1. 1 Pertemuan 24 Uji Kebaikan Suai Matakuliah: I0134 – Metoda Statistika Tahun: 2005 Versi: Revisi

2. 2 Learning Outcomes Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu : Mahasiswa dapat menghasilkan uji suatu sebaran (distribusi).

3. 3 Outline Materi Statistik uji khi-kuadrat Uji sebaran seragam Uji sebaran binomial Uji sebaran normal

4. 4 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

5. 5 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, f i, for each of the k categories. 3. Assuming H 0 is true, compute the expected frequency, e i, in each category by multiplying the category probability by the sample size. continued

6. 6 Goodness of Fit Test: A Multinomial Population 4. Compute the value of the test statistic. 5. Reject H 0 if (where  is the significance level and there are k - 1 degrees of freedom).

7. 7 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 30 20 35 15

8. 8 Contoh Soal: Finger Lakes Homes Multinomial Distribution Goodness of Fit Test –Notation p C = popul. proportion that purchase a colonial p R = popul. proportion that purchase a ranch p S = popul. proportion that purchase a split-level p A = popul. proportion that purchase an A-frame –Hypotheses H 0 : p C = p R = p S = p A =.25 H a : The population proportions are not p C =.25, p R =.25, p S =.25, and p A =.25

9. 9 Contoh Soal: Finger Lakes Homes Multinomial Distribution Goodness of Fit Test –Expected Frequencies e 1 =.25(100) = 25 e 2 =.25(100) = 25 e 3 =.25(100) = 25 e 4 =.25(100) = 25 –Test Statistic = 1 + 1 + 4 + 4 = 10

10. 10 Multinomial Distribution Goodness of Fit Test –Rejection Rule With  =.05 and k - 1 = 4 - 1 = 3 degrees of freedom 22 22 7.81 Do Not Reject H 0 Reject H 0 Contoh Soal: Finger Lakes Homes

11. 11 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.

12. 12 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, f i, for each of the k values of the Poisson random variable. b. Compute the mean number of occurrences, . 3. Compute the expected frequency of occurrences, e i, for each value of the Poisson random variable.continued

13. 13 Goodness of Fit Test: Poisson Distribution 4. Compute the value of the test statistic. 5. Reject H 0 if (where  is the significance level and there are k - 2 degrees of freedom).

14. 14 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.

15. 15 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

16. 16 Contoh Soal: Troy Parking Garage Poisson Distribution Goodness of Fit Test –Hypotheses H 0 : Number of cars entering the garage during a one-minute interval is Poisson distributed. H a : Number of cars entering the garage during a one-minute interval is not Poisson distributed

17. 17 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,

18. 18 Contoh Soal: Troy Parking Garage Poisson Distribution Goodness of Fit Test –Expected Frequencies x f (x ) xf (x ) 0.0025.25 7.138913.89 1.0149 1.49 8.104110.41 2.0446 4.46 9.06946.94 3.0892 8.9210.04174.17 4.133913.3911.02272.27 5.162016.2012.01551.55 6.160616.06 Total1.0000100.00

19. 19 Contoh Soal: Troy Parking Garage Poisson Distribution Goodness of Fit Test –Observed and Expected Frequencies i f i e i f i - e i 0 or 1 or 256.20-1.20 3108.921.08 41413.39.61 52016.203.80 61216.06-4.06 71213.89-1.89 8910.41-1.41 986.941.06 10 or more107.992.01

20. 20 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 H 0 if  2 > 14.07 –Conclusion We cannot reject H 0. There’s no reason to doubt the assumption of a Poisson distribution. Contoh Soal: Troy Parking Garage

21. 21 Goodness of Fit Test: Normal Distribution 4. Compute the value of the test statistic. 5. Reject H 0 if (where  is the significance level and there are k - 3 degrees of freedom).

22. 22 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.

23. 23 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. 33 43 44 45 52 52 56 58 63 64 64 65 66 68 70 72 73 73 74 75 83 84 85 86 91 92 94 98 102 105 (mean = 71, standard deviation = 18.54)

24. 24 Normal Distribution Goodness of Fit Test –Hypotheses H 0 : The population of number of units sold has a normal distribution with mean 71 and standard deviation 18.54. H a : 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

25. 25 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

26. 26 Contoh Soal: Victor Computers Normal Distribution Goodness of Fit Test –Interval Definition Areas = 1.00/6 =.1667 Areas = 1.00/6 =.1667 71 53.02 63.03 78.97 88.98 = 71 +.97(18.54)

27. 27 Normal Distribution Goodness of Fit Test –Observed and Expected Frequencies i f i e i f i – e i Less than 53.02651 53.02 to 63.0335-2 63.03 to 71.00651 71.00 to 78.97550 78.97 to 88.9845-1 More than 88.98651 Total3030 Contoh Soal: Victor Computers

28. 28 Normal Distribution Goodness of Fit Test –Test Statistic –Rejection Rule With  =.05 and k - p - 1 = 6 - 2 - 1 = 3 d.f., Reject H 0 if  2 > 7.81 –Conclusion We cannot reject H 0. There is little evidence to support rejecting the assumption the population is normally distributed with  = 71 and  = 18.54. Victor Computers

29. 29 Selamat Belajar Semoga Sukses.