怎么对如下的数据进行最小二乘拟合

0.62 525.75
0.62 525.78
0.62 525.81
0.62 525.88
0.62 525.94
0.62 526.00
0.62 526.06
0.62 526.19
0.62 526.25
0.62 526.28
0.54 526.31
0.54 526.38
0.54 526.44
0.54 526.50
0.54 526.56
0.54 526.69
0.54 526.72
0.62 525.75
0.62 525.78
0.62 525.81
0.62 525.88
0.62 525.94
0.62 526.00
0.62 526.06
0.62 526.19
0.62 526.25
0.62 526.28
0.54 526.31
0.54 526.38
0.54 526.44
0.54 526.50
0.54 526.56
0.54 526.69
0.54 526.72
数据是这样的，那么怎么用最小方差无偏估计量，不用polyfit函数，进行最小二乘拟合呢？请高手指点啊

不用matlab里的polyfit函数，可以用spss 之类的软件

POLYFIT Fit polynomial to data.
P = POLYFIT(X,Y,N) finds the coefficients of a polynomial P(X) of
degree N that fits the data Y best in a least-squares sense. P is a
row vector of length N+1 containing the polynomial coefficients in
descending powers, P(1)*X^N + P(2)*X^(N-1) +...+ P(N)*X + P(N+1).

[P,S] = POLYFIT(X,Y,N) returns the polynomial coefficients P and a
structure S for use with POLYVAL to obtain error estimates for
predictions. S contains fields for the triangular factor (R) from a QR
decomposition of the Vandermonde matrix of X, the degrees of freedom
(df), and the norm of the residuals (normr). If the data Y are random,
an estimate of the covariance matrix of P is (Rinv*Rinv')*normr^2/df,
where Rinv is the inverse of R.

[P,S,MU] = POLYFIT(X,Y,N) finds the coefficients of a polynomial in
XHAT = (X-MU(1))/MU(2) where MU(1) = MEAN(X) and MU(2) = STD(X). This
centering and scaling transformation improves the numerical properties
of both the polynomial and the fitting algorithm.

Warning messages result if N is >= length(X), if X has repeated, or
nearly repeated, points, or if X might need centering and scaling.

Class support for inputs X,Y:
float: double, single

用polyfit（）语句可以 polyfit(x,y,拟合次数n); x,y是你的数据，n是你要进行几次拟合，填1的话为一次也就是最小二乘法拟合 你可以这样写 x=[ 0.25 0.5 0.75 1 1.5 2 2.5 3 3.5 4 4.5 5 6 7 8 9 10 11 12 13 14 15 16]; y=[30 68 75 82 82 77 68 68 58 51 50 41 38 35 28 25 18 15 12 10 7 7 4]; p=polyfit(x,y,1); plot(x,y,'b',x,polyval(p,x),'r'); %如果作图的话，蓝线为原数据，红线为拟合数据 不知道对你有没有帮助