teknikpengolahan data& - istiarto.staff.ugm.ac.idistiarto.staff.ugm.ac.id/docs/tpd/tpd8 uji...

Teknik Pengolahan Data Uji Hipotesis (Hypothesis Tes/ng)

15-‐Oct-‐15

h8p://is/

arto.staff.ugm.ac.id

1

Universitas Gadjah Mada Jurusan Teknik Sipil dan Lingkungan Prodi Magister Teknik Pengelolaan Bencana Alam

Uji Hipotesis •  Model Matema/ka vs Pengukuran •  komparasi garis teore/k (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 /dak cukup jelas untuk menyatakan bahwa model diterima atau ditolak

•  uji hipotesis sebagai alat analisis dalam komparasi tersebut

15-‐Oct-‐15

h8p://is/


2

Prosedur Uji Hipotesis •  Rumuskan hipotesis •  Rumuskan hipotesis alterna/f •  Tetapkan sta/s/ka uji •  Tetapkan distribusi sta/s/ka uji •  Tentukan nilai kri/k sebagai batas sta/s/ka uji harus ditolak •  Kumpulkan data untuk menyusun sta/s/ka uji •  Kontrol posisi sta/s/ka uji terhadap nilai kri/s

15-‐Oct-‐15

h8p://is/


3

Kemungkinan Kesalahan

pilihan keadaan nyata

hipotesis benar hipotesis salah

menerima tak salah kesalahan /pe II

menolak kesalahan type I tak salah

15-‐Oct-‐15

h8p://is/


4

Notasi

15-‐Oct-‐15

h8p://is/


5

H0 = hipotesis (yang diuji)

Ha = hipotesis alterna/f

1−α = /ngkat keyakinan (confidence level)

Uji Hipotesis Nilai Rata-‐rata

15-‐Oct-‐15

h8p://is/


6

H0 : µ =µ1

Ha : µ =µ2

X <µ1− z1−α

σX

n⇒ Z < −z1−α

Distribusi Normal σX2 diketahui

Z =

nσX

X −µ1( )Sta/s/ka uji: berdistribusi normal

Jika μ1 > μ2: H0 ditolak jika

Jika μ1 < μ2: H0 ditolak jika X <µ1+ z1−α

σX

n⇒ Z > z1−α

15-‐Oct-‐15

h8p://is/


7

luas = α

z1−α

prob Z > z1−α( ) = α


15-‐Oct-‐15

h8p://is/


8

H0 : µ =µ1

Ha : µ =µ2

X <µ1− t1−α ,n−1

sX

n

Distribusi Normal σX2 /dak diketahui

T =

nsX

X −µ1( )Sta/s/ka uji: berdistribusi t

H0 ditolak jika: jika μ1 > μ2

X >µ1+ t1−α ,n−1

sX

njika μ1 < μ2


15-‐Oct-‐15

h8p://is/


9

H0 : µ =µ0

Ha : µ ≠µ0

Z =n

σX

X −µ0( ) > z1−α 2

Distribusi Normal σX2 diketahui

Z =

nσX

X −µ0( )Sta/s/ka uji: berdistribusi normal

H0 ditolak jika:


15-‐Oct-‐15

h8p://is/


10

H0 : µ =µ0

Ha : µ ≠µ0

t =n

sX

X −µ0( ) > t1−α 2,n−1

Distribusi Normal σX2 /dak diketahui

T =

nsX

X −µ0( )Sta/s/ka uji: berdistribusi t

H0 ditolak jika:

Uji Hipotesis Nilai Rata-‐rata •  Hasil uji hipotesis adalah •  menolak H0, atau •  /dak menolak H0

•  Ar/nya •  H0: μ = μ0

•  Tidak menolak H0 à “menerima” H0 berar/ bahwa μ /dak berbeda secara signifikan dengan μ0.

•  Tetapi /dak dikatakan bahwa μ benar-‐benar sama dengan μ0 karena kita /dak membuk/kan bahwa μ = μ0.

15-‐Oct-‐15

h8p://is/


11

Uji hipotesis beda nilai rata-‐rata dua buah distribusi normal

15-‐Oct-‐15

h8p://is/


12

H0 : µ1−µ2 = δ

Ha : µ1−µ2 ≠ δ

z =n

σX

X −µ0( ) > z1−α 2

Z =X1−X2 −δ

σ12 n1+σ2

2 n2( )1 2Sta/s/ka uji: berdistribusi normal

H0 ditolak jika:

var X1( ) dan var X2( ) diketahui

Uji hipotesis beda nilai rata-‐rata dua buah distribusi normal

15-‐Oct-‐15

h8p://is/


13

H0 : µ1−µ2 = δ

Ha : µ1−µ2 ≠ δ

t =n

sX

X −µ0( ) > t1−α 2,n1+n2−2

T =X1−X2 −δ

n1+ n2( ) n1−1( ) s12 + n2 −1( ) s2

2#$

%&

n1n2 n1+ n2 −2( )#$ %&

'()

*)

+,)

-)

1 2Sta/s/ka uji:

berdistribusi t dengan (n1+n2–2) degrees of freedom

H0 ditolak jika:

var X1( ) dan var X2( ) tidak diketahui

Uji Hipotesis Nilai Varian

15-‐Oct-‐15

h8p://is/


14

H0 : σ2 = σ02

Ha : σ2 ≠ σ02

χα 2,n−1

2 < χc2 < χ1−α 2,n−1

2

Distribusi Normal

χc

2 =Xi −X( )σ0

2i=1

n

∑Sta/s/ka uji: berdistribusi chi-‐kuadrat

H0 diterima (/dak ditolak) jika:


15-‐Oct-‐15

h8p://is/


15

H0 : σ12 = σ2

2

Ha : σ12 ≠ σ2

2

Fc > F1−α ,n1−1,n2−1

2 Distribusi Normal

Fc =

s12

s22Sta/s/ka uji: berdistribusi F dengan

H0 ditolak jika:

n1−1( ) dan n2 −1( ) degrees of freedom

s12 > s2

2


15-‐Oct-‐15

h8p://is/


16

H0 : σ12 = σ2

2 = ... = σ k2

Ha : σ12 ≠ σ2

2 ≠ ... ≠ σ k2

Qh> χ1−α ,k−1

2

Distribusi Normal

Qh

Sta/s/ka uji: berdistribusi chi-‐kuadrat dengan (k – 1) degrees of freedom

H0 ditolak jika:

Q = n−1( ) lnni −1( ) si

2

N − ki=1

k

∑#

$%%

&

'((i=1

k

∑ − n−1( ) ln si2

i=1

k

∑

h =1+1

3 k −1( )1

ni −1−

1N − k

)

*+

,

-.

i=1

k

∑

N = nii=1

k

∑

Uji Hipotesis •  La/han •  Lihat kembali data debit puncak tahunan Sungai XYZ.

•  Uji hipotesis yang menyatakan bahwa debit puncak tahunan rerata adalah 650 m3/s dan varians adalah 45.000 m6/s2.

•  Contoh uji hipotesis.pdf •  Exercises on hypothesis thesis.pdf

15-‐Oct-‐15

h8p://is/


17

CDF PLOT ON PROBABILITY PAPER Goodness of Fit Test

15-‐Oct-‐15

h8p://is/


18

Testing The Goodness of Fit of Data to Probability Distributions •  Graphical (and visual) methods to judge whether or not a par/cular distribu/on adequately describes a set of observa/ons: •  plot and compare the observed rela/ve frequency curve with the theore/cal rela/ve frequency curve

•  plot the observed data on appropriate probability paper and judge as to whether or not the resul/ng plot is a straight line

•  Sta/s/cal tests: •  chi-‐square goodness of fit test •  the Kolmogorov-‐Smirnov test

15-‐Oct-‐15

h8p://is/


19

Annual Peak Discharge of XYZ River

0.00

0.05

0.10

0.15

0.20

Rela%v

e freq

uency

Discharge (m3/s)

observed data theore/cal distribu/on

15-‐Oct-‐15

h8p://is/


20

15-‐Oct-‐15

h8p://is/


21

markers: observed data line: theoretical distribution

15-‐Oct-‐15

h8p://is/


22

Normal Distribution Paper

Chi-‐square Goodness of Fit Test

•  Method of test •  Comparison between the actual number of observa/ons and the expected number of observa/ons (expected according to the distribu/on under test) that fall in the class intervals.

•  The expected numbers are calculated by mul/plying the expected rela/ve frequency by the total number of observa/ons.

•  The test sta/s/c is calculated from the following rela/onship:

χc

2 =Oi − Ei( )2

Eii=1

k

∑

15-‐Oct-‐15

h8p://is/


23

Chi-‐square Goodness of Fit Test ˃  The test sta/s/c is calculated from the following rela/onship:

χc

2 =Oi − Ei( )2

Eii=1

k

∑where: k is the number of class intervals Oi is the number of observa/ons in the ith class interval Ei is the expected number of observa/ons in the ith class interval

according to the distribu/on being tested χc2 has a distribu/on of chi-‐square with (k – p – 1) degrees of freedom,

where p is the number of parameters es/mated from the data

15-‐Oct-‐15

h8p://is/


24

Chi-‐square Goodness of Fit Test ˃  The test sta/s/c is calculated from the following rela/onship:

χc

2 =Oi − Ei( )2

Eii=1

k

∑

˃  The hypothesis that the data are from the specified distribu/on is rejected if:

χc2 > χ1−α ,k−p−1

2

1−α α

χ1−α ,k−p−12

15-‐Oct-‐15

h8p://is/


25

The Kolmogorov-‐Smirnov Test

•  Steps in the Kolmogorov-‐Smirnov test: •  Let PX(x) be the completely specified theore/cal cumula/ve distribu/on func/on under the null hypothesis.

•  Let Sn(x) be the sample comula/ve density func/on based on n observa/ons. For any observed x, Sn(x) = k/n where k is the number of observa/ons less than or equal to x.

•  Determine the maximum devia/on, D, defined by: D = max |PX(x) – Sn(x)|

•  If, for the chosen significance level, the observed value of D is greater than or equal to the cri/cal tabulated of the Kolmogorov-‐Smirnov sta/s/c, the hypothesis is rejected. Table of Kolmogorov-‐Smirnov test sta/s/c is available in many books on sta/s/cs.

15-‐Oct-‐15

h8p://is/


26

The Kolmogorov-‐Smirnov Test

•  Notes on the Kolmogorov-‐Smirnov test: •  The test can be conducted by calcula/ng the quan//es PX(x) and Sn(x) at each observed point or

•  By plotng the data on the probability paper and and selec/ng the greatest devia/on on the probability scale of a point from the theore/cal line. •  The data should not be grouped for this test, i.e. plot each point of the data on the probability paper.

15-‐Oct-‐15

h8p://is/


27

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 distribu/on.

15-‐Oct-‐15

h8p://is/


28


•  Notes on both tests when tes/ng hydrologic frequency distribu/ons. •  Both tests are insensi/ve in the tails of the distribu/ons. •  On the other hand, the tails are important in hydrologic frequency distribu/ons.

•  To increase sensi/vity of chi-‐square test •  The expected number of observa/ons 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 observa/ons in each class interval is the same. •  The class intervals will be of unequal width. •  The interval widths will be a func/on of the distribu/on being tested.

15-‐Oct-‐15

h8p://is/


29


•  Exercise •  Redo the chi-‐square goodness of fit test and the Kolmogorov-‐Smirnov test to the annual peak discharge of XYZ River against normal distribu/on. •  Define the class intervals so that the expected number of observa/ons in each class interval is the same.

15-‐Oct-‐15

h8p://is/


30

15-‐Oct-‐15

h8p://is/


31

teknikpengolahan data& - istiarto.staff.ugm.ac.idistiarto.staff.ugm.ac.id/docs/tpd/tpd8 uji...

Documents