Reg linear solution
1. Given (x1 , y1 ), ( x2 , y 2 ),............, ( xn , y n ), best fitting data to y = f ( x ) by least squares requires minimization of (A) ∑ [ y i − f ( xi )] i =1 n n
(B) (C)
∑y i =1
i
− f ( xi )
2
∑ [yi − f (xi )] i =1 n 2
n
(D) ∑ [ yi − y ] , y = i =1
∑y i =1
n
i
n
Solution The correct answer is (C).
A measure of goodness of fit, that is, how f (x ) predicts the response variable y is the magnitude of the residual Ei at each of the n data points. Ei = yi − f ( xi ), i = 1,2,....., n − 1, n Ideally, if all the residuals Ei are zero, one may have found an equation in which all the points lie on the model. Thus, minimization of the residual is an objective of obtaining regression coefficients. The most popular method to minimize the residual is the least squares method, where the estimates of the constants of the models are chosen such that the sum of the squared residuals is minimized, that is minimize ∑ E i .
2 n i =1
Thus, best fitting data to y = f ( x ) by least squares requires minimization of
∑ [y i =1
n
i
− f (xi )]
2
2. The following data x 1 20 30 40 y 1 400 800 1300 is regressed with least squares regression to y = a0 + a1 x . The value of a1 most nearly is A) 27.480 B) 28.956 C) 32.625 D) 40.000
Solution The correct answer is (C). a1 = n∑ xi yi −∑ xi ∑ yi i =1 i =1 i =1 n n n
⎛ ⎞ n ∑ x − ⎜ ∑ xi ⎟ i =1 ⎝ i =1 ⎠ n 2 i n n n n i =1 i =1 n i =1
2
a0 =
∑ xi2 ∑ yi − ∑ xi ∑ xi yi
⎛ ⎞ n∑ xi2 − ⎜ ∑ xi ⎟ i =1 ⎝ i =1 ⎠ n i =1 2
n
Since
n=4
∑x i =1 4
4
i
yi = 1×1 + 20 × 400 + 30 × 800 + 40 ×1300 = 84001 = 1 + 20 + 30 + 40 = 91 = 1 + 400 + 800 + 1300 = 2501 = 12 + 20 2 + 30 2 + 40 2 = 2901
∑x i =1 4 i =1 4
i
∑y ∑x i =1
i
2 i
then a1 = = 4 × 84001 − 91× 2501 2 4 × 2901 − (91)
108413 3323 = 32.625
3. The following data x 1 20 30 40 y 1 400 800 1300 is regressed with least squares regression to y = a1 x . The value of a1 most nearly is A) 27.480 B)