Regresi Kuadratik

TUGAS #1 ANALISIS STATISTIK

ANALISIS REGRESI KUADRATIK

PENGAJAR: TOTO WARSA, Ir., M.S.

PENYUSUN: Ade Setiawan 150220060003

PROGRAM PASCASARJANA UNIVERSITAS PADJADJARAN 2006

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Soal: Berikut ini adalah data hasil percobaan pemupukan nitrogen pada tanaman padi: Dosis Pupuk N (Xi) 0 30 60 90 120

Hasil Panen (Yi) 2.1 3.4 4.2 7.1 6.3

Tentukan persamaan regresi kuadratiknya!

Jawab: a. Model Regresi Kuadratik Y = b0 + b1 X i + b2 X i + ε 2

X0 1 1 1 1 1

X1 0 30 60 90 120

X2 0 900 3600 8100 14400

Y 2.1 3.4 4.2 7.1 6.3

b. Definisikan Matriks-matriks Berikut: 0 ⎞ ⎛1 0 ⎛ 2.1 ⎞ ⎜1 30 ⎜ 3.4 ⎟ ⎛ b0 ⎞ 900 ⎟ ⎟ ⎜ ⎟ ⎜ b = ⎜⎜ b1 ⎟⎟ X = 1 60 3600 Y = 4.2 ⎜1 90 8100 ⎟ ⎜ 7.1 ⎟ ⎜b ⎟ ⎝ 2⎠ ⎜1 120 14400 ⎟ ⎜ 6.3 ⎟ ⎠ ⎝ ⎝ ⎠ 1 1 1 ⎞ ⎛1 1 X ′ = ⎜ 0 30 60 90 120 ⎟ ⎜ 0 900 3600 8100 14400 ⎟ ⎠ ⎝

0 ⎞ ⎛1 0 1 1 1 ⎞ ⎜1 30 900 ⎟ ⎛1 1 X ′X = ⎜ 0 30 60 90 120 ⎟ ⎜1 60 3600 ⎟ ⎜ 0 900 3600 8100 14400 ⎟ ⎜1 90 8100 ⎟ ⎝ ⎠⎜ ⎟ ⎝1 120 14400 ⎠ 300 2700 ⎛5 ⎞ 27000 2700000 ⎟ = ⎜ 300 ⎜ 27000 2700000 286740000 ⎟ ⎝ ⎠

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⎛ 2.1 ⎞ 1 1 1 ⎛1 1 ⎞ ⎜ 3.4 ⎟ X ′Y = ⎜ 0 30 60 90 120 ⎟ ⎜ 4.2 ⎟ ⎜ 0 900 3600 8100 14400 ⎟ ⎜ 7.1 ⎟ ⎝ ⎠⎜ ⎟ ⎝ 6.3 ⎠ ⎛ 23.1 ⎞ = ⎜ 1749 ⎟ ⎜166410 ⎟ ⎝ ⎠

c. Persamaan Regresi dalam bentuk Matriks: Y = Xb + ε b = ( X ′X ) −1 X ′Y

300 2700 ⎡5 ⎤ ( X ′X ) = ⎢300 27000 2700000 ⎥ ⎢⎣27000 2700000 286740000⎥⎦

−1

−1

d. Untuk mendapatkan nilai invers X’X , digunakan cara kofaktor dan determinan: 1. Minor:

M 11 =

27000

2700000

2700000 286740000

M 12 =

= 451980000000

M 21 =

300

27000

2700000 286740000

300

27000

27000 2700000

2700000

27000 286740000

M 13 =

= 13122000000

M 22 =

= 13122000000

M 31 =

300

5

27000

27000 286740000

= 81000000

5

27000

300 2700000

27000

27000 2700000

= 81000000

M 23 =

= 704700000

M 32 =

300

5

300

27000 2700000

= 5400000

M 33 =

= 5400000

5

300

300 27000

= 45000

2. Determinan: Diambil dari baris 1: Misalkan Matriks X’X = Matriks A, Maka nilai deteminannya:

A = a11M 11 − a12 M 12 + a13 M 13 atau A = a11K11 + a12 K12 + a13 K13

dimana K ij = (−1)i = j M ij

= 5(451980000000) + 300(-13122000000) + 27000(81000000) = 510300000000 3. Matriks Kofaktor: K ij = (−1)i = j M ij

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⎛ 451980000000 - 13122000000 81000000 ⎞ ⎜ ⎟ - 5400000 ⎟ K ij = ⎜ - 13122000000 704700000 ⎜ 81000000 ⎟ - 5400000 45000 ⎝ ⎠

4. Adjoint A:

[ ]

adj. A = K ij

⎛ 451980000000 - 13122000000 81000000 ⎞ - 5400000 ⎟ = ⎜ - 13122000000 704700000 ⎜ 81000000 ⎟ - 5400000 45000 ⎝ ⎠ ⎛ 451980000000 - 13122000000 81000000 ⎞ - 5400000 ⎟ = ⎜ - 13122000000 704700000 ⎜ 81000000 ⎟ - 5400000 45000 ⎝ ⎠

5. Invers: adj. A A−1 = ; A

'

A ≠0

⎛ 451980000000 - 13122000000 81000000 ⎞ ⎜ - 13122000000 704700000 - 5400000 ⎟ ⎟ ⎜ 81000000 5400000 45000 ⎠ =⎝ 510300000000 ⎛ 0.8857142857 - 0.0257142857 0.0001587302 ⎞ = ⎜ - 0.0257142857 0.0013809524 - 0.0000105820 ⎟ ⎜ 0.0001587302 - 0.0000105820 0.0000000882 ⎟ ⎠ ⎝ 6. Dengan demikian: ⎛ 0.8857142857 - 0.0257142857 0.0001587302 ⎞ ⎜ ⎟ ( X ' X ) = ⎜ - 0.0257142857 0.0013809524 - 0.0000105820 ⎟ ⎜ 0.0001587302 - 0.0000105820 0.0000000882 ⎟ ⎝ ⎠ −1

Sehingga, didapat nilai b:

b = ( X ′X ) −1 X ′Y ⎛ b 0 ⎞ ⎛ 0.8857142857 - 0.0257142857 0.0001587302 ⎞ ⎛ 23.1 ⎞ ⎜ b ⎟ = ⎜ - 0.0257142857 0.0013809524 - 0.0000105820 ⎟ ⎜ 1749 ⎟ ⎜⎜ 1 ⎟⎟ ⎜ ⎟⎜ ⎟ ⎝ b 2 ⎠ ⎝ 0.0001587302 - 0.0000105820 0.0000000882 ⎠ ⎝166410 ⎠ 1.9 ⎛ ⎞ = ⎜ 0.0603333 ⎟ ⎜ - 0.000167 ⎟ ⎝ ⎠ e. Persamaan Regresinya:

Yˆ = b0 + b1 X + b2 X 2 Yˆ = 1.9 + 0.0603333 X − 0.000167 X 2

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