Uji Goodness of Fit : Distribusi Multinomial

1. Uji Goodness of Fit : Distribusi Multinomial

2. Distribusi Multinomial Distribusi Multinomial merupakangeneralisasidaridistribusi binomial yaitudenganmelonggarkankriteriabanyaknyahasil (outcome) yang mungkinjadilebihdari 2. Dalamhalinimakapercobaannyadisebutpercobaan multinomial sedangkandistribusiprobabilitasnyadisebutdistribusi multinomial.

3. Definisi: Misalkantiappercobaanbisamenghasilkankhasil yang berbedayaitu E1, E2, …,Ekdanmasing-masingdenganprobabiliitas p1, p2, …,pk. Distribusi multinomial f(x1,x2,…,xk; n, p1,p2, ..,pk) akanmemberikanprobabilitasbahwa E1akanmunculsebanyak x1 kali, E2akanmunculsebanyak x2 kali, dstdalampengambilanindependensebanyakn kali, jadi x1+ x2+ ….+ xk=n dengan p1+p2+ …+ pk =1 dan

4. Sebuah airport memiliki 3 buahlandaspacu (runway), danprobabilitassebuah runway dipiliholehpesawatygakanmendaratadalah: runway -1 : 2/9 runway -2 : 1/6 runway -3 : 11/18 Berapakahprobabilitas 6 pesawatygdatangsecaraacakdidistribusikankedalam runway-runway tsbsptberikut: runway -1 : 2 pesawat runway -2 : 1 pesawat runway -3 : 3 pesawat Jawab. Pemilihan runway acakdanindependen, dengan p1=2/9, p2=1/6 dan p3=11/18. Probabilitasuntuk x1=2, x2= 1 dan x3=3 adalah

5. Contoh • Seorangdoktermelakukanpengobatansebanyak 6 kali terhadap 6 orangpenderitagagaljantungdenganhasilsembuhsempurna, sembuhdengangejalasisa, danmeninggal. • Berapabesarprobabilitasdari 6 kali pengobatantersebutmenghasilkan 2 orangsembuhsempurna, 2 orangsembuhdengangejalasisa, dan 2 orangmeninggal. P = 0,123 = 12,3%

6. Contoh • Berdasarkanteorigenetika, perbandinganseekorhamster betinaakanmelahirkananakdgnwarnabulumerah,hitamdanputihadalah 8:4:4. Hitungpeluangakanlahiranakdgnwarnamerah 5 ekor, hitam 2 ekor, putih 1 ekordarikelahiran 8 ekor.

7. Uji Goodness of Fit Uji Goodness of Fit Bagaimanadekathasilpengamatan/sampelsesuaidengan yang diharapkan ? Example: In tossing a coin, you expect half heads and half tails. You tossed a coin 100 times. You expected 50 heads and 50 tails. However, you obtained 48 heads and 52 tails. Are 48 heads and 52 tails close enough to call the coin fair?

8. UjiHipotesisuntukproporsidariPopulasi Multinomial 1.NyatakanHipotesisnoldanhipotesisalternatifnya. 2.Ambilsampel random dantentukanfrekuensipengamatan, fi, untukmasing-masingkkategori. 3.DenganmenganggapH0benar, frekuensiharapaneidihitunguntuktiapkategoriyaitudenganmengalikantiapkategoridenganprobabilitastiapkategoridenganukuransampel (sample size).

9. UjiHipotesisuntukproporsidariPopulasi Multinomial 4.Hitungstatistikuji dengan fi = frekuensipengamatanuntukkategorii ei = frekuensiharapanuntuki k = banyakkategori Catatan : Statistikmempunyaidistribusi chi-kuadratdenganderajatbebask– 1 asalkanfrekuensiharapanuntuksemuakategorilebihdari 5.

10. Reject H0 if UjiHipotesisuntukproporsidariPopulasi Multinomial 5.AturanPenolakan Reject H0 if p-value <a p-value approach: Critical value approach: denganadalahtingkatsignifikansidandistribusinyaadalahdistribusi chi-kuadratdenganderajatbebask– 1.

11. Multinomial Distribution Goodness of Fit Test • Example: Finger Lakes Homes (A) Finger Lakes Homes manufactures four models of prefabricated homes, a two-story colonial, a log cabin, a split-level, and an A-frame. To help in production planning, management would like to determine if previous customer purchases indicate that there is a preference in the style selected.

12. Multinomial Distribution Goodness of Fit Test • Example: Finger Lakes Homes (A) The number of homes sold of each model for 100 sales over the past two years is shown below. Split- A- Model Colonial Log Level Frame # Sold 30 20 35 15

13. Multinomial Distribution Goodness of Fit Test • Hypotheses H0: pC = pL = pS = pA = .25 Ha: The population proportions are not pC = .25, pL = .25, pS = .25, and pA = .25 where: pC = population proportion that purchase a colonial pL = population proportion that purchase a log cabin pS = population proportion that purchase a split-level pA = population proportion that purchase an A-frame

14. Hypotheses Ho : There is no preference in the home styles or all home styles have equal preferences. Ha : All home styles do not have equal preferences.

15. Multinomial Distribution Goodness of Fit Test • Rejection Rule Reject H0 if p-value < .05 or c2 > 7.815. With  = .05 and k - 1 = 4 - 1 = 3 degrees of freedom Do Not Reject H0 Reject H0 2 7.815

16. Multinomial Distribution Goodness of Fit Test • Expected Frequencies • Test Statistic • e1 = .25(100) = 25 e2 = .25(100) = 25 e3 = .25(100) = 25 e4 = .25(100) = 25 = 1 + 1 + 4 + 4 = 10

17. Multinomial Distribution Goodness of Fit Test • Conclusion Using the p-Value Approach Area in Upper Tail .10 .05 .025 .01 .005 c2 Value (df = 3) 6.251 7.815 9.348 11.345 12.838 Because c2 = 10 is between 9.348 and 11.345, the area in the upper tail of the distribution is between .025 and .01. The p-value <a . We can reject the null hypothesis. Note: A precise p-value can be found using R.

18. Multinomial Distribution Goodness of Fit Test • Conclusion Using the Critical Value Approach c2 = 10 > 7.815 We reject, at the .05 level of significance, the assumption that there is no home style preference.

19. Solusidengan SPSS : Data

20. Solusidengan SPSS • Sesudahmengimputkan data dalambentukfrekuensipengamatansepertidisampingselanjutnyadigunakanperintah Analyze  Non ParametrikStatistik  Chi-square

21. Langkahpengerjaandengan SPSS

22. Output SPSS : • Dari output SPSS diperoleh X2 = 10 dengannilai-p = 0.019 sehingga Ho ditolakartinyadistribusinyatidakseragamdiskrit (homogen).

23. Soal 1

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27. Soal 5

28. Soal 6

29. Soal 7

30. TERIMA KASIH