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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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.
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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.
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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).
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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 =
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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?
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Hypothesis Test Hypothesis Test
State the null and alternate hypotheses
State the null and alternate hypotheses
Step 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
α = 0.05
The test statistic is the F distribution
State the decision ruleState the decision ruleStep 4Step 4
Compute the test statistic and make
a decision
Compute the test statistic and make
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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 ?
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
…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…
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
• 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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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=
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
≠
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
• SS Total is the total sum of squares
nX
X2
2 )(TotalSS
Σ−Σ=
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
• 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
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•SSE is the sum of squares error
SST - totalSS SSE =
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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%.
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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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
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…continued
• SS Total (is the total sum of squares)
= 8613
= 2634 -
)( TotalSS22 Σ−Σ= n
XX
(182)2
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
…continued
( )nX
nT
SSTc
c22 Σ−
Σ=
•SST is the treatment sum of squares
( ) ( ) ( )
= 76.2513
)182(5
854
464
51 222 2
−
++=
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•SSE is the sum of squares error…continued
SSE = SS Total - SST
86 – 76.25
= 9.75
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Hypothesis Test Hypothesis Test
State the null and alternate hypotheses
State the null and alternate hypotheses
Step 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
α = 0.05The test statistic is the
F distribution
State the decision ruleState the decision ruleStep 4Step 4
Compute the test statistic and make
a decision
Compute the test statistic and make
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Inferences About
Treatment Means
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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Confidence Interval for the Difference Between Two Means
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+
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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?
Confidence Interval for the Difference Between Two Means
Confidence Interval for the Difference Between Two Means
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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)
Confidence Interval for the Difference Between Two Means
Confidence Interval for the Difference Between Two Means
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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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
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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
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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
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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
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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
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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
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– 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
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Hypothesis Test Hypothesis Test
State the null and alternate hypotheses
State the null and alternate hypotheses
Step 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
α = 0.05
The test statistic is the F distribution
State the decision ruleState the decision ruleStep 4Step 4
Compute the test statistic and make
a decision
Compute the test statistic and make
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?
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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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Hypothesis Test Hypothesis Test
State the null and alternate hypotheses
State the null and alternate hypotheses
Step 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
α = 0.05
The test statistic is the F distribution
State the decision ruleState the decision ruleStep 4Step 4
Compute the test statistic and make
a decision
Compute the test statistic and make
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
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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.
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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
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Using
See…See…
Highlight ANOVA: TWO FACTOR WITHOUT REPLICATION
…Click OK
SelectSelect
INPUT DATA INPUT DATA
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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.
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Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
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
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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
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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!
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This completes Chapter 12