2. Uji Kebaikan Suai (Uji Kecocokan) Pertemuan 23
Matakuliah : Statistika Psikologi
Tahun : 2008
Uji Kebaikan Suai (Uji Kecocokan) Pertemuan 23

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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5. 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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6. 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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7. 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).
Bina Nusantara University

8. 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 Colonial Ranch Split-Level A-Frame
# Sold
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
Hypotheses
H0: 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) = e2 = .25(100) = 25
e3 = .25(100) = e4 = .25(100) = 25
Test Statistic =
= 10
Bina Nusantara University

11. Contoh Soal: Finger Lakes Homes
Multinomial Distribution Goodness of Fit Test
Rejection Rule
With  = .05 and
k - 1 = = 3
degrees of freedom
Do Not Reject H0
Reject H0 2
7.81
Bina Nusantara University

12. Contoh Soal: Finger Lakes Homes
Multinomial Distribution Goodness of Fit Test
Conclusion
c2 = 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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14. 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).
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
Frequency
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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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18. Contoh Soal: Troy Parking Garage
Poisson Distribution Goodness of Fit Test
Estimate of Poisson Probability Function
otal Arrivals = 0(0) + 1(1) + 2(4) (1) = 600
Total Time Periods = 100
Estimate of  = 600/100 = 6
Hence,
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19. Contoh Soal: Troy Parking Garage
Poisson Distribution Goodness of Fit Test
Expected Frequencies
x f (x ) xf (x ) x f (x ) xf (x )
Total
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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
10 or more
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21. Contoh Soal: Troy Parking Garage
Poisson Distribution Goodness of Fit Test
Test Statistic
Rejection Rule
With  = .05 and k - p - 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.
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 level
and there are k - 3 degrees of freedom).
Bina Nusantara University

23. 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.
Bina Nusantara University

24. 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.
(mean = 71, standard deviation = 18.54)
Bina Nusantara University

25. Contoh Soal: Victor Computers
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
Ha: The population of number of
units sold does not have a
normal distribution with mean 71
and standard deviation
Bina Nusantara University

26. 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.
Bina Nusantara University

27. Contoh Soal: Victor Computers
Normal Distribution Goodness of Fit Test
Interval Definition
Areas
= 1.00/6
= .1667
53.02 71
88.98 = (18.54)
63.03
78.97
Bina Nusantara University

28. Contoh Soal: Victor Computers
Normal Distribution Goodness of Fit Test
Observed and Expected Frequencies
i fi ei fi – ei
Less than
53.02 to
63.03 to
71.00 to
78.97 to
More than
Total 30 30
Bina Nusantara University

29. Victor Computers Normal Distribution Goodness of Fit Test
Test Statistic
Rejection Rule
With  = .05 and k - p - 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  =
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30. Selamat Belajar Semoga Sukses.
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