Solution :
The independent variable is X, and the dependent variable is Y. In order to compute the regression coefficients, the following table needs to be used:
X ‾ = 1 n ∑ i = 1 n X i = 56 8 = 7 \overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_{i}=\frac{56}{8}=7 X = n 1 ∑ i = 1 n X i = 8 56 = 7
Y ‾ = 1 n ∑ i = 1 n Y i = 40 8 = 5 \overline{Y}=\frac{1}{n}\sum_{i=1}^{n}Y_{i}=\frac{40}{8}=5 Y = n 1 ∑ i = 1 n Y i = 8 40 = 5
S S X X = ∑ i = 1 n X i 2 − 1 n ( ∑ i = 1 n X i ) 2 = 524 − 5 6 2 8 = 132 SS_{XX}=\sum_{i=1}^{n}X_{i}^2 -\frac{1}{n}(\sum_{i=1}^{n}X_{i})^2= 524- \frac{56^2}{8}=132 S S XX = ∑ i = 1 n X i 2 − n 1 ( ∑ i = 1 n X i ) 2 = 524 − 8 5 6 2 = 132
S S Y Y = ∑ i = 1 n Y i 2 − 1 n ( ∑ i = 1 n Y i ) 2 = 256 − 4 0 2 8 = 56 SS_{YY}=\sum_{i=1}^{n}Y_{i}^2 -\frac{1}{n}(\sum_{i=1}^{n}Y_{i})^2= 256- \frac{40^2}{8}=56 S S YY = ∑ i = 1 n Y i 2 − n 1 ( ∑ i = 1 n Y i ) 2 = 256 − 8 4 0 2 = 56
S S X Y = ∑ i = 1 n X i Y i − 1 n ∑ i = 1 n X i ∑ i = 1 n Y i = 364 − 56 ∗ 40 8 = 84 SS_{XY}=\sum_{i=1}^{n}X_{i}Y_{i} -\frac{1}{n}\sum_{i=1}^{n}X_{i}\sum_{i=1}^{n}Y_{i}= 364- \frac{56*40}{8}=84 S S X Y = ∑ i = 1 n X i Y i − n 1 ∑ i = 1 n X i ∑ i = 1 n Y i = 364 − 8 56 ∗ 40 = 84
Therefore, based on the above calculations, the regression coefficients (the slope m, and the y-intercept n are obtained as follows:
m = S S X Y S S X X = 84 132 = 0.6364 m=\frac{SS_{XY}}{SS_{XX}}=\frac{84}{132}=0.6364 m = S S XX S S X Y = 132 84 = 0.6364
n = Y ‾ − X ‾ ∗ m = 5 − 7 ∗ 0.6364 = 0.5455 n=\overline{Y}-\overline{X}*m = 5-7*0.6364=0.5455 n = Y − X ∗ m = 5 − 7 ∗ 0.6364 = 0.5455
Therefore, we find that the regression equation is:
Y =0.5455+0.6364X
b)1 n ∑ i = 1 n e i ^ = 0 \frac{1}{n}\sum_{i=1}^{n}\widehat{e_{i}}=0 n 1 ∑ i = 1 n e i = 0 because the intercept of the model absorbs the mean of the residuals .
a)By definition Y i = Y i ^ + ϵ i ^ Y_{i}=\widehat{Y_{i}}+\widehat{\epsilon _{i}} Y i = Y i + ϵ i
Since
∑ i = 1 n e i ^ = 0 \sum_{i=1}^{n}\widehat{e_{i}}=0 ∑ i = 1 n e i = 0
1 n ∑ i = 1 n Y i = 1 n ∑ i = 1 n Y ^ i + 1 n ∑ i = 1 n e i ^ \frac{1}{n}\sum_{i=1}^{n}Y_{i}= \frac{1}{n}\sum_{i=1}^{n}\widehat{Y}_{i}+\frac{1}{n}\sum_{i=1}^{n}\widehat{e_{i}} n 1 ∑ i = 1 n Y i = n 1 ∑ i = 1 n Y i + n 1 ∑ i = 1 n e i
1 n ∑ i = 1 n Y i = 1 n ∑ i = 1 n Y ^ i + 0 \frac{1}{n}\sum_{i=1}^{n}Y_{i}= \frac{1}{n}\sum_{i=1}^{n}\widehat{Y}_{i}+0 n 1 ∑ i = 1 n Y i = n 1 ∑ i = 1 n Y i + 0
= = = 1 n ∑ i = 1 n Y ^ i \frac{1}{n}\sum_{i=1}^{n}\widehat{Y}_{i} n 1 ∑ i = 1 n Y i
Hence the proof.
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