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Cara Manual Korelasi Kendall Parsial

๐Ÿ“‹ Daftar Isi

Berikut adalah cara manual Korelasi Kendall Parsial.

Hipotesis

H0 : ฯ„xy,z = 0 (Tidak ada korelasi antara x dan y jika z konstan)
H1 : ฯ„xy,z โ‰  0 (Ada korelasi antara x dan y jika z konstan)

Perangkingan

Misalkan :
X โ†’ Tinggi Badan
Y โ†’ Berat Badan
Z โ†’ Lingkar Dada

N = 16, sampel kecil, dan tidak ada observasi bernilai sama

Untuk nilai ฯ„xy

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Subjek Tinggi Rank Tinggi Berat Rank Berat Pasangan Berat (Konkordan) Pasangan Berat (Diskordan)
O 140 1 37 1 15 0
K 145 2 38 2 14 0
L 152 3 46 5 11 2
F 153 4 47 6 10 2
I 155 5 43 4 10 1
M 158 6 41 3 10 0
C 160 7 50 8 8 1
B 161 8 59 12 4 4
D 163 9 56 10 5 2
N 165 10 65 14 2 4
E 168 11 58 11 3 2
G 170 12 54 9 3 1
A 171 13 49 7 3 0
H 173 14 60 13 2 0
J 180 15 73 15 1 0
P 181 16 85 16 0 0
P = 101 Q = 19
\[ S = P – Q \] \[ S = 101 – 19 \] \[ S = 82 \] \[ \tau_{xy} = \frac{S}{\frac{1}{2} N(N-1)} \] \[ \tau_{xy} = \frac{82}{\frac{1}{2} (16)(16-1)} \] \[ \tau_{xy} = 0,683 \]

Untuk nilai ฯ„xz

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Subjek Tinggi Rank Tinggi Lingkar Rank Lingkar Pasangan Berat (Konkordan) Pasangan Berat (Diskordan)
K 145 2 92 1 14 1
M 158 6 93 2 10 4
O 140 1 95 3 13 0
I 155 5 96 4 10 2
C 160 7 98 5 9 2
A 171 13 99 6 3 7
F 153 4 100 7 8 1
L 152 3 101 8 8 0
G 170 12 102 9 3 4
B 161 8 103 10 6 0
E 168 11 104 11 3 2
D 163 9 105 12 4 0
H 173 14 106 13 2 1
N 165 10 107 14 2 0
J 180 15 108 15 1 0
P 181 16 109 16 0 0
P = 96 Q = 24
\[ S = P – Q \] \[ S = 96 – 24 \] \[ S = 72 \] \[ \tau_{xz} = \frac{S}{\frac{1}{2} N(N-1)} \] \[ \tau_{xz} = \frac{72}{\frac{1}{2} (16)(16-1)} \] \[ \tau_{xz} = 0,6 \]

Untuk nilai ฯ„yz

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Subjek Berat Rank Berat Lingkar Rank Lingkar Pasangan Berat (Konkordan) Pasangan Berat (Diskordan)
K 38 2 92 1 14 1
M 41 3 93 2 13 1
O 37 1 95 3 13 0
I 43 4 96 4 12 0
C 50 8 98 5 8 3
A 49 7 99 6 8 2
F 47 6 100 7 8 1
L 46 5 101 8 8 0
G 54 9 102 9 7 0
B 59 12 103 10 4 2
E 58 11 104 11 4 1
D 56 10 105 12 4 0
H 60 13 106 13 3 0
N 65 14 107 14 2 0
J 73 15 108 15 1 0
P 85 16 109 16 0 0
P = 109 Q = 11
\[ S = P – Q \] \[ S = 109 – 11 \] \[ S = 98 \] \[ \tau_{yz} = \frac{S}{\frac{1}{2} N(N-1)} \] \[ \tau_{yz} = \frac{98}{\frac{1}{2} (16)(16-1)} \] \[ \tau_{yz} = 0,816 \]

Statistik Uji

\[ \tau_{xy,z} = \frac{\tau_{xy} – \tau_{xz} \tau_{yz}}{\sqrt{(1-\tau_{xz}^2)(1-\tau_{yz}^2)}} \] \[ \tau_{xy,z} = \frac{0,683 – (0,6)(0,816)}{\sqrt{(1-0,6^2)(1-0,816^2)}} \] \[ \tau_{xy,z} = 0,41821 \]

Keputusan

Untuk n = 16 dan merupakan uji dua arah

Sehingga akan Tolak H0 jika
– ฯ„hitung > ฯ„(1-ฮฑ/2;n) atau
– ฯ„hitung < -ฯ„(1-ฮฑ/2;n)

Nilai ฯ„(0,975;16) = 0.361 (Tabel S dibuku Castellan)

Maka Tolak H0

Kesimpulan

Dengan tingkat signifikansi sebesar 5%, maka terdapat cukup bukti untuk mengatakan bahwa ada hubungan antara variabel tinggi dan berat badan, apabila lingkar dada konstan.


Materi Lengkap

Untuk lebih memperdalam materi Korelasi Kendall Parsial, maka Anda bisa baca juga bagian berikut:


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