OLEH :

FAKULTAS PERTANIANUNIVERSITAS SWADAYA GUNUNG JATI CIREBON

2011

WIJAYA

ANALISIS REGRESI

REGRESI NON LINEAR

III. REGRESI NON LINEAR

Regresi Kuadratik : Y = b0 + b1 X + b2 X2.

n ∑ X ∑ X2 b0 ∑ Y∑ X ∑ X2 ∑ X3 b1 = ∑ XY∑ X2 ∑ X3 ∑ X4 b2 ∑ X2Y

∑ Y = b0 n + b1 ∑ X + b2 ∑ X2

∑ XY = b0 ∑ X + b1 ∑ X2 + b2 ∑ X3

∑ X2Y = b0∑ X2 + b1 ∑ X3 + b2 ∑ X4

(X’X) (b) (X’Y)

b0 n ∑ X ∑ X2 ∑ Yb1 = ∑ X ∑ X2 ∑ X3 ∑ XYb2 ∑ X2 ∑ X3 ∑ X4 ∑ X2Y

–1

Misal telah dilakukan sebuah penelitian tentangPengaruh Kadar Air Gabah Terhadap Mutu Fisik BerasGiling. Salah satu respon yang diamati yaituPersentase Butir Patah. Hasil pengamatannyadisajikan pada tabel berikut :

(b) (X’X)–1 (X’Y)

No PerlakuanButir Patah (%)

I II III IV1 k1 (8 %) 27,40 26,56 29,52 27,702 k2 (10 %) 19,40 16,88 18,28 17,783 k3 (12 %) 6,68 6,24 7,56 5,904 k4 (14 %) 3,46 3,20 4,00 2,925 k5 (16 %) 13,12 15,04 12,02 13,846 k6 (18 %) 16,76 18,32 23,64 21,42

Pengamatan Persentase Butir Patah :

Untuk memudahkan perhitungan, taraf Faktor atauVariabel Bebas Kadar Air diubah menjadi :

Xi – (Rata-rata)Ki =

2Xi = Taraf Kadar AirRata-rata = Rata-rata seluruh taraf Kadar Air = 13 %2 = Selisih antar taraf Kadar Air

Kadar Air (Xi) 8 % 10 % 12 % 14 % 16 % 18 %Ki –2,5 –1,5 –0,5 0,5 1,5 2,5

No Y X X2 X3 X4

1 27,40 -2,5 6,25 -15,625 39,06252 19,40 -1,5 2,25 -3,375 5,0625… … … … … …6 16,76 2,5 6,25 15,625 39,06257 26,56 -2,5 6,25 -15,625 39,0625… … … … … …12 18,32 2,5 6,25 15,625 39,062513 29,52 -2,5 6,25 -15,625 39,0625… … … … … …18 23,64 2,5 6,25 15,625 39,062519 27,70 -2,5 6,25 -15,625 39,0625… … … … … …24 21,42 2,5 6,25 15,625 39,0625

∑ X = 0 ∑ X2 = 70

∑ XY = –111,480 ∑ X2Y = 1490,050∑ Y = 357,640

∑ X3 = 0 ∑ X4 = 354

b0 24 0 70 357,640b1 = 0 70 0 –111,480b2 70 0 354 1490,050

–1

b0 n ∑ X ∑ X2 ∑ Yb1 = ∑ X ∑ X2 ∑ X3 ∑ XYb2 ∑ X2 ∑ X3 ∑ X4 ∑ X2Y

–1

(b) (X’X)–1 (X’Y)

b0 0,0986 0,0000 –0,0195 357,640b1 = 0,0000 0,0143 0,0000 –111,480b2 –0,0195 0,0000 0,0067 1490,050

b0 6,1725b1 = –1,5926b2 2,9929

Regresi : Y = 6,1725 – 1,5926 X + 2,9929 X2.

(b) (X’X)–1 (X’Y)

Pengujian Regresi Non Linear :

1. FK = (∑Y)2 / n = (357,640)2 / 24 = 5329,432 2. JKT = ∑ Y2 – FK = 6996,305 – 5329,432 = 1667,873 3. JKR = b1 [ (∑ XY – (∑X)(∑Y)/n ] +

b2 [ (∑ X2Y – (∑X2)(∑Y)/n ] = –1,5926 [ (–111,480 – (0)(357,640)/24 ] +

2,9929 [ (1490,050 – (70)(357,640)/24 ] = 177,540 + 1337,608 = 1515,147

4. JKG = JKT – JKR = 1667,873 – 1515,147 = 152,725

∑ X = 0 ∑ X2 = 70∑ XY = –111,480 ∑ X2Y = 1490,050∑ Y = 357,640

∑ X3 = 0 ∑ X4 = 354 ∑ Y2 = 6997,305

No Variasi DB JK KT F F5%1 Regresi 2 1515,147 757,574 104,168 3,467

R (b1) 1 177,540 177,540 24,412 4,325R (b2) 1 1337,608 1337,608 183,923 4,325

2 Galat 21 152,725 7,273Total 23 1667,873

Keterangan : 1. Regresi (b1) ≡ (F = 24,412) > (F0,05 = 4,325) Sig2. Regresi (b2) ≡ (F = 182,923) > (F0,05 = 4,325) Sig3. R2 = 0,9084 R = 0,9531

Penggunaan Metode Doolitle :

BarisMatriks (X'X) Matriks

Matriks (X'X)-1

b0 b1 b2 (X'Y)(0) 24 0 70 357,640 1 0 0(1) 70 0 -111,480 0 1 0(2) 354 1490,050 0 0 1(3) 24 0 70 357,640 1 0 0(4) 1,00 0,00 2,917 14,902 0,0417 0,0000 0,0000(5) 70,00 0,00 -111,480 0,0000 1,0000 0,0000(6) 1,00 0,00 -1,593 0,0000 0,0143 0,0000(7) 149,333 446,933 -2,9167 0,0000 1,0000(8) 1,00 2,993 -0,0195 0,0000 0,0067

Baris (3) = Baris (0)Baris (4) = Baris (3)/24

Baris (5) = (70)– (0)(Baris 4)Baris (6) = Baris (5) /70

Baris (7) = (354) – (70)(Baris 4) – (0,00)(Baris 6)Baris (8) = Baris (7) /149,33

Penentuan Koefisien Regresi :

Baris (8) 1,00 (b2) = 2,993 b2 = 2,993Baris (6) 1,00 (b1) + 0,00 (b2) = –1,593 b1 = –1,593Baris (4) 1,00 (b0) + 0,00 (b1) + 2,917 (b2) = 14,902

b0 = 6,173

Baris Matriks (X'X) MatriksMatriks (X'X)-1

b0 b1 b2 (X'Y)(3) 24 0 70 357,640 1 0 0(4) 1,00 0,00 2,917 14,902 0,0417 0,0000 0,0000(5) 70,00 0,00 -111,480 0,0000 1,0000 0,0000(6) 1,00 0,00 -1,593 0,0000 0,0143 0,0000(7) 149,333 446,933 -2,9167 0,0000 1,0000(8) 1,00 2,993 -0,0195 0,0000 0,0067

Pengujian Koefisien Regresi : Matrik (X’X)–1 Baris (3), (5), (7)

Baris Matriks (X'X) MatriksMatriks (X'X)-1

b0 b1 b2 (X'Y)(3) 24 0 70 357,640 1 0 0(4) 1,00 0,00 2,917 14,902 0,0417 0,0000 0,0000(5) 70,00 0,00 -111,480 0,0000 1,0000 0,0000(6) 1,00 0,00 -1,593 0,0000 0,0143 0,0000(7) 149,333 446,933 -2,9167 0,0000 1,0000(8) 1,00 2,993 -0,0195 0,0000 0,0067

1,0000 0,0000 0,0000 1,0000 0,0000 –2,91670,0000 1,0000 0,0000 0,0000 1,0000 0,0000

–2,9167 0,0000 1,0000 0,0000 0,0000 1,0000Matriks : T Matriks : T 1

Pengujian Koefisien Regresi : Matrik (X’X)–1 Baris (4), (6), (8)

Baris Matriks (X'X) MatriksMatriks (X'X)-1

b0 b1 b2 (X'Y)(3) 24 0 70 357,640 1 0 0(4) 1,00 0,00 2,917 14,902 0,0417 0,0000 0,0000(5) 70,00 0,00 -111,480 0,0000 1,0000 0,0000(6) 1,00 0,00 -1,593 0,0000 0,0143 0,0000(7) 149,333 446,933 -2,9167 0,0000 1,0000(8) 1,00 2,993 -0,0195 0,0000 0,0067

0,0417 0,0000 0,00000,0000 0,0143 0,0000

–0,0195 0,0000 0,0067Matriks : t

1,0000 0,0000 –2,9167 0,0417 0,0000 0,00000,0000 1,0000 0,0000 0,0000 0,0143 0,00000,0000 0,0000 1,0000 –0,0195 0,0000 0,0067

0,0986 0,0000 –0,01950,0000 0,0143 0,0000

–0,0195 0,0000 0,0067

Matriks : ( T 1 t ) = ( X’X)–1

Matriks T 1 Matriks t

BarisMatriks (X'X) Matriks

Matriks (X'X)-1

b0 b1 b2 (X'Y)(3) 24 0 70 357,640 1 0 0(4) 1,00 0,00 2,917 14,902 0,0417 0,0000 0,0000(5) 70,00 0,00 -111,480 0,0000 1,0000 0,0000(6) 1,00 0,00 -1,593 0,0000 0,0143 0,0000(7) 149,333 446,933 -2,9167 0,0000 1,0000(8) 1,00 2,993 -0,0195 0,0000 0,0067

Menghitung JKR (b1) dan JKR (b2) dari kolom Matrik (X’Y) :

1. JKR (b1) = (baris 5)(baris 6) = (–111,480)(–1,593) = 177,540

2. JKR (b2) = (baris 7)(baris 8) = (446,933)(2,993) = 1337,608

0,0986 0,0000 –0,01950,0000 0,0143 0,0000–0,0195 0,0000 0,0067

Matriks : ( T 1 t ) = ( X’X)–1

bi KTG Cii KTG.Cii Sb t6,173 7,273 0,0986 0,7173 0,847 7,288-1,593 7,273 0,0143 0,1039 0,322 -4,9412,993 7,273 0,0067 0,0487 0,221 13,562