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


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