多元统计分析方法练习题 下载本文

它们不能被测量,但对可测量的变量产生影响,或者说通过这些变量体现出来。因子分析的任务是通过原变量提供的错综复杂的关系,寻找潜在的公共因子,当初始因子不好解释时,常对其作旋转变换。提取公共因子的方法很多,主成分是最常用的提取公共因子的方法之一。因此,很多应用者将主成分和因子分析看成一回事,这事片面的。

5-2

TITLE’因子分析’;

OPTION LINESIZE=120; DATA ex5_2; INPUT x1-x12; CARDS;

46 55 126 51 75.0 25 72 6.8 489 27 8 360 ..........

48 68 100 45 53.6 23 70 7.2 522 28 9 352 ;

/*从原始数据出发,进行因子分析*/

PROC FACTOR METHOD=P N=3 ROTATE=VARIMAX; RUN;

/*从协方差矩阵出发,进行因子分析*/

PROC FACTOR COV METHOD=P N=3 ROTATE=VARIMAX; RUN;

练习5.4

TITLE’因子分析与因子旋转’; DATA ex5_4(TYPE=CORR); _TYPE_=”CORR”;

INPUT _name_$ x1—x12; CARDS;

x1 1 x2 0.69

0 1 x3 0.590.65

6 5

1

x4 0.510.550.60

5 7 0 1

x5 0.420.390.380.25

1 7 6 5

1

x6 0.350.300.250.200.610 0 2 0 1 1

x7 0.370.340.320.250.640.61

6 9 9 8 2 1

1 x8 0.400.440.350.310.66

0.640.73

1 5 8 1 0.340.380.28x9

2 1 4

x10.320.370.320 5 7 4 x10.260.380.251 0 5 5 x10.160.200.142 5 0 6

;

0 0 2 8

0.240.400.660.431 7 0 5 0.280.350.400.396 9 7 2 0.250.320.370.402 1 0 8 0.140.160.230.305 2 6 3

0.47

1 8

0.380.46

1

5 0 0.370.400.38

1

9 6 4

0.280.270.210.39

1

5 8 3 8

/*方差最大正交旋转*/

PROC FACTOR METHOD=P N=4 ROTATE=VARIMAX RES; RUN;

/*斜交旋转*/

PROC FACTOR METHOD=P N=3 ROTATE=PROMAX RES; RUN; 6-1解:

不妨假设回归方程为:logitP????x。

?p0/1?p0?(1)ln(OR)?ln???logitP0?logitP1???(???)???

p/1?p?11???ln1可解释为优势比导数的对数值。 OR对例6.1,当x=0时,logitP0?lnp055?ln??,??1.0629 1?p019 当x=1时,logitp?lnp1128?ln????,???1.3107 1?p1164所以相应的logistic回归方程为logitP=1.0629-1.3107x (2)ln(OR)?ln??p1/1?p1???logitP?1?(???)?(???)?2?

p/1?p??1?1???lnOR可解释为优势比平方根的对数值。

当x=1时,logitp1?lnp155?ln???? 1?p119p?1128?ln????

1?p?1164当x=-1时,logitp?1?ln联列上面两式,解得:??0.4075,??0.6554

所以相应的logistic回归方程为logitP=0.4075+0.6554x 6-2 解:

Title ‘logistic回归’; Data ex6-2;

Input f y x1 x2 @@ ; x12=x1*x2;

lable x1=”吸烟否”; lable x2=”用药否”; cards;

14 1 2 0 ;

Proc logistic descending; Weight f;

Model y=x1 x2 x12; Run;

Proc sort ;by x2 ;

Proc logistic descending;

Weight f; Model y=x1; By x2; Run; 6 -3 解:

Data ex6-3; Do y= 0 to 1; Do cho 1 to 4; Do sbp 1 to 4; Input f @@; Output ; End; End; End; Cards;

2 3 8 11 117 121 119

209 ;

Data temp ; Set ex6-3;

1 1 1 1 3 4 6 6 47 22 68 43 7 1 22 0 3 2 7 12 85 98 67 99 1 0 1 0 0 3 11 11 43 20 46 33

8

12 Y=y-1 ;

s1=0; If sbp=2 then s1=1; s2=0; If sbp=3 then s2=1; s3=0; If sbp=4 then s2=1; c1=0; If cho=2 then c1=1; c2=0; If cho=3 then c2=1; c3=0; If cho=4 then c2=1; run;

Proc logistic descending; Weight f;

Model y=sbp cho ;; Run;

Proc logistic descending; Weight f;

Model y=s1-s3 c1-c3; Test1: test s2-s1=s1; Test2: test s3-s2=s2-s1; Test3: test c2-c1=c1; Test4: test c3-c2=c2-c1; Run; 6-4解: Data ex6-4;

Input id chd age agrp@@; Cards;

1

0 20

… …34

0

38

; Run;

Proc gplot; Plot chd *age; Run;

/*计算条件均数p*/ Proc sort out =temp; By agrp;

Proc univariate data =temp noprint; By agrp; Var chd;

Output out=temp2 N=n sum=n1 mean=p; Run;

Proc print data =temp2; Run;

data =temp3; set temp2;

1 35

3

0 38

3 68