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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 ;

proc anova data =examp3_4; /* μ÷ó?·?2?·???1y3ì */ 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=E A MS MS / 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4

3210

:μμμμ===H

，31.1/==E A

MS MS

.03002.0))20,3((>=≥=f F P p 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 ;

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 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=E A MS MS / 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3

210

:μμμ==H

，72.15/==E A

MS MS

0001

.0))24,3((<≥=f F P p 较小，因此认为在显著水平

α0.05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 i

n j ij i n y y i

∑=?= )1/()(21

--=?=?

∑i i n j ij i n y y s

Level of -------------tgl------------- kyjf N Mean Std Dev

因素kyjf 的水平观测次数i n 各总体均值

i y 各总体样本标准差i

a1 9 6.87777778 0.81359968

a2 12 8.13333333 0.75718778 a3 6 9.20000000 0.86717934 给出i μ置信度α-1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 a n -= 24

均方误差 Error Mean Square =E

MS 0.640093

检验t 值 Critical Value of t )327(975.0-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j i

μμ-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,

8778.6,

1333.8,2000.9321===??

y y y 由表3.6知，0.64009259=E

MS ，)24()(975

.02

1t a n t

=--

=2.06390，

6,12,9321==n n n ，，

μ置信

度

-1的置信区间

???

-+---

?i E i i E i n MS a n t

y n MS a n t y /)(,/)(2121αα

故得生产能力增高量的均值3

μμμ，，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）

i μμ-的置信度95%的置信区间为

???

? ?

+-+-+----

??-??E j i E j i MS n n a n t

y y MS n n a n t y y )1

1()(,)11()(2

121αα

故得生产能力增高量的均值3

μμμ，，的两两之

差置信度95%的置信区间分别为

21μμ-：（-1.9837 ，-0.5274）31μμ-：（-3.1925 ， -1.4519）32μμ-：（-1.8923 ，

-0.2410

） 1μ显著大于3μ和2μ， 2μ显著大于3μ．

（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 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j i

μμ-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：