| Week | (X_1) | (X_2) | (Y) | (X_1Y) | (X_2Y) | (X_1^2) | (X_2^2) | (X_1X_2) | |------|---------|---------|-------|----------|----------|----------|----------|------------| | 1 | 1 | 2 | 5 | 5 | 10 | 1 | 4 | 2 | | 2 | 2 | 3 | 8 | 16 | 24 | 4 | 9 | 6 | | 3 | 3 | 4 | 11 | 33 | 44 | 9 | 16 | 12 | | 4 | 4 | 5 | 14 | 56 | 70 | 16 | 25 | 20 | | | 10 | 14 | 38 | 110 | 148 | 30 | 54 | 40 |
| ( Y ) | ( X_1 ) | ( X_2 ) | ( X_1^2 ) | ( X_2^2 ) | ( X_1 X_2 ) | ( X_1 Y ) | ( X_2 Y ) | |--------|-----------|-----------|-------------|-------------|---------------|-------------|-------------| | 7 | 2 | 3 | 4 | 9 | 6 | 14 | 21 | | 11 | 4 | 5 | 16 | 25 | 20 | 44 | 55 | | 13 | 6 | 4 | 36 | 16 | 24 | 78 | 52 | | 17 | 8 | 6 | 64 | 36 | 48 | 136 | 102 | | 21 | 10 | 7 | 100 | 49 | 70 | 210 | 147 | | | | | | | | | | | ΣY=69 | ΣX₁=30 | ΣX₂=25 | ΣX₁²=220 | ΣX₂²=135 | ΣX₁X₂=168 | ΣX₁Y=482 | ΣX₂Y=377 |
Con: [ \mathbfX = \beginbmatrix 1 & 2 & 3 \ 1 & 4 & 5 \ 1 & 6 & 4 \ 1 & 8 & 6 \ 1 & 10 & 7 \endbmatrix, \quad \mathbfY = \beginbmatrix 7 \ 11 \ 13 \ 17 \ 21 \endbmatrix ]
, en miles de dólares) basándose en la inversión en publicidad de TV ( X1cap X sub 1 , en miles) y la inversión en publicidad en Radio ( X2cap X sub 2 , en miles). X1cap X sub 1 X2cap X sub 2 Objetivo: Encontrar la ecuación Paso 1: Definir las matrices Necesitamos añadir una columna de unos (1s) a la matriz para representar la intersección ( β0beta sub 0 regresion lineal multiple ejercicios resueltos a mano
b1=16.875+1.625(-2.734)b sub 1 equals 16.875 plus 1.625 open paren negative 2.734 close paren
94=20β1+13(6−1.3β1)94 equals 20 beta sub 1 plus 13 open paren 6 minus 1.3 beta sub 1 close paren
Interesting result: Sleep hours ((X_2)) has no effect in this sample. The model simplifies to simple linear regression. | Week | (X_1) | (X_2) | (Y)
Multiplicamos (1) por 6 y restamos a (2): (2) - 6×(1): ( (30-30)b_0 + (220-180)b_1 + (168-150)b_2 = 482 - 414 ) ( 0b_0 + 40b_1 + 18b_2 = 68 ) → (A) ( 40b_1 + 18b_2 = 68 )
|XTX|=4(150−121)−7(70−66)+6(77−90)the absolute value of cap X to the cap T-th power cap X end-absolute-value equals 4 open paren 150 minus 121 close paren minus 7 open paren 70 minus 66 close paren plus 6 open paren 77 minus 90 close paren
β1=163.1≈5.161beta sub 1 equals 16 over 3.1 end-fraction is approximately equal to 5.161 Con el valor de β1beta sub 1 , hallamos β2beta sub 2 Multiplicamos (1) por 6 y restamos a (2):
) interactúan a través de la inversión de matrices para encontrar los coeficientes óptimos. Siempre añadir la columna de unos para el intercepto. El orden de las matrices es vital:
Ŷ=49.33+1.58(90)−1.33(6)cap Y hat equals 49.33 plus 1.58 open paren 90 close paren minus 1.33 open paren 6 close paren
C₁₁ = +det([102,161; 161,255]) = 89 C₁₂ = -det([22,161; 35,255]) = - (22 255 - 161 35) = -(-25) = 25 C₁₃ = +det([22,102; 35,161]) = -28 C₂₁ = -det([22,35; 161,255]) = - (22 255 - 35 161) = - (5610 - 5635) = -(-25) = 25 C₂₂ = +det([5,35; 35,255]) = (5 255 - 35 35) = 1275 - 1225 = 50 C₂₃ = -det([5,22; 35,161]) = - (5 161 - 22 35) = - (805 - 770) = -35 C₃₁ = +det([22,35; 102,161]) = (22 161 - 35 102) = 3542 - 3570 = -28 C₃₂ = -det([5,35; 22,161]) = - (5 161 - 35 22) = - (805 - 770) = -35 C₃₃ = +det([5,22; 22,102]) = (5 102 - 22 22) = 510 - 484 = 26
Determinante = 15 (no singular, bien).
[ \begincases 375 = 5b_0 + 20b_1 + 32b_2 \quad (1) \ 1550 = 20b_0 + 90b_1 + 123b_2 \quad (2) \ 2375 = 32b_0 + 123b_1 + 210b_2 \quad (3) \endcases ]