X Y X Y X 2 Y 2 4 4 16 16 16 5 6 30 25 36 3 5 15 9 25 6 7 42 36 49 10 7 70 100 49 S u m = 28 29 173 186 175 \def\arraystretch{1.5}
\begin{array}{c:c:c:c:c:c}
& X & Y & XY & X^2 & Y^2\\
\hline
& 4 & 4 & 16 & 16 & 16\\
& 5 & 6 & 30 & 25 & 36\\
& 3 & 5 & 15 & 9 & 25\\
& 6 & 7 & 42 & 36 & 49\\
& 10 & 7 & 70 & 100 & 49\\
Sum= & 28 & 29 & 173 & 186 & 175\\
\end{array} S u m = X 4 5 3 6 10 28 Y 4 6 5 7 7 29 X Y 16 30 15 42 70 173 X 2 16 25 9 36 100 186 Y 2 16 36 25 49 49 175 X ˉ = 1 n ∑ i X i = 28 5 = 5.6 \bar{X}=\dfrac{1}{n}\sum_iX_i=\dfrac{28}{5}=5.6 X ˉ = n 1 i ∑ X i = 5 28 = 5.6
Y ˉ = 1 n ∑ i Y i = 29 5 = 5.8 \bar{Y}=\dfrac{1}{n}\sum_iY_i=\dfrac{29}{5}=5.8 Y ˉ = n 1 i ∑ Y i = 5 29 = 5.8
S S X X = ∑ i ( X i − X ˉ ) 2 = 29.2 SS_{XX}=\sum_i(X_i-\bar{X})^2=29.2 S S XX = i ∑ ( X i − X ˉ ) 2 = 29.2
S S Y Y = ∑ i ( Y i − Y ˉ ) 2 = 6.8 SS_{YY}=\sum_i(Y_i-\bar{Y})^2=6.8 S S YY = i ∑ ( Y i − Y ˉ ) 2 = 6.8
S S X Y = ∑ i ( X i − X ˉ ) ( Y i − Y ˉ ) = 10.6 SS_{XY}=\sum_i(X_i-\bar{X})(Y_i-\bar{Y})=10.6 S S X Y = i ∑ ( X i − X ˉ ) ( Y i − Y ˉ ) = 10.6
m = s l o p e = S S X Y S S X X = 10.6 29.2 = 0.363014 m=slope=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{10.6}{29.2}=0.363014 m = s l o p e = S S XX S S X Y = 29.2 10.6 = 0.363014
n = Y ˉ − m X ˉ n=\bar{Y}-m\bar{X} n = Y ˉ − m X ˉ
= 5.8 − 0.363014 ( 5.6 ) =5.8-0.363014(5.6) = 5.8 − 0.363014 ( 5.6 )
= 3.767123 =3.767123 = 3.767123 Therefore, we find that the regression equation is:
Y = 3.767123 + 0.363014 X Y=3.767123+0.363014X Y = 3.767123 + 0.363014 X
Correlation coefficient:
r = S S X Y S S X X S S Y Y r=\dfrac{SS_{XY}}{\sqrt{SS_{XX}}\sqrt{SS_{YY}}} r = S S XX S S YY S S X Y
= 10.6 29.2 6.8 = 0.752246 =\dfrac{10.6}{\sqrt{29.2}\sqrt{6.8}}=0.752246 = 29.2 6.8 10.6 = 0.752246 We have strong positive correlation.
The regression equation is:
Y = 3.767123 + 0.363014 X Y=3.767123+0.363014X Y = 3.767123 + 0.363014 X
Comments