Post on 21-Feb-2015
Nonparametric Statistics
Chapter 12
Understandable Statistics Ninth EditionBy Brase and Brase Prepared by Yixun ShiBloomsburg University of Pennsylvania
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Nonparametric Situations
• At times, we will not know anything about the distributions of the populations from which we are sampling.
• Recall that all of our inference techniques thus far have assumed either a normal or binomial distribution from the populations of interest.
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Nonparametric Tests
• Advantages:– Easy to apply– Quite general in nature
• Disadvantages:– Wastes information– Accept the null hypothesis more often than
with other tests– Less sensitive
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Runs Test for Randomness
• Definitions:
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Runs Test for Randomness Hypotheses
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Conducting the Test
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Constructing a Runs Test
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Constructing a Runs Test
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Constructing a Runs Test
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Constructing a Runs Test
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Contoh
Pada taraf signifikansi 0,05, uji hipotesis keacakan barisan bilangan, jika sampel adalah5 2 2 1 6 5 3 3 1 6 5 2 1 4 4
Hipotesis H0 : Barisan bilangan adalah acak
H1 : Barisan bilangan tidak acak
Sampel Median bilangan ini adalah 3,27Runtun + + + + + + + sehingga r = 7 n+ = 7 dan n = 8
Beberapa buku menyatakan bahwa 0 sebaiknya diabaikan saja (+++0++ = 1 runtun)
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Contoh
Distribusi Probabilitas PensampelanSampel kecil sehingga pengujian dilakukan melalui tabel nilai kritis
Kriteria pengujianDari tabel nilai kritis untuk = 0,05 diperoleh bahwa hipotesis null H0 diterima pada
4 r 13
KeputusanPada taraf signifikansi 0,05, terima H0
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Sampel Besar
Uji hipotesis pada sampel besar (ada n > 20)
•Runtun di antara data X dan Y•Distribusi probabilitas pensampelan adalah distribusi probabilitas normal•Rerata dan simpangan baku r dan r adalah
)()(
)(
1
22
12
2
YXYX
YXYXYXr
YX
YXr
nnnn
nnnnnn
nn
nn
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Sampel Besar
• Statistik uji adalah
• Hipotesis diuji dengan taraf signifikansi pada dua sisi
Ujung bawah untuk runtun yang terlalu sedikitUjung atas untuk runtun yang terlalu banyak
Tolak H0 jika z < z(½) atau z > z(1-½)
Gagal Tolak H0 jika z(½) z z(1-½)
r
rrz
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Contoh
Pada taraf signifikansi 0,05, uji keacakan data jika sampel adalah1 8 4 9 5 6 2 9 7 6 3 2 5 8 7 3 6 9 3 74 8 9 5 7 6 9 8 4 8 7 6 4 9 6 5 8 5 9 9
Hipotesis H0 : data adalah acak
H1 : data tidak acak
Sampel Dari perhitungan median = 6,33 sehingga + + + + + + + + + + + + + + + + + + + r = 26 n = 21 n+ = 19
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Contoh
• Distribusi probabilitas pensampelanDistribusi probabilitas pensampelan adalah distribusi probabilitas normal dengan rerata dan simpangan baku
•Statistik uji
95,1911921
)19)(21)(2(1
2
nn
nnr
2 2
(2)(21)(19) (2)(21)(19) 21 19
(21 19) (21 19 1)
2 (2 )
( ) ( 1)
3,113
r
n n n n n n
n n n n
943,1113,3
95,1926
r
rrz
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Contoh
• Kriteria pengujianTaraf signifikansi = 0,05 ; Pengujian pada dua sisi nilai kritis
Sisi kiri z(0,025) = 1,96
Sisi kanan z(0,975) = 1,96
Tolak H0 jika z < 1,96 atau z > 1,96
Gagal tolak H0 jika 1,96 z 1,96
• Keputusan
Pada taraf signifikansi 0,05 Gagal menolak H0