《非参数统计》与MATLAB编程第五章两样本问题.doc

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2850030650397003370031000350503325036300228003560031800372503235026900382003395030450313503080037750382002895032250367003410032900380503610030150313003480026550335503135032750392002735035700388004100025200359002990040400320503520037400355002655030450x = 28500 31000 22800 32350 30450 38200 34100 30150 33550 27350 25200 32050 26550 30650 35050 35600 26900 31350 28950 32900 31300 31350 35700 35900 35200 30450 y = 39700 33250 31800 38200 30800 32250 38050 34800 32750 38800 29900 37400 33700 36300 37250 33950 37750 36700 36100 26550 39200 41000 40400 35500me=median(x;y)me = 33400length(find(xme);length(find(yme)ans = 18 8 7 17年收入me合计女职工18826男职工71724合计252550N11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 99 101 104 107 108 113 118 119 124 129 130 132 134 135 Columns 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 9 Columns 18 through 19 2 6l=d=10l = Columns 1 through 17 0 0 0 0 0 0 1 0 1 0 1 0 1 1 1 1 1 Columns 18 through 19 1 1e=1:19e = Columns 1 through 17 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Columns 18 through 19 18 19sum(e(find(l=1)ans = 139sum(e(find(l=0)ans =51当ranksum为51时，n=9 N=19,9*(19+1)/2=90,139+51=190=19*(19+1)/2P的计算：zh=nchoosek(1:19,9);2*sum(sum(zh,2)=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123.53.5567.57.5910数据1116112011251125113011471149114911511152秩号1112131415161718.518.520数据1155116011661169117111761182118411841194红色字体为12月的数据方法一：Mood 检验：n=10,N=20,m=10 g=17,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r123.53.5567.57.5910ar123.53.5567.57.5910r1112131415161718.518.520ar10987654321Ay=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123.53.5567.57.5910数据1116112011251125113011471149114911511152ar201716131298541秩号1112131415161718.518.520数据1155116011661169117111761182118411841194ar2367101114151819Sy=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123.53.5567.57.5910数据1116112011251125113011471149114911511152a2.78351.71390.93590.93590.50760.32030.1340.1340.03240.003