ANOVA实际应用中碰到的问题
现在正碰到一个实例,对工件的表面情况(用A度量)进行评估,是取相等段(距离)相邻段内的A均值进行ANOVA的,可是距离范围的大小取得不同,结果也不同,我用的方法有错吗?图如下
结果分别为
//
Source DF SS MS F P
C19 1 0.008 0.008 0.08 0.781
Error 18 1.811 0.101
Total 19 1.819
S = 0.3172 R-Sq = 0.44% R-Sq(adj) = 0.00%
Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev --------+---------+---------+---------+-
1 10 43.554 0.271 (----------------*-----------------)
2 10 43.594 0.357 (-----------------*----------------)
--------+---------+---------+---------+-
43.44 43.56 43.68 43.80
Pooled StDev = 0.317
//
和
//
Source DF SS MS F P
C20 3 1.0628 0.3543 7.49 0.002
Error 16 0.7563 0.0473
Total 19 1.8191
S = 0.2174 R-Sq = 58.43% R-Sq(adj) = 50.63%
Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev ----+---------+---------+---------+-----
1 5 43.692 0.177 (-----*------)
2 5 43.416 0.294 (------*------)
3 5 43.888 0.151 (------*------)
4 5 43.300 0.220 (------*------)
----+---------+---------+---------+-----
43.20 43.50 43.80 44.10
Pooled StDev = 0.217
//
结果分别为
//
Source DF SS MS F P
C19 1 0.008 0.008 0.08 0.781
Error 18 1.811 0.101
Total 19 1.819
S = 0.3172 R-Sq = 0.44% R-Sq(adj) = 0.00%
Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev --------+---------+---------+---------+-
1 10 43.554 0.271 (----------------*-----------------)
2 10 43.594 0.357 (-----------------*----------------)
--------+---------+---------+---------+-
43.44 43.56 43.68 43.80
Pooled StDev = 0.317
//
和
//
Source DF SS MS F P
C20 3 1.0628 0.3543 7.49 0.002
Error 16 0.7563 0.0473
Total 19 1.8191
S = 0.2174 R-Sq = 58.43% R-Sq(adj) = 50.63%
Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev ----+---------+---------+---------+-----
1 5 43.692 0.177 (-----*------)
2 5 43.416 0.294 (------*------)
3 5 43.888 0.151 (------*------)
4 5 43.300 0.220 (------*------)
----+---------+---------+---------+-----
43.20 43.50 43.80 44.10
Pooled StDev = 0.217
//
没有找到相关结果
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nela (威望:0) (上海 闵行区) 其它行业 其它
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