《非参数统计》与MATLAB编程 第七章 区组设计问题.doc

第七章 区组设计问题7.1p = friedman(X,reps) %reps是指区组重复的次数，详细说明请看matlab的anova2帮助文件。p = friedman(X,reps,off) %不显示方差分析表p = friedman(X,reps,on) %显示方差分析表p,table = friedman(.)p,table,stats = friedman(.)B=2B=3B=1 处理：A=1 A=2此时，rep=3，即：p = friedman(X,3)idea=73 75 67 61 69 79;83 81 99 82 85 87;73 60 73 77 68 74;58 64 64 71 77 74;77 75 73 59 85 82%每一列代表一个处理idea = 73 83 73 58 77 75 81 60 64 75 67 99 73 64 73 61 82 77 71 59 69 85 68 77 85 79 87 74 74 82p,table,stats=friedman(idea,1)%每一个兔子没有重复试验p = 0.0056table = Source SS df MS Chi-sq ProbChi-sq Columns 35.0000 4 8.7500 14.6087 0.0056 Error 22.5000 20 1.1250 Total 57.5000 29 stats = source: friedman n: 6 meanranks: 2.5000 4.9167 2.2500 1.9167 3.4167 sigma: 1.5478按p114（7.7）式，根据书上表7.3，计算Chi-sq值：12/30*35/(1-1/24)ans = 14.6087p=1-chi2cdf(14.6087,4)p =0.0056因此，有差异。compares,means=multcompare(stats)%参见统计学原理中的方差分析Note: Intervals can be used for testing but are not simultaneous confidence pares = 1.0000 2.0000 -4.8543 -2.4167 0.0210 1.0000 3.0000 -2.1877 0.2500 2.6877 1.0000 4.0000 -1.8543 0.5833 3.0210 1.0000 5.0000 -3.3543 -0.9167 1.5210 2.0000 3.0000 0.2290 2.6667 5.1043 2.0000 4.0000 0.5623 3.0000 5.4377 2.0000 5.0000 -0.9377 1.5000 3.9377 3.0000 4.0000 -2.1043 0.3333 2.7710 3.0000 5.0000 -3.6043 -1.1667 1.2710 4.0000 5.0000 -3.9377 -1.5000 0.9377means = 2.5000 0.6319 4.9167 0.6319 2.2500 0.6319 1.9167 0.6319 3.4167 0.6319 习题4用Friedman方法检验c=15.54 20.5 21.31 21.89 21.29 27.28; 21.34 17.52 19.25 20.56 23.00 24.04c = 15.5400 21.3400 20.5000 17.5200 21.3100 19.2500 21.8900 20.5600 21.2900 23.0000 27.2800 24.0400p,table,stats=friedman(c,2)p = 0.6547table = Source SS df MS Chi-sq ProbChi-sq Columns 0.3333 1 0.3333 0.2000 0.6547 Interaction 4.6667 2 2.3333 Error 10 6 1.6667 Total 15 11 stats = source: friedman n: 3 meanranks: 2.6667 2.3333 sigma: 1.2910P值为0.6547，接受原假设。两冷冻方法相同普通的双因素方差分析p,table,stats=anova2(c,2)p = 0.8202 0.0730 0.6933table = Source SS df MS F ProbF Columns 0.3675 1 0.3675 0.0564 0.8202 Rows 54.4530 2 27.2265 4.1775 0.0730 Interaction 5.0788 2 2.5394 0.3896 0.6933 Error 39.1041 6 6.5174 Total 99.0035 11 stats = source: anova2 sigmasq: 6.5174 colmeans: 21.3017 20.9517 coln: 6 rowmeans: 18.7250 20.7525 23.9025 rown: 4 inter: 1 pval: 0.6933 df: 6P值为0.8202，接受原假设。两冷冻方法相同。Hodges-Lehmann检验方法 残差的秩蔬菜秩和1 2 3 冷冻方法1 1，108，92，1242.00 2 11，43，65，736比36小的秩和的置换有：10+4+3+6+5+71+4+3+6