广东科学技术职业学院
广州学院
实 验 报 告
专业计算机应用 班级 成绩评定______
学号 姓名 (合作者____号____) 教师签名
实验 6 题目可视化建模基础 第 6 周星期 第 节
第二篇:实验报告6 单因素方差分析
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 0.88
a1 0.85
a1 0.79
a1 0.86
a1 0.85
a1 0.83
a2 0.87
a2 0.92
a2 0.85
a2 0.83
a2 0.90
a2 0.80
a3 0.84
a3 0.78
a3 0.81
a3 0.80
a3 0.85
a3 0.83
a4 0.81
a4 0.86
a4 0.90
a4 0.87
a4 0.78
a4 0.79
;
run;
procanova data=examp3_4; /* µ÷Ó÷½²î·ÖÎö¹ý³Ì */
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24
The SAS System 18:50 Saturday, December 4, 2012 2
The ANOVA Procedure
Dependent 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 7.6
a1 8.2
a1 6.8
a1 5.8
a1 6.9
a1 6.6
a1 6.3
a1 7.7
a1 6.0
a2 6.7
a2 8.1
a2 9.4
a2 8.6
a2 7.8
a2 7.7
a2 8.9
a2 7.9
a2 8.3
a2 8.7
a2 7.1
a2 8.4
a3 8.5
a3 9.7
a3 10.1
a3 7.8
a3 9.6
a3 9.5
;
run;
procanova 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 27
The SAS System 19:03 Saturday, December 4, 2012 2
The ANOVA Procedure
Dependent 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 <.0001
Error 24 15.36222222 0.64009259
Corrected Total 26 35.48740741
R-Square Coeff Var Root MSE tgl Mean
0.567108 10.06128 0.800058 7.951852
Source DF Anova SS Mean Square F Value Pr > 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)
procanova 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 7.4282
The 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)
procanova 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 Tukey's 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 1.0667 0.0371 2.0962 ***
a3 - a1 2.3222 1.2370 3.4074 ***
a2 - a3 -1.0667 -2.0962 -0.0371 ***
a2 - a1 1.2556 0.3476 2.1635 ***
a1 - a3 -2.3222 -3.4074 -1.2370 ***
a1 - a2 -1.2556 -2.1635 -0.3476 ***
原理:
设计思想:
对应程序:
实验结果:
实验体会: