实验报告6 单因素方差分析.doc

学号： 100111114 班级： 信息101 姓名： 赵永光 实验六 单因素方差分析实验目的：1.掌握单因素方差分析的理论与方法；2. 掌握利用SAS进行模型的建立与显著性检验，解决有关实际应用问题.实验要求:编写程序，结果分析实验内容：3.4 3.5（选作）3.4程序：data examp3_4;input chj $ delv ; cards;a1 0.88a1 0.85a1 0.79a1 0.86a1 0.85a1 0.83a2 0.87a2 0.92a2 0.85a2 0.83a2 0.90a2 0.80a3 0.84a3 0.78a3 0.81a3 0.80a3 0.85a3 0.83a4 0.81a4 0.86a4 0.90a4 0.87a4 0.78a4 0.79;run;proc anova data=examp3_4; /* */class chj; model delv=chj; run;The SAS System 18:50 Saturday, December 4, 2012 1 The ANOVA Procedure Class Level Information Class Levels Values chj 4 a1 a2 a3 a4表示一个因素chj,四个水平 Number of observations 24样本值个数24The SAS System 18:50 Saturday, December 4, 2012 2 The ANOVA ProcedureDependent Variable: delv Sum of Source DF Squares Mean Square F Value Pr F方差来源 自由度 平方和 均方 f= p值 Model 3 0.00584583 0.00194861 1.31 0.3002 Error 20 0.02985000 0.00149250 Corrected Total 23 0.03569583 R-Square Coeff Var Root MSE delv Mean 0.163768 4.601436 0.038633 0.839583 Source DF Anova SS Mean Square F Value Pr F chj 3 0.00584583 0.00194861 1.31 0.3002由计算可知检验假设，该值较大，因此认为这四种不同催化剂对该化工产品的得率无显著影响3.5（1）程序：data examp3_5;input kyjf $ tgl ; cards;a1 7.6a1 8.2a1 6.8a1 5.8a1 6.9a1 6.6a1 6.3a1 7.7a1 6.0a2 6.7a2 8.1a2 9.4a2 8.6a2 7.8a2 7.7a2 8.9a2 7.9a2 8.3a2 8.7a2 7.1a2 8.4a3 8.5a3 9.7a3 10.1a3 7.8a3 9.6a3 9.5;run;proc anova data=examp3_5; class kyjf; model tgl=kyjf; run; The SAS System 19:03 Saturday, December 4, 2012 1 The ANOVA Procedure Class Level Information Class Levels Values kyjf 3 a1 a2 a3表示一个因素kyjf,三个水平 Number of observations 27The SAS System 19:03 Saturday, December 4, 2012 2 The ANOVA ProcedureDependent Variable: tgl Sum of Source DF Squares Mean Square F Value Pr F方差来源 自由度 平方和 均方 f= p值 Model 2 20.12518519 10.06259259 15.72 F kyjf 2 20.12518519 10.06259259 15.72 .0001由计算可知检验假设，较小，因此认为在显著水平0.05下过去三年科研经费投入的不同对当年生产力的提高有显著影响。（2）proc anova data=examp3_5; class kyjf; model tgl=kyjf; means kyjf; means kyjf/t clm alpha=0.05; means kyjf/t cldiff alpha=0.05; run;The SAS System 19:03 Saturday, December 4, 2012 7 The ANOVA Procedure Level of -tgl- kyjf N Mean Std Dev因素kyjf的水平 观测次数 各总体均值 各总体样本标准差 a1 9 6.87777778 0.81359968 a2 12 8.13333333 0.75718778 a3 6 9.20000000 0.86717934给出置信度的置信区间 The ANOVA Procedure t Confidence Intervals for tgl Alpha 0.05 Error Degrees of Freedom 24 Error Mean Square 0.640093 Critical Value of t 2.06390 95% Confidence kyjf N Mean Limits a3 6 9.2000 8.5259 9.8741 a2 12 8.1333 7.6567 8.6100 a1 9 6.8778 6.3274 7.4282The SAS System 19:03 Saturday, December 4, 2012 9 The ANOVA Procedure t Tests (LSD) for tgl NOTE: This test controls the Type I comparisonwise error rate, not the experimentwise error rate. Alpha 0.05 误差平方自由度 Error Degrees of Freedom = 24 均方误差 Error Mean Square 0.640093 检验t值 Critical Value of t =2.06390*表示显著差异 Comparisons significant at the 0.05 level are indicated by *. Difference kyjf Between 95% Confidence Comparison Means Limits各因素比较 均值差估计 95%的均值差的置信区间 a3 - a2 1.0667 0.2410 1.8923 * a3 - a1 2.3222 1.4519 3.1925 * a2 - a3 -1.0667 -1.8923 -0.2410 * a2 - a1 1.2556 0.5274 1.9837 * a1 - a3 -2.3222 -3.1925 -1.4519 * a1 - a2 -1.2556 -1.9837 -0.5274 *估计结果求得由表3.6知，=2.06390，置信度的置信区间故得生产能力增高量的均值的置信度95%的置信区间分别为（8.5259 ，9.8741）（7.6567 ，8.6100）（6.3274 ，7.4282）的置信度95%的置信区间为故得生产能力增高量的均值的两两之差置信度95%的置信区间分别为：（-1.9837 ，-0.5274）：（-3.1925 ， -1.4519）：（-1.8923 ， -0.2410）显著大于和， 显著大于（3）proc anova data=examp3_5;class kyjf;model tgl=kyjf;means kyjf/bon cldiff alpha=0.05; run;下面给出均值差的同时置信区间The SAS System 19:03 Saturday, December 4, 2012 19 The ANOVA Procedure Bonferroni (Dunn) t Tests for tgl NOTE: This test controls the Type I experimentwise error rate, but it generally has a higher Type II error rate than Tukeys for all pairwise comparisons. Alpha 0.05 Error Degrees of Freedom 24 Error Mean Square 0.640093 Critical Value of t 2.57364 Comparisons significant at the 0.05 level are indicated by *. Difference kyjf Between Simultaneous 95% Comparison Means Confidence Limits各因素比较 均值差估计 95%均值差的同时置信区间 a3 - a2 1.06