Linear RegressionThis program fits a straight line to a given set of coordinates using the method of least squares ( linear regression ).
easier for all of us, but we'd like to show a step-by-step way to do it, to understand the inner concepts): function [y0, a, b, r2, r, k2] = lin_reg(x, y, x0) % Number of known points n = length(x); % Initialization j = 0; k = 0; l = 0; m = 0; r2 = 0; % Accumulate intermediate sums j = sum(x); k = sum(y); l = sum(x.^2); m = sum(y.^2); r2 = sum(x.*y); % Compute curve coefficients b = (n*r2 - k*j)/(n*l - j^2); a = (k - b*j)/n; % Compute regression analysis j = b*(r2 - j*k/n); m = m - k^2/n; k = m - j; % Coefficient of determination r2 = j/m; % Coefficient of correlation r = sqrt(r2); % Std. error of estimate k2 = sqrt(k/(n-2)); % Interpolation value y0 = a + b*x0; If we have the following data available (where every yi has its correspondent xi): x = [71 73 64 65 61 70 65 72 63 67 64]; y = [160 183 154 168 159 180 145 210 132 168 141]; we can call the function above in this manner (to obtain interpolated values for x = 70 and x = 72): [y0, a, b, r2, r, k2] = lin_reg(x, y, 70) [y0] = lin_reg(x, y, 72) And the Matlab response is: y0 = 176.5139 a = -106.7917 b = 4.0472 r2 = 0.5563 r = 0.7458 k2 = 15.4135 y0 = 184.6083 We can use the 'polyfit' and 'polyval' instructions in Matlab for this purpose, like this: a = polyfit(x, y, 1) y0 = polyval(a, 70) y0 = polyval(a, 72) Fitting a straight line through the data means thet we want to find the polynomial coefficients of a first order polynomial such that a1xi + a0 gives the best approximation for yi. We find the coefficients with 'polyfit' and evaluate any xi with 'polyval'. The Matlab results is a = 4.0472 -106.7917 y0 = 176.5139 y0 = 184.6083 confirming our previous numbers. From 'Linear Regression' to home From 'Linear Regression' to 'Matlab Cookbook'
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