Rewrite the regression equations uniformly
Y = . 0399 X + . 6934 Y = .0 399X + .6 934 Y = .0399 X + .6934
X = . 1212 Y − . 2461 X = .1 212Y - .2 461 X = .1212 Y − .2461
Denote mean values as x ‾ \overline{x} x and y ‾ \overline{y} y , correlation coefficient as r r r , standard deviations as s x s_x s x and s y s_y s y . Then the basic formulas are
Y = r s y s x X − r s y s x x ‾ + y ‾ Y=r\dfrac{s_y}{s_x}X-r\dfrac{s_y}{s_x}\overline{x}+\overline{y} Y = r s x s y X − r s x s y x + y
X = r s x s y Y − r s x s y y ‾ + x ‾ X=r\dfrac{s_x}{s_y}Y-r\dfrac{s_x}{s_y}\overline{y}+\overline{x} X = r s y s x Y − r s y s x y + x
So find correlation coefficient from following
{ r s y s x = . 0399 r s x s y = . 1212 \left\{\begin{array}{rr}r\dfrac{s_y}{s_x}=.0 399\\ \\ r\dfrac{s_x}{s_y}= .1 212\end{array}\right. ⎩ ⎨ ⎧ r s x s y = .0399 r s y s x = .1212
Multiply the equations
r s y s x ⋅ r s x s y = . 0399 ⋅ . 1212 r\dfrac{s_y}{s_x}\cdot r\dfrac{s_x}{s_y}=.0 399 \cdot .1 212 r s x s y ⋅ r s y s x = .0399 ⋅ .1212 ⇒ \Rightarrow ⇒ r 2 = . 00483588 r^2=.00483588 r 2 = .00483588
Thus
r = . 00483588 = . 069540492 r=\sqrt{.00483588}=.069540492 r = .00483588 = .069540492
Further, find mean values from
− r s y s x x ‾ + y ‾ = . 6934 -r\dfrac{s_y}{s_x}\overline{x}+\overline{y}= .6 934 − r s x s y x + y = .6934
− r s x s y y ‾ + x ‾ = − . 2461 -r\dfrac{s_x}{s_y}\overline{y}+\overline{x}=- .2 461 − r s y s x y + x = − .2461
or
− . 0399 x ‾ + y ‾ = . 6934 - .0 399\overline{x}+\overline{y}= .6 934 − .0399 x + y = .6934
− . 1212 y ‾ + x ‾ = − . 2461 -.1 212\overline{y}+\overline{x}=- .2 461 − .1212 y + x = − .2461
Construct the system of linear equations and solve it using Cramer's rule
{ − . 0399 x ‾ + y ‾ = . 6934 x ‾ − . 1212 y ‾ = − . 2461 \left\{\begin{array}{rrr}- .0 399\ \overline{x}&+\ \overline{y}=& .6 934\\
\overline{x}&-.1 212\ \overline{y}=&- .2 461\end{array}\right. { − .0399 x x + y = − .1212 y = .6934 − .2461
det ( A ) = ∣ − . 0399 1 1 − . 1212 ∣ = \det(A)= \begin{vmatrix}
- .0 399 & 1 \\
1 & -.1 212
\end{vmatrix}= det ( A ) = ∣ ∣ − .0399 1 1 − .1212 ∣ ∣ = − . 0399 ⋅ ( − . 1212 ) − 1 ⋅ 1 = -.0 399\cdot( -.1 212)-1\cdot1= − .0399 ⋅ ( − .1212 ) − 1 ⋅ 1 = − . 99516412 -.99516412 − .99516412
det ( A x ) = ∣ . 6934 1 − . 2461 − . 1212 ∣ = \det(A_x)= \begin{vmatrix}
.6 934 & 1 \\
- .2 461 & -.1 212
\end{vmatrix}= det ( A x ) = ∣ ∣ .6934 − .2461 1 − .1212 ∣ ∣ = . 6934 ⋅ ( − . 1212 ) − 1 ⋅ ( − . 2461 ) = .6934\cdot( -.1 212)-1\cdot(-.2461)= .6934 ⋅ ( − .1212 ) − 1 ⋅ ( − .2461 ) = . 16205992 .16205992 .16205992
det ( A y ) = ∣ − . 0399 . 6934 1 − . 2461 ∣ = \det(A_y)= \begin{vmatrix}
- .0 399 & .6 934 \\
1 & - .2 461
\end{vmatrix}= det ( A y ) = ∣ ∣ − .0399 1 .6934 − .2461 ∣ ∣ = − . 0399 ⋅ ( − . 2461 ) − . 6934 ⋅ 1 = -.0 399\cdot( -.2461)-.6934\cdot1= − .0399 ⋅ ( − .2461 ) − .6934 ⋅ 1 = − . 68358061 -.68358061 − .68358061
x ‾ = det ( A x ) det ( A ) = . 16205992 − . 99516412 = − . 162847431 \overline{x}=\dfrac{\det(A_x)}{\det(A)}=\dfrac{.16205992}{-.99516412}=-.162847431 x = det ( A ) det ( A x ) = − .99516412 .16205992 = − .162847431
y ‾ = det ( A y ) det ( A ) = − . 68358061 − . 99516412 = . 686902388 \overline{y}=\dfrac{\det(A_y)}{\det(A)}=\dfrac{-.68358061}{-.99516412}=.686902388 y = det ( A ) det ( A y ) = − .99516412 − .68358061 = .686902388
Answer
correlation coefficient
r = . 0695 r=.0695 r = .0695
mean values
x ‾ = − . 1628 \overline{x}=-.1628 x = − .1628
y ‾ = . 6869 \overline{y}=.6869 y = .6869
Remark. Weight must be positive. Perhaps condition is incorrect.
Comments