SAS例题及程序输出6

实用标准文档文案大全地质勘探中，在A,B,C三个地区采集了一些岩石，测量其部分化学成分，其数据见表3.5。假定这三个地区掩饰的成分遵从。（1）检验不全（2）检验；（3）检验。表3.5岩石部分化学成分数据SiO2FeOK2OA地区47.225.060.1047.454.350.1547.526.850.1247.864.190.1747.317.570.18B地区54.336.220.1256.173.310.1554.402.430.2252.625.920.12C地区43.1210.330.0542.059.670.0842.509.620.0240.779.680.04解：(1)检验假设，在H0成立时，取近似检验统计量为统计量：。由样本值计算三个总体的样本协方差阵：进一步计算可得对给定显著性水平，利用软件SAS9.3进行检验时，首先计算p值：p=P{ξ≥13.896916}=0.3073394。因为p值=0.3073394>0.05，故接收，即认为方差阵之间无显著性差异。prociml;n1=5;n2=4;n3=4;n=n1+n2+n3;k=3;p=3;x1={47.225.060.1,47.454.350.15,47.526.850.12,47.864.19 0.17,47.317.57 0.18};x2={54.336.220.12,56.173.31 0.15,54.42.43 0.22,52.625.92 0.12};x3={43.1210.330.05,42.059.67 0.08,42.59.620.02,40.779.68 0.04};xx=x1//x2//x3;/*三组样本纵向拼接*/mm1=i(5)-j(5,5,1)/n1;mm2=i(4)-j(4,4,1)/n2;mm=i(n)-j(n,n,1)/n;a1=x1`*mm1*x1;printa1;a2=x2`*mm2*x2;printa2;a3=x3`*mm2*x3;printa3;tt=xx`*mm*xx;printtt;/*总离差阵*/a=a1+a2+a3;printa;/*组内离差阵*/da=det(a/(n-k));/*合并样本协差阵*/da1=det(a1/(n1-1));/*每个总体的样本协差阵阵*/da2=det(a2/(n2-1));da3=det(a3/(n3-1));m=(n-k)*log(da)-(4*log(da1)+3*log(da2)+3*log(da3));dd=(2*p*p+3*p-1)*(k+1)/(6*(p+1)*(n-k));df=p*(p+1)*(k-1)/2;/*卡方分布自由度*/kc=(1-dd)*m;/*统计量值*/printdada1da2da3mdddf;p0=1-probchi(kc,df);/*显著性概率*/printkcp0;quit;(2)提出假设。取检验统计量为，由样本值计算得：进一步计算得：对给定显著性水平，利用软件SAS9.3进行检验时，首先计算p值：p=P{F≥32.098939}=0.0010831。因为p值=0.0010831<0.05，故否定，即认为A，B两地岩石化学成分数据存在显著性差异。在这种情况下，可能犯第一类错误，且犯第一类错误的概率为0.05。SAS程序及结果如下：prociml;n=5;m=4;p=3;x={47.225.060.1,47.454.350.15,47.526.850.12,47.864.19 0.17,47.317.57 0.18};ln={[5]1};x0=(ln*x)`/n;printx0;mx=i(n)-j(n,n,1)/n;a1=x`*mx*x;printa1;y={54.336.220.12,56.173.31 0.15,54.42.43 0.22,52.625.92 0.12};lm={[4]1};y0=(lm*y)`/m;printy0;my=i(m)-j(m,m,1)/m;a2=y`*my*y;printa2;a=a1+a2;xy=x0-y0;ai=inv(a);printaai;dd=xy*ai*xy`;d2=(m+n-2)*dd;t2=n*m*d2/(n+m);f=(n+m-1-p)*t2/((n+m-2)*p);fa=finv(0.95,p,m+n-p-1);beta=probf(f,p,m+n-p-1,t2);printd2t2fbeta;pp=1-probf(f,p,m+n-p-1);printpp;quit;(3)检验假设；因似然比统计量，本题中k-1=2，可以利用统计量与F统计量的关系，去检验统计量为F统计量：由样本值计算得：及，进一步计算得：对给定显著性水平，利用软件SAS9.3进行检验时，首先计算p值：p=P{F≥18.390234}=2.3451×10-6。因为p值=2.3451×10-6<0.05，故否定，即认为A，B，C三地岩石化学成分数据存在显著性差异。在这种情况下，可能犯第一类错误，且犯第一类错误的概率为0.05。prociml;n1=5;n2=4;n3=4;n=n1+n2+n3;k=3;p=3;x1={47.225.060.1,47.454.350.15,47.526.850.12,47.864.19 0.17,47.317.57 0.18};x2={54.336.220.12,56.173.31 0.15,54.42.43 0.22,52.625.92 0.12};x3={43.1210.330.05,42.059.67 0.08,42.59.620.02,40.779.68 0.04};xx=x1//x2//x3;/*三组样本纵向拼接*/ln={[5]1};lnn{[4]1};lnnn={[13]1};x10=(ln*x1)`/n1;x20=(lnn*x2)`/n2;x30=(lnn*x3)`/n3;xx0=(lnnn*x1)`/n1;mm1=i(5)-j(5,5,1)/n1;mm2=i(4)-j(4,4,1)/n2;mm=i(n)-j(n,n,1)/n;a1=x1`*mm1*x1;a2=x2`*mm2*x2;a3=x3`*mm2*x3;tt=xx`*mm*xx;printtt;/*总离差阵*/a=a1+a2+a3;printa;/*组内离差阵*/da=det(a);/*合并样本协差阵*/dt=det(tt);a0=da/dt;printdadta0;b=sqrt(a0);printb;f=(n-k-p+1)*(1-b)/(b*p);df1=2*p;df2=2*(n-k-p+1)