Magister Pengelolaan Air dan Air Limbah Universitas Gadjah Mada. 18-Aug-17. 1http://istiarto.staff.ugm.ac.id. Statistika Teknik.

Transkripsi

1 Magister Pengelolaan Air dan Air Limbah Universitas Gadjah Mada Statistika Teknik Uji Hipotesis 1

2 Uji Hipotesis Model Matematika vs Pengukuran komparasi garis teoretik (prediksi menurut model) dan data pengukuran jika prediksi model sesuai dengan data pengukuran, maka model diterima jika prediksi model menyimpang dari data pengukuran, maka model ditolak Dalam sejumlah kasus, yang terjadi adalah hasil komparasi prediksi model dan data pengukuran tidak cukup jelas untuk menyatakan bahwa model diterima atau ditolak uji hipotesis sebagai alat analisis dalam komparasi tersebut 2

3 Prosedur Uji Hipotesis Rumuskan hipotesis Rumuskan hipotesis alternatif Tetapkan statistika uji Tetapkan distribusi statistika uji Tentukan nilai kritik sebagai batas statistika uji harus ditolak Kumpulkan data untuk menyusun statistika uji Kontrol posisi statistika uji terhadap nilai kritis 3

4 Kemungkinan Kesalahan keadaan nyata pilihan hipotesis benar hipotesis salah menerima tak salah kesalahan tipe II menolak kesalahan tipe I tak salah 4

5 Notasi H 0 = hipotesis (yang diuji) H a = hipotesis alternatif 1 α = tingkat keyakinan (confidence level) 5

6 Uji Hipotesis Nilai Rata-rata H 0 : µ = µ 1 H a : µ = µ 2 Distribusi Normal σ X2 diketahui Statistika uji: Z = n σ X X µ 1 ( ) Jika μ 1 > μ 2 : H 0 ditolak jika Jika μ 1 < μ 2 : H 0 ditolak jika berdistribusi normal σ X < µ 1 z X 1 α n Z < z 1 α σ X < µ 1 + z X 1 α n Z > z 1 α 6

7 7 z 1 α luas = α prob( Z > z ) 1 α = α

8 Uji Hipotesis Nilai Rata-rata H 0 : µ = µ 1 H a : µ = µ 2 Distribusi Normal σ X2 tidak diketahui Statistika uji: T = n s X X µ 1 ( ) berdistribusi t s H X < µ 1 t X 0 ditolak jika: 1 α,n 1 jika μ n 1 > μ 2 X > µ 1 + t 1 α,n 1 s X n jika μ 1 < μ 2 8

9 Uji Hipotesis Nilai Rata-rata H 0 : µ = µ 0 H a : µ µ 0 Distribusi Normal σ X2 diketahui Statistika uji: H 0 ditolak jika: Z = Z = n σ X X µ 0 ( ) n ( X µ σ 0 ) > z 1 α 2 X berdistribusi normal 9

10 Uji Hipotesis Nilai Rata-rata H 0 : µ = µ 0 H a : µ µ 0 Distribusi Normal σ X2 tidak diketahui Statistika uji: H 0 ditolak jika: T = t = n s X X µ 0 ( ) n ( X µ s 0 ) > t 1 α 2,n 1 X berdistribusi t 10

11 Uji Hipotesis Nilai Rata-rata Hasil uji hipotesis adalah menolak H 0, atau tidak menolak H 0 Artinya H 0 : μ = μ 0 Tidak menolak H 0 à menerima H 0 berarti bahwa μ tidak berbeda secara signifikan dengan μ 0. Tetapi tidak dikatakan bahwa μ benar-benar sama dengan μ 0 karena kita tidak membuktikan bahwa μ = μ 0. 11

12 Uji hipotesis beda nilai rata-rata dua buah distribusi normal H 0 : H a : µ 1 µ 2 = δ µ 1 µ 2 δ var( X 1 ) dan var( X 2 ) diketahui Statistika uji: H 0 ditolak jika: Z = z = X 1 X 2 δ ( σ 2 1 n 1 + σ 2 2 n ) n ( X µ σ 0 ) > z 1 α 2 X berdistribusi normal 12

13 Uji hipotesis beda nilai rata-rata dua buah distribusi normal H 0 : H a : µ 1 µ 2 = δ µ 1 µ 2 δ var( X 1 ) dan var( X 2 ) tidak diketahui Statistika uji: T = ' ) ( *) X 1 X 2 δ n 1 + n # 2 ( )s ( n 2 1)s % + $ 2 &), # $ n 1 n 2 ( n 1 + n 2 2) % & -) ( ) n 1 1 berdistribusi t dengan (n 1 +n 2 2) degrees of freedom 1 2 H 0 ditolak jika: t = n s X X µ 0 ( ) > t 1 α 2,n1 +n

14 Uji Hipotesis Nilai Varian H 0 : σ 2 = σ 0 2 H a : σ 2 σ 0 2 Distribusi Normal Statistika uji: χ c 2 = n i=1 ( X i X) 2 σ 0 H 0 diterima (tidak ditolak) jika: berdistribusi chi-kuadrat 2 χ α 2,n 1 < χ 2 2 c < χ 1 α 2,n 1 14

15 Uji Hipotesis Nilai Varian H 0 : σ 1 2 = σ 2 2 H a : σ 1 2 σ Distribusi Normal Statistika uji: F c = s s 2 berdistribusi F dengan ( n 1 1) dan ( n 2 1) degrees of freedom s 1 2 > s 2 2 H 0 ditolak jika: F c > F 1 α,n1 1,n

16 Uji Hipotesis Nilai Varian H 0 : σ 1 2 = σ 2 2 =... = σ k 2 H a : σ 1 2 σ σ k 2 Distribusi Normal Statistika uji: H 0 ditolak jika: Q h berdistribusi chi-kuadrat dengan (k 1) degrees of freedom k # k 2 ( n Q = ( n 1) ln i 1)s & k % i ( n 1 i=1 $ % i=1 N k '( i=1 k ) 1 h =1+ n i 1 1, +. * N k - N = k i=1 1 3 k 1 ( ) n i Q h > χ 2 1 α,k 1 i=1 ( ) lns i 2 16

17 Uji Hipotesis Latihan Lihat kembali data debit puncak tahunan Sungai XYZ. Uji hipotesis yang menyatakan bahwa debit puncak tahunan rerata adalah 650 m 3 /s dan varians adalah m 6 /s 2. Contoh uji hipotesis.pdf Exercises on hypothesis thesis.pdf 17

18 Goodness of Fit Test CDF PLOT ON PROBABILITY PAPER 18

19 Testing The Goodness of Fit of Data to Probability Distributions Graphical (and visual) methods to judge whether or not a particular distribution adequately describes a set of observations: plot and compare the observed relative frequency curve with the theoretical relative frequency curve plot the observed data on appropriate probability paper and judge as to whether or not the resulting plot is a straight line Statistical tests: chi-square goodness of fit test the Kolmogorov-Smirnov test 19

20 Relative frequency Annual Peak Discharge of XYZ River 0.20 theoretical distribution observed data Discharge (m 3 /s) 20

21 markers: observed data line: theoretical distribution 21

22 Normal Distribution Paper 22

23 Chi-square Goodness of Fit Test Method of test Comparison between the actual number of observations and the expected number of observations (expected according to the distribution under test) that fall in the class intervals. The expected numbers are calculated by multiplying the expected relative frequency by the total number of observations. The test statistic is calculated from the following relationship: χ c 2 = k ( O i E ) 2 i i=1 E i 23

24 Chi-square Goodness of Fit Test The test statistic is calculated from the following relationship: χ c 2 = k ( O i E ) 2 i i=1 E i where: k is the number of class intervals O i is the number of observations in the ith class interval E i is the expected number of observations in the ith class interval according to the distribution being tested χ c2 has a distribution of chi-square with (k p 1) degrees of freedom, where p is the number of parameters estimated from the data 24

25 Chi-square Goodness of Fit Test The test statistic is calculated from the following relationship: χ c 2 = k ( O i E ) 2 i i=1 E i The hypothesis that the data are from the specified distribution is rejected if: χ 2 2 c > χ 1 α,k p 1 1 α α 2 χ 1 α,k p 1 25

26 The Kolmogorov-Smirnov Test Steps in the Kolmogorov-Smirnov test: Let P X (x) be the completely specified theoretical cumulative distribution function under the null hypothesis. Let S n (x) be the sample comulative density function based on n observations. For any observed x, S n (x) = k/n where k is the number of observations less than or equal to x. Determine the maximum deviation, D, defined by: D = max P X (x) S n (x) If, for the chosen significance level, the observed value of D is greater than or equal to the critical tabulated of the Kolmogorov- Smirnov statistic, the hypothesis is rejected. Table of Kolmogorov- Smirnov test statistic is available in many books on statistics. 26

27 The Kolmogorov-Smirnov Test Notes on the Kolmogorov-Smirnov test: The test can be conducted by calculating the quantities P X (x) and S n (x) at each observed point or By plotting the data on the probability paper and and selecting the greatest deviation on the probability scale of a point from the theoretical line. The data should not be grouped for this test, i.e. plot each point of the data on the probability paper. 27

28 Chi-square Goodness of Fit Test and The Kolmogorov-Smirnov Test Exercise Do the chi-square goodness of fit test and the Kolmogorov- Smirnov test to the annual peak discharge of XYZ River against normal distribution. 28

29 Chi-square Goodness of Fit Test and The Kolmogorov-Smirnov Test Notes on both tests when testing hydrologic frequency distributions. Both tests are insensitive in the tails of the distributions. On the other hand, the tails are important in hydrologic frequency distributions. To increase sensitivity of chi-square test The expected number of observations in a class shall not be less than 3 (or 5). Define the class interval so that under the hypothesis being tested, the expected number of observations in each class interval is the same. The class intervals will be of unequal width. The interval widths will be a function of the distribution being tested. 29

30 Chi-square Goodness of Fit Test and The Kolmogorov-Smirnov Test Exercise Redo the chi-square goodness of fit test and the Kolmogorov- Smirnov test to the annual peak discharge of XYZ River against normal distribution. Define the class intervals so that the expected number of observations in each class interval is the same. 30

31 31