Stats Lec9

7/28/2019 Stats Lec9

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2 andFDistributions

Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition Pearson Education Limited 2006


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2 Distribution

The

2

distributionis

used

to:

constructconfidenceintervalsforavariance

frequencies

table

freedom



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FDistribution

TheFdistribution

is

used

to

testthehypothesisofequalityoftwovariances

con uct

an

ana ysis

o

variance ANOVA ,

comparing

means

acrossseveralsamples

sasymme r can epen son e egreeso ree om



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Tailsof

the

219 Distribution

2.5% 2.5%

0 10 20 30 40 50

8.91 32.85



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CriticalValues

of

the

Chi

squared

Distribution

NB

chi

squared

is

not

symmetric

so

table

will

give

different

values

for

ExcerptfromTableA4:

0.990 0.975 0.050 0.025 0.0101 0.000 0.001 3.841 5.024 6.635

2 0.020 0.051 5.991 7.378 9.210

3 0.115 0.216 7.815 9.348 11.345

: : : : : :

. . . . .

19 7.633 8.907 30.144 32.852 36.191

20 8.260 9.591 31.410 34.170 37.566



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Case1:

Estimating

aVariance

Arandom

sample

of

size

n =

20

yields

astandard

deviation

of

s

= . ow owees ma e epopu a onvar ance

Pointestimate:uses2 =252 =625whichisunbiased(E(s2)=2 )

Intervalestimate:weneedthesamplingdistributionofs2...



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TheSampling

Distribution

of

s2

212

2

~

1

n

sn

n-1 gives the degrees of freedom for the 2 distribution,.



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Limitsto

the

Confidence

Interval

Fort e95%CI,wenee t e va uescuttingo 2.5%ineac

tail

of

the

distribution

ExcerptfromTableA4:

0.990 0.975 0.050 0.025 0.010

1 0.000 0.001 3.841 5.024 6.6352 0.020 0.051 5.991 7.378 9.210

. . . . .

: : : : : :

187.015 8.231 28.869 31.526 34.805. . . . .

20 8.260 9.591 31.410 34.170 37.566



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Tailsof

the

219 Distribution(cont.)

Wecan

be

95%

confident

that

(n

1)s2/2 lies

between

8.91

and

32.85

orn = 85.32

191.8

2

2

sn

11 222 snsn

Substitutings2 =625andn =20:

..

8.332,15.3612



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Case2:

Comparing

Actual

vs

72ro so a iceyie :

Score on dice 1 2 3 4 5 6

Frequency 6 15 15 7 15 14

times.



11/35

TestStatistic

0:t e ce s a rH

1:thediceisbiased

Thiscanbetestedusing

EO2

2

E

,

inthiscasebecausewehave6outcomes.



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Calculatingthe

Test

Statistic

Score Observed Expected O E (O E)2 (O E)2

frequency (O) frequency (E) E

1 6 12 6 36 3.00

2 15 12 3 9 0.75

3 15 12 3 9 0.75

.

5 15 12 3 9 0.75

.

Totals 72 72 0 7.66



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Calculatingthe

Test

Statistic

(cont.)

Thetest

statistic,

7.66,

is

less

than

the

critical

value

of

2 with

=5,11.1

observedand

expected

outcomes

is

random

.

thedistribution.OnlylargevaluesoftheteststatisticrejectH0



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Case3:

Contingency

Tables

Theassociationbetweentwovariablescanalsobeanalysedvia

the

2 distribution

Votingbehaviourbasedonasampleof200:

S o c i a l c l a s s L a b o u r C o n s e r v a t i v e L i b e r al

D e m o c r a t

T o t a l

B 40 35 25 100

C 30 20 10 60

Total 80 70 50 200



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AreSocial

Class

and

Voting

H0:noassociationbetweensocialclassandvotingbehaviourH1:someassociation

Expectedvalues are

calculated,

based

on

the

null

of

no

association

E.g.ifthereisnoassociation:

40%

(80/200)

of

every

social

class

should

vote

Labour,

i.e.

16

from

classA,40fromBand24fromC



16/35

Observed(and

Expected)

Values

Social class Labour Conservative Liberal Democrat Total

A 10(16) 15(14) 15(10) 40

C 30(24) 20(21) 10(15) 60

Total 80 70 50 200



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Calculatingthe

Test

Statistic

101514151610 222

252535354040 222

04.8

151021202430222

For

=

(rows

1)

(columns

1)

=

4,

the

critical

value

of

the

2

distributionis9.50,sothenullofnoassociationisnotrejected

atthe5%significancelevel.



18/35

Distribution

Dotwo

sam les

have

e ual

variances i.e.

come

from

populationswiththesamevariance)?

n1 =30 s1 =25

n =30 s =20



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TestingTwo

Variances

the

F

.

H0:1 =2

H1:12 =2

2

or,equivalently

H0:12/2

2 =1

2 21 1 2



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TheTest

Statistic

T eteststatisticis

s12

~

Evaluatingthis:s

n n

2

2 1 2,

5625.120

2 F

F*29,29 =2.09>1.5625,sothenullisnotrejected. Thevariancesmaybeconsideredequal



21/35

xcerpt

rom

a e

:

1 1 2 24 30 40

21 647.79 799.48 997.27 1001.40 1005.60

. . . . .

: : : : : :28 5.61 4.22 2.17 2.11 2.05

29 5.59 4.20 2.15 2.09 2.03

30 5.57 4.18 2.14 2.07 2.01

(Using 1 = 30 (rather than 29) makes little practicaldifference.)



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Oneor

Two

Tailed

Test?

Aslongasthelargervarianceismadethenumeratorofthetest ,

.

ThesmallestpossiblevalueofFis1,whichoccursifthesamplevar ancesareequa . 0 s ou no ere ec e n scase.

So,despitetheinH1,thisisaonetailedtest.



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Case5:

Analysis

of

Variance

(ANOVA)

A

test

for

the

equality

of

several

means,

not

just

two

as e ore

Inourexamplewetestfortheequalityofoutputof

threefactories,

i.e.

are

they

equally

productive,

on

average,ornot?



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Example:Daily

Output

of

Three

Factories

Observation Factory 1 Factory 2 Factory 3

1 415 385 408

2 430 410 415

4 399 403 440

5 408 405 425

6 418 400

7 399



25/35

Chartof

Output

Factory 1

Factory 2Factory 3

380 390 400 410 420 430 440 450



26/35

HypothesisTest

H0:1 =2 =3

H :

Principleofthetest:breakdownthetotalvarianceofall

o serva ons n o ew n ac ory var ancean e e ween

factory variance

Ifthebetweenvariancecomponentislargerelativetothe

withinvariance

component,

reject

H0



27/35

Sumsof

Squares

at ert anvar ances,wor w t sumso squares

2xx

Sum of squares

1 ns

Variance



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ThreeSums

of

Squares

Total sum

of

squares

(TSS) Sumofsquaresofalldeviationsfromtheoverallaverage

Between sumofsquares(BSS)

Sum

of

squares

of

deviations

of

factory

means

from

overall

average

Within sumofsquares(WSS)

Sum

of

squares

of

deviations

within

each

factory,

from

factoryaverage



29/35

TestStatistic

kBSSF

1

TheFstatisticistheratioofBSStoWSS eachad ustedb their

degreesof

freedom

(k

1and

nk)

factoriesdeviationslarge rejectH0



30/35

Calculations

TSS=

2

i xx

(jindexesfactories,iindexesobservations)

j i

=(415

410.11)

2

+(430

410.11)

2

+

+(440

410.11)

2

+(425 410.11)2 =2,977.778

410.11istheoverall,orgrand,average



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Calculations(cont.)

BSS= j i

i xx

where istheaverageoutputoffactoryi

=6 (410.83 410.11)2 +7 (401.57 410.11)2 +5 (421.2ix

410.11 =1,128.43

410.83,401.57,421.11arethethreeaverages,respectively.



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Calculations(cont.)

WSS=TSS BSS=2,977.778 1,128.430= , .

2

erna ve y, = j i

iij

- . - . -401.57)2 + + (399-401.57)2 + (408-421.2)2 +

- 2 .= 1,849.348



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Resultof

the

Test

1343.11281 kBSS

.318348.1849

knWSS

F* 2,15 =3.682

(5%

significance

level)

F>F*hencewerejectH0.Therearesignificantdifferencesbetweenthefactories.

EasiesttodoinEXCEL.....



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ANOVATable

(Excel

Format)

SUMMARY

Groups Count Sum Average Variance

Factory 1 6 2465 410.833 166.967

Factory 2 7 2811 401.571 70.6191

. .

Source ofVariation SS df MS F P-value F crit

Between Groups 1128.430 2 564.215 4.576 0.028 3.68

Within Groups 1849.348 15 123.290


Total 2977.778 17


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Summary

Usethe2 distributionto

a cu a e e oravar ance

Compareactualandexpectedvalues

Analyse

a

contingency

table

Testfortheequalityoftwovariances

es

or

eequa y

o

severa

means


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