11+Regression

I1 example 1 suppose we want to model the monthly

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Unformatted text preview: First equation meas that the sum of the errors (SE ) equals to zero, n SE = εi = 0. i=1 The solution of the system is ˆ β0 = y − β1 x ¯ ˆ¯ SSxy ˆ β1 = SSxx 2 A. Zhensykbaev Regression where 1 x= ¯ n n 1 y= ¯ n xi , i=1 n n SSxy = (xi − x)(yi − y ) = ¯ ¯ i=1 i=1 n SSxx = x2 i (xi − x) = ¯ i=1 i=1 xi , i=1 1 xi y i − n n 2 n 1 − n n n xi i=1 yi , i=1 n 2 xi . i=1 Example 1. Suppose we want to model the monthly sales revenue as a function of the monthly advertising expenditure. We base on the following table Advertising Sales Month expenditure Revenue xi ($100) yi ($1000) 1 1 1 2 2 1 3 3 2 4 4 2 5 5 4 Total xi = 15 yi = 10 SSxy = 37 − x2 i xi yi 1 2 6 8 20 xi yi = 37 1 · 15 · 10 = 7 5 SSxx = 55 − x2 i 1 4 9 16 25 1 · 152 = 10 5 SSxy 7 = = 0.7 SSxx 10 ˆ β1 = ˆ β0 1 1 = y − β1 x = · 10 − 0.7 · · 15 = −0.1 ¯ ˆ¯ 5 5 3 A. Zhensykbaev Regression and fitted line is y = 0.7x − 0.1 6 • " " " " " " " " " " " " " " " " " " " " " " " •""" • " " " " " " " " " " " " " " " " " " "• " " • "" " " " " " " " " " " " " - Pic. 1 We make four assumptions about the random error ε 1. The mean of the probability distribution of ε equals to 0. It implies that the mean value is equal to the deterministic component Ey = β0 + β1 x. 2.The variance σ 2 of the probability distribution of ε is constant for all x. 3.The probability distribution of ε is normal. 4. The values of ε associated with any two observed values of y are independent. Estimation of σ 2 . In most practical situations σ 2 is unknown and we estimate it using given data. The estimated standard error of the regression model is the sample standard deviation s. SSE s2 = n−2 n (yi − yi )2 , ˆ SSE = yi = β0 + β1 xi . ˆ i=1 ˆ SSE = SSyy − β1 · SSxy , n 2 yi SSyy = i=1 4 1 − n n 2 yi i=1 . A. Zhensykbaev Regression Ex...
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