# Homework Help: Least Squares Regression

1. Jun 28, 2010

1. The problem statement, all variables and given/known data

[PLAIN]http://img683.imageshack.us/img683/4744/leastsquares.jpg [Broken]
[PLAIN]http://img149.imageshack.us/img149/4793/graphwd.jpg [Broken]

2. Relevant equations

3. The attempt at a solution

So would these be the points?
(-41,51),(-22,62),(23,63),(44,24)

I'm not too sure how to evaluate the sigmas because it shows n as being the subscript up above in the directions but in the problem it shows i as being the subscript, and I don't understand how i can equal 1. When clearly the subscript is not always 1.

Last edited by a moderator: May 4, 2017
2. Jun 28, 2010

### LCKurtz

Your xi's are x1=-4, x2=-2, x3= 2, x4= 4 and your yi's are the corresponding y values. i goes from 1 to 4.

3. Jun 28, 2010

So I would have...

nc+0b+40a=19
0c+40b+144a=-12
40c+144b+544a=160

Is this correct for the system of equations? What would n be?

Last edited: Jun 28, 2010
4. Jun 28, 2010

### LCKurtz

They look correct. n is the number of points.

5. Jun 29, 2010

The system of equations I have created is wrong. I started to solve the system of equations

4c+0b+40a=19
0c+40b+144a=-12
40c+144b+544a=160

I used linear combinations to start to solve it and I got -9\52 for b. The equation for the regression is suppose to be
y=(-5/14)x2-(3/10)x+(41/6) and since the b I got doesn't match up with the b in the answer if I solved for a and c those values would also be wrong. I know I'm solving the system correctly because I checked my solution with a matrix on my calculator and my solution matches the one calculated using matrices.

What's wrong with my system of equations?

Last edited: Jun 29, 2010
6. Jun 29, 2010

### ehild

I got the same result as you. Try to plug in the x values into the equation given as solution. You will see, they do not match to the given curve. Try the same with your parameters.

ehild

7. Jun 30, 2010

### TheoMcCloskey

Check your term for "sum-of-x-cubes" again. Since the ordinates x_i are equally disposed about zero, this term should be zero as well. I have not check any other values.

8. Jun 30, 2010

### ehild

Oops... That is right.

ehild

9. Jun 30, 2010

ohhh... ok, thanks. haha that was a stupid mistake.

4c+0b+40a=19
0c+40b+0a=-12
40c+0b+544a=160

10. Jul 1, 2010

I'm stuck on another one of these problems. This time I have the system of equations and I know It's right, but I'm not sure how I can solve it.

4c+9b+29a=20
9c+29b+99a=70
29c+99b+353a=254

I know that the solution is a=1 b=-1 c=0 but I'm not sure how to solve the equation to get those variables.

I'm trying to get it into row echelon form. The only thing that I can do to eliminate a vairable is subtract the third equation from 11 times the first equation. That yields:

-11c+0b+63a=254
9c+29b+99a=70
29c+99b+353a=254

I don't see how I can eliminate any other variables.

11. Jul 3, 2010

### ehild

You should do Gauss elimination systematically, but before it, try to make the equations a bit more "friendly". At start, subtract 2 times the first equation from the second, and 3 times the second equation from the third.

ehild