单因素方差分析示例1

下表是 6 种溶液及对照组的雌激素活度鉴定，指标是小鼠子宫重。对表中的数据做方差分析，若差异是显著的，则需做多重比较。

鼠  号	溶  液  种  类
	Ⅰ(ck)	Ⅱ	Ⅲ	Ⅳ	Ⅴ	Ⅵ	Ⅶ
1	89.9	84.4	64.4	75.2	88.4	56.4	65.6
2	93.8	116.0	79.8	62.4	90.2	83.2	79.4
3	88.4	84.0	88.0	62.4	73.2	90.4	65.6
4	112.6	68.6	69.4	73.8	87.8	85.6	70.2

答：所用程序及结果如下：

options linesize=76 nodate;

data uterus;

    infile 'e:\data\exr8-2e.dat';

        do solution=1 to 7;

           do repetit=1 to 4;

              input weight @@;

              output;

           end;

       end;

run;

proc anova;

    class solution;

    model weight=solution;

    means solution/duncan;

run;

 

The SAS System                          Analysis of Variance Procedure                           Class Level Information                         Class    Levels    Values                         SOLUTION      7    1 2 3 4 5 6 7                       Number of observations in data set = 28                                  The SAS System                          Analysis of Variance Procedure   Dependent Variable: WEIGHT                                    Sum of          Mean Source                  DF        Squares        Square   F Value     Pr > F   Model                    6     2419.10500     403.18417      2.77     0.0385   Error                   21     3061.30750     145.77655   Corrected Total         27     5480.41250                     R-Square           C.V.      Root MSE          WEIGHT Mean                     0.441409       15.03118       12.0738              80.3250     Source                  DF       Anova SS   Mean Square   F Value     Pr > F   SOLUTION                 6     2419.10500     403.18417      2.77     0.0385                                  The SAS System                          Analysis of Variance Procedure                Duncan's Multiple Range Test for variable: WEIGHT           NOTE: This test controls the type I comparisonwise error rate, not               the experimentwise error rate                        Alpha= 0.05  df= 21  MSE= 145.7765               Number of Means     2     3     4     5     6     7             Critical Range  17.75 18.64 19.20 19.60 19.89 20.12           Means with the same letter are not significantly different.                Duncan Grouping              Mean      N  SOLUTION                              A            96.175      4  1                            A                    B       A            88.250      4  2                    B       A                    B       A            84.900      4  5                    B       A                    B       A            78.900      4  6                    B                    B                    75.400      4  3                    B                    B                    70.200      4  7                    B                    B                    68.450      4  4

溶液种类的显著性概率P＝0.038 5，P <0.05，不同种类的溶液影响显著。其中1、2、5、6间差异不显著；2、5、6、3、7、4间差异不显著。以上结果可以归纳成下表：

变差来源	平方和	自由度	均方	F	P
溶液间	2 419.105 00	6	403.184 17	2.77	0.038 5
重复间	3 061.307 50	21	145.776 55
总  和	5 480.412 50	27

 

1(ck)	2	5	6	3	7	4