非参数统计与matlab编程第三章符号检验法 1.docx

1、非参数统计与matlab编程第三章符号检验法 1第三章 符号检验法3.1 符号检验函数 signtest格式 p = signtest(x) 原假设为x的中位数为0，显著性水平为0.05的双侧检验。 p = signtest(x,m) 原假设为x的中位数为m，显著性水平为0.05的双侧检验。 p = signtest(x,m，alpha) 原假设为x的中位数为m，显著性水平为alpha的双侧检验。 p,h,stats = signtest(x,m,alpha)，当样本小于100时，stats只会显示sign,取x小于假设中位数m的个数与大于假设中位数m的个数的较小值，当样本容量大于或等于100

2、时，stats还将显示zval，即正态统计量值。 k为大于中位数的个数，当k大于n/2时，取负号，当k小于n/2时，号取正号。p,h = signtest(.,alpha, alpha)例：p,h = signtest(.,alpha, 0.01)p,h,stats = signtest(.,method, exact)用精确的方法p,h = signtest(.,method, approximate)用正态近似的方法p,h,stats=signtest(x,y,alpha,0.01,method,exact)p,h,stats=signtest(y1,y2,0.01,method, app

3、roximate )例3.1某市劳动和社会保障部门的资料说明，1998年高级技师的年收入的中位数为21700，该市某个行业有一个由50名高级技师组成的样本，这些高级技师的年收入如下:23072243702032724296222561914025669224042674426744234062043824890248152455618472215142251625112234802652224074180642259025261211802618821625243332314618324359826040208462043819474192142307226744234432463026893

4、2648518138201792674423554257062158817990建立一个50行和1列的向量x，即上面那个数据。a=; %把上表数据粘到a中b=size(a);x=reshape(a,b(1)*b(2),1);format longp,h,stats=signtest(x,21700)p = 0.06490864707227h = 0stats = sign: 18sign值的计算：length(x(find(x21700)ans = 32 length(x(find(x25000)ans =13 length(x(find(x25000)ans = 37 P值等于： binoc

5、df(13,50,0.5)ans =4.681114554259028e-004 也可用：1-binocdf(36,50,0.5)ans =4.681114554372368e-004 对于双侧：p,h,stats=signtest(x,25000)p = 9.362229108518057e-004h = 1stats = sign: 13 P/2得：4.681114554259028e-004与左侧计算的P值相等。下面我们分析大样本情况：23072243702032724296222561914025669224042674426744234062043824890248152455618

6、47221514225162511223480265222407418064225902526121180261882162524333231461832435982604020846204381947419214230722674423443246302689326485181382017926744235542570621588179902307224370203272429622256191402566922404267442674423406204382489024815245561847221514225162511223480XX文库 - 让每个人平等地提升自我2652224074

7、180642259025261211802618821625243332314618324359826040208462043819474192142307226744234432463026893264851813820179267442355425706215881799023072243702032724296222561914025669224042674426744234062043824890248152455618472215142251625112234802652224074180642259025261211802618821625243332314618324359826

8、0402084620438194741921423072267442344324630268932648518138201792674423554257062158817990根据以上数据建立1列150行的向量bb length(find(bb21700)ans = 96 p,h,stats=signtest(bb,21700)p = 8.150399118268979e-004h = 1stats = zval: 3.34763598180368sign: 54zval值的计算：zval=(96-75-0.5)/(150/4)0.5 （96大于75，减去0.5）zval=3.34763598

9、180368p值的计算：2*(1-normcdf(zval)ans = 8.150399118269647e-004对于右侧检验：P值的计算：上面双侧检验所得P值除以2，即得右侧检验的P值。对于：length(bb(find(bb24000)ans = 63 length(bb(find(bb0)ans = 23 length(z(find(z6500)ans = 13 length(find(x6500)ans = 26左侧检验的P值为：p/2ans = 0.02662595704896在0.05显著性水平，接受备择假设，中位数低于6500。区间估计： 中位数me的各层的区间估计第一层：置信

10、度1-0.538ans = 0.99999999999636或1-binocdf(0,39,0.5)/0.5ans = 0.99999999999636第二层：1-0.538-39*0.538ans = 0.99999999985448或：1-binocdf(1,39,0.5)/0.5ans = 0.99999999985448第十四层：1-binocdf(13,39,0.5)/0.5ans = 0.94674808590207 x1=sort(x)x1(14)x1(14)ans = 5800x2=wrev(x1);x2(14)ans = 6500因此，上限6400置信度为0.94674808

11、590207，区间估计为：5800，65002双侧检验：精确算法：2*binocdf(30,70,0.5)ans = 0.281978921793662*(1-binocdf(39,70,0.5)ans = 0.28197892179364正态近似：2*normcdf(35-40+0.5)/sqrt(70/4)ans = 0.28205887576156接受原假设，新建住宅的房价中位数是每平方米6500元。3精确算法：binocdf(527,2079,0.5)ans = 6.355204377699287e-1171-binocdf(1551,527+1552,0.5)ans = 0n很大时，

12、binocdf不方便算近似算法：normcdf(2079/2-1552+0.5)/sqrt(2079/4)ans = 5.334462242411411e-112因此，相信孩子过上更好的日子的成年人较多。点估计：p=1552/(1552+527)p = 0.7465127465127595%的置信区间： p-1.96*sqrt(p*(1-p)/(1552+527) p+1.96*sqrt(p*(1-p)/(1552+527)ans = 0.72781343987309 0.765212053152404. p=normcdf(157860*0.094-18154+0.5)/sqrt(157860*0.094*0.906)p = 4.8162e-180因此，拒绝原假设。90.6%的分位数超过65岁。引深思考：157860*0.094ans = 1.4839e+004若原题中的18154改为15000,则：p=normcdf(157860*0.094-15000+0.5)/sqrt(157860*0.094*0.906)p = 0.0829则在0.1显著性水平下，接受原假设。