协方差分析例题

1、协方差分析例子RZ6proc gplot data=sasuser.rz6; plot y*x=t; symbol1 i=r1 v=star; symbol2 i=r1 v=plus;proc reg data=sasuser.rz6; model y=x; by t;proc glm data=sasuser.rz6; class t; model y=x t/solution; means t/duncan; lsmeans t/stderr pdiff;run;例1：新培育的4个水稻品种（v），测量其株高（x）和产量（y），试分析4个水稻品种的产量是否相同？data;do v =1 to

2、 4; do i=1 to 4; input x y ; output; end;end;cards;47 54 58 66 53 63 46 5149 56 56 66 54 61 44 5052 54 53 53 64 67 58 6259 62 61 63 63 64 66 69;proc means; var x y; class v;proc glm; class v; model y=x v /solution; means v /duncan; lsmeans v /stderr pdiff;run;结果：协变量X df=1 回归平方和SS=436.34723386 均方MS=4

3、36.34723386 误差均方MSe=1.8548 F=235.25 Pr F = 0.0001X的回归系数=1.2096回归方程： = ()v1：63.5650 = 58.501.2096(51.0055.188)v2：63.6174 = 58.251.2096(50.7555.188)v3：57.1101 = 59.001.2096(56.7555.188)v4：55.9575 = 64.501.2096(62.2555.188)例2: 10种不同的药剂( 包括CK )防治咖啡绿蚧效果比较试 , X 为处理前的产量, Y 为处理后的产量。1.sas程序data;input block t

4、 x y ;cards;1 1 342 3241 2 200 2621 3 350 3481 4 210 3911 5 370 4681 6 235 4061 7 260 3761 8 250 2741 9 210 3181 10 189 2702 1 381 3522 2 245 2622 3 395 3882 4 254 3892 5 394 4842 6 246 4242 7 241 3492 8 258 2802 9 208 2942 10 194 2643 1 366 3393 2 275 2853 3 364 3683 4 222 3943 5 386 4913 6 270 430

5、3 7 243 3613 8 270 2883 9 230 2963 10 196 2974 1 383 3414 2 248 2754 3 439 4364 4 246 3984 5 406 5014 6 229 3884 7 224 3504 8 266 2864 9 172 2884 10 181 265proc glm;class block t ;model y=block t x /solution ;means block t /Duncan;lsmeans block t / stderr pdiff;run;例3：研究3种蛋白质饲料的营养价值，实验将同种系的24只白鼠随机分为

6、3组（Group），每组8只，记录白鼠的初始周龄（Age）进食量（Food）与体重增重（Weight）,试分析这几种饲料的增重效果是否相同。序号Group 1Group 2Group 3agefoodweightagefoodweightagefoodweight16281.7375309.8248259.382210274.0476317.8435241.26638253.83710326.1606248.57445261.4348322.1507242.87957272.8427323.5478255.78265272.2276321.2427254.37676272.3325311.83

7、95244.67387293.24410324.55310243.890data;do i=1 to 8; do group=1 to 3; input age food weight ; output; end;end;cards; 6 281.7 37 5 309.8 24 8 259.3 8210 274.0 47 6 317.8 43 5 241.2 66 8 253.8 37 10 326.1 60 6 248.5 74 5 261.4 34 8 322.1 50 7 242.8 79 7 272.8 42 7 323.5 47 8 255.7 82 5 272.2 27 6 321

8、.2 42 7 254.3 76 6 272.3 32 5 311.8 39 5 244.6 73 7 293.2 44 10 324.5 53 10 243.8 90;proc means; var age food weight; class group;proc glm; class group; model weight=group age food/solution; means group/duncan; lsmeans group/stderr pdiff;run;结果：协变量AGE df=1 回归平方和SS=868.74897252 均方MS=868.74897252 误差均方

9、MSe=15.03387498 F=57.79 Pr F = 0.0001AGE的回归系数=3.7376协变量FOOD df=1 回归平方和SS=69.15070809 均方MS=69.15070809 误差均方MSe=15.03387498 F=4.60 Pr F = 0.0451FOOD 的回归系数=0.22054回归方程： = ()()Group1：39.971237.503.7376(6.756.958)0.22054(272.675280.35)Group2：35.471044.753.7376(7.1256.958)0.22054(319.60280.35)Group3：84.55

10、7777.753.7376(7.006.958)0.22054(248.775280.35)例4 下表为运动员(A)与大学生(S)的身高（cm）与肺活量（cm3）的数据，考虑到身高与肺活量有关，而一般运动员的身高高于大学生，为进一步分析肺活量的差异是否由于体育锻炼所致，试作控制身高变量的协方差分析。身份G身高H肺活量L身份G身高H肺活量LA184.94300S168.73450A167.93850S170.84100A1714100S1653800A1714300S169.73300A1884800S171.53450A1794000S166.53250A1775400S1653600A179

11、.54000S1653200A1874800S1733950A1874800S1694000A1694500S173.84150A1884780S1743450A176.73700S170.53250A1795250S1764100A1834250S169.53650A180.54800S176.33950A1795000S1633500A1783700S172.53900A1643600S1773450A1744050S1733850Data;Input G H L;Cards;A 184.9 4300A 167.9 3850A 171 4100A 171 4300A 188 4800A 1

12、79 4000A 177 5400A 179.5 4000A 187 4800A 187 4800A 169 4500A 188 4780A 176.7 3700A 179 5250A 183 4250A 180.5 4800A 179 5000A 178 3700A 164 3600A 174 4050S 168.7 3450S 170.8 4100S 165 3800S 169.7 3300S 171.5 3450S 166.5 3250S 165 3600S 165 3200S 173 3950S 169 4000S 173.8 4150S 174 3450S 170.5 3250S 1

13、76 4100S 169.5 3650S 176.3 3950S 163 3500S 172.5 3900S 177 3450S 173 3850;proc gplot; plot L*H=G; symbol1 i=r1 v=star; symbol2 i=r1 v=plus;proc reg; model L=H; by G;proc glm; class G; model L=G H /solution; means G/duncan; lsmeans H/stderr pdiff;run;一个协变量、二个因子的协方差分析例5某园艺家研究鲜花的种类（因子A：花种LP（a=1）和花种WB）a=2）和湿度（因子B：湿度低（b=1）和湿度高（b