sas习题大全带程序编码

1、P265 1今有某种型号的电池三批，它们分别是A、B、C三个工厂所生产的，为评比其质量，各随机抽取5只电池为样品，经试验得其寿命（h）如下：ABC40 4248 453826 2834 323039 5040 5043试在显著性水平0.05下检验电池的平均寿命有无显著的差异，若差异是显著的，试求均差A-B，A-C和B-C的置信水平为95%的置信区间。代码：data l1; do b=1 to 5;do a=1 to 3;input x;output;end;end;cards;40 26 39 42 28 50 48 34 40 45 32 50 38 30 43proc anova;clas

2、s a;model x=a;run;结果输出：The SAS System 19:15 Friday, April 9, 2012 5 The ANOVA Procedure Class Level Information Class Levels Values a 3 1 2 3 Number of observations 15The SAS System 19:15 Friday, April 9, 2012 6 The ANOVA ProcedureDependent Variable: x Sum of Source DF Squares Mean Square F Value Pr

3、 > F Model 2 615.6000000 307.8000000 17.07 0.0003 Error 12 216.4000000 18.0333333 Corrected Total 14 832.0000000 R-Square Coeff Var Root MSE x Mean 0.739904 10.88863 4.246567 39.00000 Source DF Anova SS Mean Square F Value Pr > F a 2 615.6000000 307.8000000 17.07 0.0003结论：结论：在显著水平为0.05下0.0003&

4、lt;0.05,所以各个总体均值间有显著差异。代码：data l1;p265 1 (ua-ub)input lei n;do rep= 1 to n;input x;output;end;cards;1 540 42 48 45 382 526 28 34 32 30;proc ttest;class lei;var x;run;结果输出：The SAS System 19:15 Friday, April 9, 2012 25 The TTEST Procedure Statistics Lower CL Upper CL Lower CL Upper CL Variable lei N M

5、ean Mean Mean Std Dev Std Dev Std Dev Std Err x 1 5 37.664 42.6 47.536 2.3815 3.9749 11.422 1.7776 x 2 5 26.074 30 33.926 1.8946 3.1623 9.087 1.4142 x Diff (1-2) 7.3618 12.6 17.838 2.426 3.5917 6.8808 2.2716 T-Tests Variable Method Variances DF t Value Pr > |t| x Pooled Equal 8 5.55 0.0005 x Satt

6、erthwaite Unequal 7.62 5.55 0.0006 Equality of Variances Variable Method Num DF Den DF F Value Pr > F x Folded F 4 4 1.58 0.6685 代码：data l1;（p265 1 ub-uc）input lei n;do rep= 1 to n;input x;output;end;cards;1 526 28 34 32 302 539 50 40 50 43;proc ttest;class lei;var x;run;结果输出：The SAS System 19:08

7、 Friday, April 23, 2012 1 The TTEST Procedure Statistics Lower CL Upper CL Lower CL Upper CL Variable lei N Mean Mean Mean Std Dev Std Dev Std Dev Std Err x 1 5 26.074 30 33.926 1.8946 3.1623 9.087 1.4142 x 2 5 37.795 44.4 51.005 3.1873 5.3198 15.287 2.3791 x Diff (1-2) -20.78 -14.4 -8.018 2.9558 4.

8、3761 8.3835 2.7677 T-Tests Variable Method Variances DF t Value Pr > |t| x Pooled Equal 8 -5.20 0.0008 x Satterthwaite Unequal 6.51 -5.20 0.0016 Equality of Variances Variable Method Num DF Den DF F Value Pr > F x Folded F 4 4 2.83 0.3378 代码：data l1;(p265 1 ua-uc)input lei n;do rep= 1 to n;inp

9、ut x;output;end;cards;1 540 42 48 45 382 539 50 40 50 43;proc ttest;class lei;var x;run;结果输出：The SAS System 19:15 Friday, April 9, 2012 28 The TTEST Procedure Statistics Lower CL Upper CL Lower CL Upper CL Variable lei N Mean Mean Mean Std Dev Std Dev Std Dev Std Err x 1 5 37.664 42.6 47.536 2.3815

10、3.9749 11.422 1.7776 x 2 5 37.795 44.4 51.005 3.1873 5.3198 15.287 2.3791 x Diff (1-2) -8.648 -1.8 5.0485 3.1718 4.6957 8.996 2.9698 T-Tests Variable Method Variances DF t Value Pr > |t| x Pooled Equal 8 -0.61 0.5613 x Satterthwaite Unequal 7.41 -0.61 0.5626 Equality of Variances Variable Method

11、Num DF Den DF F Value Pr > F x Folded F 4 4 1.79 0.5862结论：在置信水平为95%的置信区间。ua-uc、ua-ub、ub-uc分别为（-7.65，4.05）、（6.75，18.45）、（-20.25，-8.55）.p 265 2为了寻找飞机控制面板上仪器表的最佳布置，试验了三个方案，观察领航在紧急情况的反应时间（以1/10秒记），随机地选择28名领航员，得到他们对于不同的布局方案的反应时间如下：方案14 13 9 15 11 13 14 11方案10 12 7 11 8 12 9 10 13 9 10 9方案11 5 9 10 6 8

12、 8 7试在显著性水平0.05下检验各个方案的反应时间有无显著的差异，若有差异，试求1-2，1-3，2-3的置信水平为0.95的置信区间。代码:data l1;(p 265 2)input type$ n;do i=1 to n;input x;output;end;cards;M1 814 13 9 15 11 13 14 11M2 1210 12 7 11 8 12 9 10 13 9 10 9M3 811 5 9 10 6 8 8 7;proc anova;class type;model x = type;run;结果输出：The SAS System 19:10 Friday, Ap

13、ril 16, 2012 1 The ANOVA Procedure Class Level Information Class Levels Values type 3 M1 M2 M3 Number of observations 28The SAS System 19:10 Friday, April 16, 2012 2 The ANOVA ProcedureDependent Variable: x Sum of Source DF Squares Mean Square F Value Pr > F Model 2 81.4285714 40.7142857 11.31 0.

14、0003 Error 25 90.0000000 3.6000000 Corrected Total 27 171.4285714 R-Square Coeff Var Root MSE x Mean 0.475000 18.70643 1.897367 10.14286 Source DF Anova SS Mean Square F Value Pr > F type 2 81.42857143 40.71428571 11.31 0.0003结论：在显著水平为0.05下0.0003<0.05，所以各个方案的反应时间有着明显的差异。代码：data l1;（p265 2 u1-u

15、2）input lei n;do rep= 1 to n;input x;output;end;cards;1 814 13 9 15 11 13 14 112 1210 12 7 11 8 12 9 10 13 9 10 9;proc ttest;class lei;var x;run;结果输出：The SAS System 19:08 Friday, April 23, 2012 2 The TTEST Procedure Statistics Lower CL Upper CL Lower CL Upper CL Variable lei N Mean Mean Mean Std Dev

16、 Std Dev Std Dev Std Err x 1 8 10.828 12.5 14.172 1.3223 2 4.0705 0.7071 x 2 12 8.883 10 11.117 1.2454 1.7581 2.985 0.5075 x Diff (1-2) 0.7203 2.5 4.2797 1.4024 1.8559 2.7446 0.8471 T-Tests Variable Method Variances DF t Value Pr > |t| x Pooled Equal 18 2.95 0.0085 x Satterthwaite Unequal 13.7 2.

17、87 0.0125 Equality of Variances Variable Method Num DF Den DF F Value Pr > F x Folded F 7 11 1.29 0.6738结论：u1-u2，u1-u3，u2-u3的置信水平为0.95的置信区间为（0.72，4.28），（2.55，6.45），（0.22，3）代码：data l1;(p265 2 u1-u3)input lei n;do rep= 1 to n;input x;output;end;cards;1 814 13 9 15 11 13 14 112 811 5 9 10 6 8 8 7;pr

18、oc ttest;class lei;var x;run;结果输出：The SAS System 19:08 Friday, April 23, 2012 3 The TTEST Procedure Statistics Lower CL Upper CL Lower CL Upper CL Variable lei N Mean Mean Mean Std Dev Std Dev Std Dev Std Err x 1 8 10.828 12.5 14.172 1.3223 2 4.0705 0.7071 x 2 8 6.328 8 9.672 1.3223 2 4.0705 0.7071

19、x Diff (1-2) 2.3552 4.5 6.6448 1.4643 2 3.1542 1 T-Tests Variable Method Variances DF t Value Pr > |t| x Pooled Equal 14 4.50 0.0005 x Satterthwaite Unequal 14 4.50 0.0005 Equality of Variances Variable Method Num DF Den DF F Value Pr > F x Folded F 7 7 1.00 1.0000结论：代码：data l1;(p265 2 u2-u3)i

20、nput lei n;do rep= 1 to n;input x;output;end;cards;1 1210 12 7 11 8 12 9 10 13 9 10 92 811 5 9 10 6 8 8 7;proc ttest;class lei;var x;run;结果输出：The SAS System 19:08 Friday, April 23, 2012 4 The TTEST Procedure Statistics Lower CL Upper CL Lower CL Upper CL Variable lei N Mean Mean Mean Std Dev Std Dev

21、 Std Dev Std Err x 1 12 8.883 10 11.117 1.2454 1.7581 2.985 0.5075 x 2 8 6.328 8 9.672 1.3223 2 4.0705 0.7071 x Diff (1-2) 0.2203 2 3.7797 1.4024 1.8559 2.7446 0.8471 T-Tests Variable Method Variances DF t Value Pr > |t| x Pooled Equal 18 2.36 0.0297 x Satterthwaite Unequal 13.7 2.30 0.0378 Equal

22、ity of Variances Variable Method Num DF Den DF F Value Pr > F x Folded F 7 11 1.29 0.6738结论：p265 3某防治站对4个林场的松毛虫密度进行调查，每个林场调查5块地得资料如下表：地点松毛虫密度（头/标准地）A1192 189 176 185 190A2190 201 187 196 200A3188 179 191 183 194A4187 180 188 175 182判断4个林场松毛从密度有无显著差异，取显著性水平=0.05.代码：data l1;（p265 3）do b=1 to 5;do a

23、=1 to 4;input x;output;end;end;cards;192 190 188 187 189 201 179 180 176 187 191 188 185 196 183 175 190 200 194 182proc anova;class a;model x=a;run;结果输出：The SAS System 19:15 Friday, April 9, 2012 7 The ANOVA Procedure Class Level Information Class Levels Values a 4 1 2 3 4 Number of observations 20

24、The SAS System 19:15 Friday, April 9, 2012 8 The ANOVA ProcedureDependent Variable: x Sum of Source DF Squares Mean Square F Value Pr > F Model 3 403.3500000 134.4500000 3.77 0.0321 Error 16 571.2000000 35.7000000 Corrected Total 19 974.5500000 R-Square Coeff Var Root MSE x Mean 0.413883 3.184091

25、 5.974948 187.6500 Source DF Anova SS Mean Square F Value Pr > F a 3 403.3500000 134.4500000 3.77 0.0321结论：P265 4一试验用来比较4种不同药品解除外科手术后疼痛的延长时间（h），结果如下表：药品时间长度（h）A8642B6644C810101012D442试在显著性水平=0.05下检验各种药品对解除疼痛的延续时间有无显著差异。代码：data l1;(P265 4)input type$ n;do i=1 to n;input x;output;end;cards;M1 48 6

26、4 2M2 46 6 4 4 M3 58 10 10 10 12M4 34 4 2;proc anova;class type;model x = type;run;结果输出：The SAS System 19:15 Friday, April 9, 2012 35 The ANOVA Procedure Class Level Information Class Levels Values type 4 M1 M2 M3 M4 Number of observations 16The SAS System 19:15 Friday, April 9, 2012 36 The ANOVA Pr

27、ocedureDependent Variable: x Sum of Source DF Squares Mean Square F Value Pr > F Model 3 108.3333333 36.1111111 12.50 0.0005 Error 12 34.6666667 2.8888889 Corrected Total 15 143.0000000 R-Square Coeff Var Root MSE x Mean 0.757576 27.19477 1.699673 6.250000 Source DF Anova SS Mean Square F Value P

28、r > F type 3 108.3333333 36.1111111 12.50 0.0005结论：P266 5将抗生素注入人体会产生抗生素与血浆蛋白质结合的现象。以致减少了药效，下表列出5种常用的抗生素注入到牛的体内时，抗生素与血浆蛋白质结合的百分比。试在水平=0.05下检验这些百分比的均值有无显著的差异。青霉素四环素链霉素红霉素氯霉素29.627.35.821.629.224.332.66.217.432.828.530.811.018.325.032.034.88.319.024.2代码：data l1;do b=1 to 4;do a=1 to 5;input x;output

29、;end;end;cards;29.6 27.3 5.8 21.6 29.2 24.3 32.6 6.2 17.4 32.8 28.5 30.8 11.0 18.3 25.0 32.0 34.8 8.3 19.0 24.2proc anova;class a;model x=a;run;结果输出：The SAS System 19:15 Friday, April 9, 2012 9 The ANOVA Procedure Class Level Information Class Levels Values a 5 1 2 3 4 5 Number of observations 20The

30、 SAS System 19:15 Friday, April 9, 2012 10 The ANOVA ProcedureDependent Variable: x Sum of Source DF Squares Mean Square F Value Pr > F Model 4 1480.823000 370.205750 40.88 <.0001 Error 15 135.822500 9.054833 Corrected Total 19 1616.645500 R-Square Coeff Var Root MSE x Mean 0.915985 13.12023 3.009125 22.93500 Source DF Anova SS Mean Square F Value Pr > F a 4 1480.823000 370.205750 40.88 <.0001结论：P266 6下表给出某种化工过程在三种浓度、四种温度水平下得率的数据：温度（因素B）10243852浓度（因素A）2%14 1011 1113 910 124% 9 710 8 7 11 6 106% 5 1113 1412 1314 10试在显著性