多项式回归分析R平方肿么求？ 用matlab肿么编程？

t=[0 4 8 9 10 18 21 23 24 25 29 37 38 39 40 41 44 46];c=[85.90 133.54 152.33 173.28 149.00 160.00 170.40 206.85 181.79 172.48 184.67 267.67 167.50 213.76 196.24 182.59 164.60 101.33];[p,S,mu] =polyfit(t,c,2)结果为:p = -0.131131380060804 7.404136387597593 93.846288472422543分别对应beta2,beta1,beta0的系数具体参考doc polyfit

楼主的多项式回归是怎么定义的？

程序如下：t0=ones(18,1);t1=[0 4 8 9 10 18 21 23 24 25 29 37 38 39 40 41 44 46]';t2=t1.*t1;c=[85.90 133.54 152.33 173.28 149.00 160.00 170.40 206.85 181.79 172.48 184.67 267.67 167.50 213.76 196.24 182.59 164.60 101.33]';T=[t0 t1 t2];[b,bint,r,rint,s]=regress(c,T,0.05);disp(['c =' num2str(b(3)) '+' num2str(b(2)) 't+' num2str(b(1)) 't^2'])%输出方程b%方程系数bint%b的置信区间r%残差rint%r的置信区间s%检验回归模型的统计量（对应4个参数值：相关系数r^2，F值，与F对应的概率P， 误差估计）结果：c =-0.13113+7.4041t+93.8463t^2b = 93.8463 7.4041 -0.1311bint = 49.0695 138.6230 3.0080 11.8003 -0.2206 -0.0417r = -7.9463 12.1753 7.6430 23.4181 -5.7745 -24.6342 -21.1042 12.0771 -14.2239 -24.5126 -13.6148 79.3895 -18.3498 30.6032 16.0385 5.6060 -1.1579 -55.6326rint = -56.6714 40.7788 -46.0597 70.4102 -54.7131 69.9992 -38.1024 84.9386 -68.9365 57.3874 -86.0124 36.7440 -82.5767 40.3683 -50.1051 74.2593 -76.2751 47.8273 -85.5679 36.5427 -76.3083 49.0788 33.9153 124.8637 -81.4707 44.7712 -30.6668 91.8733 -46.5710 78.6479 -56.9761 68.1881 -60.5541 58.2383 -101.2318 -10.0333s = 0.5007 7.5203 0.0055 933.2755

clc,clear,clfx=[19.1,25.0,30.1 36 40 45.1 50 ];y=[76.3 77.8 79.75 80.8 82.35 83.9 85.1 ];plot(x,y,'or')title('电阻与温度之间的关系')n=length(x);B1=(n*sum(x.*y)-sum(x)*sum(y))/(n*sum(x.^2)-sum(x)^2);B0=mean(y)-B1*mean(x);sprintf('y=%f+%f*x',B0,B1) %y=70.762374+0.288069*x%%下面求R值，即求相关性Rfenzi=sum((x-mean(x)).*(y-mean(y)));Rfenmu=sqrt(sum((x-mean(x)).^2).*sum((y-mean(y)).^2));R=Rfenzi/Rfenmu %R =0.9978%%求拟合后的曲线hold on x=19:0.1:50;plot(x,B0+B1.*x)这是我之前在课堂上做的小作业，里面涉及到 R的求法，然后，你就可以求R^2了希望对你有所帮助