x y x y x 2 y 2 4 6300 25200 16 39690000 4 5800 23200 16 33640000 5 5700 28500 25 32490000 5 4500 22500 25 20250000 7 4500 31500 49 20250000 7 4200 29400 49 17640000 8 4100 32800 64 16810000 9 3100 27900 81 9610000 10 2100 21000 100 4410000 11 2500 27500 121 6250000 12 2200 26400 144 4840000 S u m = 82 45000 265900 690 205880000 \begin{matrix}
& x & y & xy & x^2 & y^2 \\
& 4 & 6300 & 25200 & 16 & 39690000 \\
& 4 & 5800 & 23200 & 16 & 33640000 \\
& 5 & 5700 & 28500 & 25 & 32490000 \\
& 5 & 4500 & 22500 & 25 & 20250000 \\
& 7 & 4500 & 31500 & 49 & 20250000 \\
& 7 & 4200 & 29400 & 49 & 17640000 \\
& 8 & 4100 & 32800 & 64 & 16810000\\
& 9 & 3100 & 27900 & 81 & 9610000 \\
& 10 & 2100 & 21000 & 100 & 4410000 \\
& 11 & 2500 & 27500 & 121 & 6250000 \\
& 12 & 2200 & 26400 & 144 & 4840000 \\
Sum= & 82 & 45000 & 265900 & 690 & 205880000
\end{matrix} S u m = x 4 4 5 5 7 7 8 9 10 11 12 82 y 6300 5800 5700 4500 4500 4200 4100 3100 2100 2500 2200 45000 x y 25200 23200 28500 22500 31500 29400 32800 27900 21000 27500 26400 265900 x 2 16 16 25 25 49 49 64 81 100 121 144 690 y 2 39690000 33640000 32490000 20250000 20250000 17640000 16810000 9610000 4410000 6250000 4840000 205880000
x ˉ = 1 n ∑ i = 1 n x i = 82 11 ≈ 7.454545 \bar{x}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nx_i=\dfrac{82}{11}\approx7.454545 x ˉ = n 1 i = 1 ∑ n x i = 11 82 ≈ 7.454545
y ˉ = 1 n ∑ i = 1 n y i = 45000 11 ≈ 40.909091 \bar{y}=\dfrac{1}{n}\displaystyle\sum_{i=1}^ny_i=\dfrac{45000}{11}\approx40.909091 y ˉ = n 1 i = 1 ∑ n y i = 11 45000 ≈ 40.909091
S S x x = ∑ i = 1 n x i 2 − 1 n ( ∑ i = 1 n x i ) 2 = SS_{xx}=\displaystyle\sum_{i=1}^nx_i^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nx_i)^2= S S xx = i = 1 ∑ n x i 2 − n 1 ( i = 1 ∑ n x i ) 2 =
= 690 − 8 2 2 11 ≈ 78.727273 =690-\dfrac{82^2}{11}\approx78.727273 = 690 − 11 8 2 2 ≈ 78.727273
S S y y = ∑ i = 1 n y i 2 − 1 n ( ∑ i = 1 n y i ) 2 = SS_{yy}=\displaystyle\sum_{i=1}^ny_i^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^ny_i)^2= S S yy = i = 1 ∑ n y i 2 − n 1 ( i = 1 ∑ n y i ) 2 =
= 205880000 − 4500 0 2 11 ≈ 21789090.909091 =205880000 -\dfrac{45000^2}{11}\approx21789090.909091 = 205880000 − 11 4500 0 2 ≈ 21789090.909091
S S x y = ∑ i = 1 n x i y i − 1 n ( ∑ i = 1 n x i ) ( ∑ i = 1 n y i ) = SS_{xy}=\displaystyle\sum_{i=1}^nx_iy_i-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nx_i)(\displaystyle\sum_{i=1}^ny_i)= S S x y = i = 1 ∑ n x i y i − n 1 ( i = 1 ∑ n x i ) ( i = 1 ∑ n y i ) =
= 265900 − 82 ⋅ 45000 11 ≈ − 39554.545455 =265900 -\dfrac{82\cdot45000}{11}\approx-39554.545455 = 265900 − 11 82 ⋅ 45000 ≈ − 39554.545455 Therefore, based on the above calculations, the regression coefficients (the slope m , m, m ,
and the y-intercept n n n ) are obtained as follows:
m = S S x y S S x x ≈ − 502.424942 m=\dfrac{SS_{xy}}{SS_{xx}}\approx-502.424942 m = S S xx S S x y ≈ − 502.424942
n = y ˉ − m ⋅ x ˉ ≈ n=\bar{y}-m\cdot\bar{x}\approx n = y ˉ − m ⋅ x ˉ ≈
≈ 40.909091 − ( − 502.424942 ) ⋅ 7.454545 ≈ 7836.25866 \approx40.909091-(-502.424942)\cdot7.454545\approx7836.25866 ≈ 40.909091 − ( − 502.424942 ) ⋅ 7.454545 ≈ 7836.25866 Therefore, we find that the regression equation is:
y = 7836.25866 − 502.424942 x y=7836.25866-502.424942x y = 7836.25866 − 502.424942 x y ( 4 ) = 7836.25866 − 502.424942 ( 4 ) = 5826.56 y(4)=7836.25866-502.424942(4)=5826.56 y ( 4 ) = 7836.25866 − 502.424942 ( 4 ) = 5826.56
y ( 5 ) = 7836.25866 − 502.424942 ( 5 ) = 5324.13 y(5)=7836.25866-502.424942(5)=5324.13 y ( 5 ) = 7836.25866 − 502.424942 ( 5 ) = 5324.13
y ( 7 ) = 7836.25866 − 502.424942 ( 7 ) = 4319.28 y(7)=7836.25866-502.424942(7)=4319.28 y ( 7 ) = 7836.25866 − 502.424942 ( 7 ) = 4319.28
y ( 8 ) = 7836.25866 − 502.424942 ( 8 ) = 3816.86 y(8)=7836.25866-502.424942(8)=3816.86 y ( 8 ) = 7836.25866 − 502.424942 ( 8 ) = 3816.86
y ( 9 ) = 7836.25866 − 502.424942 ( 9 ) = 3314.43 y(9)=7836.25866-502.424942(9)=3314.43 y ( 9 ) = 7836.25866 − 502.424942 ( 9 ) = 3314.43
y ( 10 ) = 7836.25866 − 502.424942 ( 10 ) = 2812.01 y(10)=7836.25866-502.424942(10)=2812.01 y ( 10 ) = 7836.25866 − 502.424942 ( 10 ) = 2812.01
y ( 11 ) = 7836.25866 − 502.424942 ( 11 ) = 2309.58 y(11)=7836.25866-502.424942(11)=2309.58 y ( 11 ) = 7836.25866 − 502.424942 ( 11 ) = 2309.58
y ( 12 ) = 7836.25866 − 502.424942 ( 12 ) = 1807.16 y(12)=7836.25866-502.424942(12)=1807.16 y ( 12 ) = 7836.25866 − 502.424942 ( 12 ) = 1807.16
6300 − 5826.56 + 5800 − 5826.56 + 6300-5826.56+5800-5826.56+ 6300 − 5826.56 + 5800 − 5826.56 + + 5700 − 5324.13 + 4500 − 5324.13 + +5700-5324.13+4500-5324.13+ + 5700 − 5324.13 + 4500 − 5324.13 + + 4500 − 4319.28 + 4200 − 4319.28 + +4500-4319.28+4200-4319.28+ + 4500 − 4319.28 + 4200 − 4319.28 + + 4100 − 3816.86 + 3100 − 3314.43 + +4100-3816.86+3100-3314.43+ + 4100 − 3816.86 + 3100 − 3314.43 + + 2100 − 2812.01 + 2500 − 2309.58 + +2100-2812.01+2500-2309.58+ + 2100 − 2812.01 + 2500 − 2309.58 + + 2200 − 1807.16 = 0.02 ≈ 0 +2200-1807.16=0.02\approx0 + 2200 − 1807.16 = 0.02 ≈ 0
R S S = ∑ i = 1 n ( y i o b s − y i p r e d ) 2 = S S y y − m S S x y = RSS=\displaystyle\sum_{i=1}^n(y_{iobs}-y_{ipred})^2=SS_{yy}-mSS_{xy}= RSS = i = 1 ∑ n ( y i o b s − y i p re d ) 2 = S S yy − m S S x y =
= S S y y ( 1 − S S x y 2 S S x x S S y y ) = =SS_{yy}(1-\dfrac{SS_{xy}^2}{SS_{xx}SS_{yy}})= = S S yy ( 1 − S S xx S S yy S S x y 2 ) =
= 21789090.909091 ( 1 − ( − 39554.545455 ) 2 78.727273 ( 21789090.909091 ) ) = =21789090.909091(1-\dfrac{(-39554.545455)^2}{78.727273(21789090.909091)})= = 21789090.909091 ( 1 − 78.727273 ( 21789090.909091 ) ( − 39554.545455 ) 2 ) =
= 1915896.5844 =1915896.5844 = 1915896.5844 The sum of square of residuals is minimum for points lying on the regression line and so cannot be less than 1915896.5844 1915896.5844 1915896.5844 for any other line.
s e s t = R S S n − 2 ≈ 1915896.5844 11 − 2 ≈ 461.39 s_{est}=\sqrt{\dfrac{RSS}{n-2}}\approx\sqrt{\dfrac{1915896.5844}{11-2}}\approx461.39 s es t = n − 2 RSS ≈ 11 − 2 1915896.5844 ≈ 461.39
r = S S x y S S x x S S y y = − 39554.545455 78.727273 21789090.909091 ≈ r=\dfrac{SS_{xy}}{\sqrt{SS_{xx}}\sqrt{SS_{yy}}}=\dfrac{-39554.545455}{\sqrt{78.727273}\sqrt{21789090.909091}}\approx r = S S xx S S yy S S x y = 78.727273 21789090.909091 − 39554.545455 ≈
≈ − 0.955024 \approx-0.955024 ≈ − 0.955024 Strong negarive correlation.
r 2 ≈ 0.912071 r^2\approx0.912071 r 2 ≈ 0.912071
Comments
Dear Jeco, please use the panel for submitting a new question.
2. You have to examine the relationship between the age and price for used cars sold in the last year by a car dealership company with the given table of data. Car Age (in years) Price (in dollars) 4 6300 4 5800 5 5700 5 4500 7 4500 7 4200 8 4100 9 3100 10 2100 11 2500 12 2200 What is the estimated cost value of car ages 2 and 15?
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