如何作（X1，X2）的Chi-square图

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data awt5_2;
input x1 x2 @@;
cards;
3   2.30
5   1.90
5   1.00
7   0.70
7   0.30
7   1.00
8   1.05
9   0.45
10  0.70
11  0.30
;
请问如何作（X1，X2）的Chi-square图？
同时判断（X1，X2）的二元正态性？请老鸟踩踩.....小鸟在此谢过了

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关键词：Square ARE cards Input Data 如何 小鸟

"请问如何作（X1，X2）的Chi-square图？"

I don't understand what you are asking?

"同时判断（X1，X2）的二元正态性？请老鸟踩踩.....小鸟在此谢过了"

The bi-variate normal distribution has five parameters. mean1, std1,mean2, std2, and correlation.

Since a joint normal distribution has normal marginal distributions. You may test them separately and then test the correlation.

You can use histogram q-q plot, p - p plot to eye check your data. If it is not passed your eye-check, you can reject it.

bobguy 发表于 2012-4-12 10:23
"请问如何作（X1，X2）的Chi-square图？"

I don't understand what you are asking?

哦，忘了告诉您，是通过相关性检验（ correlation）这个参量来检测二元（x1,x2）的正态性，我只画得出他们的置信椭圆

Son_Of_The_Sun 发表于 2012-4-12 13:29
哦，忘了告诉您，是通过相关性检验（ correlation）这个参量来检测二元（x1,x2）的正态性，我只画得出他们 ...

You really need to read some basic books before do any statistic analysis and inference.

https://bbs.pinggu.org/thread-793528-1-1.html不知道这个能帮你吗？

bobguy 发表于 2012-4-13 11:08
You really need to read some basic books before do any statistic analysis and inference.

ok thank you

（4）计算每一对观测值到样本均值向量的统计距离平方。


  
程序：proc iml;
n=10;p=2;
xx={x1 x2};
use awt5_2;
read all var xx into x;
e={[10] 1};
x0=(e*x)/n;
mm=i(10)-j(10,10,1)/n;
a=x`*mm*x;
s=a/(n-1);
si=inv(s);print x0 s si; /*si为s的逆矩阵*/
use awt5_2(obs=1);
read all var xx into xx1;
d1=(xx1-x0)*si*(xx1-x0)`; /*d 为马氏距离*/
use awt5_2(firstobs=2 obs=2);
read all var xx into xx2;
d2=(xx2-x0)*si*(xx2-x0)`;
use awt5_2(firstobs=3 obs=3);
read all var xx into xx3;
d3=(xx3-x0)*si*(xx3-x0)`;
use awt5_2(firstobs=4 obs=4);
read all var xx into xx4;
d4=(xx4-x0)*si*(xx4-x0)`;
use awt5_2(firstobs=5 obs=5);
read all var xx into xx5;
d5=(xx5-x0)*si*(xx5-x0)`;
use awt5_2(firstobs=6 obs=6);
read all var xx into xx6;
d6=(xx6-x0)*si*(xx6-x0)`;
use awt5_2(firstobs=7 obs=7);
read all var xx into xx7;
d7=(xx7-x0)*si*(xx7-x0)`;
use awt5_2(firstobs=8 obs=8);
read all var xx into xx8;
d8=(xx8-x0)*si*(xx8-x0)`;
use awt5_2(firstobs=9 obs=9);
read all var xx into xx9;
d9=(xx9-x0)*si*(xx9-x0)`;
use awt5_2(firstobs=10 obs=10);
read all var xx into xx10;
d10=(xx10-x0)*si*(xx10-x0)`;
print d1 d2 d3 d4 d5 d6 d7 d8 d9 d10;
run;
     通过d1=(xx1-x0)*si*(xx1-x0)`可计算各样本到达样本均值的马氏距离。
（5）计算样本点落在二元正态50％置信区域内的比率。（χ22（0.5）=1.39）
     
先算dj*dj
d1*d1=3.4101>1.39   d2*d2=1.4915>1.39  d3*d3=0.8642
d4*d4=0.2344       d5*d5=1.2746      d6*d6=0.0052
d7*d7=0.1873       d8*d8=0.5593      d9*d9=1.0701
d10*d10=1.8592>1.39
所以有七组数据时小于1.39的，落在二元正态50%置信区域内的比率是70%。
程序：proc lifetest method=pl width=2;
time  x2*x2(0);
freq x2;
run;
proc lifetest method=pl width=2;
time  x1*x2(0);
freq x2;
run;
      运行模块lifetest，我们可以得到x1,x2的点估计和区间估计，同时又相对应的75%、50%、25%三种情况下的置信区间比率。
（6）作（X1，X2）的Chi-square图。

程序：data md;
input n d @@;
cards;
1  4.0586824
2  2.1095808
3  2.1074318
4  0.6361144
5  3.2654794
6  0.0079034
7  0.5218616
8  0.6479336
9  2.0590803
10 2.5859323
;
run;
proc sort data=md;
by d;
run;
proc print data=md;
run;
proc means  data=md noprint;
var d;
output out=chiqn n=totn;
run;
data chiqq;
if (_n_=1) then set chiqn;
set md;
novar=2;
chisq=cinv(((_n_-0.5)/totn),novar);
prop=0;
d0=cinv(0.5,novar);
if d <=d0 then prop=1;
proc univariate data=chiqq;
var prop;
run;
proc gplot;
plot d*chisq;
label d='Mahalanobis Distance'
      chisq='Chi-Square Quantile';
          symbol1 v=star;
          *symbol2 i=join v=+;
run;
（7）判断（X1，X2）的二元正态性。
由(6)中图可知,(x1,x2)符合二元正态分布