0.50002.7750df:s: comparison=------means--''[--0.6530]p204《统计手册》 ans- - comparison=--------means'scheffe3.85740.6926]p204《统 ans- -或设检验F=ans接受原假设，即1（安慰剂）与2（新药）无差异。 comparison=-------- comparison=------means--间是怎么计算《商务统计中国财政p181-ans=-3.4823-新0其实也可看comparison--------如果[-3.4823，-1.0677]012则没有差0，那么存在差异。第二节两向分类模型（无交互效应BAx=[58.256.265.3;;60.170.939.2;75.858.2p= anovatab [ [[ [ statssource:sigmasq:colmeans:[60.800059.8500coln:rowmeans:[59.900051.600056.7333rown:inter:0pval:NaNdf:compares--means-x=[63.163.965.666.8;65.166.467.869;67.27171.9p= anovatab [[ [[[[6.9271e- [[statssource:sigmasq:colmeans:[65.133367.100068.4333coln:rowmeans:[64.850067.0750rown:inter:0pval:NaNdf:A[compares compares------means---AB[compares compares-------------means--可得:1与34不同2与4不同，3与不同 与不同。型型21x=[1303420;1554070;1748082;1807558;15013625;18870;15910658;12611545;13817496;110120104;16882;160139p= anovtab[[1.5183e-[[ [[[8.0678e- statssource:'anova2'sigmasq:colmeans:[153.1667107.5833coln:rowmeans:[91.5000108.3333rown:inter:pval:8.0678e-004df:27[compares atthelevelyouspecified. Testingmaineffectsundertheseconditionsisparesmeans[compares Note:Yourmodelincludesaninctiontermthatissignificantatthelevelyouspecified.Testingmaineffectsunderconditionsiscompares-----means--Clickonthegroupyouwantto123 Therowmeansofgroups1and3aresignificantly13

第四节x=[8589828465616771;7270918859627578;706785605685y=[sum(x(:,[12]),2)sum(x(:,[34]),2)sum(x(:,[56]),2)8]),2)ssb=ssa=ssesst=ans=p=1.0e-014 A、B第五节》参计建与R件编著社 p34-1》AA1、A2、A3、A4A1=[160016101650168017001700A2=[1500164014001700A3=[1640155016001620164016001740A4=[15101520153015701640lillietesth=0plcllillietest统计量值，c为临界值。PP值接受原假设。A1正态h=0A2h=1plcA3在0.05显著性下，原假设h0A4qqplot(x)normplot(x)1bartlett（）si2=[var(A1var(A2var(A3var(A4)]%每个水平的样本方差s2=si2*(ni-1)/(sum(ni)-r);r=4;n=sum(ni);k2=5.8056p=2、最大F比法（Hartley法，也可用于非平衡情形si21.0e+004 ans7.11A1=[232521141520171126211619131624202127A2=[283019172226242125261918192025262629A3=[181523181021251212221923221413221322A1h=0plu[h,p,l,u]=lillietest(A2),BA=[A1;A2;A3];B1=A(:,[1:5]);B2=A(:,[6:10]);B3=A(:,[11:15]);Abartlett齐性检验A1=[232521141520171126211619131624202127A2=[283019172226242125261918192025262629A3=[181523181021251212221923221413221322si2=[var(A1var(A2var(A3每个水平的样本方差s2=si2*(ni-1)/(sum(ni)-3);r=3;n=sum(ni);k2=1/c*((n-r)*log(s2)-log(si2)*(ni-1))k2p=AA因素的最大F比法（Hartley法Fmax=临界值为：r=3,n0=202.95，接受原假设。PA因素的最大方差检验法(Cochran法)，n较小时，经常C=Bbartlettsi2=[var(B1(var(B2(var(B3(var(B4(每个水平的样本s2=si2*(ni-1)/(sum(ni)-k2=1/c*((n-r)*log(s2)-log(si2)*(ni-1))k2pB以下参考书目《线性模型引论》菊密霞编著科学2004年 A1=[107310581071103710661026105310491065A2=[101610581038104210201045104410611034A3=[108410691106107810751090107910941111p=[lillietest(A1)lillietest(A2)lillietest(A3)]p= Levene检验法（只适合平衡的情形e1=(A1-mean(A1)).^2;e2=(A2-mean(A2)).^2;e3=(A3-mean(A3)).^2;L=3*(10-1)*ssb/((3-1)*ssa)%nssassbL=PHarleyFmsi=[var(A1)var(A2)var(A3)]msi FmaxBartlettsi2=[var(A1var(A2var(A3每个水平的样本方差s2=si2*(ni-1)/(sum(ni)-3);r=3;n=sum(ni);k2=1/c*((n-r)*log(s2)-log(si2)*(ni-1))k2pCochranmsi=[var(A1)var(A2)var(A3)]C第六节anova1、anova2略有不同。其中有相乘之两项或三项者，即为其间之交互作用 之行归纳在一个细胞数组group中。其内可有Ｎ层之，以此指认Ｘ资料之归属。在group细胞中之可以是一向量、p=p=anovan(x,group,'Param1',val1,'Param2',val2,...)[p,table]=anovan(...)[p,table,stats]=[p,table,stats,terms]=1,2,3（默认值）1,2,3（默认值）组别名称预设为'X1','X2','X3','XN''disANOVA，'on'|信任水平（0.05决定采用何种函数，如'linearp（默认值）；'inction'除主效应外尚计算两者间'full'计算所有主效应及交互效应。若为义，如[100]A[010]B[001C[110]AB[101]AC应、[011]BC[111]ABC三、在输出项当中，TERMSANOVAMODELMODEL参数不采用随T中之字段将为TERMS、ＳTERMS之型态（即固定或随机）、期输出参数STATS的结构包括使用 PARE函数之 变异项估计值（每一随机项一个） 六、实际上anova2anovan也可以通用，只是其安排的语法略有不同而已。anova2是将重复的观察值置于列向，用repsanovan则必须另假设属anova2X=[251721;281865;45560;4210Xp= m2：=123123123123=111111222222其次，再将Ｘ，m1，m2[Xans11121332anovananovan(X,{m1m2},2)ans=anova2naovan好像比anova2之型式复杂的绿故，若naovan之需要处理，则会简单许多。指2即为'model'=2的意思，表示分析结果包括anova1相同。下面是一个 变异分析之例子，设ｙ为观察之料，其余三组g1、g2、g3为对应之组合项，各组内仅有两种内容，g1为[12]、g2['hi''lo']、g3['june''may']。由十五、双因素方差分析（非平衡情形资料来源：威分析系统的自带数据，略做修y=[74.11 0.962.41.6541.50.350.30.9]';%sw=[ones(4,1);2*ones(4,1)ones(4,1);2*ones(4,1)];%水温定剂''水温'})panvoatab 'Sum 'Mean '稳定剂 [ [ [ '水温 [ '稳定剂*水温[[[[[[[[[stats[[source:'anovan'resid:[1x16double]coeffs:[9x1Rtr:[4x4double]rowbasis:[4x9dfe:mse:1.1225nullproject:[9x4double]terms:[3x2double]nlevels:[2x1double]termcols:[4x1double]coeffnames:{9x1cell}vars:[9x2double]varnames:{2x1cell}grpnames:{2x1ems:denom:[]dfdenom:[]msdenom:[]varest:[]varci:[]txtdenom:txtems:[]rtnames:水温稳定剂'})p=anvoatab 'Sum 'Mean '水温[[[[4.6089e-'稳定剂[[[[[ '水温*稳定剂[[[[[ [[[stats[[source:'anovan'resid:[1x16double]coeffs:[9x1Rtr:[4x4double]rowbasis:[4x9dfe:mse:1.1225nullproject:[9x4double]terms:[3x2double]nlevels:[2x1double]termcols:[4x1double]coeffnames:{9x1cell}vars:[9x2double]varnames:{2x1cell}grpnames:{2x1ems:denom:[]dfdenom:[]msdenom:[]varest:[]varci:[]txtdenom:txtems:[]rtnames:y=[52.757.545.944.553.057.045.9g1=[1212121g2=g3={'may';'may';'may';'may';'june';'june';'june';利用上述数据进 变异分析如下>>p=anovan(y,{g1g2g3})p=x1、x2、x3对应g1、g2、g3等三组。由于2之参数在最后一项，或令'model'2。三个Ｐ值则对应个别零假设(NullHypothesis)Ho1、Ho2、Ho3之机率值。基本上Ｐ值趋近零，对于零假设的成分愈小。例如，Ho20.0028已足够小到可以认0.005或0.001，依问题的>>[p,table]=anovan(y,{g1g2p=table 'SumSq.''d.f.''Singular?''MeanSq.' 3.7812] 1] [199.0013] 1] 0.0612] 1] [18.3012] 1] [ 0.2112] 1] 0.0113] 1] 0.0113][ 1][ [221.3788][ 7][ >>p=anovan(y,{g1g2g3},'varnames',{'time''limit'把威的方差分析打开，把所有数据粘到 中的y.mes',{'稳定剂','水浴温度','浸泡口表','加盖与否'},'model',[1000;0100;0010;0001;1100;100panovantab'Sum'Mean'稳定剂[[[[ '水浴温度[[['浸泡口表[[[[[ '加盖与否[[[[[ '稳定剂*水浴温度[[[[[ '稳定剂*加盖与否[[[[[ [[[[[stats

source:resid:[16x1double]coeffs:[17x1double]Rtr:[7x7double]rowbasis:[7x17double]dfe:9mse:nullproject:[17x7double]terms:[6x4double]nlevels:[4x1double]termcols:[7x1double]coeffnames:{17x1vars:[17x4double]varnames:{4x1cell}grpnames:{4x1cell}ems:[]denom:dfdenom:[]msdenom:[]varest:[]varci:txtdenom:[]txtems:[]rtnames: compares= means mes',{'水浴温度','稳定剂','浸泡口表','加盖与否'},'model',[1000;0100;0010;0001;1100;010panovantab'Sum'Mean'水浴温度[[[[6.2310e-'稳定剂[[[[ '浸泡口表[[[[[ '加盖与否[[[[[ '水浴温度*稳定剂[[[[[ '稳定剂*加盖与否[[[[[ [[[[[statssource:resid:[16x1double]coeffs:[17x1double]Rtr:[7x7double]rowbasis:[7x17double]dfe:9mse:1.3249nullproject:[17x7double]terms:[6x4double]nlevels: