第五章 两样本问题5.1a=hygepdf(0,22,11,12) hygepdf(1,22,11,12) hygepdf(2,22,11,12) hygepdf(3,22,11,12)a =0 0.00001701085292 0.00093559691083 0.01403395366244sum(a)ans =0.01498656142619hygecdf(3,22,11,12)ans =0.014986561426195.2.5P72x1=20.6 19.9 18.6 18.9 18.8 20.2 21 20.5 19.8 19.8 19.2 20.5;x2=21.3 17.6 17.4 18.5 19.7 21.1 17.3 18.8 17.8 16.9 18 20.1;format longp,h,stats=ranksum(x1,x2)p =0.04317169310436h =1stats = zval: 2.02204509752701ranksum: 1.855000000000000e+002zval的算法：n=12;m=12;N=24;me=n*(N+1)/2,de=n*m*(N+1)/12-n*m*3*(23-2)/(12*N*(N-1)me =150de =2.996086956521739e+002z=(185.5-me-0.5)/sqrt(de)z =2.02204509752701P值的计算：p=2*(1-normcdf(z)p =0.04317169310436精算算法：p,h,stats=ranksum(x1,x2,alpha,0.05,method,exact)p =0.04004280818118h =1stats = ranksum: 1.855000000000000e+002习题五1.女职工 男职工28500 30650 39700 3370031000 35050 33250 3630022800 35600 31800 3725032350 26900 38200 3395030450 31350 30800 3775038200 28950 32250 3670034100 32900 38050 3610030150 31300 34800 2655033550 31350 32750 3920027350 35700 38800 4100025200 35900 29900 4040032050 35200 37400 3550026550 30450 x =285003100022800323503045038200341003015033550273502520032050265503065035050356002690031350 2895032900313003135035700359003520030450y =397003325031800382003080032250380503480032750388002990037400 337003630037250339503775036700361002655039200410004040035500me=median(x;y)me =33400length(find(xme);length(find(yme)ans =18 87 17年收入me 合计女职工 18 8 26男职工 7 17 24合计 25 25 50N11取值范围为：1，25hygecdf(0,50,25,26)ans=0单侧检验：p=1-hygecdf(17,50,25,26)p =0.00506098777511在显著性水平 0.05下，拒绝原假设。即女职工年收入低于男职工。对双侧检验：p=2*(1-hygecdf(17,50,25,26)p =0.01012197555021在显著性水平 0.05下，拒绝原假设。即女职工年收入不同于男职工。大样本用正态近似算：E(N11)=25*26/50=13; D(N11)=26*24*25*25/(50*50*49)e=25*26/50;d=26*24*25*25/(50*50*49);p=1-normcdf(18,e,sqrt(d)p = 0.00253743404897拒绝原假设。对于双侧检验：u2=49*(18*17-8*7)2/(25*25*26*24)u2 =7.852564102564101-chi2cdf(u2,1)ans =0.00507486809794在显著性水平下，拒绝原假设。p,h,stats=ranksum(x,y)p =3.396221501665853e-004h =1stats = zval: 3.58303725316353ranksum: 797对于单侧检验：P 值为p/2ans =1.698110750832926e-004所以男职工工资高用精确算法：p,h,stats=ranksum(x,y,method,exact)p =2.034089051793435e-004h =1stats = ranksum: 797对于单侧检验：P 值为p/2ans =1.017044525896717e-004拒绝原假设。所以男性工资高。2a=134 146 130 113 119 161 107 132 135 129;b=70 118 101 104 108 83 94 124 99;精确算法：p,h,stats=ranksum(a,b)p =6.495052934681418e-004h =1stats = ranksum: 51ranksum的计算：c,d=sort(a b)c =Columns 1 through 17 70 83 94 99 101 104 107 108 113 118 119 124 129 130 132 134 135Columns 18 through 19 146 161d =Columns 1 through 17 11 16 17 19 13 14 7 15 4 12 5 18 10 3 8 1 9Columns 18 through 19 2 6l=d=139)/nchoosek(19,10)ans =6.495052934681418e-004对于双侧检验，拒绝原假设，有差异。对于单侧检验，matlab 没有现存程序。把把相应的双侧检验所得 p值除以 2，即得单侧检验的 p值。7x=1149 1152 1176 1149 1155 1169 1182 1160 1120 1171;y=1116 1130 1184 1194 1184 1147 1125 1125 1166 1151;秩号 1 2 3.5 3.5 5 6 7.5 7.5 9 10数据 1116 1120 1125 1125 1130 1147 1149 1149 1151 1152秩号 11 12 13 14 15 16 17 18.5 18.5 20数据 1155 1160 1166 1169 1171 1176 1182 1184 1184 1194红色字体为 12月的数据方法一：Mood 检验：n=10,N=20,m=10 g=17, ，02231 其 它em=n*(N2-1)/12-3*(23-2)/(12*N)em =3.317500000000000e+002em也可这样计算：n*sum(d-(N+1)/2).2)/Nans =3.317500000000000e+002d=1 2 3.5 3.5 5 6 7.5 7.5 9 10 11 12 13 14 15 16 17 18.5 18.5 20;de=n*m/(N*(N-1)*(sum(d-(N+1)/2).4)-N*(N2-1)/144)de =1.032775219298246e+004r=1 3.5 3.5 5 6 9 13 18.5 18.5 20;ar=(r-(N+1)/2).2;my=sum(ar)my =4.655000000000000e+002p=2*(1-normcdf(my-0.5,em,sqrt(de)p =0.18979492105955在显著性水平 0.05下，接受原假设，两月份相同。方法二：Ansari-Bradley 检验N=20，k=10,为偶数r 1 2 3.5 3.5 5 6 7.5 7.5 9 10ar 1 2 3.5 3.5 5 6 7.5 7.5 9 10r 11 12 13 14 15 16 17 18.5 18.5 20ar 10 9 8 7 6 5 4 3 2 1Ay=1+3.5+3.5+5+6+9+8+3+2+1=42，d=1 2 3.5 3.5 5 6 7.5 7.5 9 10 11 12 13 14 15 16 17 18.5 18.5 20;对于双侧检验，在显著性水平 0.05下，查表，P（A41）=0.025, Ay=42，接受原假设。用大样本近似，ea=n*(N+2)/4,da=n*m/(N*(N-1)*(sum(d.2)-N*(N+2)2/16)da =5.956578947368421e+002p=2*normcdf(42+0.5,ea,sqrt(da)p =0.60853416186003对于双侧检验，在显著性水平 0.05下,接受原假设，两月份相同。方法三：Siegel-Turkey 检验秩号 1 2 3.5 3.5 5 6 7.5 7.5 9 10数据 1116 1120 1125 1125 1130 1147 1149 1149 1151 1152ar 20 17 16 13 12 9 8 5 4 1秩号 11 12 13 14 15 16 17 18.5 18.5 20数据 1155 1160 1166 1169 1171 1176 1182 1184 1184 1194ar 2 3 6 7 10 11 14 15 18 19Sy=20+16+13+12+9+4+6+15+18+19Sy =132查 Wilcoxon秩和检验表，双侧检验，临界值为 78，接受原假设。同样也可用大样本来近似。方法四：Klotz 检验ai=(norminv(d/(N+1).2;ea=n*sum(ai)/N%期望ea =7.48858756774053da=n*m/(N*(N-1)*sum(ai-ea).2)%方差da =2.428193625138611e+002秩号 1 2 3.5 3.5 5 6 7.5 7.5 9 10数据 1116 1120 1125 1125 1130 1147 1149 1149 1151 1152a 2.7835 1.7139 0.9359 0.9359 0.5076 0.3203 0.134 0.134 0.0324 0.0036秩号 11 12 13 14 15 16 17 18.5 18.5 20数据 1155 1160 1166 1169 1171 1176 1182 1184 1184 1194a 0.0036 0.0324 0.0918 0.1855 0.3203 0.5076 0.7676 1.3918 1.3918 2.7835Ky=sum (ai(1 3 4 5 6 9 13 18 19 2