24-korelasi-point-biserial.docx
This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA
Overview
Download & View 24-korelasi-point-biserial.docx as PDF for free.
More details
- Words: 1,607
- Pages: 11
STATYSTIC NON-PARAMETRIC
NAMA
:
NURUL CHAIRUNNISA UTAMI PUTRI
NIM
:
1620070008
FAK / JUR :
SAINS & TEKNOLOGI / MATEMATIKA
http://roelcup.wordpress.com
UNIVERSITAS ISLAM AS-SYAFI’IYAH JAKARTA TIMUR
2010
KORELASI POINT BISERIAL
Rumus : 𝛤𝑝𝑏𝑖 =
𝑥𝑖 − 𝑥𝑡 𝑝 √ 𝑆𝑡 1−𝑝
Keterangan : 𝑥𝑖 = Mean Butir yang Menjawab Benar 𝑥𝑡 = Mean Skor Total 𝑆𝑡 = Simpangan Baku Total 𝑝 = Proposi yang Menjawab Benar
CONTOH SOAL : Multiple Choice (𝐴, 𝐵, 𝐶, 𝐷, 𝐸) o Jika jawab Benar → Skor 1 o Jika jawab Salah → Skor 0 Uji Validitas & Reliabilitas Data Uji Validitas yaitu Mencari Korelasi antara Butir Soal & Totalnya Case : Soal = 7 Buah Peserta =10 B = Benar / Proposi
Datanya Adalah : BUTIR Peserta
Total (X) 1
2
3
4
5
6
7
1
1
1
1
1
0
0
0
4
2
1
1
0
1
1
1
0
5
3
0
1
1
1
0
0
0
3
4
1
1
0
0
0
0
0
2
5
0
1
0
0
0
0
0
1
6
1
1
1
1
1
1
1
7
7
1
1
1
1
1
1
0
6
8
0
0
0
0
0
0
0
0
9
1
1
0
0
1
0
0
3
10
1
1
1
1
1
0
0
5
B
7
9
5
6
5
3
1
36
36 = 3,6 10 2 1 = (𝑛 ∙ ∑ 𝑥 2 − (∑ 𝑥) ) 𝑛 ∙ (𝑛 − 1)
𝑥𝑡 = 𝑆𝑡 2
𝑆𝑡 2 =
1 (10 ∙ 174 − (36)2 ) 10 ∙ (10 − 1)
1 (1740 − 1296) 90 1 (444) = 90
𝑆𝑡 2 = 𝑆𝑡 2
𝑆𝑡 2 = 4,9333 𝑆𝑡 = √4,9333 𝑆𝑡 = 2,2211
1. Antara Butir Ke-1 dengan Total Peserta
Butir ke-1
Total (X)
𝑥2
1 2 3 4 5 6 7 8 9 10
1 1 0 1 0 1 1 0 1 1
4 5 3 2 1 7 6 0 3 5
16 25 9 4 1 49 36 0 9 25
B
7
36
174
7 = 0,7 10 (4 ∙ 1) (5 ∙ 1) (3 ∙ 0) (2 ∙ 1) (1 ∙ 0) (7 ∙ 1) (6 ∙ 1) (0 ∙ 0) (3 ∙ 1) 𝑥𝑖 = + + + + + + + + 7 7 7 7 7 7 7 7 7 (5 ∙ 1) + 7 4 5 0 2 0 7 6 0 3 5 𝑥𝑖 = + + + + + + + + + 7 7 7 7 7 7 7 7 7 7 32 𝑥𝑖 = 7 𝑝=
𝑥𝑖 = 4,5714 𝛤𝑝𝑏𝑖 =
𝑥𝑖 − 𝑥𝑡 𝑝 √ 𝑆𝑡 1−𝑝
𝛤𝑝𝑏𝑖 =
4,5714 − 3,6 0,7 √ 2,2211 1 − 0,7
𝛤𝑝𝑏𝑖 =
0,97 0,7 √ 2,2211 0,3
𝛤𝑝𝑏𝑖 = (0,4374)√2,3333 𝛤𝑝𝑏𝑖 = (0,4374)(1,5275) 𝛤𝑝𝑏𝑖 = 0,6681
2. Antara Butir Ke-2 dengan Total Peserta
Butir ke-2
Total (X)
𝑥2
1 2 3 4 5 6 7 8 9 10
1 1 1 1 1 1 1 0 1 1
4 5 3 2 1 7 6 0 3 5
16 25 9 4 1 49 36 0 9 25
B
9
36
174
9 = 0,9 10 (4 ∙ 1) (5 ∙ 1) (3 ∙ 1) (2 ∙ 1) (1 ∙ 1) (7 ∙ 1) (6 ∙ 1) (0 ∙ 0) (3 ∙ 1) 𝑥𝑖 = + + + + + + + + 9 9 9 9 9 9 9 9 9 (5 ∙ 1) + 9 4 5 3 2 1 7 6 0 3 5 𝑥𝑖 = + + + + + + + + + 9 9 9 9 9 9 9 9 9 9 36 𝑥𝑖 = 9 𝑝=
𝑥𝑖 = 4 𝛤𝑝𝑏𝑖 =
𝑥𝑖 − 𝑥𝑡 𝑝 √ 𝑆𝑡 1−𝑝
𝛤𝑝𝑏𝑖 =
4 − 3,6 0,9 √ 2,2211 1 − 0,9
𝛤𝑝𝑏𝑖 =
0,4 0,9 √ 2,2211 0,1
𝛤𝑝𝑏𝑖 = (0,1801)√9,00 𝛤𝑝𝑏𝑖 = (0,1801)(3,00) 𝛤𝑝𝑏𝑖 = 0,5403
3. Antara Butir Ke-3 dengan Total Peserta
Butir ke-3
Total (X)
𝑥2
1 2 3 4 5 6 7 8 9 10
1 0 1 0 0 1 1 0 0 1
4 5 3 2 1 7 6 0 3 5
16 25 9 4 1 49 36 0 9 25
B
5
36
174
5 = 0,5 10 (4 ∙ 1) (5 ∙ 0) (3 ∙ 1) (2 ∙ 0) (1 ∙ 0) (7 ∙ 1) (6 ∙ 1) (0 ∙ 0) (3 ∙ 0) 𝑥𝑖 = + + + + + + + + 5 5 5 5 5 5 5 5 5 (5 ∙ 1) + 5 4 0 3 0 0 7 6 0 0 5 𝑥𝑖 = + + + + + + + + + 5 5 5 5 5 5 5 5 5 5 25 𝑥𝑖 = 5 𝑝=
𝑥𝑖 = 5 𝛤𝑝𝑏𝑖 =
𝑥𝑖 − 𝑥𝑡 𝑝 √ 𝑆𝑡 1−𝑝
𝛤𝑝𝑏𝑖 =
5 − 3,6 0,5 √ 2,2211 1 − 0,5
𝛤𝑝𝑏𝑖 =
1,4 0,5 √ 2,2211 0,5
𝛤𝑝𝑏𝑖 = (0,6303)√1,00 𝛤𝑝𝑏𝑖 = (0,6303)(1,00) 𝛤𝑝𝑏𝑖 = 0,6303
4. Antara Butir Ke-4 dengan Total Peserta
Butir ke-4
Total (X)
𝑥2
1 2 3 4 5 6 7 8 9 10
1 1 1 0 0 1 1 0 0 1
4 5 3 2 1 7 6 0 3 5
16 25 9 4 1 49 36 0 9 25
B
6
36
174
6 = 0,6 10 (4 ∙ 1) (5 ∙ 1) (3 ∙ 1) (2 ∙ 0) (1 ∙ 0) (7 ∙ 1) (6 ∙ 1) (0 ∙ 0) (3 ∙ 0) 𝑥𝑖 = + + + + + + + + 6 6 6 6 6 6 6 6 6 (5 ∙ 1) + 6 4 5 3 0 0 7 6 0 0 5 𝑥𝑖 = + + + + + + + + + 6 6 6 6 6 6 6 6 6 6 30 𝑥𝑖 = 6 𝑝=
𝑥𝑖 = 5 𝛤𝑝𝑏𝑖 =
𝑥𝑖 − 𝑥𝑡 𝑝 √ 𝑆𝑡 1−𝑝
𝛤𝑝𝑏𝑖 =
5 − 3,6 0,6 √ 2,2211 1 − 0,6
𝛤𝑝𝑏𝑖 =
1,40 0,6 √ 2,2211 0,4
𝛤𝑝𝑏𝑖 = (0,6303)√1,5 𝛤𝑝𝑏𝑖 = (0,6303)(1,2247) 𝛤𝑝𝑏𝑖 = 0,7720
5. Antara Butir Ke-5 dengan Total Peserta
Butir ke-5
Total (X)
𝑥2
1 2 3 4 5 6 7 8 9 10
0 1 0 0 0 1 1 0 1 1
4 5 3 2 1 7 6 0 3 5
16 25 9 4 1 49 36 0 9 25
B
5
36
174
5 = 0,5 10 (4 ∙ 0) (5 ∙ 1) (3 ∙ 0) (2 ∙ 0) (1 ∙ 0) (7 ∙ 1) (6 ∙ 1) (0 ∙ 0) (3 ∙ 1) 𝑥𝑖 = + + + + + + + + 5 5 5 5 5 5 5 5 5 (5 ∙ 1) + 5 0 5 0 0 0 7 6 0 3 5 𝑥𝑖 = + + + + + + + + + 5 5 5 5 5 5 5 5 5 5 26 𝑥𝑖 = 5 𝑝=
𝑥𝑖 = 4 𝛤𝑝𝑏𝑖 =
𝑥𝑖 − 𝑥𝑡 𝑝 √ 𝑆𝑡 1−𝑝
𝛤𝑝𝑏𝑖 =
5,20 − 3,6 0,5 √ 2,2211 1 − 0,5
𝛤𝑝𝑏𝑖 =
1,6 0,5 √ 2,2211 0,5
𝛤𝑝𝑏𝑖 = (0,7204)√1,00 𝛤𝑝𝑏𝑖 = (0,7204)(1,00) 𝛤𝑝𝑏𝑖 = 0,7204
6. Antara Butir Ke-6 dengan Total Peserta
Butir ke-6
Total (X)
𝑥2
1 2 3 4 5 6 7 8 9 10
0 1 0 0 0 1 1 0 0 0
4 5 3 2 1 7 6 0 3 5
16 25 9 4 1 49 36 0 9 25
B
3
36
174
3 = 0,3 10 (4 ∙ 0) (5 ∙ 1) (3 ∙ 0) (2 ∙ 0) (1 ∙ 0) (7 ∙ 1) (6 ∙ 1) (0 ∙ 0) (3 ∙ 0) 𝑥𝑖 = + + + + + + + + 3 3 3 3 3 3 3 3 3 (5 ∙ 0) + 3 0 5 0 0 0 7 6 0 0 0 𝑥𝑖 = + + + + + + + + + 3 3 3 3 3 3 3 3 3 3 18 𝑥𝑖 = 3 𝑝=
𝑥𝑖 = 6 𝛤𝑝𝑏𝑖 =
𝑥𝑖 − 𝑥𝑡 𝑝 √ 𝑆𝑡 1−𝑝
𝛤𝑝𝑏𝑖 =
6 − 3,6 0,3 √ 2,2211 1 − 0,3
𝛤𝑝𝑏𝑖 =
2,4 0,3 √ 2,2211 0,7
𝛤𝑝𝑏𝑖 = (1,0805)√0,4286 𝛤𝑝𝑏𝑖 = (0,1801)(0,6547) 𝛤𝑝𝑏𝑖 = 0,7074
7. Antara Butir Ke-7 dengan Total Peserta
Butir ke-7
Total (X)
𝑥2
1 2 3 4 5 6 7 8 9 10
0 0 0 0 0 1 0 0 0 0
4 5 3 2 1 7 6 0 3 5
16 25 9 4 1 49 36 0 9 25
B
1
36
174
1 = 0,1 10 (4 ∙ 0) (5 ∙ 0) (3 ∙ 0) (2 ∙ 0) (1 ∙ 0) (7 ∙ 1) (6 ∙ 0) (0 ∙ 0) (3 ∙ 0) 𝑥𝑖 = + + + + + + + + 1 1 1 1 1 1 1 1 1 (5 ∙ 0) + 1 0 0 0 0 0 7 0 0 0 0 𝑥𝑖 = + + + + + + + + + 1 1 1 1 1 1 1 1 1 1 7 𝑥𝑖 = 1 𝑝=
𝑥𝑖 = 7 𝛤𝑝𝑏𝑖 =
𝑥𝑖 − 𝑥𝑡 𝑝 √ 𝑆𝑡 1−𝑝
𝛤𝑝𝑏𝑖 =
7 − 3,6 0,1 √ 2,2211 1 − 0,1
0,1 𝛤𝑝𝑏𝑖 = 3,4√ 0,9 𝛤𝑝𝑏𝑖 = (0,1801)√0,1111 𝛤𝑝𝑏𝑖 = (1,5308)(0,3333) 𝛤𝑝𝑏𝑖 = 0,5103
Nurul Chairunnisa Utami Putri : http://roelcup.wordpress.com [email protected] [email protected]
NAMA
:
NURUL CHAIRUNNISA UTAMI PUTRI
NIM
:
1620070008
FAK / JUR :
SAINS & TEKNOLOGI / MATEMATIKA
http://roelcup.wordpress.com
UNIVERSITAS ISLAM AS-SYAFI’IYAH JAKARTA TIMUR
2010
KORELASI POINT BISERIAL
Rumus : 𝛤𝑝𝑏𝑖 =
𝑥𝑖 − 𝑥𝑡 𝑝 √ 𝑆𝑡 1−𝑝
Keterangan : 𝑥𝑖 = Mean Butir yang Menjawab Benar 𝑥𝑡 = Mean Skor Total 𝑆𝑡 = Simpangan Baku Total 𝑝 = Proposi yang Menjawab Benar
CONTOH SOAL : Multiple Choice (𝐴, 𝐵, 𝐶, 𝐷, 𝐸) o Jika jawab Benar → Skor 1 o Jika jawab Salah → Skor 0 Uji Validitas & Reliabilitas Data Uji Validitas yaitu Mencari Korelasi antara Butir Soal & Totalnya Case : Soal = 7 Buah Peserta =10 B = Benar / Proposi
Datanya Adalah : BUTIR Peserta
Total (X) 1
2
3
4
5
6
7
1
1
1
1
1
0
0
0
4
2
1
1
0
1
1
1
0
5
3
0
1
1
1
0
0
0
3
4
1
1
0
0
0
0
0
2
5
0
1
0
0
0
0
0
1
6
1
1
1
1
1
1
1
7
7
1
1
1
1
1
1
0
6
8
0
0
0
0
0
0
0
0
9
1
1
0
0
1
0
0
3
10
1
1
1
1
1
0
0
5
B
7
9
5
6
5
3
1
36
36 = 3,6 10 2 1 = (𝑛 ∙ ∑ 𝑥 2 − (∑ 𝑥) ) 𝑛 ∙ (𝑛 − 1)
𝑥𝑡 = 𝑆𝑡 2
𝑆𝑡 2 =
1 (10 ∙ 174 − (36)2 ) 10 ∙ (10 − 1)
1 (1740 − 1296) 90 1 (444) = 90
𝑆𝑡 2 = 𝑆𝑡 2
𝑆𝑡 2 = 4,9333 𝑆𝑡 = √4,9333 𝑆𝑡 = 2,2211
1. Antara Butir Ke-1 dengan Total Peserta
Butir ke-1
Total (X)
𝑥2
1 2 3 4 5 6 7 8 9 10
1 1 0 1 0 1 1 0 1 1
4 5 3 2 1 7 6 0 3 5
16 25 9 4 1 49 36 0 9 25
B
7
36
174
7 = 0,7 10 (4 ∙ 1) (5 ∙ 1) (3 ∙ 0) (2 ∙ 1) (1 ∙ 0) (7 ∙ 1) (6 ∙ 1) (0 ∙ 0) (3 ∙ 1) 𝑥𝑖 = + + + + + + + + 7 7 7 7 7 7 7 7 7 (5 ∙ 1) + 7 4 5 0 2 0 7 6 0 3 5 𝑥𝑖 = + + + + + + + + + 7 7 7 7 7 7 7 7 7 7 32 𝑥𝑖 = 7 𝑝=
𝑥𝑖 = 4,5714 𝛤𝑝𝑏𝑖 =
𝑥𝑖 − 𝑥𝑡 𝑝 √ 𝑆𝑡 1−𝑝
𝛤𝑝𝑏𝑖 =
4,5714 − 3,6 0,7 √ 2,2211 1 − 0,7
𝛤𝑝𝑏𝑖 =
0,97 0,7 √ 2,2211 0,3
𝛤𝑝𝑏𝑖 = (0,4374)√2,3333 𝛤𝑝𝑏𝑖 = (0,4374)(1,5275) 𝛤𝑝𝑏𝑖 = 0,6681
2. Antara Butir Ke-2 dengan Total Peserta
Butir ke-2
Total (X)
𝑥2
1 2 3 4 5 6 7 8 9 10
1 1 1 1 1 1 1 0 1 1
4 5 3 2 1 7 6 0 3 5
16 25 9 4 1 49 36 0 9 25
B
9
36
174
9 = 0,9 10 (4 ∙ 1) (5 ∙ 1) (3 ∙ 1) (2 ∙ 1) (1 ∙ 1) (7 ∙ 1) (6 ∙ 1) (0 ∙ 0) (3 ∙ 1) 𝑥𝑖 = + + + + + + + + 9 9 9 9 9 9 9 9 9 (5 ∙ 1) + 9 4 5 3 2 1 7 6 0 3 5 𝑥𝑖 = + + + + + + + + + 9 9 9 9 9 9 9 9 9 9 36 𝑥𝑖 = 9 𝑝=
𝑥𝑖 = 4 𝛤𝑝𝑏𝑖 =
𝑥𝑖 − 𝑥𝑡 𝑝 √ 𝑆𝑡 1−𝑝
𝛤𝑝𝑏𝑖 =
4 − 3,6 0,9 √ 2,2211 1 − 0,9
𝛤𝑝𝑏𝑖 =
0,4 0,9 √ 2,2211 0,1
𝛤𝑝𝑏𝑖 = (0,1801)√9,00 𝛤𝑝𝑏𝑖 = (0,1801)(3,00) 𝛤𝑝𝑏𝑖 = 0,5403
3. Antara Butir Ke-3 dengan Total Peserta
Butir ke-3
Total (X)
𝑥2
1 2 3 4 5 6 7 8 9 10
1 0 1 0 0 1 1 0 0 1
4 5 3 2 1 7 6 0 3 5
16 25 9 4 1 49 36 0 9 25
B
5
36
174
5 = 0,5 10 (4 ∙ 1) (5 ∙ 0) (3 ∙ 1) (2 ∙ 0) (1 ∙ 0) (7 ∙ 1) (6 ∙ 1) (0 ∙ 0) (3 ∙ 0) 𝑥𝑖 = + + + + + + + + 5 5 5 5 5 5 5 5 5 (5 ∙ 1) + 5 4 0 3 0 0 7 6 0 0 5 𝑥𝑖 = + + + + + + + + + 5 5 5 5 5 5 5 5 5 5 25 𝑥𝑖 = 5 𝑝=
𝑥𝑖 = 5 𝛤𝑝𝑏𝑖 =
𝑥𝑖 − 𝑥𝑡 𝑝 √ 𝑆𝑡 1−𝑝
𝛤𝑝𝑏𝑖 =
5 − 3,6 0,5 √ 2,2211 1 − 0,5
𝛤𝑝𝑏𝑖 =
1,4 0,5 √ 2,2211 0,5
𝛤𝑝𝑏𝑖 = (0,6303)√1,00 𝛤𝑝𝑏𝑖 = (0,6303)(1,00) 𝛤𝑝𝑏𝑖 = 0,6303
4. Antara Butir Ke-4 dengan Total Peserta
Butir ke-4
Total (X)
𝑥2
1 2 3 4 5 6 7 8 9 10
1 1 1 0 0 1 1 0 0 1
4 5 3 2 1 7 6 0 3 5
16 25 9 4 1 49 36 0 9 25
B
6
36
174
6 = 0,6 10 (4 ∙ 1) (5 ∙ 1) (3 ∙ 1) (2 ∙ 0) (1 ∙ 0) (7 ∙ 1) (6 ∙ 1) (0 ∙ 0) (3 ∙ 0) 𝑥𝑖 = + + + + + + + + 6 6 6 6 6 6 6 6 6 (5 ∙ 1) + 6 4 5 3 0 0 7 6 0 0 5 𝑥𝑖 = + + + + + + + + + 6 6 6 6 6 6 6 6 6 6 30 𝑥𝑖 = 6 𝑝=
𝑥𝑖 = 5 𝛤𝑝𝑏𝑖 =
𝑥𝑖 − 𝑥𝑡 𝑝 √ 𝑆𝑡 1−𝑝
𝛤𝑝𝑏𝑖 =
5 − 3,6 0,6 √ 2,2211 1 − 0,6
𝛤𝑝𝑏𝑖 =
1,40 0,6 √ 2,2211 0,4
𝛤𝑝𝑏𝑖 = (0,6303)√1,5 𝛤𝑝𝑏𝑖 = (0,6303)(1,2247) 𝛤𝑝𝑏𝑖 = 0,7720
5. Antara Butir Ke-5 dengan Total Peserta
Butir ke-5
Total (X)
𝑥2
1 2 3 4 5 6 7 8 9 10
0 1 0 0 0 1 1 0 1 1
4 5 3 2 1 7 6 0 3 5
16 25 9 4 1 49 36 0 9 25
B
5
36
174
5 = 0,5 10 (4 ∙ 0) (5 ∙ 1) (3 ∙ 0) (2 ∙ 0) (1 ∙ 0) (7 ∙ 1) (6 ∙ 1) (0 ∙ 0) (3 ∙ 1) 𝑥𝑖 = + + + + + + + + 5 5 5 5 5 5 5 5 5 (5 ∙ 1) + 5 0 5 0 0 0 7 6 0 3 5 𝑥𝑖 = + + + + + + + + + 5 5 5 5 5 5 5 5 5 5 26 𝑥𝑖 = 5 𝑝=
𝑥𝑖 = 4 𝛤𝑝𝑏𝑖 =
𝑥𝑖 − 𝑥𝑡 𝑝 √ 𝑆𝑡 1−𝑝
𝛤𝑝𝑏𝑖 =
5,20 − 3,6 0,5 √ 2,2211 1 − 0,5
𝛤𝑝𝑏𝑖 =
1,6 0,5 √ 2,2211 0,5
𝛤𝑝𝑏𝑖 = (0,7204)√1,00 𝛤𝑝𝑏𝑖 = (0,7204)(1,00) 𝛤𝑝𝑏𝑖 = 0,7204
6. Antara Butir Ke-6 dengan Total Peserta
Butir ke-6
Total (X)
𝑥2
1 2 3 4 5 6 7 8 9 10
0 1 0 0 0 1 1 0 0 0
4 5 3 2 1 7 6 0 3 5
16 25 9 4 1 49 36 0 9 25
B
3
36
174
3 = 0,3 10 (4 ∙ 0) (5 ∙ 1) (3 ∙ 0) (2 ∙ 0) (1 ∙ 0) (7 ∙ 1) (6 ∙ 1) (0 ∙ 0) (3 ∙ 0) 𝑥𝑖 = + + + + + + + + 3 3 3 3 3 3 3 3 3 (5 ∙ 0) + 3 0 5 0 0 0 7 6 0 0 0 𝑥𝑖 = + + + + + + + + + 3 3 3 3 3 3 3 3 3 3 18 𝑥𝑖 = 3 𝑝=
𝑥𝑖 = 6 𝛤𝑝𝑏𝑖 =
𝑥𝑖 − 𝑥𝑡 𝑝 √ 𝑆𝑡 1−𝑝
𝛤𝑝𝑏𝑖 =
6 − 3,6 0,3 √ 2,2211 1 − 0,3
𝛤𝑝𝑏𝑖 =
2,4 0,3 √ 2,2211 0,7
𝛤𝑝𝑏𝑖 = (1,0805)√0,4286 𝛤𝑝𝑏𝑖 = (0,1801)(0,6547) 𝛤𝑝𝑏𝑖 = 0,7074
7. Antara Butir Ke-7 dengan Total Peserta
Butir ke-7
Total (X)
𝑥2
1 2 3 4 5 6 7 8 9 10
0 0 0 0 0 1 0 0 0 0
4 5 3 2 1 7 6 0 3 5
16 25 9 4 1 49 36 0 9 25
B
1
36
174
1 = 0,1 10 (4 ∙ 0) (5 ∙ 0) (3 ∙ 0) (2 ∙ 0) (1 ∙ 0) (7 ∙ 1) (6 ∙ 0) (0 ∙ 0) (3 ∙ 0) 𝑥𝑖 = + + + + + + + + 1 1 1 1 1 1 1 1 1 (5 ∙ 0) + 1 0 0 0 0 0 7 0 0 0 0 𝑥𝑖 = + + + + + + + + + 1 1 1 1 1 1 1 1 1 1 7 𝑥𝑖 = 1 𝑝=
𝑥𝑖 = 7 𝛤𝑝𝑏𝑖 =
𝑥𝑖 − 𝑥𝑡 𝑝 √ 𝑆𝑡 1−𝑝
𝛤𝑝𝑏𝑖 =
7 − 3,6 0,1 √ 2,2211 1 − 0,1
0,1 𝛤𝑝𝑏𝑖 = 3,4√ 0,9 𝛤𝑝𝑏𝑖 = (0,1801)√0,1111 𝛤𝑝𝑏𝑖 = (1,5308)(0,3333) 𝛤𝑝𝑏𝑖 = 0,5103
Nurul Chairunnisa Utami Putri : http://roelcup.wordpress.com [email protected] [email protected]
More Documents from "wirdayanti"
Patofisiologi Ulkus Diabetik.docx
April 2020 10
24-korelasi-point-biserial.docx
April 2020 17
01-gdl-dewiwuland-674-1-ktidewi-i.pdf
April 2020 8
Sepsis.xlsx
April 2020 13