 # Uji Kebaikan Suai (Uji Kecocokan) Pertemuan 23

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Uji Kebaikan Suai (Uji Kecocokan) Pertemuan 23
Matakuliah : Statistika Psikologi Tahun : 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. Bina Nusantara University

Outline Materi Statistik uji Khi-kuadrat Uji kenormalan
Uji sebaran binomial Bina Nusantara University

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Selamat Belajar Semoga Sukses.
Bina Nusantara University