Matakuliah
Tahun
: Statistika Psikologi
: 2008
Uji Kebaikan Suai (Uji Kecocokan)
Pertemuan 23
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Mahasiswa akan dapat menghasilkan simpulan dari hasil
uji kenormalan suatu data.
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Outline Materi
• Statistik uji Khi-kuadrat
• Uji kenormalan
• Uji sebaran binomial
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Tests of Goodness of Fit and
Independence
• Goodness of Fit Test: A Multinomial Population
• Tests of Independence: Contingency Tables
• Goodness of Fit Test: Poisson and Normal Distributions
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Goodness of Fit Test:
A Multinomial Population
1. Set up the null and alternative hypotheses.
2. Select a random sample and record the observed
frequency, fi , for each of the k categories.
3. Assuming H0 is true, compute the expected
frequency, ei , in each category by multiplying the
category probability by the sample size.
continued
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Goodness of Fit Test:
A Multinomial Population
4. Compute the value of the test statistic.
( f ii  eii ) 22
 
eii
ii11
22
kk
 2   2
5. Reject H0 if
(where  is the significance level and there are k - 1
degrees of freedom).
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Contoh Soal: Finger Lakes Homes
• Multinomial Distribution Goodness of Fit Test
The number of homes sold of each model for 100
sales over the past two years is shown below.
Model
# Sold
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Colonial Ranch
30 20
Split-Level A-Frame
35
15
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Contoh Soal: Finger Lakes Homes
• Multinomial Distribution Goodness of Fit Test
– Notation
pC = popul. proportion that purchase a colonial
pR = popul. proportion that purchase a ranch
pS = popul. proportion that purchase a split-level
pA = popul. proportion that purchase an A-frame
– Hypotheses
H0: pC = pR = pS = pA = .25
Ha: The population proportions are not
pC = .25, pR = .25, pS = .25, and pA = .25
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Contoh Soal: Finger Lakes Homes
• Multinomial Distribution Goodness of Fit Test
– Expected Frequencies
e1 = .25(100) = 25
e2 = .25(100) = 25
e3 = .25(100) = 25
e4 = .25(100) = 25
– Test Statistic
( 30  25) 2 ( 20  25) 2 ( 35  25) 2 (15  25) 2
 



25
25
25
25
=1+1+4+4
= 10
2
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Contoh Soal: Finger Lakes Homes
• Multinomial Distribution Goodness of Fit Test
– Rejection Rule
With  = .05 and
k-1=4-1=3
degrees of freedom
Do Not Reject H0
Reject H0
2
7.81
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Contoh Soal: Finger Lakes Homes
• Multinomial Distribution Goodness of Fit Test
– Conclusion
2 = 10 > 7.81, so we reject the assumption there is
no home style preference, at the .05 level of
significance.
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Goodness of Fit Test: Poisson
Distribution
• 1. Set up the null and alternative hypotheses.
• 2. Select a random sample and
a. Record the observed frequency, fi , for each of the
k values of the Poisson random variable.
b. Compute the mean number of occurrences, μ.
• 3. Compute the expected frequency of occurrences,
ei , for each value of the Poisson random variable.
continued
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Goodness of Fit Test: Poisson
Distribution
4. Compute the value of the test statistic.
2
(
f

e
)
2   i i
ei
i 1
k
 2   2
5. Reject H0 if
(where  is the significance level and there are k - 2
degrees of freedom).
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Contoh Soal: Troy Parking Garage
• Poisson Distribution Goodness of Fit Test
In studying the need for an additional entrance to a
city parking garage, a consultant has recommended
an approach that is applicable only in situations where
the number of cars entering during a specified time
period follows a Poisson distribution.
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Contoh Soal: Troy Parking Garage
• Poisson Distribution Goodness of Fit Test
A random sample of 100 one-minute time intervals
resulted in the customer arrivals listed below. A
statistical test must be conducted to see if the
assumption of a Poisson distribution is reasonable.
# Arrivals 0 1 2 3 4 5 6 7 8 9 10 11
12
Frequency 0 1 4 10 14 20 12 12 9 8 6 3
1
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Contoh Soal: Troy Parking Garage
• Poisson Distribution Goodness of Fit Test
– Hypotheses
H0: Number of cars entering the garage during
a one-minute interval is Poisson distributed.
Ha: Number of cars entering the garage during a
one-minute interval is not Poisson
distributed
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Contoh Soal: Troy Parking Garage
• Poisson Distribution Goodness of Fit Test
– Estimate of Poisson Probability Function
otal Arrivals = 0(0) + 1(1) + 2(4) + . . . + 12(1) = 600
Total Time Periods = 100
Estimate of  = 600/100 = 6
Hence,
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6 x e 6
f ( x) 
x!
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Contoh Soal: Troy Parking Garage
• Poisson Distribution Goodness of Fit Test
– Expected Frequencies
x
f (x ) xf (x )
x
f (x )
0.0025 .25
7
1.0149 1.49
8
2.0446 4.46
9
3.0892 8.92
10
4.133913.39
11
5.162016.20
12
6.160616.06 Total
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.1389
.1041
.0694
.0417
.0227
.0155
1.0000
xf (x )
13.89
10.41
6.94
4.17
2.27
1.55
100.00
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Contoh Soal: Troy Parking Garage
• Poisson Distribution Goodness of Fit Test
– Observed and Expected Frequencies
i
fi
ei
fi - ei
0 or 1 or 2
5
6.20
-1.20
3
10
8.92
1.08
4
14
13.39
.61
5
20
16.20
3.80
6
12
16.06
-4.06
7
12
13.89
-1.89
8
9
10.41
-1.41
9
8
6.94
1.06
10 or more 10
7.99
2.01
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Contoh Soal: Troy Parking Garage
• Poisson Distribution Goodness of Fit Test
– Test Statistic
2
2
2
(

1
.
20
)
(
1
.
08
)
(
2
.
01
)
2 

 ... 
 3. 42
6. 20
8. 92
7. 99
2
 .05
 14 . 07
– Rejection Rule
With  = .05 and k - p - 1 = 9 - 1 - 1 = 7 d.f. (where
= number of categories and p = number of population
parameters estimated),
Reject H0 if 2 > 14.07
– Conclusion
We cannot reject H0. There’s no reason to doubtthe
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assumption of a Poisson distribution.
k
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Goodness of Fit Test: Normal
Distribution
4. Compute the value of the test statistic.
k ( f  e )2
2   i i
ei
i 1
 
2

 
5. Reject H0 if
(where  is the significance level
and there are k - 3 degrees of freedom).
2
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Contoh Soal: Victor Computers
• Normal Distribution Goodness of Fit Test
Victor Computers manufactures and sells a
general purpose microcomputer. As part of a study to
evaluate sales personnel, management wants to
determine if the annual sales volume (number of
units sold by a salesperson) follows a normal
probability distribution.
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Contoh Soal: Victor Computers
• Normal Distribution Goodness of Fit Test
A simple random sample of 30 of the
salespeople was taken and their numbers of units
sold are below.
33
64
83
43
65
84
44
66
85
45
68
86
52
70
91
52
72
92
56
73
94
58 63 64
73 74 75
98 102 105
(mean = 71, standard deviation = 18.54)
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Contoh Soal: Victor Computers
• Normal Distribution Goodness of Fit Test
– Hypotheses
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H0: The population of number of
units sold has a normal
distribution with mean 71 and
standard deviation 18.54.
Ha: The population of number of
units sold does not have a
normal distribution with mean 71
and standard deviation 18.54.
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Contoh Soal: Victor Computers
• Normal Distribution Goodness of Fit Test
– Interval Definition
To satisfy the requirement of an expected
frequency of at least 5 in each interval we
will divide the normal distribution into 30/5 = 6
equal probability intervals.
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Contoh Soal: Victor Computers
• Normal Distribution Goodness of Fit Test
– Interval Definition
Areas
= 1.00/6
= .1667
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53.02
71
88.98 = 71 + .97(18.54)
63.03 78.97
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Contoh Soal: Victor Computers
• Normal Distribution Goodness of Fit Test
– Observed and Expected Frequencies
i
fi
ei
fi – ei
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Less than 53.02 6
53.02 to 63.03 3
63.03 to 71.00 6
71.00 to 78.97 5
78.97 to 88.98 4
More than 88.98
6
Total 30
5
5
5
5
5
5
30
1
-2
1
0
-1
1
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Victor Computers
• Normal Distribution Goodness of Fit Test
– Test Statistic
2
2
2
2
2
2
(
1
)
(

2
)
(
1
)
(
0
)
(

1
)
(
1
)
2 





 1. 60
5
5
5
5
5
5
– Rejection Rule
With  = .05 and k - p - 1 = 6 - 2 - 1 = 3 d.f.,
 .205  7. 81
Reject H0 if 2 > 7.81
– Conclusion
We cannot reject H0. There is little evidence to
support rejecting the assumption the population
is normally distributed with  = 71 and  = 18.54.
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• Selamat Belajar Semoga Sukses.
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