1. y=a+bx Linear Regression: Method of Least Squares slope y intercept
The Method of Least Squares is a procedure to determine the best fit line to data; the proof uses simple calculus and linear algebra. The basic problem is to find the best fit straight line y = a + bx given that, for n ϵ {1,…,N}, the pairs (xn; yn) are observed.
The form of the fitted curve is
Sum of squares of errors
slope
y=a+bx
y intercept

2. Example 1: Find a 1st order polynomial y=a+bx for the values given in the table.-5 -2 2 4 7
3.5
a=1.188 b=0.484
y= x
With Matlab:
clc;clear
x=[-5,2,4];
y=[-2,4,3.5];
p=polyfit(x,y,1)
x1=-5:0.01:7;
yx=polyval(p,x1);
plot(x,y,'ro',x1,yx,'b')
xlabel('x value')
ylabel ('y value')
Data point
Fitted curve

3. Example 2: x y 200 3 230 5 240 8 270 10 290 y=a+bx y=200.13 + 8.82x
200 3
230 5
240 8
270 10
290
y=a+bx
y= x
clc;clear
x=[0,3,5,8,10];
y=[200,230,240,270,290];
p=polyfit(x,y,1)
x1=-1:0.01:12;
yx=polyval(p,x1);
plot(x,y,'ro',x1,yx,'b')
xlabel('x value')
ylabel ('y value')
Data point
Fitted curve

4. Example 3: Intercept Slope
Tensile tests were performed for a composite material having a crack in order to calculate the fracture toughness. Obtain a linear relationship between the breaking load F and crack length a.
Method of Least Squares
Slope
Intercept

5. with Matlab: clc;clear x=[10,9.25,9.1,9.4,8.5];
y=[0.5,0.4,0.35,0.45,0.28];
p=polyfit(x,y,1)
F=8:0.01:12;
a=polyval(p,F);
plot(x,y,'ro‘,F,a,'b')
xlabel('x value')
ylabel ('y value')

6. Example 4:
The change in the interior temperature of an oven with respet to time is given in the Figure. It is desired to model the relationship between the temperature (T) and time (t) by a first order polynomial as T=c1t+c2. Determine the coefficients c1 and c2.
T (°C)
t (min.)
175
204
200
212
Slope
Intercept

7. clc;clear
x=[0,5,10,15];
y=[175,204,200,212];
p=polyfit(x,y,1)
t=0:0.01:15;
T=polyval(p,t);
plot(x,y,'ro',t,T,'b')
xlabel('x value')
ylabel ('y value‘)
with Matlab: