#$&* course mth 272 Question: `qQuery problem 7.7.4 points (1,0), (2,0), (3,0), (3,1), (4,1), (4,2), (5,2), (6,2)Give the equation of the least squares regression line and explain how you obtained the equation.
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Given Solution: `a 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. STUDENT QUESTION I am very clear on this concept, but I am having some issues calculating my sum of the squared errors. Can you provide some clarity here; I am obviously confusing myself. Aside from that point, I am confident with the remaining steps here. INSTRUCTOR RESPONSE Your solution includes S = (1a + b - 0)^2 + (2a + b - 0)^2 + (3a + b - 0)^2 + (3a + b -1)^2 + (4a + b -1)^2 + (4a + b -2)^2 + (5 a + b - 2)^2 + (6 a + b - 2)^2, which seems to agree with the data and with my expression ( (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 You need to expand your squares. All it takes is the Distributive Law. (4a + b -1)^2 = (4a + b -1) * (4a + b -1) = 4a * (4a + b -1) + b * (4a + b -1) + (-1) * (4a + b -1). This is a straightforward application of the distributive law. 4a * (4a + b -1) + b * (4a + b -1) + (-1) * (4a + b -1) = (16 a^2 + 4 a b - 4 a) + (4 a b + b^2 - b) + (-4a - b + 1) = 16 a^2 + b^2 + 8 a b - 8 a - 2 b - 1. The other terms can also be squared. You collect all the a^2, b^2, ab, a and b terms, and the pure numbers, and you should get 116•a^2 + 2•a•(28•b - 37) + 8•b^2 - 16•b + 14. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qQuery problem 7.7.6 (was 7.7.16) use partial derivatives,etc., to find least-squares line for (-3,0), (-1,1), (1,1), (3,2) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: | (a * -3 + b) - 0 |, | (a * -1 + b) - 1 |, | (a * 1 + b) - 1 | | (a * 3 + b) - 2 | ( (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 40 a - 12 = 0 if a = 12/40 = .3. 8b - 8 = 0 if b = 1 critical point gives a min confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: What was your expression for the sum of the squared errors? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 20·a^2 - 12·a + 4·b^2 - 8·b + 6 a = .3 and b = 1 1.8 - 3.6 + 4 - 8 + 6 = .2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Right, for the values of a and b you correctly obtained above. The expression for squared errors is 20·a^2 - 12·a + 4·b^2 - 8·b + 6. For a = .3 and b = 1 this expression gives 1.8 - 3.6 + 4 - 8 + 6 = .2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!