Extensive study I assume the mean age of this class is not 23.5!
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Extensive study
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I assume the mean age of this class is not 23.5!
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Ho: u=80, but the value of u is really equal to 82 score
80 82
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Decision-making always exists risk
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Prob. of sample ?
X
2.5%2.5%
Prob.=Significance level
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Judgment in a relaxed manner
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n
x
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ZZ
a/2
z
X0 :H
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ZZ
p/2
a/2
z
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n
Xz
X1
X0
:H
:H
X1
X0
:H
:H
n
Xz
X1
X0
:H
:H
X1
X0
:H
:H
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Critical limit Critical limit
368:H X0
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When n≥30, t distribution is almost equal to standard normal distribution. So we also use z-test.
)30(n Z
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s
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EXCEL
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Introduction to analysis of variance
Suppose you are interested in determining whether certain situations produce differing amounts of stress. You know the amount of the hormone corticosterone (HC) circulating in the blood is a good measure of how stressed a person is. You randomly assign 15 students into three groups of 5 each. The students in group 1 have their HC levels measured immediately after returning from vacations (low stress). The students in group 2 have
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their HC levels measured after they have been in class for a week (moderate stress). The students in group 3 are measured immediately before final exam week (high stress). All measurements are taken at the same time of day. You record the data shown in Table 7.1. Scores are in milligrams of HC per 100 milliliters of blood.
Null hypothesis: the different situations affect stress equally. (H0: u1=u2=u3)
Can we develop a technique to solve the problem easily? ----F-test
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Table 7.1----sample data
group1 group2 group3
X1 X2 X3
2 10 10
3 8 13
7 7 14
2 5 13
6 10 15
3.8X
jX 4 8 13
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The distribution of statistic F
• F fundamentally is the ratio of two independent variance estimates of the same population variance σ2
22
21
s
s
of σstimate 2 variance e
of σstimate 1 variance eF 2
2
Like the t distribution, F distribution varies with degrees of freedom . However, F distribution has two values for degrees of freedom, one for the numerator and the other for the denominator.
df1=n1-1 df2=n2-1
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FF ((nn11-1-1, , nn22-1-1))
00 FF1
The distribution of statistic F
critF
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The analysis of variance ------ANOVA
• It is a statistical technique employed to analyze multigroup experiments.
• Using the F-test allows us to make one overall comparison that tells whether there is a significant difference between the means of the groups.
• If the groups are divided by a factor (for example: situation) to be investigated in the experiment, then the analysis of variance is called as one-way’s.
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The assumption of ANOVA
• The population of each group is of normal distribution
• The population of each group is of equal variance, σ1
2 = σ22 = …… = σk
2 = σ2
• The sampling from different group is of independent
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SST SSW and SSB
group jth of number then
k)j(1 group jth of mean the
samples the of mean total the
:where ).squares( of sum
groups-between the called groups, each between exists that
yvariabilit the and ),
(SSW squares of sum groups-within
the called group, each within exists thaty variabilit the
:sources two into dpartitione is )
(SST data of squares of sum total the
j
j
2j
k
2
k
1
2
12ij
2jij
2
ij2ij
2ij
X
X
SSWSST)XX(SSB
n
X
n
XX
)XX(
N
XX
)XX(
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H retain FF If H reject FF If
becomes. ) :(H
hypothesis null the leunreasonab more the ratio, F the larger the
Thus, .unaffected is SW the whereas ect,factor)eff (observed
svariable' tindependen the of magnitude the with increases SBSWSB
F
)1-kSSBBestimate(S variance
groups-between the called isy variabilit groups-between the
on based which and ),k-nSSWWestimate(S variance
groups-within the called isy variabilit groups-within the
on based )( variance population the of estimate The
0crit0crit
0
2
2
2
2
2
2
2
,,321
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Solution of Table 7.1
• K=3, n=15, nj=5
• SSB=203.333 SSW=54
• dfB=k-1=2 dfW=n-k=12
• SB2=SSB/dfB=203.333/2=101.667
• SW2=SSW/dfW=54/12=4.5
• F=SB2/SW2=22.59 Fcrit=3.88 (α=0.05)
• Therefore, H0 is rejected
FFFF
FF
FF
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Multiple comparisons
• We concluded that the three situations were not the same in the stress levels they produced. This means that at least one condition differs from at least one of the other. To determine which conditions differ, multiple comparisons between pairs of group means are usually made. Here we shall discuss a priori comparisons
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A priori comparisons
• Definition of statistic t
0H reject then t
,tIf
knt
n1
n1
SW
XXt
:H:H
ji
2
ji
ji1ji0
∽
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Solution of Table 7.1(α=0.05)
0
2
21
0
2
32
0
2
31
crit
H reject
SW
XXt
H reject
SW
XXt
H reject
SW
XXt
(12)t 3)-(15tt
98.2
52
5.4
84
n1
n1
73.3
52
5.4
138
n1
n1
71.6
52
5.4
134
n1
n1
18.2
21
32
31
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Two-way analysis of variance
• The two-way analysis of variance allows us in one experiment to evaluate the effect of two independent variable (factors) and the interaction between them
• suppose a professor in physical education conducts an experiment to compare the effects on nighttime sleep of different intensities of exercise and the time of day when the exercise is done. For this example, let’s assume that there are two levels of exercise (light and heavy) and two times of
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day (morning and evening). The experiment is shown diagrammatically in Figure 7.1.
Figure
7.1
Factor B, exercise intensity
b1,light b2,heavy
Facto
r A, tim
e of d
ay
a1,m
orn
ing
a1b1 ; sleep scores of subjects who do
light exercise in the morning
a1b2 ; sleep scores of subjects who do heavy exercise in
the morning
a2,e
ve
nin
g
a2b1 ; sleep scores of subjects who do
light exercise in the evening
a2b2 ; sleep scores of subjects who do heavy exercise in
the evening
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Here there are two factors and each factor has two levels. Thus, the design is referred to as a 2×2 design. For example, if factor A has three levels, then the experiment would be called a 3×2 design.There are three analyses done in this design: First, we want to determine whether factor A has a significant effect on sleep, disregarding the effect of factor B. Second, we want to determine whether factor B has a significant effect on sleep, without considering the effect of factor A. Finally, we want to determine whether there is an interaction between factors A and B in their effect on sleep.
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Intensity of exercise
light heavy
No significant time of day or intensity of exercise effects
sleep
morning
evening
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Intensity of exercise
light heavy
sleep
morning
evening
Significant time of day effect; no other effect
Intensity of exercise
light heavy
sleep
morning
evening
Significant time of day effect; no other effect
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light heavy
Intensity of exercise
sleep
morning
evening
Significant intensity of exercise effect; no other effect
light heavy
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light heavy
Intensity of exercise
sleep
morning
evening
Significant intensity of exercise and time of day effects; no interaction effect
light heavy
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light heavy
Significant interaction effect; no other effect
Intensity of exercise
sleep
morning
evening
light heavy
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light heavy
Intensity of exercise
sleep
morning
evening
Significant intensity of exercise and time of day effects; no interaction effect
light heavy
Intensity of exercise
light heavy
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B∑
b1 b2 bc
A
a1 cell11 cell12 … cell1c ∑row1X
a2 cell21 cell22 … cell2c ∑row2X
… … … … …
ar cellr1 cellr2 … cellrc ∑rowrX
∑ ∑col1X ∑col2X ∑colcX
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SSW SSR SSC and SSRC
1prc
SSW
df
SSW
rc
XXX
XXXXXX
w
2
cellrc
2
12cell
2
11cell
2
cellrccellrc
2
12cell12cell
2
11cell11cell
2
2ij
rc1211
SW estimate variance cell-within
X
SSSSSSSSW squares of sum cells-within the
Not affected by A or BWhere: p= the number of data in each cell
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SSW SSR SSC and SSRC
data of number total theN
row each in data of number the n :where
SR estimate variancerow
SSR
squares of sumrow the
row
2
1
2
22
2
2
1
22
2
2
1
r
SSR
df
SSR
N
X
n
XXX
XXXXXXn
R
row
rowrrowrow
rowrrowrowrow
Only affected by A
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SSW SSR SSC and SSRC
data of number total theN
column each in data of number the n :where
SC estimate variancerow
SSC
squares of sum column the
col.
2
1
2
.
2
.
2
2.
2
1.
2
.
2
2.
2
1..
c
SSC
df
SSC
N
X
n
XXX
XXXXXXn
c
col
ccolcolcol
ccolcolcolcol
Only affected by B
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SSW SSR SSC and SSRC
estimate variance columnrow
squares of sum columnrow the
2
1c1r
SSRC
df
SSRCSRC
SSCSSRN
X
rc
XXX
SSCSSR
XXXXXXpSSRC
RC
2
2
cellrc
2
12cell
2
11cell
2
cellrc
2
12cell
2
11cell
Only affected by interaction
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effect ninteractio tsignifican a is there FF if SWSRC
F
:effect) column(row B and A factors of ninteractio the test To
j)i,( H
B factor for effect main tsignifican a is there FF if SWSC
F
:effect) (column B factor of effect main test To
H
A factor for effect main tsignifican a is there FF if SWSR
F
:effect)(row A factor of effect main test To
H
crit2
2
0
crit2
2
0
crit2
2
0
,
:
,
:
,
:
ji
c21
r21
ba
bbb
aaa
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Data light moderate heavy
Morn-ing
6.5 7.4
7.3 7.2
6.6 6.8
7.4 7.3
6.8 7.6
6.7 7.4
8.0 7.6
7.7 6.6
7.1 7.2
∑x=129.20
∑x2=930.50
n=18
=6.97 =7.20 =7.37 =7.18
Even-ing
7.1 7.2
7.6 7.5
8.2 7.6
7.4 8.0
8.1 7.6
8.2 8.0
8.2 8.7
8.5 9.6
9.5 9.4
∑x=147.20
∑x2=1212.68
n=18
=7.67 =7.88 =8.98 =8.18
∑x=90.50
∑x2=645.30
n=12
=7.32
∑x=90.50
∑x2=685.07
n=12
=7.54
∑x=98.10
∑x2=812.81
n=12
=8.18
∑x=276.4
∑x2=2143.18
N=36
x x x x
x
xx x
x x x
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577.532
9.533.470.462.442.438.4118.2143
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754.436
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1298.190.587.8
SSC
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653.947.346.044.243.241.8
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186.030
577.5
df
SSWSW
856.02
712.1
df
SSRCSRC
377.22
754.4
df
SSCSC
000.91
000.9
df
SSRSR
212)1c)(1r(df
2131cdf
1121rdf
3056)1p(rcdf
W
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w
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1.For the row effect F=SR2/SW2=9.000/0.186=48.42>F
crit(0.05,1,30)=4.17
We reject H0 with respect to the A variable, which in thi
s experiment is time of day.
2.For the column effect F=SC2/SW2=2.377/0.186=12.78
>Fcrit(0.05,2,30)=3.32
We reject H0 with respect to the B variable, which in thi
s experiment is amount of exercise.
3.For the row×column interaction effect F=SRC2/SW2=
0.856/0.186=4.60>Fcrit(0.05,2,30)=3.32
We reject H0 regarding the interaction of variables A an
d B.Homework: S402 12.11, 12.12, 12.22, 12.36EXCEL