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1、1、考察温度x对产量y的影响,测得下列10组数据:x=20253035404550556065y=13.215.116.417.117.918.719.621.222.524.3求y关于x的线性回归方程,检验回归效果是否显著,并预测x=42℃时产量的估值及预测区间(置信度95%).解答:法①:回归:x=20:5:65;X=[ones(10,1)x'];Y=[13.215.116.417.117.918.719.621.222.524.3]';[b,bint,r,rint,stats]=regress(Y,X);b,bint,stats结果:b=9.12120.2230bint=8.
2、021110.22140.19850.2476stats=0.9821439.83110.00000.2333即=9.1212,=0.2230;的置信区间为[8.021110.2214],的置信区间为0.19850.2476;r2=0.9281,F=439.8311,p=0.0000p<0.05,可知回归模型y=9.1212+0.2230x成立.法②:拟合:x=20:5:65;y=[13.215.116.417.117.918.719.621.222.524.3];[A,B]=polyfit(x,y,1)Y=polyval(A,42)[Y,DELTA]=polyconf(A,42,
3、B,0.05)结果:A=0.22309.1212B=R:[2x2double]df:8normr:1.3660Y=18.4885Y=18.4885DELTA=1.1681在42度时的预估计值为Y=18.4885,Y的显著性为1-0.05,其置信区间为18.4885+1.1681,18.4885-1.1681.1、某零件上有一段曲线,为了在程序控制机床上加工这一零件,需要求这段曲线的解析表达式,在曲线横坐标xi处测得纵坐标yi共11对数据如下:x=02468101214161820y=0.62.04.47.511.817.123.331.239.649.761.7求这段曲线的纵坐标y
4、关于横坐标x的二次多项式回归方程法①:拟合:x=0:2:20;y=[0.62.04.47.511.817.123.331.239.649.761.7];a=polyfit(x,y,2)结果:a=0.14030.19711.0105既有方程:y=0.1403*x^2+0.1971*x+1.0105法②:回归:x=0:2:20;X=[ones(11,1)x'(x.^2)'];Y=[0.62.04.47.511.817.123.331.239.649.761.7]';[b,bint,r,rint,stats]=regress(Y,X);b结果:b=1.01050.19710.1403既有
5、方程:y=0.1403*x^2+0.1971*x+1.0105法③:非线性:functionf=tier03(beta,x)f=beta(1)*x.^2+beta(2)*x+beta(3);后x=0:2:20;y=[0.62.04.47.511.817.123.331.239.649.761.7];beta0=[123]';[beta,r,J]=nlinfit(x',y','tier03',beta0);beta结果:beta=0.14030.19711.0105既有方程:y=0.1403*x^2+0.1971*x+1.01053、在研究化学动力学反应过程中,建立了一个反应速度和反
6、应物含量的数学模型,形式为其中是未知参数,是三种反应物(氢,n戊烷,异构戊烷)的含量,y是反应速度.今测得一组数据如表4,试由此确定参数,并给出置信区间.的参考值为(1,0.05,0.02,0.1,2).序号反应速度y氢x1n戊烷x2异构戊烷x318.554703001023.79285801034.8247030012040.024708012052.754708010614.391001901072.54100806584.3547019065913.0010030054108.50100300120110.05100801201211.3228530010133.132851
7、901201.建立M文件:functionf=tisan(beta,x)x1=x(:,1);x2=x(:,2);x3=x(:,3);f=(beta(1)*x2-x3/beta(5))./(1+beta(2)*x1+beta(3)*x2+beta(4)*x3);2.输入数据:x=[470300102858010470300120470801204708010100190101008065470190651003005410030012010080120285300102