010 statistika-analisis-korelasi
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Transcript of 010 statistika-analisis-korelasi
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ANALISIS KORELASI
OLEH :
WIJAYA
FAKULTAS PERTANIAN
UNIVERSITAS SWADAYA GUNUNG JATI CIREBON
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2010
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ANALISIS KORELASI
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II. ANALISIS KORELASI
1. Koefisien Korelasi Pearson¾ Koefisien Korelasi Moment Product¾ Korelasi Data Berskala Interval dan Rasio
2. Koefisien Korelasi Spearman¾ Korelasi Data Berskala Ordinal (Rank)
3. Koefisien Kontingensi¾ Korelasi Data yang Disusun dalam Baris - Kolom
4. Koefisien Korelasi Phi¾ Korelasi Data Berskala Nominal
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II. ANALISIS KORELASI
Analisis Korelasi merupakan studi yang membahastentang derajat keeratan hubungan antar peubah, yang dinyatakan dengan Koefisien Korelasi. Hubungan antarapeubah X dan Y dapat bersifat :
a. Positif, artinya jika X naik (turun) maka Y naik(turun).
b. Negatif, artinya jika X naik (turun) maka Y turun(naik).
c. Bebas, artinya naik turunnya Y tidak dipengaruhioleh X.
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II. ANALISIS KORELASI
Positif Negatif Bebas (Nol)
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1. KORELASI PEARSON
Rumus umum Koefisien Korelasi :r2 _ _JKG _ JKT - JKG _ JKR
- 1 JKT - JKT -JKT
r2 = Koefisien Determinasi (Koefisien Penentu)r = √ r2 = Koefisien KorelasiJKG = Jumlah Kuadrat Galat JKT = Jumlah Kuadrat Total JKR = Jumlah Kuadrat Regresi
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Rumus Koefisien Korelasi Pearson :nL xy - CL X) CL y)
Y = Variabel Terikat (Variabel Tidak Bebas)X = Variabel Bebas (Faktor)
Nilai r : – 1 ≤ r ≤ 1 Æ …. ≤ r2 ≤ ….
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5I0,21 0,40
I 6 0,07 0,207 0,50 0,90
\ 8 1,00 2,009 0,70 1,2010 0,14 0,3511 0,35 0,7012 0,28 0,65
-
- - •,
Data keuntungan usahatani (Y) pada berbagai luas lahan (X) :l
No Petani Luas Lahan (X) Keuntungan (Y)1 0,212 0,50
)
. I ,• ,
0,501,10
3 0,14 0,25l •
i 4 1,00 1,80•
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No X Y X2 Y2 XY1 0,21 0,50 0,0441 0,2500 0,10502 0,50 1,10 0,2500 1,2100 0,55003 0,14 0,25 0,0196 0,0625 0,03504 1,00 1,80 1,0000 3,2400 1,80005 0,21 0,40 0,0441 0,1600 0,08406 0,07 0,20 0,0049 0,0400 0,01407 0,50 0,90 0,2500 0,8100 0,45008 1,00 2,00 1,0000 4,0000 2,00009 0,70 1,20 0,4900 1,4400 0,840010 0,14 0,35 0,0196 0,1225 0,049011 0,35 0,70 0,1225 0,4900 0,245012 0,28 0,65 0,0784 0,4225 0,1820
JumlahRata-rata
n
5,100,4312
10,050,84
-
3,3232--
12,2475--
6,3540--
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∑ X = 5,10 ; ∑ Y = 10,05 ; ∑ X2 = 3,3232 ;
∑Y2 =12,2475 ; ∑ XY = 6,3540 ; n = 12nL xy - (L X)(Ly)
r = ------------J[n L X2 - CL x)2][n Ly2 - CLy)2]
12(6,3540) - (5,10)(10,05)r~----------------------------
)[12(3,3232) - (5,10)2][12(12,2475) - (10,05)2]
76,2480 - 51,2550r~~~~~~~~~~~~~~~
J[39,8784 - 26,0100] [146, 9700 - 101,0025]
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76,2480 - 51,2550r~------------------------------
) [39,8784 - 26,0100] [146, 9700 - 101,0025]
24,9930 24,9930r= ------
)[13,8684][45,9675] 25,2487
r = 0,9899 r2 = 0,9798 = 97,98 0/0
Nilai r2 = 97,98 % artinya sebesar 97,98 % variasi besarnya keuntungan (nilai Y) diperngaruhi oleh variasi besarnya luas lahan (nilai X).
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tPenr
g••
ujian Koefisien Korelasi Pearson :
1. H0 ≡ r = 0 lawan H1 ≡ r ≠ 0
2. Taraf Nyata α = 5 % = 0,053. Uji Statistik = Uji- t4. Wilayah Kritik (Daerah Penolakan H0) :
t < –tα/2(n-2) atau t > tα/2(n-2)
t < –t0,025(10) atau t > t0,025(10)
t < –2,228 atau t > 2,228
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5. Perhitungan :n-2
t=r 1- r2
t = 0,989912 - 2
1- 0,9798
t = 0,989910
0,0202
t = 0,9899 (22,2772) = 22,052
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6. Kesimpulan :
Karena nilai ( t = 22,052) > ( t0,025(10) = 2,228) maka disimpulkan untuk menolak H0, artinya terdapat hubungan yang signifikan antara
keuntungan usahatani (Y) dengan luas lahan garapan (X)
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6. Kesimpulan :
Nilai t = 22,052 dan t0,025(10) = 2,228.
Tolak H0•I-a.
Terima H0
Tolak H0
–2,228 2,228
22,052
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d
2. KORELASI SPEARMAN
1. Jika tidak6
aL
darnilai pengamatan yang sama
:
rs = 1 - n(n2 _ 1)
2. Jika ada nilai pengamatan yang sama :
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r
2. KORELASI SPEARMAN
L X2 + L y2 - L dr=-------
s 2~ CL X2)(L y2)
L T y
t3-t=--12
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No X • Y1 12 852 10 743 10 784 13 905 11 856 14 877 13 948 14 989 11 81
10 14 9111 10 7612 8 74
No X Rank1 8 12 10 33 10 34 10 35 11 5,56 11 5,57 12 .. 78 13 8,59 13 8,5
10 1411 14 1112 14 11
No X Rank1 74 1,52 74 1,53 76 34 78 45 81 56 85 6,57 85 6,58 87 89 90 9
10 91 1011 94 1112 98 12
-Data Pengalaman Usahatani (X) dan Penerapan Teknologi(Y) dari 12 petani :
-
•
11-•
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No X Y Rank-X Rank-Y d 2i1 12 85 7 6,5 0,252 10 74 3 1,5 2,253 10 78 3 4 1,004 13 90 8,5 9 0,255 11 85 5,5 6,5 1,006 14 87 11 8 9,007 13 94 8,5 11 6,258 14 98 11 12 1,009 11 81 5,5 5 0,2510 14 91 11 10 1,0011 10 76 3 3 0,0012 8 74 1 1,5 0,25
Jml 22,50
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i∑ d 2 = 22,50 n = 12
6(22,50)rs = 1- 12 (144 - 1)
135Ts = 1- 1716 = 1 - 0,0787
Ts = 0,9213
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5,5 28,5 211 3
Jml
RUMUS II :
Rank-X t Tx Rank-Y t .. Ty3 3 2,0 1,5 2 0,5
0tJ,I5f-p·~.....t0,52,0
6,5 2 0,5
•~ 2 123 -12
5,0 Jml 1,0
L x = 12 - 5, 0 = 138
~ 2 123 -12L Y = 12 - 1,0 = 142
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r L X2 + L y2 - L dr
s 2J (L X2)(Ly2)
138 + 142 - 22, 50r =
s 2.) (138)(142)
257,50rs = 279,9 71 = 0, 9197
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Pengujian Koefisien Korelasi Spearman :
1. H0 ≡ rs = 0 lawan H1 ≡ rs ≠ 0
2. Taraf Nyata α = 5 % = 0,05
3. Uji Statistik = Uji- t
4. Wilayah Kritik (Daerah Penolakan H0) :
t < –tα/2(n-1) atau t > tα/2(n-1)
t < –t0,025(10) atau t > t0,025(10)
t < –2,228 atau t > 2,228
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5. Perhitungan :n-2
t=r s 1- r2s
12 - 2t=O,9197 1-(0,9197)2
10t = 0,9197 0,1541
t = 0,9197(8,0560) = 7,409
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.•
6. Kesimpulan :..
Karena nilai ( t = 7,409) > ( t0,025(10) = 2,228)maka disimpulkan untuk menolak H0, artinya
.k.
0:..- terdapat hubungan yang signifikan antarapengalaman usahatani (X) dengan p
...
enerapan
/ teknologi (Y)\ ~.~
,..'/'I
./
..:..-I"o _
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3. KORELASI PHI
Koefisien korelasi phi rφ merupakan ukuran derajat keeratan hubungan antara dua variabel dengan skalanominal yang bersifat dikotomi (dipisahduakan).
Kolom JumlahBaris A B (A+B)
C D (C+D)Jumlah (A+C)AD-BC
(B+D) N
r (J = ---;J==(A==+====B==) (==C==+==D==)==(A==+====C)==(==B==+==D==)
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Uji signifikansi rφ dengan statistik χ2 Pearson :
Atau dengan rumus :2 _ N[ I(AD - Be)1 - OJ 5N]2
X - (A + B)(C + D) (A + C)(B + D)
Derajat Bebas χ2 = (b – 1)(k –1)
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Contoh : Data banyaknya petani tebu berdasarkanpenggunaan jenis pupuk dan cara tanam.
PupukTunggal
PupukMajemuk
Jumlah
Tanam Awal 5 9 14Keprasan 9 7 16Jumlah 14 16 30
Tentukan nilai Koefisien Korelasinya dan Ujilah padataraf nyata 1% apakah penggunaan jenis pupuk tergantung dari cara tanamnya ?
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PupukTunggal
PupukMajemuk
Jumlah
Tanam Awal 5 9 14Keprasan 9 7 16Jumlah 14 16 30
J
Jawab :
AD-BCr6=~~~~~~~~~
)(A + B)(C + D)(A + C)(B + D)
(5) (7) - (9) (9) -46 -46r - - ---
8 - (14)(16)(14)(16) - V50176 - 224
ro = -0,2054
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X >
Uji Koefisien Korelasi phi :
1. H0 ≡ rφ = 0 lawan H1 ≡ rφ ≠ 0
2. Taraf Nyata α = 5 % = 0,05
3. Uji Statistik = Uji- X2
4. Wilayah Kritik (Daerah Penolakan H0) :
2 20,05(1)
5. Perhitungan :
atau X2 > 3,841
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PupukTunggal
PupukMajemuk
Jumlah
oi ei oi eiTanam Awal 5 6,53 9 7,47 14Keprasan 9 7,47 7 8,53 16Jumlah 14 16 30
2 _ [(5 - 6J 53) - OJ 5]2 [(7 - 8J 53) - OJ 5]2X - 6 53 + ...+ 8 53
J J
x2 == 0,571
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0,05(1)
6. Kesimpulan
Karena nilai (X2 = 0,571) < (X2 = 6,635)maka H0 diterima artinya penggunaan jenispupuk tidak tergantung pada cara tanam.
_. ', -.. IIIII!IIII'.· ···liIel'.:._~
.j.. ••
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PupukTunggal
PupukMajemuk
Jumlah
Tanam Awal 5 9 14Keprasan 9 7 16Jumlah 14 16 302 _ N[ I(AD - Be)1 - 0, 5N]2
X -(A+B)(C+D)(A+C)(B+D)
2 30[1(35-81)1-15]2X = (14)(16)(14)(16)
2 30[ 1-461 - 15]2X = 50176 = 0,575
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PupukTunggal
PupukMajemuk
Jumlah
Tanam Awal 5 9 14Keprasan 9 7 16Jumlah 14 16 30
4. KORELASI ClARDAM-BEelRv=-----------------
.j(A + B)(C + D)(A + C)(B + D)
V = 1(5)(7) - (9) (9) I = 0, 2054J (14)(16)(14)(16)
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4. KORELASI KONTINGENSI
Koefisien kontingensi C merupakan ukuran korelasi antara dua variabel kategori yang disusun dalam tabel kontingensi berukuran ( b x k ).
Pengujian koefisien kontingensi C digunakan sebagai Uji Kebebasan (Uji Independensi) antara dua variabel. Jadi apabila hipotesis nol dinyatakan sebagai C = 0 diterima, berarti kedua variabel tersebut bersifat bebas.
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4. KORELASI KONTINGENSI
c=
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Contoh : (Ada anggapan bahwa pelayanan bank swasta terhadappara nasabahnya lebih memuaskan dari pada bank pemerintah. Untuk mengetah,ui hal tersebut, makadilakukan wawancara terhadap nasabah bank swasta dan bank pemerintah masing-masing sebanyak 40 orang. Hasil wawancara yang tercatat adalah :
Swasta PemerintahTidak Puas 16 10Netral 9 5Puas 15 25
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X >
Pengujian Hipotesis :
1. H0 ≡ C = 0 lawan H1 ≡ C ≠ 0
2. Taraf Nyata α = 5 % = 0,05
3. Uji Statistik = Uji- X2
4. Wilayah Kritik (Daerah Penolakan H0) :2 2
0,05(2)
5. Perhitungan :
atau X2 > 5,991
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Swasta PemerintahJumlahoi ei oi ei
Tidak Puas 16 13 10 13 26Netral 9 7 5 7 14Puas 15 20 25 20 40Jumlah 40 40 80
L
Pengujian Hipotesis :
X2 = (0't
- e·)2t
e·t
2 (16 - 13)2 (5 - 7)2X = + 000 + = 5 02713 7'
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.J
c=
c = 5,0275, 027 + 80 = 0, 0591 = 0, 243
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0,05(2)
6. Kesimpulan :
Karena nilai (X2 = 5,027) < (X2 = 5,991) makaH0 diterima artinya hubungan antara keduavariabel tersebut bersifat tidak nyata (tingkat kepuasan nasabah terhadap pelayanan bank swasta tidak berbeda nyata dengan bank pemerintah).
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5. KORELASI BISERI
Koefisien korelasi biseri merupakan ukuran derajat keeratan hubungan antara Y yang kontinu (kuantitatif) dengan X yang diskrit bersifat dikotomi.
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5. KORELASI BISERI(Yl - YZ)pq
Tb =
rb = Koefisien Korelasi BiseriY1Y2p
===
Rata-rata Variabel Y untukRata-rata Variabel Y untukProporsi kategori ke-1
kategori ke-1 kategori ke-2
q = 1 – pu = Tinggi ordinat kurva z dengan peluang p dan qSy = Simpangan Baku Variabel Y
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Data berikut merupakan hasil nilai ujian statistika dari145 mahasiswa yang belajar dan tidak belajar.
Nilai UjianJumlah Mahasiswa
TotalBelajar Tidak Belajar
55 – 59 1 31 3260 – 64 0 27 2765 – 69 1 30 3170 – 74 2 16 1875 – 79 5 12 1780 – 84 6 3 985 – 89 6 5 11Total 21 124 145
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Interval Y1 F FY1 Y2 F FY255 – 5960 – 6465 – 6970 – 7475 – 7980 – 8485 – 89
57626772778287
1012566
570
67144385492522
57626772778287
312730161235
1767167420101152924246435
JumlahRata-rata
21 166779,38
124 820866,19
Yl = 79,38; Yz = 66,19 ; P = 21/45 = 0,14
q = 0,86 ; Sy = 9,26 ; U = 0,223
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(79,38 - 66, 19) (0,14) (0,86)r, = (0, 223) (9,26)
Tb =(13,19)(0,120)
2,065 = 0'769
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2 JKRr = JKT =
6. KORELASI LINEAR GANDA DAN PARSIAL
1. Korelasi Linear Ganda
Untuk regresi linier ganda Y = b0 + b1 X1 + b2 X2 +… + bk Xk , maka koefisien korelasi ganda dihitung dari Koefsisien Determinasi dengan rumus :
b1x1y + b2X2Y + ...+ bkXkYLy2
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1. Korelasi Linear Ganda
rZ = JKR JKT
= b1x1y + bz
zY + .
..+ bkXkY
x
JKR = Jumlah Kuadrat RegresiJKT = Jumlah Kuadrat Total
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Skor tes (X1) Frek. Bolos (X2) Nilai Ujian (Y)65 1 8550 7 7455 5 7665 2 9055 6 8570 3 8765 2 9470 5 9855 4 8170 3 9150 1 7655 4 74
∑ X1 = 725
∑ X2 = 43
2∑ X1 = 44.475
2 = 195∑ X2
∑ X1X2 = 2.540
∑ Y = 1.011
∑ X1Y = 61.685
∑ X2Y = 3.581
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∑ X1Y = b0 ∑ X1 + b1 ∑
∑ X2Y = b0 ∑ X2 + b1 ∑
b1 = ∑ X1Yb2 ∑ X2Y
Regresi Dugaan : Y = b0 + b1 X1 + b2 X2. Kemudian persamaan normal yang dapat dibentuk yaitu :
∑ Y = b0 n + b1 ∑ X1 + b2 ∑ X2
X12 + b2 ∑ X1X2
X1X2 + b2 ∑ X22
Matrik dari persamaan normal diatas :
n ∑ X1 ∑ X2 b0 ∑ Y∑ X1 ∑ X1
2 ∑ X1X2
∑ X2 ∑ X1X2 ∑ X22
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Nilai b0 , b1 dan b2 dapat dihitung melalui :
1. Matriks :a. Determinan Matriks, b. Invers Matriks
2. Substitusi, dan (b) Eliminasi
Melalui salah satu cara diatas diperoleh nilaib0 = 27,254b1 = 0,922b2 = 0,284
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∑ X1 = 725 ∑ X12 = 44.475 ∑ Y = 1.011
∑ X2 = 43 ∑ X22 = 195 ∑ X1X2 = 2.540
b∑o X=1Y2=7,6215.648;5
b1 =∑ 0X,29Y2=2 3.;58b12
= 0,2∑8Y42 = 85.905
Analisis Ragam :
FK = (∑Y)2 / n = (1,011)2 / 12 = 85.176,75JKT = ∑ Y2 – FK = 85.905 – 85,175,75 = 728,25JKR = b1 [ (∑ X1Y – (∑X1)(∑Y)/n ] + b2 [ (∑ X2Y – (∑X2)(∑Y)/n]
= 0,922 [ (61.685 – (725)(1.011)/12 ] +
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0,284 [ (3.581 – (43)(1.011)/12 ]= 556,463 – 11.867 = 544,596
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No Variasi DB JK KT F F5%
12
RegresiGalat
29
544,596183,654
272,29820,406
13,344 4,256
Total 11 728,250
Analisis Ragam :
JKG = JKT – JKR = 728,25 – 544,596 = 183,654
2 JKR 544,596r = JKT = 728,250 = 0,7478
r = .jO, 7478 = 0,8648
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Pengujian Korelasi Ganda :
(r2)j(k)F - --_..:: ....:, -----::-
- (1 - r2)j(n - k - 1)
Tolak H 0 jika F > F 0,05(k; n-k-l)
Tolak H 0 jika F > F 0,05(2; 9)
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r2 = 0, 7478 ,· k = 2
,· n - k - 1 = 9
(r2)j(k)F=------(1 - r2)j(n - k - 1)
(0,7478)/2F = (0,2522)/9 = 13,343
F0,05(2 ; 9) = 4,2565
Karena nilai ( F = 13,343) > ( F0,05(2 ; 9) = 4,2565) artinya koefisien korelasi ganda tersebut bersifat nyata.
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2. Koefisien Korelasi Parsial :
A. Korelasi X1 dengan Y jika X2 tetap :Tyl - Ty2T12
B. Korelasi X2 dengan Y jika X1 tetap :
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2. Koefisien Korelasi Parsial :n LXIY - (LXI)(L Y)
rI=y j[nLXi - (LXI)Z][nLYZ - (LY)Z]
nLXZY - (LXZ)(LY)r Z = ----;::::::==============================
y j[nLX~ - CLXZ)Z][nLYZ - CLY)Z]
nLX1XZ - (LX1)(LX2)r12 = ----;::::::=========================================:
J[nLXi - (LX1)Z][nLX~ - (LXZ)Z]
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y
)
2. Koefisien Korelasi Parsial :
ry1 = 0,862 ; r 2 = 0,743 ; ry2 = –0,2422rY2
2 = 0,059 ; r12 = –0,349 ; r12 = 0,122
A. Korelasi TXyl1 d-enTgy2aTn12Y jika X2 tetap :
T yl /2 = -;::::::============j(l - r;z)(l - Ti2)
0,862 - [(-0,242)(-0,349)]r 1/2 = --;::=================---
y 0,059)(1 - 0,122)
0,778ryl/2 = 0, 909 = 0, 855
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y
j(1
B. Korelasi X2 dengan Y jika X1 tetap :
ry1 = 0,862 ; r 2 = 0,743 ; ry2 = –0,2422rY2
2 = 0,059 ; r12 = –0,349 ; r12
ry2jl = ,.....--------
r;1)(1 - riz)
= 0,122
-0,242 - [(0,862)(-0,349)]r 2 j 1 = --;:::::::======================--
y J(1 - 0,941)(1- 0,122)
0,059ryZ/l = 0,475 = 0,124
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Pengujian Koefisien Korelasi Parsial :n-3
t = Tyi/j 1- T;i/j
A. Korelasi X1 dengan Y jika X2 tetap (ry1/2) :
B. Korelasi X2 dengan Y jika X1 tetap (ry2/1) :n-3t = Ty2/1 1- Ty22/1
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= 0,855 ;= 0,124 ;
r
/2
r
A. Korelasi X1 dengan Y jika X2 tetap (ry1/2) :
ry1/22 y1.../2 = 0,731 ;
ry2/1
t = ry1n-31- Ty2l/2
2Y2/1 = 0,015
12 - 3 = 4,9491 t = 0, 855 1 _ 0, 731
t0,025(9) = 2,262 Æ Korelasi Signifikan
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y1/
Y2/
B. Korelasi X2 dengan Y jika X1 tetap (ry2/1) :
ry1/2 = 0,855 ; r 2
ry2/1 = 0,124 ; r 2
n-3
= 0,731 ;= 0,015
t = ry2/1 1- r 2 /y2 1
12 - 3 = 0,374t = 0, 124 1 _ 0,015
t0,025(9) = 2,262 Æ Korelasi Tidak Signifikan
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7. KORELASI DATA DIKELOMPOKKAN
nI/i xy - (L/x x)CI/y y)r=~=======================
j[nI/Xx2 - (L/xX)2] [nI/yy2 - CI/yy)2]
Atau :nI/i c.c, - (L/x Cx)CI/y Cy)r=~========================
j[nI/xci - cu. Cx)2] [nI/yC; - CI/y Cy)2]
•
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Out Put(Y)
Jml (fy )1 – 20 21 – 40 41 – 60 J 61 – 80 81 – 100
1 – 2021 – 4041 – 6061 – 8081 – 100
1 241
13521
2732
234
491587
Jml (fx) 1 7 12 14 9 n = 43
-Pendapatan (X) dan Pengeluaran (Y) Bulanan (ribu(rupiah) karyawan sebuah pabrik :
:/
In Put (X)
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YX 10,5 30,5 50,5 70,5 90,5
Cy .Cx – 2 – 1 0 1 2 fy fy.Cy fy.Cy2 fi CxCy
10,5 – 2 •1 2 1 4 – 8 16 8
30,5 – 1 4 3 2 9 – 9 9 2
50,5• 0 1 5 7 2 15 0 0 0
70,5 1 2 3 3 8 8 8 9
90,5 2 1 2 4 7 14 28 20
fx 1 7 12 14 9 43 5 61 39
fx.Cx – 2 – 7 0 14 18 23
fx.Cx2 4 7 0 14 36
fi Cx.Cy 4 8 0 5 22 39
-
-
..
6-1
•
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Mencari •fi Cx.Cy = 8 pada titik tengah (X) = 30,5adalah : 8 = (2)(–2)(–1) + (4)(–1)(–1) + (1)(0)(–1)
nL.fi c.c; - (Lfx Cx)(L.fy cy)r=~========================
[nL.fxCi - (Lfx Cx)2] [nL.fyc~ - (L.fy Cy)2]
. 43 (39) - (23) (5)r=~-----------------------
.)[(43)(61) - (23)2][(43)(61) - (5)2]..
r = 0,67
... . .. -