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Chapter 17Nonparametric Methods:

Chi-Square Applications

GOALSWhen you have completed this chapter, you will be able to:1Karakteristik distribusi chi-square.

2 Melakukan uji hipotesis dengan membandingkan data frekuensi dengan frekuensi harapan.

3Melakukan uji hipotesis apakah dua kriteria klasifikasi saling terkait

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Characteristics of the Chi-Square Distribution

• Ciri-ciri :· Derajat kemiringan posistif/positively

skewed· Non-negatif· Berbasis degrees of freedom df· Ketika degrees of freedom berubah,

distribusi baru terbentuk.

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CHI-SQUARE DISTRIBUTION CHI-SQUARE DISTRIBUTION CHI-SQUARE DISTRIBUTION CHI-SQUARE DISTRIBUTION

df = 3

df = 5

df = 10

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Goodness-of-Fit Test: Equal Expected Frequencies

• Jika adalah data pengamatan dan frekuensi harapan.

• : tidak ada perbedaan• : ada perbedaan antara• Uji statistika:

• Nilai kritis adalah nilai chi-square (x2), dengan (k-1)=df degrees of freedom, dimana k adalah

jumlah kategori.

f and fe0

H0 eff and 0

1H eff and 0

x

f f

fe

e

2 0

2

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EXAMPLE 1

• Data berikut adalah absensi karyawan suatu pabrik. Pada level signifikan 0,05, ujilah apakah ada perbedaan dalam absensi dari hari ke hari dalam seminggu.

Hari Frekuensi

Senin 120

Selasa 45

Rabu 60

Kamis 90

Jumat 130

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EXAMPLE 1 continued

• Asumsikan frekuensi harapan adalah sama: (120+45+60+90+130)/5=89.

• Hitung uji statistik adalah .• Nilai degrees of freedom adalah (5-

1)=4.• Maka, nilai kritis adalah 9,488 (dari

tabel)

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x

f f

fe

e

2 0

2

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Hari Frekuensi (fo)

F harapan

(fe)

Fo-fe (f0-fe)2 (fo-fe)2 fe

Senin 120 89

Selasa 45 89

Rabu 60 89 -

Kamis 90 89

Jumat 130 89

0 X

x

f f

fe

e

2 0

2

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EXAMPLE 1 continued

1. : tidak ada perbedaan antara data hasil pengamatan dan frekuensi harapan dlm absensi. : ada perbedaan antara hasil pengamatan dengan frekuensi harapan dlm absensi.

2. Tingkat signifikan =0,053. Pilih Uji statistika: chi-square4. Decision Rule: tolak jika uji statistika lebih besar

dibanding critical value. Hitung nilai dalam tabel chi-square df=4 , sig.=0.05

5. Hitung chi-square=60,8 6. Conclusion: tolak dan disimpulkan bahwa terdapat

perbedaan antara data hasil pengamatan dan frekuensi harapan dalam absensi .

H0

H1

H0

H0

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Karakteristik chi-square

• Nilai tidak pernah negatif• Terdapat sebuah keluarga distribusi

· Tidak tergantung ukuran sampeljumlah kategori

• Distribusi chi-square kemiringannya positif

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Goodness-of-Fit Test for NormalityUji ketepatan Perkiraaan: frekuensi harapan tdk

setara

• Kartu Baseball : fe = 20, dalam kasus berikut uji chi-square dipakai untuk fe tidak setara

• Kegunaan: untuk menguji apakah frekuensi data yang diamati dalam distribusi frekuensi sesuai dengan teori distribusi normal .

• Contoh:

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Juml. Rawat (1th)

% AHAA dr total

Juml. Warga Bartow (f0)

Juml. Warga yg diharapkan (fe)

0 40 55 60

1 30 50 45

2 20 32 30

3 10 13 15

100 150 150

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Steps

1. Ho= tidak ada perbedaan kondisi lokal dan nasional utk jumlah perawatan di RS

2. H1= ada perbedaan kondisi lokal dan nasional utk jumlah perawatan di RS

3. Level sign. 0,054. Uji statistika chi-square5. Decision Rule: tolak Ho jika uji statistika lebih besar

(>)dibanding critical value. Hitung nilai dalam tabel chi-square df=3, sig.=0.05 7,815

6. Hitung chi-square (x2)=1,37237. Conclusion: terima ho dan disimpulkan bahwa tidak

terdapat perbedaan antara data hasil pengamatan lokal dan nasional.

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Goodness-of-Fit Test: Unequal Expected FrequenciesEXAMPLE 2

• The U.S. Bureau of the Census indicated that 63.9% of the population is married, 7.7% widowed, 6.9% divorced (and not re-married), and 21.5% single (never been married). A sample of 500 adults from the Philadelphia area showed that 310 were married, 40 widowed, 30 divorced, and 120 single. At the .05 significance level can we conclude that the Philadelphia area is different from the U.S. as a whole?

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EXAMPLE 2 continued

Status

Married 310 319.5 .2825Widowed 40 38.5 .0584

Divorced 30 34.5 .5870

Single 120 107.5 1.4535

Total 500 2.3814

f0 fe ( ) /f f fe e02

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EXAMPLE 2 continued

• Step 1: : The distribution has not changed.

• : The distribution has changed.• Step 2: The significance level is .05.• Step 3: choose chi-square test • Step 4: is rejected if• Step 5:• Step 6: is accepted. The distribution has

not changed.

H0

H1

H0 05.,3,815.72 df

3824.22 H0

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Keterbatasan Chi-square

• Jika hanya 2 cell, fe dalam tiap sel minimal 5• Jika lebih dari 2 cell, chi-square seharusnya tidak

digunakan jika lebih dari 20% cell fe memiliki frekuensi harapan kurang dari 5

tingkat Fo fe

Petugas 30 32

Supervisor 110 113

Manajer 86 87

Manajer mid 23 24

Ass. VP 5 2

VP 5 4

Senior VP 4 1

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tingkat Fo fe 0-e (0-e)2/e % of chisq

Petugas 30 32 -2 0,125 0,89

Supervisor

110 113 -3 0,079 0,57

Manajer 86 87 -1 0,011 0,08

Manajer mid

23 24 -1 0,041 0,30

Ass. VP 5 2 3 4,5 32,12

VP 5 4 1 0.25 1,78

Senior VP 4 1 3 9 64,25

263 263 14,008 100

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tingkat Fo fe 0-e (0-e)2/e % of chisq

Petugas 30 32 -2 0,125

Supervisor

110 113 -3 0,079

Manajer 86 87 -1 0,011

Manajer mid

23 24 -1 0,041

VP 14 7 7 7

263 263

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Analisis Tabel Kemungkinan

• Bab 4. data bivariat• Bab. 5. tabel kemungkinan• Chi-square dapat dipakai utk uji dua

variabel skala nominal. (2 variabel bebas)Pilihan tinggal

penyesuaian

Luar biasa

baik cukup Tidak puas

total

Kota asal

27 35 33 25 120

Bukan kota asal

13 15 27 25 80

40 50 60 50 200

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• Df= (jumlah baris -1)(jumlah kolom-1)

= (r-1)(c-1) = (2-1)(4-1)=3• Sign. 0,01• Nilai kritis 11,345• Tolak ho jika x2 > 11,345

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• Fe =(juml.baris)(juml. Kolom)/(juml.seluruh)• = ((120)(40))/(200)=24

Pilihan tinggal

penyesuaian

Luar biasaF0-fe

Baik

F0-fe

cukup

F0-fe

Tidak puasF0-fe

TotalF0 –fe

Kota asal

27 - 24

35 – 30

33 - 36

25 - 30 120-120

Bukan kota asal

13 – 16

15 – 20

27 – 24

25 – 20

80-80

40 – 40

50 - 50 60 – 60

50 - 50 200-200

x

f f

fe

e

2 0

2

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• X2= ((27-24)2)/24 + ((35-30)2)/30 + dst

• =5,729• terima ho

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• Prosedur: · tentukan rata-rata dan standar deviasi dari

distribusi frekuensi. (f0-fe)2/fe· Hitung nilai-z untuk batas bawah kelas dan

batas atas kelas. · Tentukan untuk setiap kategori .· Gunakan chi-square goodness-of-fit test

untuk menentukan jika berkoinsiden dgn .

fe

f0fe

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EXAMPLE 3

• A sample of 500 donations to the Arthritis Foundation is reported in the following frequency distribution. Is it reasonable to conclude that the distribution is normally distributed with a mean of $10 and a standard deviation of $2? Use the .05 significance level.

• Note: To compute for the first class, first compute the probability for this class. P(x<6)=P[z<(6-10)/2]=.0228. Thus is (.0228)(500)=11.4

fe

fe

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Example 3 continued

Amount fo Area fe e

eof

ff 2

Less than $6 20 .02 11.40 6.49

$6 up to 8 60 .14 67.95 0.93

$8 up to 10 140 .34 170.65 5.50

$10 up to 12 120 .34 170.65 15.03

$12 up to 14 90 .14 67.95 7.16

More than $14 70 .02 11.40 301.22

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EXAMPLE 3 continued

• Step 1: : The distribution is normal. : The distribution is not

normal.• Step 2: The significance level is .05.• Step 3: is rejected if• Step 4:• Step 5: is rejected . The

distribution is not normal

H0

H1

H0 x df2 1107 5 05 . , , .x2 336 33 .

H0

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Contingency Table Analysis

• Contingency table analysis dgunakan untuk menguji apakah dua traits atau variables terkait.

• Setiap observasi diklasifikasikan menyesuaikan dua variabel. The usual hypothesis testing procedure is used.

• degrees of freedom /df sama dengan: (jumlah baris-1)(jumlah-kolom-1).

• Frekuensi harapan/expected frequency dihitung sebagai berikut:

(row total)(column total)/grand total

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EXAMPLE 4

• Is there a relationship between the location of an accident and the gender of the person involved in the accident? A sample of 150 accidents reported to the police were classified by type and gender. At the .05 level of significance, can we conclude that gender and the location of the accident are related?

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EXAMPLE 4 continued

Sex Work Home Other Total

Male 60 20 10 90

Female 20 30 10 60

Total 80 50 20 150

Note: The expected frequency for the work-male intersection is computed as (90)(80)/150=48.Similarly, you can compute the expected frequencies for the other cells.

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EXAMPLE 4 continued

• Step 1: : Gender and location are not related. : Gender and location are related.

• Step 2: The significance level is.05.• Step 3: is rejected if• Step 4:• Step 5: is rejected. Gender and

location are related.

H0

H1

H0 x df2 5991 2 05 . , , .

x2 16 667 .H0

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