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参数估计与假设检验的SAS分析参数估计的SAS分析参数估计1、点估计2、区间估计区间估计SAS程序:①根据原始变量值,计算均数的双侧95%可信区间,用MEANS过程,加clm选项完成即可。②后面t-test也可以计算③如果只知道样本均数,S,求95%可信区间。可以用下列程序。X①求某市120名12岁健康男孩的平均身高Dataprg5_1;Inputx@@;Datalines;142.3156.6142.7145.7138.2141.6142.5130.5132.1135.5134.5148.8134.4148.8137.9151.3140.8149.8143.6149.0145.2141.8146.8135.1150.3133.1142.7143.9142.4139.6151.1144.0145.4146.2143.3156.3141.9140.7145.9144.4141.2141.5148.8140.1150.6139.5146.4143.8150.0142.1143.5139.2144.7139.3141.9147.8140.5138.9148.9142.4134.7147.3138.1140.2137.4145.1145.8147.9146.7143.4150.8144.5137.1147.1142.9134.9143.6142.3143.3140.2125.9132.7152.9147.9141.8141.4140.9141.4146.7138.7160.9154.2137.9139.9149.7147.5136.9148.1144.0137.4134.7138.5138.9137.7138.5139.6143.5142.9146.5145.4129.4142.5141.2148.9154.0147.7152.3146.6139.2139.9;Procmeansnmeanstdstderrcvclm;Varx;Run;②见后面t检验③知道样本均数,标准差S,求95%可信区间。可以用下列程序。Dataprg5_2;N=120;Mean=143.05;Std=5.82;T=tinv(0.975,n-1);In=t*std/sqrt(n);Lclm=mean-in;Uclm=mean+in;Run;Procprint;Varlclmuclm;Run;两总体均数相差的可信区间两样本均数比较,总体均数差值的95%可信区间与假设检验Dataprg5_3;N1=10;N2=10;M1=10.2;M2=9.4;S1=3.58;S2=4.27;Sc2=(s1**2*(n1-1)+s2**2*(n2-1))/(n1+n2-2);St=sqrt(sc2*(1/n1+1/n2));T=tinv90.975,n1+n2-2);In=t*st;Lclm=abs(m1-m2)-in;Uclm=abs(m1-m2)+in;Procprint;Varlclmuclm;Run;总体率的置信区间估计例题6-6n=120,x=94,p=78.3%估计95%的可信区间。Dataprg6_6;N=120;P=0.783;sp=sqrt(p*(1-p)/n);u=probit(0.975);usp=u*sp;Lclm=p-usp;Uclm=p+usp;Procprint;Varnpsplclmuclm;Run;t检验单样本均数的t检验某医生测量了36名男性铅作业工人的血红蛋白含量,均数为130.8g/L,标准差为25.74g/L,正常男性血红蛋白含量一般为140g/L,铅作业工人的血红蛋白含量与正常人有无不同?Dataprg5_4;N=36;S_m=130.8;P_m=140.0;Std=25.74;Df=n-1;T=(s_m-p_m)/(std/sqrt(n));p=(1-probt(abs(t),df))*2;Procprint;Vartp;Run;某医生又测量了另外30名男性铅作业工人的血红蛋白含量,分别是:1717913578118175122105111140138132142140168113131145128124134116129155135134136113119132,问这批工人与正常男性血红蛋白含量140g/L有无不同?Dataprg5_5;Inputx@@;Cards;1717913578118175122105111140138132142140168113131145128124134116129155135134136113119132;Procttesth0=140;Varx;Run;配对设计资料的t检验DATAprg6_2;INPUTx1x2@@;d=x2-x1;CARDS;2.792.693.062.892.342.243.413.373.483.503.232.932.272.242.482.553.032.823.073.053.613.582.692.663.093.202.982.922.652.60;PROCMEANSNMEANstdstderrtPRT;VARd;TITLE'paired-comparisonsttest';RUN;DATAprg6_2;INPUTx1x2@@;d=x2-x1;CARDS;2.792.693.062.892.342.243.413.373.483.503.232.932.272.242.482.553.032.823.073.053.613.582.692.663.093.202.982.922.652.60;Procunivariate;Vard;Run;DATAprg6_2;INPUTx1x2@@;d=x2-x1;CARDS;2.792.693.062.892.342.243.413.373.483.503.232.932.272.242.482.553.032.823.073.053.613.582.692.663.093.202.982.922.652.60;Procttest;Vard;Run;成组资料两样本均数比较例3.9测得14名慢性支气管炎病人与11名健康人的尿中17酮类固醇(mol/24h)排出量如下,试比较两组人的尿中17酮类固醇的排出量有无不同。原始调查数据如下:病人X1:n=14;10.0518.7518.9915.9413.9617.6720.5117.2214.6915.109.428.217.2424.60健康人X2:n=11;17.9530.4610.8822.3812.8923.0113.8919.4015.8326.7217.29Dataprg5_6;Inputxc@@;Cards;10.05118.75118.99115.94113.96117.67120.51117.22114.69115.1019.4218.2117.24124.60117.95230.46210.88222.38212.89223.01213.89219.40215.83226.72217.292;Procttest;Varx;Classc;Run;Dataprg5_6;Inputx@@;Ifn14thenc=1;elsec=2;Cards;10.0518.7518.9915.9413.9617.6720.5117.2214.6915.109.428.217.2424.6017.9530.4610.8822.3812.8923.0113.8919.4015.8326.7217.29;Procttest;Varx;Classc;Run;二项分布Probbnml(π,n,r)是SAS默认的二项分布的概率函数一、阳性事件发生的概率某药物治疗某传染性疾病的有效率为0.70,无效率为0.30,今用该药物治疗患者10人,试分析这10人中有6人、7人、8人有效的概率?π=0.7,n=10,r=6、7、8Dataprg6_4;Dor=6to8;D=probbnml(0.7,10,r)-probbnml(0.7,10,r-1);Output;End;Procprint;Varrd;Run;某药物治疗某传染性疾病的有效率为0.70,无效率为0.30,今用该药物治疗患者10人,试分析这10人中至少有6人、7人、8人、9、10人有效的概率?π=0.7,n=10,x=6、7、8、9、10Dataprg6_5;Dox=6to10;p=1-probbnml(0.7,10,x-1);Output;End;Procprint;Varxp;Run;样本率与总体率的比较(直接法)例题:输卵管结扎的育龄妇女实施壶腹部-壶腹部吻合手术后,受孕率为0.55,对10名妇女实施峡部-峡部吻合手术后,有9名受孕,问峡部吻合手术受孕率是否高于壶腹部吻合手术?Π=0.55,n=10,x=9Dataprg6_6;dox=8;p=1-probbnml(0.55,10,x);output;end;Procprint;Varp;Run;样本率与总体率的比较(正态近似法)应用条件:当样本含量n足够大,且样本率p和(1-p)均不太小,如np与n(1-p)均≥5时nppp)1(||||z0000根据以往经验,一般胃溃疡病患者有20%发生胃出血症状。现某医生观察65岁以上胃溃疡病人152例,其中48例发生胃出血,占31.6%。问老年胃溃疡病患者是否较一般胃溃疡病患者易发生胃出血。Π=20%,N=152,X=48,p=31.6%Dataprg6_6;N=152;X=48;Pai=0.2;P=x/n;Z=(p-pai)/sqrt(pai*(1-pai)/n);Prob=(1-probnorm(abs(z)))*2;Procprint;Varzprob;Run;两样本率比较的z检验组别观察人数发病人数发病率(%)用药组1001414对照组1203025合计2204420)11)(1(z21212121nnppppSppccpp2121nnxxpcDataprg6_7;N1=100;N2=120;X1=14;X2=30;P1=x1/n1;P2=x2/n2;Pc=(x1+x2)/(n1+n2);Sp=sqrt(pc*(1-pc)*(1/n1+1/n2));Z=(p1-p2)/sp;P=(1-probnorm(abs(z)))*2;Formatzp8.4;Procprint;Varpcspzp;Run;Poisson分布一、样本均数与总体均数比较(直接法)poisson(λ,n)是SAS默认的Poisson分布的概率函数。例题:一般人群先心病的发病率约为8‰,某研究者探讨母亲吸烟与孩子先心病之间的关系,观察了一组20-25岁的孕妇,她们生育的120名孩子中,有4例先心病,分析吸烟与先心病的关系。分析:Π=0.008,n=120,x=4Dataprg6_7;N=120;Pai=0.008;X=4;Lam=n*pai;P=1-poisson(lam,x-1);Procprint;Varp;Run;二、样本均数与总体均数比较(正态分布法)当λ≥20时,可用正态分布法。例题:根据医院消毒卫生标准,细菌总数按每立方米菌落形成单位(CFU/m3)表示。无菌间的卫生标准为细菌菌落数应不大于200(CFU/m3)。某医院引进三氧消毒机,每天自动对无菌间进行2小时消毒。对无菌间抽样调查显示,细菌总数为121CFU/m3。试问该医院无菌间的细菌总数低于国家卫生标准。本例:λ=200,x=121Dataprg6_8;X=121;Lam=200;Z=(x-lam)/sqrt(lam);P=1-probnorm(abs(z));Procprint;Varzp;Run;dataprg6_10;x1=51;x2=23;z=abs(x1-x2)/sqrt(x1+x2);p=(1-probnorm(z))*2;procprint;varzp;run;Poisson分布两样本z检验(观察单位相同时)当观
本文标题:假设检验SAS
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