regresi kuadratik
TRANSCRIPT
1
TUGAS #1
ANALISIS STATISTIK
ANALISIS REGRESI KUADRATIK
PENGAJAR:
TOTO WARSA, Ir., M.S.
PENYUSUN:
Ade Setiawan 150220060003
PROGRAM PASCASARJANA UNIVERSITAS PADJADJARAN
2006
2
Soal:
Berikut ini adalah data hasil percobaan pemupukan nitrogen pada tanaman padi:
Dosis Pupuk N (Xi)
Hasil Panen (Yi)
0 2.1 30 3.4 60 4.2 90 7.1
120 6.3 Tentukan persamaan regresi kuadratiknya!
Jawab:
a. Model Regresi Kuadratik
ε+++= 2210 ii XbXbbY
X0 X1 X2 Y
1 0 0 2.1 1 30 900 3.4 1 60 3600 4.2 1 90 8100 7.1 1 120 14400 6.3
b. Definisikan Matriks-matriks Berikut:
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
=
14400120181009013600601900301
001
X ⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
=
3.61.72.44.31.2
Y ⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛=
2
1
0
bbb
b
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=′
144008100360090001209060300
11111X
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=′
28674000027000002700027000002700030027003005
14400120181009013600601900301
001
144008100360090001209060300
11111XX
3
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=′
166410174923.1
3.61.72.44.31.2
14400810036009000120906030011111
YX
c. Persamaan Regresi dalam bentuk Matriks: ε+= XbY
YXXXb ′′= −1)( 1
1
28674000027000002700027000002700030027003005
)(−
−
⎥⎥⎦
⎤
⎢⎢⎣
⎡=′XX
d. Untuk mendapatkan nilai invers X’X , digunakan cara kofaktor dan determinan:
1. Minor:
0045198000002867400002700000270000027000
11
=
=M
01312200000286740000270002700000300
12
=
=M
8100000027000002700027000300
13
=
=M
013122000002867400002700000
2700030021
=
=M
70470000028674000027000
27000522
=
=M
5400000270000027000
300523
=
=M
8100000027000002700027000300
31
=
=M
54000002700000300270005
32
=
=M
45000270003003005
33
=
=M
2. Determinan: Diambil dari baris 1: Misalkan Matriks X’X = Matriks A, Maka nilai deteminannya:
0051030000000000)27000(8100000000)300(-1312200)4519800000(5
)1(dimana131312121111
131312121111
=++=
−=++=
+−=
=ij
jiij MKK a Ka KaA
atauM a Ma MaA
3. Matriks Kofaktor:
ijji
ij MK =−= )1(
4
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛=
450005400000-810000005400000-70470000001312200000-
8100000001312200000-004519800000
ijK
4. Adjoint A: [ ]
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=
=
450005400000-810000005400000-70470000001312200000-
8100000001312200000-004519800000
450005400000-810000005400000-70470000001312200000-
8100000001312200000-004519800000
.'
ijKAadj
5. Invers:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
≠=−
820.00000008200.00001058-020.00015873200.00001058-240.00138095570.02571428-020.00015873570.02571428-570.88571428
005103000000450005400000-810000005400000-70470000001312200000-
8100000001312200000-004519800000
0;.1 AA
AadjA
6. Dengan demikian:
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛=−
820.00000008200.00001058-020.00015873200.00001058-240.00138095570.02571428-
020.00015873570.02571428-570.88571428)'( 1XX
Sehingga, didapat nilai b:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
′′= −
0.000167-0.0603333
1.9
166410174923.1
820.00000008200.00001058-020.00015873200.00001058-240.00138095570.02571428-020.00015873570.02571428-570.88571428
bbb
)(
2
1
0
1 YXXXb
e. Persamaan Regresinya:
2
2210
00016700603333091ˆ
ˆ
X.X..Y
XbXbbY
−+=
++=