Please tell me how to use sas in one sample chi square test

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Sorry cannot type Chinese

                              Here is the question:





thetrue standard deviation of thickness is 1.5 millimeters, Thickness measurements in millimeters of 10specimens produced on a particular shift resulted in the following data: 226,228, 226, 225, 232, 228, 227, 229 225, 230.  α=0.05  Dothe data substantiate the suspicion that the process variability exceeded thestated level on this particular shift?(Test at α=.05.)

suppose H0: σ<1.5    and    H1: σ>=1.5;             The rejection region is :  { (n-1)S*S/σ*σ}>χ(square)(n-1),α

s=2.27  ;   X(square){9,α=0.05}=16.9190

(n-1)S*S/σ*σ=9*2.27*2.27/1.5*1.5=20.612   

I can infer 20.612>16.919  so fail to reject the null hypothesis.

However, I can not find any information about this kind of hypothesis using SAS,  referring
σ and x square...often containing two samples


please help me guys, Thank you so much.

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关键词：Sample Square please Lease AMPL particular following produced question standard

1# anyme

1. 
   
2. data a;
   
3. infile datalines4 dlm = ',';
   
4. input x@@;
   
5. datalines4;
   
6. 226,228, 226, 225, 232, 228, 227, 229 ,225, 230
   
7. ;;;;
   
8. proc means noprint;
   
9.   var x;
   
10.   output out =_MeansOut(keep=var n std) n=n var=var std=std;
    
11. run;
    
12. data _null_;
    
13.   set _MeansOut;  
    
14. df =n-1;  sigma0 = 1.5;
    
15.    chisq =(df*var)/sigma0 **2;
    
16.    pvalue =sdf('chisq', chisq, df);
    
17. put chisq= pvalue=;
    
18. run;

复制代码

As far as I know, SAS does not provide the test in a procedure directly (I am not sure in fact). But very close to that, proc univariate computes the confidence interval for std.
JingJu

2# jingju11

Thank you very much .

You are so nice