07 statistika - regresi linear sederhana
TRANSCRIPT
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OLEH :
FAKULTAS PERTANIANUNIVERSITAS SWADAYA GUNUNG JATI CIREBON
2010
WIJAYA
ANALISIS REGRESI
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I. ANALISIS REGRESI
1. Regresi Linear : Regresi Linear SederhanaRegresi Linear Ganda
2. Regresi Non Linear Regresi Kuadratik
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Analisis Regresi merupakan studi yangmembahas tentang bentuk keeratanhubungan antar peubah.
Model atau persamaan regresi populasisecara umum dapat dituliskan dalam bentuk :
μy/x1, x2, …, xk = f (x1, x2, … , xk | β1, β2, … , βk )
I. REGRESI LINEAR SEDERHANA
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Untuk regresi Linear sederhana, yaitu regresi Yatas X bentuknya :
β0 dan β1 disebut Koefisien Regresi, yangmerupakan parameter. Regresi populasitersebut dapat diduga melalui contoh denganpersamaan :
μy/x = β0 + β1 X
Y = b0 + b1 X
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Jadi β0 diduga oleh b0 dan β1 diduga oleh b1. Nilaib0 dan b1 dapat ditentukan dengan Metode KuadratTerkecil, yaitu :
b0 = Intersep (titik potong regresi dengan sumbu Y)b1 = Koefisien Arah Regresi
Besarnya peningkatan Y apabila X meningkatsebesar satu satuan.
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(xI , yI) (xI , yI)yI
xI X
YY
X
n = ukuran sampel, k = banyaknya variabel bebas.
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Pada Regresi Linear Sederhana nilai k = 1, sehingga :
Ragam untuk konstanta b0 yaitu Sb02 dan koefisien
regresi b1 yaitu Sb12 yaitu :
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Tabel berikut menunjukkan skor tes kecerdasan (X)dan nilai ujian statistika (Y) dari 12 mahasiswa :
X 65 50 55 65 55 70 65 70 55 70 50 55Y 85 74 76 90 85 87 94 98 81 91 76 74
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Persamaan Regresi Dugaan :
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Pengujian Koefisien Regresi :
Wilayah Kritik : t <–tα/2(n-2) atau t > tα/2(n-2)
1. Uji t :
2. Uji F (Analisis Varians)
Wilayah Kritik : F > Fα (db1 ; db2)
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Pengujian Koefisien Regresi :
H0 ≡ βi = 0 Lawan H1 ≡ βi ≠ 0
1. Uji t :
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tα/2(n-2) = t0,025(10) = 2,228
Kesimpulan : H0 ditolak, artinya koefisienregresi bersifat nyata, regresi :
Pengujian Koefisien Regresi :
Uji t :
dapat digunakan untuk peramalan, karenabesarnya Y tergantung dari besarnya X.
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Uji Kelinearan Regresi :
Uji Kelinearan Regresi dapat dilakukan apabila peubahbebas X dirancang dengan adanya pengulangan(pengulangan tidak harus sama). Statistik uji yangdigunakan adalah Uji F dalam Analisis Ragam.
X 65 50 55 65 55 70 65 70 55 70 50 55Y 85 74 76 90 85 87 94 98 81 91 76 74
X 50 50 55 55 55 55 65 65 65 70 70 70Y 74 76 76 85 81 74 85 90 94 87 98 91∑Yi 150 316 269 276
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Analisis Ragam :
1. FK = (∑Y)2 / n = (1011)2 / 12 = 85176,7500
2. JKT = ∑ Y2 – FK = 85905 – 85176,7500 = 728,2500
3. JKR = b1 [ (∑ XY – (∑X)(∑Y)/n ]
= 0,8972 [ (61685 – (725)(1011)/12 ] = 541,6927
4. JKG = JKT – JKR = 728,2500 – 541,6927 = 186,5573
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Analisis Ragam :
JKG dibagi dua, yaitu JK Galat Murni (JKGM) dan JK Simpangan Dari Model (JK SDM)
X 50 50 55 55 55 55 65 65 65 70 70 70Y 74 76 76 85 81 74 85 90 94 87 98 91∑Yi 150 316 269 276
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Uji Kelinearan Regresi :
1. FK = 85176,7500
2. JKT = 728,2500
3. JKR = 541,6927
4. JKG = 186,5573
JK GM = 178,667
JK SDM = JK G – JK GM = 7,8906
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No Variasi DB JK KT F F5%1 Regresi 1 541,6927 541,6927 29,0363 4,4952 Galat 10 186,5573 18,6557
G-Murni 8 178,6667 22,3333G-SDM 2 7,8906 3,953 0,1767 4,459Total 11 728,2500
DB (G-SDM) = k–2 = 4–2 = 2 ; DB (G-Murni) = n–k = 12–4 = 8
Uji Kelinearan Regresi :
Regresi bersifat Nyata : Regresi Linear dapat diterimaR2 = JKR / JKT = 0,7438 R = 0,8625
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Penggunaan Matriks :
Persamaan Normal dari : Y = b0 + b1 X yaitu :
n ∑ X b0=
∑ Y∑ X ∑ X2 b1 ∑ XY
∑ Y = b0 n + b1 ∑ X
∑ XY = b0 ∑ X + b1 ∑ X2
Matrik dari persamaan normal diatas :
12 725 b0=
1011725 44475 b1 61685
X’X b X’Y
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b0 = 12 725 1011b1 725 44475 61685
–1
b0 = 5,508 –0,090 1011b1 –0,090 0,001 61685
b0 = 30,0433b1 0,8972
Regresi Dugaan : Y = 30,0433 + 0,8972 X
b (X’X)–1 X’Y
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bi KTG Cii KTG.Cii Sb t30,0433 18,6557 5,508 102,7509 10,1366 2,9640,8972 18,6557 0,001 0,0277 0,1665 5,389
t0,025 (10) = 2,228
b (X’X)–1 X’Y
b0 = 5,508 –0,090 1011b1 –0,090 0,001 61685