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Noorlaily Fitdiarini
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Comparing the tallies or counts of categorical responses between twoindependent groups two way cross-classification table (contingency table )
H0: there is no difference between the twopopulation proportions◦ H 0 : p 1=p 2
H1: two population proportions are not thesame◦ H 0 : p 1≠p 2
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It is never negative There is a family of chi-square distributions
◦ The shape of the chi-square distribution does notdepend on the size of the sample, but the number
of categories used (k) It is positively skewed
◦ As the number of both d.f. increases, thedistribution begins to approximate the normaldistribution
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CHI-SQUARE DISTRIBUTION
df = 3
df = 5
df = 10
c2
2-2
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Compare several proportion (Multinomial Test )
One of nonparametric or distribution-free tests of hypothesis
Data : nominal-scale or ordinal-scale
The test statistic is :
test statistic is equal to the squared difference between theobserved and expected frequencies, divided by the expectedfrequency in each cell of the table
f 0 is observed frequency in a particular cell of a contingency
table f e is theoretical or expected frequency in a particular cell if the
null hypothesis is true
2 c
6
e
e
f
f f x
2
02
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RowVariable
Column variable (group)
1 2 totals
successe
s
X 1 X 2 X
failures n1-X 1 n2 -X n-X
totals n1 n2 n
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X 1= number of successes in group 1 X 2 = number of successes in group 2 n 1-X 1= number of failures in group 1 n 2 -X 2 = number of failures in group 2
X= X 1+ X 2 is the total number of successes n-X=(n 1-X 1 )+(n 2 -X 2 ) is the total number of
failures n 1=the sample size in group 1
n 2 =the sample size in group 2 n=n 1+n 2 is the total sample size
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Choose
hotel
again?
Hotel
Sheratonlagoon
SheratonNusa Dua
Total
yes 163 154 317
no 64 108 172
Total 227 262 489
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Chi-Square Test: Lagoon, Nusa Dua
Expected counts are printed below observed counts
Lagoon Nusa Dua Total
1 163 154 317
147.16 169.84
2 64 108 172
79.84 92.16
Total 227 262 489
Chi-Sq = 1.706 + 1.478 +
3.144 + 2.724 = 9.053
DF = 1, P-Value = 0.003
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Chi-square test is used to :◦ Test whether an observed set of frequencies couldhave come from a hypothesized populationdistribution
◦ Determine whether the sample observations come
from a particular distribution such as the normaldistribution
◦ Contingency table analysis, is used to test whethertwo traits or characteristics are related (Test of
Independency )
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12
00
(1-α)
α
χ 2 Region of non-rejection
Region of rejectionCritical value
Reject H0 if χ 2>χ 2U
Otherwise do not reject H0
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If the null hypothesis is true, the computed
statistic should be close to zero because thesquared difference between what is actuallyobserved in each cell f 0 , and what is theoretically
expected f e , would be very small On the other hand, if H0 is false, and there are real
differences in the population proportions, thecomputed statistic is expected to be large. Thisis because the difference between what is actually
observed in each cell and what is theoreticallyexpected will be magnified when the difference aresquared
2 c
2 c
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The purpose of Goodness-of-Fit Test is tocompare an observed set of frequencies(fo) to an expected set of frequencies (fe).
Ho: no difference between fo and fe H1: there is a difference between fo and fe
The critical value is a chi-square value with
(k - 1) degrees of freedom, where k is thenumber of categories
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Pemain
JumlahTerjual (fo)
Jumlah yangdiharapkan Terjual
(fe)
Owen 13 20
Ronaldo 33 20
Nesta 14 20
Dida 7 20
Becham 36 20
Zidane 17 20
TOTAL 120 120
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Pemain fo fe (fo –
fe) (fo –
fe)2
Owen 13 20 -7 49 2,45
Ronaldo 33 20 13 169 8,45
Vieri 14 20 -6 36 1,80
Buffon 7 20 -13 169 8,45
Becham 36 20 16 256 12,80
Zidane 17 20 -3 9 0,45
0 34,40
16
e
eo
f f f
2
)(
2
X
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Contoh :Dosen mengharapkan distribusi nilai ujian: A = 40%, B = 40%, dan C = 20%. Hasil
ujian menunjukkan distribusi nilai sebagaiberikut :
A : 30 orang B : 20 orang C : 10 orang
Uji dengan level of significance 10%,apakah distribusi nilai tersebut sesuaidengan harapan dosen tersebut ?
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If there are only two cells, the expectedfrequency in each cell should be 5 or more
For more than two cells, Chi-Square should notbe used if more than 20% of the expected
frequency cells have expected frequency lessthan 5.
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Level of Management fo fe
Foreman 30 32
Supervisor 110 113
Manager 86 87
Middle Manager 23 24
Assistant vice president 5 2
Vice president 5 4Senior vice president 4 1
TOTAL 263 263
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Level of Management fo fe
Foreman 30 32
Supervisor 110 113
Manager 86 87
Middle Manager 23 24
Vice president 14 7
TOTAL 263 263
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Purpose: To test whether the observedfrequencies in a frequency distribution matchthe theoretical normal distribution.
Procedure:◦ Determine the mean and standard deviation of the
frequency distribution.◦ Compute the z-value for the lower class limit and
the upper class limit for each class.◦ Determine fe for each category◦ Use the chi-square goodness-of-fit test to
determine if fo coincides with fe.
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Salary ($ 000) frequency20 – 30 4
30 – 40 20
40 –
50 4150 – 60 44
60 – 70 29
70 – 80 16
80 – 90 2
90 – 100 4
TOTAL 16022
76.13
03.54
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Salary (S 000) Z Value Area fe
Under 30 Under –
1.75 0.0401 6.41630 – 40 -1.75 to -1.02 0.1138 18.208
40 – 50 -1.02 to -0.29 0.2320 37.120
50 –
60 -0.29 to 0.43 0.2805 44.88060 – 70 0.43 to 1.16 0.2106 33.696
70 – 80 1.16 to 1.89 0.0936 14.976
80 or more over 1.89 0.0294 1.704
1 160
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x Z
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Salary (S 000) fo fe (fo –
fe) (fo –
fe)2
Under 30 4 6.416 -2.416 5.837 0.910
30 – 40 20 18.208 1.792 3.211 0.176
40 – 50 41 37.120 3.880 15.054 0.406
50 – 60 44 44.880 -0.880 0.774 0.017
60 – 70 29 33.696 -4.696 22.052 0.654
70 – 80 16 14.976 1.024 1.049 0.07080 or more 6 1.704 1.296 1.680 0.357
160 160 2.590
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e
eo
f
f f 2)(
2
X
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Suppose we knew the mean and standarddeviation of population but wished to find
whether some sample data conform tothe normal distribution,d.f. = k - 1
On the other hand, if we don’t know the
mean and standard deviation of population but we wish to test whethersome sample data follow the normaldistribution,
d.f. = k – p – 1 (where p is the number of populationparameter being estimated from thesample data)
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Contingency table analysis is used to testwhether two traits or variables are related.
Two-way classification table Each observation is classified according to two
variables. d.f. : (number of rows-1)(number of columns-
1). The expected frequency (fe) is computed as:
Coefficient of Contingency :
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total Grand
total Coloumntotal Row f e _
_ _
N X
X C
2
2
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Manajer produksi meneliti tingkat kerusakanpada mesin produksi. Hasilnya pengamatanterhadap barang yang diproduksi sebagaiberikut
Apakah kerusakan tersebut disebabkanmesin atau kebetulan saja ? Uji dengan =0,05
l 27
Kondisi Mesin 1 Mesin 2 Mesin 3Rusak 12 15 6
Baik 88 105 74
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Lembaga riset meneliti apakah adahubungan antara jenis surat kabar yangdibaca dengan kelompok masyarakat.Hasilnya sebagai berikut :
Uji dengan = 0,1
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Surat Kabar Kelompok A B C
Atas 170 124 90
Menengah 120 112 100Bawah 130 90 88