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**Uji Kebaikan Suai (Uji Kecocokan) Pertemuan 23**

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

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**Learning Outcomes Pada akhir pertemuan ini, diharapkan mahasiswa**

akan mampu : Mahasiswa akan dapat menghasilkan simpulan dari hasil uji kenormalan suatu data. Bina Nusantara University

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**Outline Materi Statistik uji Khi-kuadrat Uji kenormalan**

Uji sebaran binomial Bina Nusantara University

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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

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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 Bina Nusantara University

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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). Bina Nusantara University

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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 Colonial Ranch Split-Level A-Frame # Sold Bina Nusantara University

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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 Bina Nusantara University

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**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

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**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

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**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

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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 Bina Nusantara University

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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). Bina Nusantara University

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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. Bina Nusantara University

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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 Frequency Bina Nusantara University

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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 Bina Nusantara University

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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) (1) = 600 Total Time Periods = 100 Estimate of = 600/100 = 6 Hence, Bina Nusantara University

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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 ) xf (x ) Total Bina Nusantara University

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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 10 or more Bina Nusantara University

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**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

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**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

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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. Bina Nusantara University

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

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**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

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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. Bina Nusantara University

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**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

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**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

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**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 = Bina Nusantara University

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**Selamat Belajar Semoga Sukses.**

Bina Nusantara University

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