当用K-S方法作正态性检验时，SPSS会用样本均数来估计mu，用样本标准差来估计sigma，语法是：

NPAR TESTS  /K-S(NORMAL)= x.

但问题是：如果我想知道一组n=20的数是否服从mu=0,sigma=1的正态分布，用K-S来作时，可以认为参数是已知的，不需要样本统计量来估计，这时在SPSS语法上应如何改动
我查了帮助，看到的NPAR TESTS语法是这样的：

NPAR TESTS Command Syntax
NPAR TESTS [CHISQUARE=varlist[(lo,hi)]/]
           [/EXPECTED={EQUAL      }]
                      {f1,f2,...fn}
 [/K-S({UNIFORM[,lo,hi]})=varlist]
       {NORMAL[,m,sd]  }
       {POISSON[,m]    }
       {EXPONENTIAL[,m]    }
 [/RUNS({MEAN  })=varlist]
        {MEDIAN}
        {MODE  }
        {value }
 [/BINOMIAL[({.5})]=varlist[({v1,v2})]]
             { p}            {value}
 [/MCNEMAR=varlist [WITH varlist [(PAIRED)]]]
 [/SIGN=varlist [WITH varlist [(PAIRED)]]]
 [/WILCOXON=varlist [WITH varlist [(PAIRED)]]]
 [/MH=varlist [WITH varlist [(PAIRED)]]]††
 [/COCHRAN=varlist]
 [/FRIEDMAN=varlist]
 [/KENDALL=varlist]
 [/MEDIAN[(value)]=varlist BY var (v1,v2)]
 [/M-W=varlist BY var (v1,v2)]
 [/K-S=varlist BY var (v1,v2)]
 [/W-W=varlist BY var (v1,v2)]
 [/J-T=varlist BY var (value1,value2)]††
 [/MOSES[(n)]=varlist BY var (v1,v2)]
 [/K-W=varlist BY var (v1,v2)]
 [/MISSING={ANALYSIS**}  [INCLUDE]]
           {LISTWISE  }
 [/SAMPLE]
 [/STATISTICS=[DESCRIPTIVES][QUARTILES][ALL]]
 [/METHOD={MC [CIN({99.0})] [SAMPLES({10000})]}]††
                   {value}           {value}
          {EXACT [TIMER({5    })]              }
                        {value}
 **Default if the subcommand is omitted.
 ††Available only if the Exact Tests Option is installed.

貌试是可以指定参数的，但是我还是没有能调用出来