用Matlab作线性回归

怎样用matlab做出一元线性回归图形~

>> clear
>> x=[1656,2122,2864,4033,6099];
>> y=[2112,2170,2291,2456,2759];
>> [P,S]=polyfit(x,y,1)
%P为拟合回归系数即y=P(1)*x+p(2)
P =
1.0e+003 *
0.0001 1.8666
S =
R: [2x2 double]
df: 3
normr: 9.4223

>> [Y,delta]=polyconf(P,x,S)
%给出回归Y的95%的置信区间为[Y-delta，Y+delta]
Y =
1.0e+003 *

2.1090 2.1772 2.2858 2.4569 2.7592
delta =

20.6907 19.8923 19.1147 19.2502 23.2005
>> x1=1600:100:6100;
>> f=polyval(P,x1);
>> plot(x,y,'ro',x1,f,'-')%绘图查看拟合效果
>> hold on
>> plot(x,Y+delta,'*g')
>> plot(x,Y-delta,'*g')%给出拟合的置信区间
希望我的回答能帮助你！

regress(y',[ones(numel(x),1),x'])
ans =
7.9874e+02
1.0597e+01

方法二：
>> lm = LinearModel.fit(x',y')
lm =
Linear regression model:
y ~ 1 + x1

Estimated Coefficients:
Estimate SE tStat pValue
(Intercept) 798.74 79.626 10.031 1.5448e-06
x1 10.597 4.1075 2.5799 0.02742

Number of observations: 12, Error degrees of freedom: 10
Root Mean Squared Error: 117
R-squared: 0.4, Adjusted R-Squared 0.34
F-statistic vs. constant model: 6.66, p-value = 0.0274

画图(续方法二）:
plot(lm)

x=[
1.99 2.00 1.00;
11.43 14.76 12.86;
44.29 50.00 50.00;
72.86 81.43 75.71;
87.61 89.51 87.61;
93.33 92.86 94.29];

day=[1:6]'; %天数

%这里我想，应当是想得到发芽率与天数的关系，而跟组无关
%所以这里我将它们横向求平均。求出每天的平均出芽率

xx=sum(x,2)/size(x,2);

A=[day,ones(size(day))];
c=A\xx;

k=c(1);
b=c(2);
yy=k*day+b;

plot(day,xx,'r*',day,yy)
%红色的*点，是平均出芽率与天数的关系
%线是拟合出来的

%=====================================
%那我就拿第一组做试验
%实际上就是将xx=sum(x,2)/size(x,2)
%这句改成xx=x(:,1);
%第二组，你就改成xx=x(:,2);

x=[
1.99 2.00 1.00;
11.43 14.76 12.86;
44.29 50.00 50.00;
72.86 81.43 75.71;
87.61 89.51 87.61;
93.33 92.86 94.29];

day=[1:6]'; %天数
xx=x(:,1);
A=[day,ones(size(day))];
c=A\xx;

k=c(1);
b=c(2);
yy=k*day+b;

plot(day,xx,'r*',day,yy)

sorry偶不懂~~

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用Matlab作线性回归视频