statistik 1 10 12 edited_anova

12 - 1

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.

12 - 2


1. Discuss the general idea of analysis of variance.

2. List the characteristics of the F distribution.

When you have completed this chapter, you will be able to:

Organize data into a one-way and a two-way ANOVA table.

3. Conduct a test of hypothesis to determine whether the variances of two populations are equal.

4.

12 - 3


5. Define the terms treatments and blocks.

6. Conduct a test of hypothesis to determine whether three or more treatment means are equal.

7. Develop multiple tests for difference between each pair of treatment means.

12 - 4


Karakteristik Distribusi-F

Karakteristik Distribusi-F

Terdapat ‘keluarga’ F-Distributions:Terdapat ‘keluarga’ F-Distributions:

Each member of the family is determined by two parameters:

…the numerator degrees of freedom, and the … denominator degrees of freedom

F cannot be negative, and it is a continuous distribution

The F distribution is positively skewed

Its values range from 0 to ∞ as F → ∞ , the curve approaches the X-axis

12 - 5


The F-Distribution, F(m,n)

0 1.0

Not symmetric (skewed to the right)

F

Nonnegative values only

α

Each member of the family is determined by two parameters: the numerator degrees of freedom (m) and the denominator degrees of freedom (n).

12 - 6


Test for Equal Variances Test for Equal Variances

For the two tailed test, the test statistic is given by:

The null hypothesis is rejected if the computed value of the test statistic

is greater than the critical value

22

21

s

sF =

and are the sample variances for the two samples21s 2

2s

12 - 7


Test for Equal Variances• For the two tail test, the test statistic is given by:

• where s12 and s2

2 are the sample variances for the two samples.

• The null hypothesis is rejected at α level of significance if the computed value of the test statistic is greater than the critical value with a confidence level α/2 and numerator and denominator dfs.

),(

),( arg=

22

21

22

21SSofSmaller

SSoferLF

22

21

s

sF =

12 - 8


Test for Equal Variances

• For the one tail test, the test statistic is given by:

where s12 and s2

2 are the sample variances for the two samples.

• The null hypothesis is rejected at α level of significance if the computed value of the test statistic is greater than the critical value with a confidence level α and numerator and denominator dfs.

22

2112

2

21 > :H if = σσS

SF

12 - 9


Bejo, seorang pialang di Maju Securities, mencatat rata-rata return pada sebuah sample 10 saham internet 12.6 persen dengan standar deviasi 3.9 persen.

Sedangkan rata-rata return pada sebuah sample 8 saham utilitas adalah 10.9 persen dengan standar deviasi 3.5 persen.

Pada tingkat signifikansi 0.05, dapatkah Bejo menyimpulkan bahwa terdapat variasi yang lebih besar pada saham internet?

12 - 10


Do not reject H0Do not reject H0 Reject H0 and accept H1

Reject H0 and accept H1

State the null and alternate hypothesesState the null and alternate hypothesesStep 1Step 1

Select the level of significanceSelect the level of significanceStep 2Step 2

Identify the test statisticIdentify the test statisticStep 3Step 3

State the decision ruleState the decision ruleStep 4Step 4

Step 5Step 5

Hypothesis Testing Hypothesis Testing

Compute the value of the test statistic and make a decision

Compute the value of the test statistic and make a decision

12 - 11


Hypothesis Test Hypothesis Test

State the null and alternate hypotheses


Step 1Step 1




α = 0.05

The test statistic is the F distribution


Compute the test statistic and make

a decision


a decision

Step 5Step 5

Reject H0 if F > 3.68 The df are 9 in the numerator and

7 in the denominator.

Do not reject the null hypothesis; there is insufficient evidence to show more variation in the internet stocks.

= 1.2416 = 1.2416F 22

21

s

s= =2

2

)5.3(

)9.3(

220: U IH σσ ≤

U22

1 : IH σ > σ

12 - 12


Contoh

Nilai UjianNilai Ujian

Kelas AKelas A Kelas BKelas B

5252 5959

6767 6060

5656 6161

4545 5151

7070 5656

5454 6363

6464 5757

6565

Uji apakah ada perbedaan yang signifikan varians nilai ujian kelas A dan kelas B ?

12 - 13


This this technique is called analysis of variance or ANOVA

The F distribution is also used for testing whether two or more sample means came from

the same or equal populations

The F distribution is also used for testing whether two or more sample means came from

the same or equal populations

ANOVA

12 - 14


…the populations have equal standard deviations

…the samples are randomly selected and are independent

…the sampled populations follow the normal distribution

ANOVA requires the following

conditions…

ANOVA requires the following

conditions…

12 - 15


• ANOVA requires the following conditions:– The populations being sampled are normally

distributed.– The populations have equal standard deviations.– The samples are randomly selected and are

independent– Data must be at least interval-scale.

• Type of ANOVA :– One-Way (One-Factor) ANOVA– Multi-Way (Multi-Factor) ANOVA

Two-Way ANOVA

12 - 16


One Way ANOVA

TREATMENTTREATMENT11 22 33XX1.11.1 XX1.21.2 XX1.31.3XX2.12.1 XX2.22.2 XX1.11.1XX3.13.1 XX3.23.2 XX1.11.1XX4.14.1 XX4.24.2 XX1.11.1TT11 TT22 TT33jT

jX 1X 2X 3X

: Overall Mean (Grand Mean); X

XT

XT Σ=

12 - 17


The Null Hypothesis (H0) is that the population means are the same

The Alternative Hypothesis (H1) is that

at least one of the means is different

ANOVA ProcedureANOVA Procedure

The Test Statistic is the F distribution

The Decision rule is to reject H0

if F(computed) is greater than F(table)

with numerator and denominator df

12 - 18


The hypothesis

• Suppose that we have independent samples of n1, n2, . . ., nK observations from K populations. If the population means are denoted by µ1, µ2, . . ., µK, the one-way analysis of variance framework is designed to test the null hypothesis

ji1

210

, pair one least at For:

===:

μμμμH

μμμH

ji

K

≠

12 - 19


Sample Observations from Independent Random Samples of K Populations

Population1 2 . . . K

Meanµ1 µ2 . . . µK

Varianceσ2 σ2 . . . σ2

Sample

observations

from the

population

x11

x12

.

.

.x1n1

x21

x22

.

.

.x2n2

. . .. . .

. . .

xK1

xK2

.

.

.xKnK

Sample sizen1 n2 . . . nK

Same !!

unequal !!

Unequal number of observations in the K samples in general.nT=n1+…+nK

12 - 20


12 - 21


Terminology

Total Variation …is the sum of the squared differences between each observation and

the overall mean

Random Variation …is the sum of the squared differences between each observation and

its treatment mean

Treatment Variation …is the sum of the squared differences

between each treatment mean and the overall mean

12 - 22


SSESSTF

( )kn−=

( )k −1

If there are a total of n observations the denominator df =n - k

The test statistic is computed by:

If there k populations being sampled, the numerator df = k – 1

12 - 23


• SS Total is the total sum of squares

nX

X2

2 )(TotalSS

Σ−Σ=

12 - 24


• SST is the treatment sum of squares

( )nX

nT

SSTc

c22 Σ−

Σ=

TC is the column total, nc is the number of observations in each column, ΣX the sum of all the observations, and n the total number of observations

12 - 25


•SSE is the sum of squares error

SST - totalSS SSE =

12 - 26


One-Way ANOVA Table

Source of Source of VariationVariation

Sum of Sum of SquareSquare

d.f.d.f. Mean SquareMean Square FF

TreatmentTreatment SSTSST k – 1k – 1 MST = SST/(k – 1)MST = SST/(k – 1) MST/MSEMST/MSE

ErrorError SSESSE nnT T - k- k MSE = SSE/(N – k)MSE = SSE/(N – k)

TotalTotal SSSStotaltotal nT - 1

12 - 27


D’Cost memiliki spesialisasi makanan murah meriah.

Bejo, Manager D’cost, baru saja membuat menu Nasi Goreng baru. Sebelum dimasukkan dalam menu reguler, dia memutuskan mencoba menawarkannya di beberapa restorannya.

Bejo ingin mengetahui apakah ada perbedaan jumlah rata-rata nasi goreng baru terjual per hari di restoran D’cost Kaza, Royal, dan Kayon restaurants. Gunakan tingkat signifikansi 5%.

12 - 28


Kaza Royal Kayon13 10 1812 12 1614 13 1712 11 17

17

Tc 51 46 85nc 4 4 5

Tc 51 46 85nc 4 4 5

…continued

12 - 29


KAZA ROYAL KAYON KAZA^2 ROYAL^2 KAYON^2

13 10 18 169 100 324

12 12 16 144 144 256

14 13 17 196 169 289

12 11 17 144 121 289

17 289

Tc 51 46 85 182 653 534 1447 2634

nc 4 4 5 13

12 - 30


…continued

• SS Total (is the total sum of squares)

= 8613

= 2634 -

)( TotalSS22 Σ−Σ= n

XX

(182)2

12 - 31


…continued

( )nX

nT

SSTc

c22 Σ−

Σ=

•SST is the treatment sum of squares

( ) ( ) ( )

= 76.2513

)182(5

854

464

51 222 2

−

++=

12 - 32


•SSE is the sum of squares error…continued

SSE = SS Total - SST

86 – 76.25

= 9.75

12 - 33





Step 1Step 1




α = 0.05The test statistic is the

F distribution



a decision


a decision

Step 5Step 5

Reject H0 if F > 4.10 The df are 2 in the numerator and

10 in the denominator.

= 39.10 = 39.10

1:H

0:H 1µ 2µ == 3µTreatment means are not all equal

SSE

SSTF

( )kn −=

( )k − 1

109.75 2= 76.25

12 - 34


The decision is to reject the null hypothesis

The treatment means are not the same

The mean number of meals sold at the three locations is not the same

…continued

12 - 35


Analysis of Variance

Source DF SS MS F P

Factor 2 76.250 38.125 39.10 0.000

Error 10 9.750 0.975

Total 12 86.000

Individual 95% CIs For Mean Based on Pooled St.Dev

Level N Mean St.Dev ---------+---------+---------+-------

Kaza 4 12.750 0.957 (---*---)

Royal 4 11.500 1.291 (---*---)

Kayon 5 17.000 0.707 (---*---)

---------+---------+---------+-------

Pooled St.Dev = 0.987 12.5 15.0 17.5

ANOVA TableANOVA Table

…from the Minitab system

12 - 36


One-Way ANOVA Table

Source of Source of VariationVariation

Sum of Sum of SquareSquare

d.f.d.f. Mean SquareMean Square FF

TreatmentTreatment 76.2576.25 22 MST = 38.12MST = 38.12 39.0939.09

ErrorError 9.759.75 1010 MSE = 0.975MSE = 0.975

TotalTotal 86.0086.00 12

12 - 37


Inferences About

Treatment Means

12 - 38


Ketika menolak hipothesis null bahwa rata-ratanya sama, kita mungkin juga ingin mengetahui rata-rata treatment mana yang berbeda

Salah satu prosedur yang paling sederhana adalah melalui penggunaan confidence intervals

Inferences

About Treatment

Means

Confidence IntervalConfidence Interval

12 - 39


Confidence Interval for the Difference Between Two Means


where t is obtained from the t table with degrees of freedom (n - k).

MSE = [SSE/(n - k)]

where t is obtained from the t table with degrees of freedom (n - k).

MSE = [SSE/(n - k)]

( )X X1 2− t± MSEn n1 2

1 1+

12 - 40


Develop a 95% confidence interval for the difference in the mean number

of Nasi Goreng sold in Kayon and Kaza.

Can Bejo conclude that there is a difference between the two restaurants?



12 - 41


MSEMSE

( )X X1 2− t± MSE

n n1 2

1 1+

(17-12.75) 2.228± .9751

4

1

5+

. .4 25 1 48± ⇒ ( 2.77, 5.73)



12 - 42


Contoh• Berikut adalah hsl panen padi (kuintal) dari 12 petak

sawah dengan 3 jenis pupuk yang berbeda. Tiap jenis pupuk diberikan pada masing-masing 4 petak sawah. Apakah ada perbedaan hsl panen dari ketiga jenis pupuk tsb? Gunakan α = 5%

Jenis PupukJenis Pupuk

AA BB CC

5555 6666 4747

5454 7676 5151

5959 6767 4646

5656 7171 4848

12 - 43


Latihan Soal

• Berikut adalah waktu (menit) yang dibutuhkan untuk mengerjakan 1 soal latihan dari 4 mata kuliah. Sampel random masing-masing 5 mhs untuk tiap mata kuliah. Uji apakah ada perbedaan yang signifikan ? (Kerjakan menggunakan Tabel ANOVA)

E. MakroE. Makro E.MikroE.Mikro MatematikaMatematika StatistikaStatistika

1818 2020 2020 2222

2121 2222 2424 2424

2020 2323 2525 2323

2525 2121 2828 2525

2626 2424 2828 2525

12 - 44


Two-Way ANOVA

TREATMENTTREATMENT11 22 33 TTii

BBLLOOCCKK

11 XX1.11.1 XX1.21.2 XX1.31.3 TT11

22 XX2.12.1 XX2.22.2 XX1.11.1 TT22

33 XX3.13.1 XX3.23.2 XX1.11.1 TT33

44 XX4.14.1 XX4.24.2 XX1.11.1 TT44

TTjj TT11 TT22 TT33jX

iX

TX1X 3X2X

4X3X

2X

1X

12 - 45


Untuk ANOVA dua-faktor kita menguji:

apakah ada perbedaan yang signifikan pada treatment effect dan

apakah ada perbedaan pada blocking effect.

…Let Br be the block totals (r for rows),

nr be the number of observations in each row

…Let SSB represent the sum of squares for the blocks

Two Factor ANOVATwo Factor ANOVA

SSBBn

Xn

r=

−Σ Σ2 2( )

=( )

( ))1)(1(

1

−−−

bkSSE

kSSTF

r

12 - 46


General Format of Two-Way Analysis of Variance Table

Source of Variat ion

Sums of Squares

Degrees of

Freedom

Mean Squares F Ratios

Treatments

SST k-1 MST=SST/(k-1) MST/MSE

Blocks SSB b-1 MSB=SSB/(b-1) MSB/MSEError SSE (k-1)(b-1) MSE=SSE/[(k-1)(b-

1)]Total SSTotal nT-1

k : number of treatment (column)b : number of block (row)n : number of observation

12 - 47


Pabrik Tempe Bieber beroperasi 24 jam sehari selama 5 hari dalam seminggu. Para pekerja bergantian shifts kerja tiap minggu. Pak Todd Bieber, pemilik pabrik, ingin mengetahui apakah ada perbedaan dalam jumlah produksi Tempe ketika para pegawai bekerja pada shift yang berbeda.

Sampel yang terdiri dari lima pegawai dipilih dan produksi mereka dicatat pada setiap shift. Pada tingkat signifikansi 0.05, apakah kita dapat menyimpulkan bahwa ada perbedaan rata-rata produksi pada shift dan rata-rata produksi pada pegawai?

Two Factor ANOVATwo Factor ANOVA

12 - 48


ANOVAANOVA

Employee DayOutput

EveningOutput

NightOutput

McCartney 31 25 35

Neary 33 26 33

Schoen 28 24 30

Thompson 30 29 28

Wagner 28 26 27

…continued

12 - 49


Day Evening Night Tr nr Day^2 Eve^2 Night^2

McCartney 31 25 35 91 3 961

625

1,225

Neary 33 26 33 92 3 1,089

676

1,089

Schoen 28 24 30 82 3 784

576

900

Thompson 30 29 28 87 3 900

841

784

wagner 28 26 27 81 3 784

676

729

Tc 150 130 153 433 4,518

3,394

4,727

12,639

nc 5 5 5 15

12 - 50


– Compute the various sum of squares:

Source SS df MS F p-value

Treatments 62.53 2 31.267 5.75 .0283

Blocks 33.73 4 8.433 1.55 .2762

Error 43.47 8 5.433

Total 139.73 14 　　　

12 - 51





Step 1Step 1




α = 0.05




a decision


a decision

Step 5Step 5

Reject H0 if F > 4.46. The df are 2

and 8

1:H

0:H 1µ 2µ == 3µNot all means are equal

=( )

( ))1)(1(

1

−−−

bkSSE

kSSTF

Difference between various shifts?Difference between various shifts?

12 - 52


Since 5.754 > 4.46, H0 is rejected.

( )( )( )( )151343.47

1353.62

−−−

=

ANOVAANOVA

…continued

Step 5Step 5

There is a difference in the mean number of units produced on the different shifts.

=( )

( ))1)(1(

1

−−−

bkSSE

kSSTF

= 5.754

12 - 53





Step 1Step 1




α = 0.05




a decision


a decision

Step 5Step 5

1:H

0:H 1µ 2µ == 3µNot all means are equal

=( )

( ))1)(1(

1

−−−

bkSSE

kSSTF

Difference between various shifts?Difference between various shifts?

Reject H0 if F > 3.84 The df are 4 and 8

12 - 54


ANOVAANOVA

…continued

Step 5Step 5=

( )( ))1)(1(

1

−−−

bkSSE

kSSTF

Since 1.55 < 3.84, H0 is not rejected.

= 1.55( )( )4243.47

433.73=

There is no significant difference in the mean number of units produced by the various employees.

12 - 55


Units versus Worker, ShiftAnalysis of Variance for Units Source DF SS MS F P

Worker 4 33.73 8.43 1.55 0.276

Shift 2 62.53 31.27 5.75 0.028

Error 8 43.47 5.43

Total 14 139.73

Units versus Worker, ShiftAnalysis of Variance for Units Source DF SS MS F P

Worker 4 33.73 8.43 1.55 0.276

Shift 2 62.53 31.27 5.75 0.028

Error 8 43.47 5.43

Total 14 139.73

…from the Minitab system

ANOVAANOVA

12 - 56


Using

See…See…

Highlight ANOVA: TWO FACTOR WITHOUT REPLICATION

…Click OK

SelectSelect

INPUT DATA INPUT DATA

12 - 57


SS TotalSS Total

SSTSSE

SSBFtestFtest FcriticalFcritical

Using

Since F(test) < F(critical), there is not sufficient evidence to reject H0

Since F(test) < F(critical), there is not sufficient evidence to reject H0

There is no significant difference in the average

number of units produced by the different employees.

There is no significant difference in the average

number of units produced by the different employees.

12 - 58


Latihan soal

• Berikut adalah waktu (menit) yang dibutuhkan untuk mengerjakan 1 soal latihan dari 4 mata kuliah. Sampel random masing-masing 5 mhs untuk tiap mata kuliah. Uji apakah ada perbedaan yang signifikan waktu pengerjaan antara keempat mata kuliah dan antara mahasiswa tersebut?

E. MakroE. Makro E.MikroE.Mikro MatematikaMatematika StatistikaStatistika

AA 1818 2020 2020 2222

BB 2121 2222 2424 2424

CC 2020 2323 2525 2323

DD 2525 2121 2828 2525

EE 2626 2424 2828 2525

12 - 59


Latihan Soal

• Uji dengan α = 0,1 apakah ada perbedaan hsl produksi (kg) antara ketiga mesin dan kelima karyawan tersebut?

MesinMesin

KaryawanKaryawan II IIII IIIIII

AA 2121 1717 3131

BB 2727 2525 2828

CC 2929 2020 3232

DD 2323 1515 3030

EE 2525 2323 2424

12 - 60


Test your learning…Test your learning…

www.mcgrawhill.ca/college/lindClick on…Click on…

Online Learning Centrefor quizzes

extra contentdata setssearchable glossaryaccess to Statistics Canada’s E-Stat data…and much more!

http://www.mcgrawhill.ca/college/lind

12 - 61


This completes Chapter 12

statistik 1 10 12 edited_anova

Education

test statistic

test of hypothesis

equal variances test

mcgrawhill companies

tail test

tailed test

sample variances

computed value