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.


Outline Materi

• Statistik uji Khi-kuadrat• Uji kenormalan • Uji sebaran binomial


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


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


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


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



• 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



• 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

( ) ( ) ( ) ( )


• 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




• 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.


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


Goodness of Fit Test: Poisson Distribution


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


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.



• 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



• 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



• 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



• 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


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


• 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



Goodness of Fit Test: Normal Distribution


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


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.



• 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)


• 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.



• 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.




• 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)


• 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



• 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


• Selamat Belajar Semoga Sukses.

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

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