SAS软件及其应用

1、SAS软件及其应用作业（10级研究生）学院：农学院 专业：作物栽培 姓名：王进 学号: 10090101210009 成绩 ：_ 一、问答题（10分） 1. SAS是国际通用统计软件，由众多模块组成的系统，其核心模块是什么？什么模块几乎覆盖了所有的实验数理统计分析方法？ SAS程序由哪两部分组成？（5分）答：核心模块是Base SAS；STAT模块几乎覆盖了所有的实验数理统计分析方法；SAS程序通常由DATA数据步和PROC过程步两部分组成。2. 方差分析SAS程序中，“means a|b|c/ duncan;” 语句的作用是什么？ （5分）答：means a|b|c/ duncan 语句的作

2、用是要求算出 a b c a*b a*c b*c a*b*c间的均数，用duncan测验进行多重比较。二、编写SAS程序，写出主要SAS输出结果及统计结论（90分）1. （10分）21个小麦双列杂交组合F1的单株产量y（克），每株穗数x1，每穗的粒数x2,千粒重x3（克）数据如下表：组合代号X1X2X3Y110.3729.5633.3110.52210.4734.2529.0510.072010.0339.2729.2111.33试写出只求x1、x3 ，Y的最小值，最大值,极差，均值，方差，标准差，标准误，变异系数，峰度系数，偏度系数，并计算结果保留4位的SAS程序（不要结果及解释）。答：da

3、ta aa;input x1 x2 x3 y;cards;10.37 29.56 33.31 10.5210.47 34.25 29.05 10.07.10.03 39.27 29.21 11.33;proc print;proc means min max range mean var std stderr cv kurtosis skewness maxdec=4;run;2.(10分)将14只大白鼠随机分为两组，一组做成白血病模型鼠，一组为正常鼠，两组鼠脾脏DNA含量（mg/g）如下，请编程度分析两组鼠表脾脏DNA平均含量是否不同？写出主要输出结果及统计分析结论。白血病组（x1）：12.

4、4 13.3 13.7 15.2 15.4 15.8 16.7 正常组（x2）： 10.8 11.6 12.3 12.7 13.5 12.5 14.8答：data aa;input trt $ mount ;cards;a 12.4 a 13.3 a 13.7 a 15.2 a 15.4 a 15.8 a 16.7b 10.8 b 11.6 b 12.3 b 12.7 b 13.5 b 12.5 b 14.8;proc ttest;class trt;var mount;run;StatisticsVariabletrtNLower CLMeanMeanUpper CLMeanLower CL

5、Std DevStd DevUpper CLStd DevStd ErrMinimumMaximummounta713.22114.64316.0640.99051.53723.38490.58112.416.7mountb711.40612.613.7940.83191.2912.84290.48810.814.8mountDiff (1-2) 0.38982.04293.6961.01781.41942.34310.7587    T-TestsVariableMethodVariancesDFt ValuePr

6、0;> |t|mountPooledEqual122.690.0196mountSatterthwaiteUnequal11.72.690.0200  Equality of VariancesVariableMethodNum DFDen DFF ValuePr > FmountFolded F661.420.6824分析结论：两组大白鼠脾脏DNA含量不同两者之间差异显著，且白血病组DNA含量显著高于正常组。3（15分）许多害虫的发生量都和气象条件有一定关系。某地连续测定10年7月下旬的温雨系数（X，雨量mm/平均温度)

7、和大豆第二代造桥虫发生量（Y，每百株大豆上的虫数），得结果于下表：温雨系数X1.58 9.98 9.42 1.25 0.30 2.41 11.01 1.85 6.04 5.92虫口密度Y180 28 25 117 165 175 40 160 120 80编一SAS程序，写出程序及主要输出结果，并回答下列问题：（1）写出回归方程。（2）当x=7时，写出其:1）y/x的95置信区间。2）条件总体观察值Y的95预测区间。答：data xx;input x y;cards;1.58 180 9.98 28 9.42 25 1.25 117 0.30 165 2.41 175 11.01 40 1.8

8、5 160 6.04 120 5.92 80 ;proc reg;model y=x/cli clm;run;主要输出结果如下：SAS 系统The REG ProcedureModel: MODEL1Dependent Variable: y Analysis of VarianceSourceDFSum ofSquaresMeanSquareF ValuePr > FModel1292392923944.140.0002Error85298.91104662.36388  Corrected Total934538  

9、0;    Parameter EstimatesVariableDFParameterEstimateStandardErrort ValuePr > |t|Intercept1179.2121913.3383613.44<.0001x1-14.110172.12373-6.640.0002  SAS 系统The REG ProcedureModel: MODEL1Dependent Variable: y Output StatisticsObsDependentVariablePredictedValueStd Er

10、rorMean Predict95% CL Mean95% CL PredictResidual1180.0000156.918110.8744131.8418181.994592.4895221.346723.0819228.000038.392713.38557.525769.2598-28.5027105.2882-10.3927325.000046.294412.462317.556375.0326-19.6457112.2346-21.29444117.0000161.574511.3513135.3983187.750696.7099226.4391-44.57455165.000

11、0174.979112.8395145.3713204.5870108.6553241.3030-9.97916175.0000145.20679.7946122.6204167.793081.7058208.707629.7933740.000023.859215.1806-11.147258.8657-45.044192.762616.14088160.0000153.108410.5028128.8888177.328089.0084217.20846.89169120.000093.98688.446474.5093113.464331.5240156.449526.01321080.

12、000095.68008.381976.3514115.008633.2635158.0965-15.6800  结果：（1）回归方程为：y= 179.21219-14.11017*x; （2）x=7时，1）y/x的95置信区间【59.2167，101.6654】；2）条件总体观察值Y的95预测区间【17.4117，143.4703】。 4.（10分）由于环境作用对光合速率的影响很大，要得到能反映环境对光合作用影响的数据，必须在不同的天气下测定光合作用各种指标。表1中的数据使用Li6400光合测量仪测定，其中因变量光合速率；气孔导度；胞间二氧化碳浓度；蒸腾速率；叶片水汽压亏损；叶片的

13、温度；相对湿度；试对数据作逐步回归分析。表1 环境对光合作用影响数据表观测号123456789108.378.198.038.328.388.167.447.286.507.850.09960.09870.10300.10400.09900.10100.09790.09650.08930.09882042022081991922002082082052032.802.793.113.443.483.783.883.903.853.452.782.792.993.273.453.653.883.954.203.4434.8135.0635.8136.7637.4637.8738.3938.723

14、9.6146.681063106911141162121912311288130012951193请写出该逐步回归分析的SAS程序及主要输出结果，写出最优回归方程。答：data phs;input y x1-x6;cards;8.370.09962042.82.7834.8110638.190.09872022.792.7935.0610698.030.1032083.112.9935.8111148.320.1041993.443.2736.7611628.380.0991923.483.4537.4612198.160.1012003.783.6537.8712317.440.097920

15、83.883.8838.3912887.280.09652083.93.9538.7213006.50.08932053.854.239.6112957.850.09882033.453.4446.681193;proc stepwise;model y=x1-x6;run;主要输出结果：Stepwise Selection: Step 4Variable x3 Entered: R-Square = 0.9894 and C(p) = 3.1073Analysis of VarianceSourceDFSum ofSquaresMeanSquareF ValuePr >

16、60;FModel43.296450.82411116.69<.0001Error50.035310.00706  Corrected Total93.33176     VariableParameterEstimateStandardErrorType II SSF ValuePr > FIntercept29.711255.111870.2385733.780.0021x1-119.9335151.211090.038735.480.0662x2-0.037690.005940

17、.2839240.200.0014x35.206921.339040.1067915.120.0115x4-5.906741.390210.1274918.050.0081Bounds on condition number: 592.49, 4248.6All variables left in the model are significant at the 0.1500 level.No other variable met the 0.1500 significance level for entry into the model.  Summary of Stepwise

18、SelectionStepVariableEnteredVariableRemovedNumberVars InPartialR-SquareModelR-SquareC(p)F ValuePr > F1x1 10.72970.729773.263321.590.00172x2 20.13050.860137.00356.530.03783x4 30.09720.957410.504213.680.01014x3 40.03210.98943.107315.120.0115最优方程为：Y=29.71125-119.93351X1

19、-0.03769X2+5.20692X3-5.90674X45.（15分）四种不同配合饲料（A）的饲喂试验，每组各有10头幼猪，试验猪的始重（x）与试验期的平均日增重（y）见下表：A1A2A3A4xyxyxyxy1360.89320.52280.55280.642300.8270.58330.62270.813260.74250.64260.58270.7310160.58170.51170.48200.571)试判断应用何种统计分析法为好，并编其SAS程序分析。(10分)2)其SAS程度主要输出结果如下： The GLM ProcedureLeast Squares MeanstrtY LS

20、MEANStandard ErrorPr > |t|LSMEAN Number10.748607570.02066772<.0001120.541529820.02060465<.0001230.591835460.02061050<.0001340.680027150.02065781<.00014  Least Squares Means for effect trtPr > |t| for H0: LSMean(i)=LSMean(j)Dependent Variable: yi/j12341 <.0001

21、<.00010.02532<.0001 0.0936<.00013<.00010.0936 0.004640.0253<.00010.0046 请写出4种饲料对猪日增重的差异显著性测验的结论。（5分）答：（1）用协方差分析较好，程序如下：data aa;input trt$ x y;cards;1 36 0.89 1 30 0.8 1 26 0.74.1 16 0.582 32 0.52 2 27 0.58 2 25 0.64.2 170.513 28 0.55 3 33 0.62 3 26 0.58.3 170.484 28 0.6

22、4 4 27 0.81 4 27 0.73.4 200.57;proc glm;class trt;model y=trt x/solution;means trt/t;lsmeans trt/stderr pdiff;run;（2）结论：饲料A1对猪日增重效果最好，显著的好于A4饲料，极显著的好于A2、A3组；A4效果极显著的好于A2、A3组；A3效果好于A2，但差异不明显。6（15分）比较A1 、A2 、A3、A4四种催化剂。每种催化剂温度都分为三个水平:A1：50，55，60； A2:70，80，90；A3：55，65，75； A4：90，95，100。试验重复三次，测得转化率如表2，试

23、作方差分析。表2 转化率表ABA1A2A3A4B185，89，8782，84，8665，61，6367，71，69B272，70，6991，88，8959，62，6475，78，73B370，67，7185，83，8560，56，6185，89，87 写出SAS程序、主要输出结果及统计结论。（注：转化率越大效果越好）答：data aa;do a=1 to 4;do b=1 to 3;do n=1 to 3;input y;output;end;end;end;cards;85 89 87 72 70 69 70 67 7182 84 86 91 88 89 85 83 8565 61 63 5

24、9 62 64 60 56 6167 71 69 75 78 73 85 89 87;proc glm;class a b n;model y=a b(a);means a b(a)/duncan;lsmeans b(a)/stderr pdiff;run;输出结果为：SAS 系统The GLM ProcedureClass Level InformationClassLevelsValuesa41 2 3 4b31 2 3n31 2 3  Number of observations36  SAS 系统The GLM Procedure  Dependent V

25、ariable: y SourceDFSum of SquaresMean SquareF ValuePr > FModel113987.888889362.53535487.01<.0001Error24100.0000004.166667  Corrected Total354087.888889     R-SquareCoeff VarRoot MSEy Mean0.9755372.7236732.04124174.94444  SourceDFType I SSMean S

26、quareF ValuePr > Fa32818.333333939.444444225.47<.0001b(a)81169.555556146.19444435.09<.0001  SourceDFType III SSMean SquareF ValuePr > Fa32818.333333939.444444225.47<.0001b(a)81169.555556146.19444435.09<.0001  SAS 系统The GLM Procedure  Duncan's

27、Multiple Range Test for y This test controls the Type I comparisonwise error rate, not the experimentwise error rate.Alpha0.05Error Degrees of Freedom24Error Mean Square4.166667  Number of Means234Critical Range1.9862.0862.150  Means with the same letterare not significantly different

28、.Duncan GroupingMeanNaA85.888992    B77.111194B   B75.555691    C61.222293  SAS 系统The GLM Procedure  Level ofbLevel ofaNyMeanStd Dev11387.00000002.0000000021370.33333331.5275252331369.33333332.0816660012384.00000002.0000000022389.

29、33333331.5275252332384.33333331.1547005413363.00000002.0000000023361.66666672.5166114833359.00000002.6457513114369.00000002.0000000024375.33333332.5166114834387.00000002.00000000  SAS 系统The GLM ProcedureLeast Squares Meansbay LSMEANStandard ErrorPr > |t|LSMEAN Number1187.00000001.1

30、785113<.000112170.33333331.1785113<.000123169.33333331.1785113<.000131284.00000001.1785113<.000142289.33333331.1785113<.000153284.33333331.1785113<.000161363.00000001.1785113<.000172361.66666671.1785113<.000183359.00000001.1785113<.000191469.00000001.1785113<.0001102475

31、.33333331.1785113<.0001113487.00000001.1785113<.000112  Least Squares Means for effect b(a)Pr > |t| for H0: LSMean(i)=LSMean(j)Dependent Variable: yi/j1234567891011121 <.0001<.00010.08440.17430.1227<.0001<.0001<.0001<.0001<.00011.00002<.0001 0.5541<

32、.0001<.0001<.00010.0002<.0001<.00010.43160.0062<.00013<.00010.5541 <.0001<.0001<.00010.00090.0001<.00010.84320.0014<.000140.0844<.0001<.0001 0.00380.8432<.0001<.0001<.0001<.0001<.00010.084450.1743<.0001<.00010.0038 0.0062<

33、.0001<.0001<.0001<.0001<.00010.174360.1227<.0001<.00010.84320.0062 <.0001<.0001<.0001<.0001<.00010.12277<.00010.00020.0009<.0001<.0001<.0001 0.43160.02450.0014<.0001<.00018<.0001<.00010.0001<.0001<.0001<.00010.4316 0.12

34、270.0002<.0001<.00019<.0001<.0001<.0001<.0001<.0001<.00010.02450.1227 <.0001<.0001<.000110<.00010.43160.8432<.0001<.0001<.00010.00140.0002<.0001 0.0009<.000111<.00010.00620.0014<.0001<.0001<.0001<.0001<.0001<.00010.000

35、9 <.0001121.0000<.0001<.00010.08440.17430.1227<.0001<.0001<.0001<.0001<.0001  To ensure overall protection level, only probabilities associated with pre-planned comparisons should be used.分析表明:Fa=225.47,其Pr>F的值小于0.0001，因子A不同水平间差异极显著。在固定催化剂A的前提下，温度B间的差异也极其显著F b(a)= 35.09，其Pr>F的值小于0.0001。对A多重比较，A2高于其他催化剂，A1,A4高于A3，而A1,A4差异不显著。 表中b为2，a也为2时，均值89.3333333为最大值，为最好。所以，催化剂为A2，温度为80时效果最好。7. （15分）为研究某木薯品种的栽培规律，分三种耕作方法A、B、C，每处理随机取4个样点，即重复4次，记录了三种性状，即