Assignment 31

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17:00:24

Query problem 7.7.4 points (1,0), (2,0), (3,0), (3,1), (4,1), (4,2), (5,2), (6,2)

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17:00:35

Give the equation of the least squares regression line and explain how you obtained the equation.

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sum xi = 28

sum yi = 8

sum xiyi = 37

sum xi^2 = 116

a = (8(37) - 28(8)) / ((8)116 - 28^2) = 1/2

b = 1/8(8-1/2(28)) = -3/4

y = 1/2x - 3/4

confidence assessment: 2

You need to solve this problem using the methods of the course, as opposed to the standard formulas for calculating the regression line. From your work on the subsequent problems it seems clear that you know how to do this. For your reference:

** The text gives you equations related to the sum of the x terms, sum of y values, sum of x^2, sum of y^2 etc, into which you can plug the given information.

To use partial derivatives and get the same results. The strategy is to assume that the equation is y = a x + b and write an expression for the sum of the squared errors, then minimize this expression with respect to a and b, which are treated as variables.

If y = a x + b then the errors at the four points are respectively

| (a * 1 + b) - 0 |,

| (a * 2 + b) - 0 |,

| (a * 3 + b) - 0 |,

| (a * 3 + b) - 1 |,

| (a * 4 + b) - 1 |,

| (a * 4 + b) - 2 |,

| (a * 5 + b) - 2 |, and

| (a * 6 + b) - 2 |.

The sum of the squared errors is therefore

sum of squared errors: ( (a * 1 + b) - 0 )^2+( (a * 2 + b) - 0 )^2+( (a * 3 + b) - 0 )^2+( (a * 3 + b) - 1 )^2+( (a * 4 + b) - 1 )^2+( (a * 4 + b) - 2 )^2+( (a * 5 + b) - 2 )^2+( (a * 6 + b) - 2 )^2.

It is straightforward if a little tedious to simplify this expression, but after simplifying all terms, squaring and then collecting like terms we get

116•a^2 + 2•a•(28•b - 37) + 8•b^2 - 16•b + 14.

We minimize this expression by finding the derivatives with respect to a and b:

The derivatives of this expression with respect to a and b are respectively

56•a + 16•b - 16 and 232•a + 56•b - 74.

Setting both derivatives equal to zero we get the system

56•a + 16•b - 16 = 0

232•a + 56•b - 74 = 0.

Solving this system for a and b we get

a = 1/2, b = - 3/4.

So see that this is a minimum we have to evaluate the expression f_aa * f_bb - 4 f_ab^2.

f_aa = 56 and f_bb = 56, while f_ab = 0 so f_aa * f_bb - 4 f_ab^2 is positive, telling us we have a minimum.

Thus our equation is

y = a x + b or

y = 1/2 x - 3/4. *&*&

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Good. There are some additional details in the following, for your reference:

** If y = a x + b then the errors at the four points are respectively

| (a * -3 + b) - 0 |,

| (a * -1 + b) - 1 |,

| (a * 1 + b) - 1 | and

| (a * 3 + b) - 2 |. The sum of the squared errors is therefore

( (a * -3 + b) - 0 )^2 + ( (a * -1 + b) - 1 )^2 + ( (a * 1 + b) - 1 )^2 + ( (a * 3 + b) - 2 )^2 =

[ 9 a^2 - 6 ab + b^2 ] + [ (a^2 - 2 a b + b^2) - 2 ( -a + b) + 1 ] + [ a^2 + 2 ab + b^2 - 2 ( a + b) + 1 ] + [ 9 a^2 + 6 ab + b^2 - 4 ( 3a + b) + 4 ] =

20•a^2 - 12•a + 4•b^2 - 8•b + 6.

This expression is to be minimized with respect to variables a and b.

The derivative with respect to a is 40 a - 12 and the derivative with respect to b is 8 b - 8.

40 a - 12 = 0 if a = 12/40 = .3.

8b - 8 = 0 if b = 1.

The second derivatives with respect to and and b are both positive; the derivative with respect to a then b is zero. So the test for max, min or saddle point yields a max or min, and since both derivatives are positive the critical point gives a min. **

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Good work overall. See my notes.

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Let me know if you have questions. &#