《非參數(shù)統(tǒng)計》與MATLAB編程第五章兩樣本問題

1、 第五章 兩樣本問題5.1a=hygepdf(0,22,11,12) hygepdf(1,22,11,12) hygepdf(2,22,11,12) hygepdf(3,22,11,12)a = 0 0.292 0.083 0.244sum(a)ans = 0.619hygecdf(3,22,11,12)ans = 0.6195.2.5P72x1=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

2、 longp,h,stats=ranksum(x1,x2)p = 0.436h = 1stats = zval: 2.701ranksum: 1.0000e+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 = 2.1739e+002z=(185.5-me-0.5)/sqrt(de)z = 2.701P值的計算：p=2*(1-normcdf(z)p = 0.436精算算法：p,h,stats=ranksum(x1,x2,alpha,0.05,method,exact

3、)p = 0.118h = 1stats = ranksum: 1.0000e+002習題五1.女職工男職工285003065039700337003100035050332503630022800356003180037250323502690038200339503045031350308003775038200289503225036700341003290038050361003015031300348002655033550313503275039200273503570038800410002520035900299004040032050352003740035500265503

4、0450x = 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

5、40400 35500me=median(x;y)me = 33400length(find(xme);length(find(yme)ans = 18 8 7 17年收入me合計女職工18826男職工71724合計252550N11取值范圍為：1，25hygecdf(0,50,25,26) ans=0單側(cè)檢驗：p=1-hygecdf(17,50,25,26)p = 0.511在顯著性水平0.05下，拒絕原假設。即女職工年收入低于男職工。對雙側(cè)檢驗：p=2*(1-hygecdf(17,50,25,26)p = 0.021在顯著性水平0.05下，拒絕原假設。即女職工年收入不同于男職工。大樣本用正

6、態(tài)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 = 0.897拒絕原假設。對于雙側(cè)檢驗：u2=49*(18*17-8*7)2/(25*25*26*24)u2 = 7.4101-chi2cdf(u2,1)ans = 0.794在顯著性水平下，拒絕原假設。p,h,stats=ranksum(x,y)p = 3.5853e-004h = 1stats = zval: 3.353ranksum: 797對于單側(cè)檢驗：P值

7、為p/2ans =1.2926e-004所以男職工工資高用精確算法：p,h,stats=ranksum(x,y,method,exact)p = 2.3435e-004h = 1stats = ranksum: 797對于單側(cè)檢驗：P值為p/2ans =1.6717e-004拒絕原假設。所以男性工資高。2a=134 146 130 113 119 161 107 132 135 129;b=70 118 101 104 108 83 94 124 99;精確算法：p,h,stats=ranksum(a,b)p = 6.1418e-004h = 1stats = ranksum: 51ranks

8、um的計算：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 Colu

9、mns 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當ranksum為51時，n=9 N=19,9*(19+1)/2=90,139+51=190=19*(19+1)/2P的計算：zh=nchoosek(1:19,9);2*sum(sum(zh,2)=139)/nchoosek(19,10)ans =6.1418e-0

10、04對于雙側(cè)檢驗，拒絕原假設，有差異。對于單側(cè)檢驗，matlab沒有現(xiàn)存程序。把把相應的雙側(cè)檢驗所得p值除以2，即得單側(cè)檢驗的p值。7x=1149 1152 1176 1149 1155 1169 1182 1160 1120 1171;y=1116 1130 1184 1194 1184 1147 1125 1125 1166 1151;秩號123.53.5567.57.5910數(shù)據(jù)1116112011251125113011471149114911511152秩號1112131415161718.518.520數(shù)據(jù)11551160116611691171117611821184118411

11、94紅色字體為12月的數(shù)據(jù)方法一：Mood 檢驗：n=10,N=20,m=10 g=17,em=n*(N2-1)/12-3*(23-2)/(12*N)em =3.0000e+002em也可這樣計算：n*sum(d-(N+1)/2).2)/Nans = 3.0000e+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 =1.8246e+004r=1 3.5 3.5 5 6 9 13 18.5 18.5 20;a

12、r=(r-(N+1)/2).2;my=sum(ar)my =4.0000e+002p=2*(1-normcdf(my-0.5,em,sqrt(de)p = 0.955在顯著性水平0.05下，接受原假設，兩月份相同。方法二：Ansari-Bradley檢驗N=20，k=10,為偶數(shù)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

13、 18.5 18.5 20;對于雙側(cè)檢驗，在顯著性水平0.05下，查表，P（A41）=0.025, Ay=42，接受原假設。用大樣本近似，ea=n*(N+2)/4,da=n*m/(N*(N-1)*(sum(d.2)-N*(N+2)2/16)da = 5.8421e+002p=2*normcdf(42+0.5,ea,sqrt(da)p = 0.003對于雙側(cè)檢驗，在顯著性水平0.05下,接受原假設，兩月份相同。方法三：Siegel-Turkey檢驗秩號123.53.5567.57.5910數(shù)據(jù)1116112011251125113011471149114911511152ar2017161312

14、98541秩號1112131415161718.518.520數(shù)據(jù)1155116011661169117111761182118411841194ar2367101114151819Sy=20+16+13+12+9+4+6+15+18+19Sy = 132查Wilcoxon秩和檢驗表，雙側(cè)檢驗，臨界值為78，接受原假設。同樣也可用大樣本來近似。方法四：Klotz檢驗ai=(norminv(d/(N+1).2;ea=n*sum(ai)/N%期望ea = 7.053da=n*m/(N*(N-1)*sum(ai-ea).2)%方差da =2.8611e+002秩號123.53.5567.57.5910數(shù)據(jù)1116112011251125113011471149114911511152a2.78351.71390.93590.93590.50760.32030.1340.1340.03240.0