計量不負責任解答1.14(1.15睡醒再來XD)

[AaronT (自己的路)]

a.
#include <oxstd.h>
#include <oxdraw.h>
#include <oxprob.h>
#import <maximize>
decl n=10;

main()
{
        ranseed("GM");
        decl tData = rant(n,1,3);
        decl mean = meanc(tData);
        decl Var = variance(tData);
        decl intervalA=mean+2*sqrt(Var/n);
        decl intervalB=mean-2*sqrt(Var/n);
        println("The Interval is :","( ",intervalB[0][0]," , ",
                intervalA[0][0]," )" );
        if((intervalA>0 && intervalB>0) || (intervalA<0 && intervalB<0))
        println("Reject the null Hypothesis!");
        else
        println("Don't reject the null Hypothesis!");
}

b.
#include <oxstd.h>
#include <oxdraw.h>
#include <oxprob.h>
#import <maximize>
decl n=10;
decl Rejs=0;
main()
{
        for(decl i=0;i<10000;i++)
        {
                decl tData;
                decl means,Vars;
                decl intervalA,intervalB;
                tData=rant(n,1,3);
                means=meanc(tData);
                Vars=variance(tData);
                intervalA=means+2*sqrt(Vars/n);
                intervalB=means-2*sqrt(Vars/n);
                if((intervalA>0&&intervalB>0)||(intervalA<0&&intervalB<0))
                        Rejs++;
                        println("There are ",Rejs," numbers will be rejected");
        }

}

c.直接把a.b小提的n改成100以及1000
d.請分析 分析完請跟我討論(其實是高手幫我分析拉XD)

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※ 編輯: AaronT          來自: 140.119.202.75       (03/07 13:46)