Uji Kebaikan Suai (Uji Kecocokan) Pertemuan 23 Matakuliah: Statistika Psikologi Tahun: 2008.
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Transcript of Uji Kebaikan Suai (Uji Kecocokan) Pertemuan 23 Matakuliah: Statistika Psikologi Tahun: 2008.
Uji Kebaikan Suai (Uji Kecocokan) Pertemuan 23
Matakuliah : Statistika PsikologiTahun : 2008
Bina Nusantara University 3
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu :• Mahasiswa akan dapat menghasilkan simpulan
dari hasil uji kenormalan suatu data.
Bina Nusantara University 4
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
Bina Nusantara University 6
Goodness of Fit Test:A Multinomial Population
1. Set up the null and alternative hypotheses.2. Select a random sample and record the
observedfrequency, 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.
5. Reject H0 if
(where is the significance level and there are k - 1 degrees of freedom).
22
1
( )f ee
i i
ii
k2
2
1
( )f ee
i i
ii
k
2 2 2 2
Bina Nusantara University 8
Contoh Soal: Finger Lakes Homes
•Multinomial Distribution Goodness of Fit TestThe number of homes sold of each model
for 100sales over the past two years is shown below.
Model Colonial Ranch Split-Level A-Frame
# Sold 30 20 35 15
Bina Nusantara University 9
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
– HypothesesH0: pC = pR = pS = pA = .25
Ha: The population proportions are not
pC = .25, pR = .25, pS = .25, and pA = .25
Bina Nusantara University 10
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
= 1 + 1 + 4 + 4 = 10
22 2 2 230 25
2520 25
2535 25
2515 25
25 ( ) ( ) ( ) ( )2
2 2 2 230 2525
20 2525
35 2525
15 2525
( ) ( ) ( ) ( )
Bina Nusantara University 11
• Multinomial Distribution Goodness of Fit Test– Rejection Rule
With = .05 and
k - 1 = 4 - 1 = 3 degrees of freedom
22
7.81 7.81
Do Not Reject H0Do Not Reject H0 Reject H0Reject H0
Contoh Soal: Finger Lakes Homes
Bina Nusantara University 12
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.
Bina Nusantara University 13
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.
5. Reject H0 if
(where is the significance level and there are k - 2 degrees of freedom).
22
1
( )f ee
i i
ii
k2
2
1
( )f ee
i i
ii
k
2 2 2 2
Bina Nusantara University 15
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.
Bina Nusantara University 16
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
Bina Nusantara University 17
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, f x
ex
x
( )!
6 6
f xex
x
( )!
6 6
Bina Nusantara University 19
Contoh Soal: Troy Parking Garage
• Poisson Distribution Goodness of Fit Test–Expected Frequenciesx f (x ) xf (x ) x f (x ) xf (x )
0.0025 .25 7 .1389 13.89 1.0149 1.49 8 .1041 10.41 2.0446 4.46 9 .0694 6.94 3.0892 8.92 10 .0417 4.17 4.133913.39 11 .0227 2.27 5.162016.20 12 .0155 1.55 6.160616.06 Total 1.0000 100.00
Bina Nusantara University 20
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 -1.203 10 8.92 1.084 14 13.39 .615 20 16.20 3.806 12 16.06 -4.067 12 13.89 -1.898 9 10.41 -1.419 8 6.94 1.06
10 or more 10 7.99 2.01
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• Poisson Distribution Goodness of Fit Test– Test Statistic
– Rejection Rule With = .05 and k - p - 1 = 9 - 1 - 1 = 7 d.f. (where k = 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 doubt
the assumption of a Poisson distribution.
22 2 21 20
6 201 088 92
2 017 99
3 42 ( . ).
( . ).
. . .( . )
.. 2
2 2 21 206 20
1 088 92
2 017 99
3 42 ( . ).
( . ).
. . .( . )
..
. .052 14 07 . .052 14 07
Contoh Soal: Troy Parking Garage
Bina Nusantara University 22
Goodness of Fit Test: Normal Distribution
4. Compute the value of the test statistic.
5. Reject H0 if
(where is the significance leveland there are k - 3 degrees of freedom).
22
1
( )f ee
i i
ii
k2
2
1
( )f ee
i i
ii
k
2 2 2 2
Bina Nusantara University 23
Contoh Soal: Victor Computers
• Normal Distribution Goodness of Fit Test Victor Computers manufactures and sells ageneral 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.
Bina Nusantara University 24
Contoh Soal: Victor Computers
• Normal Distribution Goodness of Fit TestA simple random sample of 30 of the
salespeople was taken and their numbers of units sold are below.
33 43 44 45 52 52 56 58 63 6464 65 66 68 70 72 73 73 74 7583 84 85 86 91 92 94 98 102 105
(mean = 71, standard deviation = 18.54)
Bina Nusantara University 25
• Normal Distribution Goodness of Fit Test– Hypotheses
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.
Contoh Soal: Victor Computers
Bina Nusantara University 26
• 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.
Contoh Soal: Victor Computers
Bina Nusantara University 27
Contoh Soal: Victor Computers
• Normal Distribution Goodness of Fit Test– Interval Definition
Areas = 1.00/6 = .1667
Areas = 1.00/6 = .1667
717153.0253.02
63.0363.03 78.9778.9788.98 = 71 + .97(18.54)88.98 = 71 + .97(18.54)
Bina Nusantara University 28
• Normal Distribution Goodness of Fit Test– Observed and Expected Frequencies
i fi ei fi – ei
Less than 53.02 6 5 1 53.02 to 63.03 3 5 -2 63.03 to 71.00 6 5 1 71.00 to 78.97 5 5 0 78.97 to 88.98 4 5 -1 More than 88.98 6 5 1 Total30 30
Contoh Soal: Victor Computers
Bina Nusantara University 29
• Normal Distribution Goodness of Fit Test– Test Statistic
– Rejection Rule With = .05 and k - p - 1 = 6 - 2 - 1 = 3 d.f., 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.
22 2 2 2 2 21
525
15
05
15
15
1 60 ( ) ( ) ( ) ( ) ( ) ( ).2
2 2 2 2 2 215
25
15
05
15
15
1 60 ( ) ( ) ( ) ( ) ( ) ( ).
. .052 7 81 . .052 7 81
Victor Computers
Bina Nusantara University 30
• Selamat Belajar Semoga Sukses.