数学建模-统计与回归

1、统计与回归1、考察温度x对产量y的影响，测得下列10组数据:温度C）20253035404550556065产量（kg）13.215.116.417.117.918.719.621.222.524.3求：(1) y关于x的线性回归方程，检验回归效果是否显著；(2) 并预测x=42 C时产量的估计值；(3) 预测x=42 C时产量置信度为 95%的预测区间(请参考本课件中多项式回归polyfit 与polyco nf，或非线性拟合命令nlinfit 或nlpredci实现区间预测)Matlab 程序：x=20 25 30 35 40 45 50 55 60 65;X=ones(10,1) x&#

2、39;y=13.2 15.1 16.4 17.1 17.9 18.7 19.6 21.2 22.5 24.3;b,b in t,r,ri nt,stats=regress(y',X);b,statsrstool(x',y','li near' )p,S=polyfit(x,y,1);Y,DELTA=polyco nf(p,x,S);plot(x,y, 'k+',x,Y,'r')Y,DELTA=polyco nf(p,42,S)b =9.12120.2230stats =0.9821 439.83110.00000.2333

3、Y =18.4885DELTA =1.1681>>Q Rredictibfi Plot of Linear ModelFile Edit View nsert Tools Desktop Window HelpPrMicted Y1佩 4885*0.4S637253036；EwtLinearCbs-g(1 ) y关于x的线性回归方程为 y=9.1212+0.223x,=0.9821, p=0.0000<0.05 ，所以回归方程成立，回归效果显著。(2 )预测 x=42 C 时产量为 18.4885( kg ).(3 )预测x=42 C时产量置信度为 95%的预测区间为(17.

4、3204,19.6566)2.某人记录了 21天每天使用空调器的时间和使用烘干器的次数，并监视电表以计算出每天的 耗电量，数据见下表，试研究耗电量(KWH，记作y)与空调器使用的小时数(AC，记作x1)和烘干器使用次数(DRYER，记作x2)之间的关系：(1)建立y与x1、x2之间的线性回归模型，并分析模型效果的显著性；(2 )如有必要，考虑引入非线性项(平方项x12, x22以及交叉项x1*x2 )，建立新的回归模型；(3)分析模型中新引入的非线性项是否都是必要的，若不是，请去掉多余项，建立新的模型， 并分析新模型的效果。(1 )x1=1.5 4.5 5.0 2.0 8.5 6.0 13.5

5、 8.0 12.5 7.5 6.5 8.0 7.5 8.0 7.5 12.0 6.0 2.5 5.0 7.5 6.0;x2=1 2 2 0 3 3 1 1 1 2 3 1 2 2 1 1 0 3 0 1 0;y=35 63 66 17 94 79 93 66 94 82 78 65 77 75 62 85 43 57 33 65 33' x=ones(21,1) x1' x2'b,b in t,r,ri nt,stats=regress(y,x);rcoplot(r,ri nt)x(21,:)=;y(21,:)=;b,b in t,r,ri nt,stats=regre

6、ss(y,x);rcoplot(r,ri nt)b,b in t,r,ri nt,stats=regress(y,x);b,bi nt,statsm iTodsJfflindow Helpm y 口Bfindcw £|elpRc-srdual Case Onfef PloflCase NumtwrRc-sidual Case Oidef Plofl6 S 1012 M 1519 Z0Case NutterEie £dit 里iew Dnsert leak Rwlrtqp d為h哄觀逗必-o£序号1234567891011KWH3563661794799366948

7、278AC1.54.55.02.08.56.013.58.012.57.56.5DRYER12203311123序号12131415161718192021kWH65777562854357:336533AC8.07.58.07.512.06.02.55.07.56.0DRYER1221103010b =9.79665.416012.5843bint =4.9528 14.64044.89125.940910.8997 14.2690stats =0.9759 343.87650.0000 12.0793>>y与x1、x2之间的线性回归模型为y 二 9.7966 5.416x1 1

8、2.5843x2, r2 二 0.9759 ,p=0.0000<0.05，回归方程成立，回归模型显著。(2)x1=1.5 4.5 5.0 2.0 8.5 6.0 13.5 8.0 12.5 7.5 6.5 8.0 7.5 8.0 7.5 12.0 6.02.5 5.0 7.5 6.0;x2=1 2 2 0 3 3 1 1 1 2 3 1 2 2 1 1 0 3 0 1 0;y=35 63 66 17 94 79 93 66 94 82 78 65 77 75 62 85 43 57 33 65 33'x=x1' x2'rstool(x,y, 'li nea

9、r' )Variables have bee n created in the curre nt workspace.beta,rmsebeta =8.10545.465913.2166rmse =3.9354故回归模型为：y =8.1054 5.4659x1 13.2166x2，剩余标准差为3.9354rstool(x,y, 'purequadratic')Variables have bee n created in the curre nt workspace.beta1,rmse1beta1 =7.06895.117321.29840.0000-2.6562rm

10、se1 =3.1943故回归模型为：y =7.0689,5.1173x, - 21.2984% -2.6562x；，剩余标准差为 3.1943rstool(x,y,'in teracti on')Variables have bee n created in the curre nt workspace.beta2,rmse2beta2 =8.70725.354712.65900.0984rmse2 =4.0427故回归模型为：y =8.7072 5.3547 12.6590x2 0.0984x,x2 ,剩余标准差为 4.0427rstool(x,y, 'quadrat

11、ic' )Variables have bee n created in the curre nt workspace.beta3,rmse3beta3 =9.47894.556319.91280.31360.0129-2.7739rmse3 =3.2108故回归模型为：y =9.4789 4.5563% 19.912&2 0.3136x； 0.0129X2 - 2.7739x；, 剩余标准差为3.2108(3 ) x1=1.5 4.5 5.0 2.0 8.5 6.0 13.5 8.0 12.5 7.5 6.5 8.0 7.5 8.0 7.5 12.06.0 2.5 5.0 7

12、.5 6.0;x2=1 2 2 0 3 3 1 1 1 2 3 1 2 2 1 1 0 3 0 1 0;y=35 63 66 17 94 79 93 66 94 82 78 65 77 75 62 85 43 57 33 65 33'X=ones(21,1) x1' x2' (x1.A2)' (x1.*x2)' (x2.A2)' stepwise(X,y)x=o nes(21,1) x1' x2' (x2.A2)'b,b in t,r,ri nt,stats=regress(y,x);rcoplot(r,ri nt)x(1

13、7,:)=;y(17,:)=;b,b in t,r,ri nt,stats=regress(y,x);rcoplot(r,ri nt)x(14,:)=;y(14,:)=;b,b in t,r,ri nt,stats=regress(y,x);rcoplot(r,ri nt)b,bi nt,statsb =5.29135.091824.2141-3.4016bint =1.58798.99474.66695.516619.5268 28.9013-4.8329 -1.9702stats =0.9892 459.75630.0000 6.4484>>AAAAA即 s =5.2913,打=5.0918,匕=24.21413 =3.4016 ； 'o 的置信区间为1.5879,8.9947,的置信区间为4.6669,5.5166,匕的置信区间为19.5268,28.9013,匕的置信区间为-4.8329,-1.97022r =0.9892 , F