proc capability得出集中分布的平均值和方差为何有差异

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使用proc capability或者proc univariate对同一组数据检验正态分布、lognormal, beta, gamma之类的分布，得出的均值和方差为何不一样。
我查了SAS的帮助，举的例子里面的结果也是不一样的，非常费解。由于第一次发贴，不能用网址，很郁闷。
大家可以到SAS网站上查询：Example 5.9 Fitting Lognormal, Weibull, and Gamma Curves

我这里贴出来统计结果，大家关注一下里面的mean和STD。有高手知道为什么不一样的，请帮忙解释一下。
The CAPABILITY Procedure
Fitted Lognormal Distribution for Gap

Parameters for Lognormal DistributionParameterSymbolEstimateThresholdTheta0ScaleZeta-0.58375ShapeSigma0.499546Mean 0.631932Std Dev 0.336436

The CAPABILITY Procedure
Fitted Weibull Distribution for Gap

Parameters for Weibull DistributionParameterSymbolEstimateThresholdTheta0ScaleSigma0.719208ShapeC1.961159Mean 0.637641Std Dev 0.339248

The CAPABILITY Procedure
Fitted Gamma Distribution for Gap

Parameters for Gamma DistributionParameterSymbolEstimateThresholdTheta0ScaleSigma0.155198ShapeAlpha4.082646Mean 0.63362Std Dev 0.313587

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关键词：Capability Ability APabi ROC cap 方差 均值 proc univariate proc capability

1# lzg10.01

Hey,

Because the distribution is different, so is the mean. I believe, it computes the mean based on estimated parameters, such as shape and scale. I just take an example of mean as follows:

1. 
   
2. ods output parameterestimates =pes; **get estimated parameters for each distribution;
   
3. proc capability data =_last_;
   
4. ...
   
5. run;
   
6. *test if estimated means were computed by formula of scale and shape;
   
7. options ls =max;
   
8. data _null_;
   
9. set pes; by distribution notsorted;
   
10. retain scale shape;
    
11. if Parameter ='Scale' then Scale =Estimate;
    
12. if Parameter ='Shape' then Shape =Estimate;
    
13. if Distribution ='Lognormal' then mean_test =exp(scale +shape**2/2);
    
14. if Distribution ='Weibull' then mean_test =scale*gamma(1+1/shape);
    
15. if Distribution ='Gamma' then mean_test =Scale *shape;
    
16. mean_dif =mean_test -Estimate;
    
17. if parameter ='Mean';
    
18. put @1 Distribution @15 Scale =F10.8  @35 shape =F10.8 @55 'mean=' Estimate F10.8 @75 'computation difference of mean =' mean_dif;
    
19. run;

复制代码

You will see there is no difference between the estimated and computed at all. JingJu

thanks for your explanation.
it's great.

2# jingju11