Extensive study I assume the mean age of this class is not 23.5!

Extensive study

I assume the mean age of this class is not 23.5!

Ho: u=80, but the value of u is really equal to 82 score

80 82

Decision-making always exists risk

Prob. of sample ?

X

2.5%2.5%

Prob.=Significance level

Judgment in a relaxed manner

ZZ

a/2

z

X0 :H

ZZ

p/2

a/2

z

n

Xz

X1

X0

:H

:H

X1

X0

:H

:H

n

Xz

X1

X0

:H

:H

X1

X0

:H

:H

Critical limit Critical limit

368:H X0

When n≥30, t distribution is almost equal to standard normal distribution. So we also use z-test.

)30(n Z

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

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

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

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

FF ((nn11-1-1, , nn22-1-1))

00 FF1

The distribution of statistic F

critF

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.

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

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(

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

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

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

A priori comparisons

• Definition of statistic t

0H reject then t

,tIf

knt

n1

n1

SW

XXt

:H:H

ji

2

ji

ji1ji0

∽

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

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

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

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.

Intensity of exercise

light heavy

No significant time of day or intensity of exercise effects

sleep

morning

evening


light heavy

sleep

morning

evening

Significant time of day effect; no other effect


light heavy

sleep

morning

evening

Significant time of day effect; no other effect

light heavy


sleep

morning

evening

Significant intensity of exercise effect; no other effect

light heavy

light heavy


sleep

morning

evening

Significant intensity of exercise and time of day effects; no interaction effect

light heavy

light heavy

Significant interaction effect; no other effect


sleep

morning

evening

light heavy

light heavy


sleep

morning

evening

Significant intensity of exercise and time of day effects; no interaction effect

light heavy


light heavy

B∑

b1 b2 bc

A

a1 cell11 cell12 … cell1c ∑row1X

a2 cell21 cell22 … cell2c ∑row2X

… … … … …

ar cellr1 cellr2 … cellrc ∑rowrX

∑ ∑col1X ∑col2X ∑colcX

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


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


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


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

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

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

577.532

9.533.470.462.442.438.4118.2143

222222

22

12

2

11

XSSW 2ij rc

XXXcellrccellcell

000.936

4.276

18

2.1472.129 222

2

2

2

2

1

SSRN

X

n

XX

row

rowrow

754.436

4.276 2

2

.

2

3.

2

2.

2

1.

1298.190.587.8

SSC

222

N

X

n

XXX

col

colcolcol

1.712

4.754-9.000- 36

276.4-

653.947.346.044.243.241.8

SSRC

2222222

SSCSSR

N

X

rc

XXXcellcellcell

2

2

23

2

12

2

11

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

2

RC

2

C

2

R

2

RC

C

R

w

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

Extensive study I assume the mean age of this class is not 23.5!

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Transcript of Extensive study I assume the mean age of this class is not 23.5!