1 power 14 goodness of fit & contingency tables. 2 ii. goodness of fit & chi square u...

11

Power 14Goodness of Fit

& Contingency Tables

22

II. Goodness of Fit & Chi Square

Rolling a Fair DieRolling a Fair Die The Multinomial DistributionThe Multinomial Distribution Experiment: 600 TossesExperiment: 600 Tosses

33

Outcome Probability Expected Frequency1 1/6 1002 1/6 1003 1/6 1004 1/6 1005 1/6 1006 1/6 100

The Expected Frequencies The Expected Frequencies

44

Outcome Expected Frequencies Expected Frequency1 100 1142 100 943 100 844 100 1015 100 1076 100 107

The Expected Frequencies & Empirical FrequenciesThe Expected Frequencies & Empirical Frequencies

Empirical FrequencyEmpirical Frequency

55

Hypothesis Test Null HNull H00: Distribution is Multinomial: Distribution is Multinomial

Statistic: (OStatistic: (Oii - E - Eii))22/E/Ei, i, : observed minus : observed minus

expected squared divided by expectedexpected squared divided by expected Set Type I Error @ 5% for exampleSet Type I Error @ 5% for example Distribution of Statistic is Chi SquareDistribution of Statistic is Chi Square

P(nP(n1 1 =1, n=1, n2 2 =0, nn3 3 =0, n =0, n4 4 =0, n=0, n5 5 =0, n=0, n6 6 =0) = n!/=0) = n!/

n

j

jnn

j

jpjn1

)(

1

)]([])(

P(nP(n1 1 =1, n=1, n2 2 =0, nn3 3 =0, n =0, n4 4 =0, n=0, n5 5 =0, n=0, n6 6 =0)= 1!/1!0!0!0!0!0!(1/6)=0)= 1!/1!0!0!0!0!0!(1/6)11(1/6)(1/6)00

(1/6)(1/6)0 0 (1/6)(1/6)0 0 (1/6)(1/6)0 0 (1/6)(1/6)00

One Throw, side one comes up: multinomial distributionOne Throw, side one comes up: multinomial distribution

66

Face Observed, Oj Expected, Ej Oj - Ej (Oj – Ej)2 /Ej

1 114 100 14 196/100 = 1.96

2 92 100 - 8 64/100 = 0.64

3 84 100 - 16 256/100 = 2.56

4 101 100 1 1/100 = 0.01

5 107 100 7 49/100 = 0.49

6 107 100 7 49/100 = 0.49

Sum = 6.15

Chi Square: xChi Square: x22 = = (O (Oii - E - Eii))2 2 = 6.15 = 6.15

0.00

0.05

0.10

0.15

0.20

0 5 10 15

CHI

DE

NS

ITY

Chi Square Density for 5 degrees of freedomChi Square Density for 5 degrees of freedom

11.0711.07

5 %5 %

88

Contingency Table Analysis

Tests for Association Vs. Independence For Tests for Association Vs. Independence For Qualitative VariablesQualitative Variables

99

Purchase Consumer Inform Cons. Not Inform . TotalsFrost FreeNot Frost FreeTotals

Does Consumer Knowledge Affect Purchases?Does Consumer Knowledge Affect Purchases?

Frost Free Refrigerators Use More ElectricityFrost Free Refrigerators Use More Electricity

1010

Purchase Consumer Inform Cons. Not Inform . TotalsFrost Free 432Not Frost Free 288Totals 540 180 720

Marginal CountsMarginal Counts

1111

Purchase Consumer Inform Cons. Not Inform . TotalsFrost Free 0.6Not Frost Free 0.4Totals 0.75 0.25 1

Marginal Distributions, f(x) & f(y)Marginal Distributions, f(x) & f(y)

1212

Purchase Consumer Inform Cons. Not Inform . TotalsFrost Free 0.45 0.15 0.6Not Frost Free 0.3 0.1 0.4Totals 0.75 0.25 1

Joint Disribution Under IndependenceJoint Disribution Under Independencef(x,y) = f(x)*f(y)f(x,y) = f(x)*f(y)

1313

Purchase Consumer Inform Cons. Not Inform . TotalsFrost Free 324 108 432Not Frost Free 216 72 288Totals 540 180 720

Expected Cell Frequencies Under IndependenceExpected Cell Frequencies Under Independence

1414

Purchase Consumer Inform Cons. Not Inform . TotalsFrost Free 314 118Not Frost Free 226 62Totals

Observed Cell CountsObserved Cell Counts

1515

Purchase Consumer Inform Cons. Not Inform . TotalsFrost Free 0.31 0.93Not Frost Free 0.46 1.39Totals

Contribution to Chi Square: (observed-Expected)Contribution to Chi Square: (observed-Expected)22/Expected/Expected

Chi Sqare = 0.31 + 0.93 + 0.46 +1.39 = 3.09Chi Sqare = 0.31 + 0.93 + 0.46 +1.39 = 3.09(m-1)*(n-1) = 1*1=1 degrees of freedom (m-1)*(n-1) = 1*1=1 degrees of freedom

Upper Left Cell: (314-324)Upper Left Cell: (314-324)22/324 = 100/324 =0.31/324 = 100/324 =0.31

0.0

0.2

0.4

0.6

0.8

1.0

0 2 4 6 8 10 12 14

Chi-Square Variable

Figure 4: Chi-Square Density, One Degree of Freedom

Density

5%5%

5.025.02

1717

Conclusion

No association between consumer No association between consumer knowledge about electricity use and knowledge about electricity use and consumer choice of a frost-free refrigeratorconsumer choice of a frost-free refrigerator

1818

Using Goodness of Fit to Choose Between Competing

Probability Models Men on base when a home run is hitMen on base when a home run is hit

1919

Men on base when a home run is hit

# 0 1 2 3 Sum

Observed 421 227 96 21 765

Fraction 0.550 0.298 0.125 0.027 1

2020

Conjecture

Distribution is binomialDistribution is binomial

2121

Average # of men on base# 0 1 2 3

fraction 0550 0.298 0.125 0.027

product 0 0.298 0.250 0.081

Sum of products = n*p = 0.298+0.250+0.081 = 0.63Sum of products = n*p = 0.298+0.250+0.081 = 0.63

21.03/63.0/ˆˆ npnp

2222

Using the binomialk=men on base, n=# of trials

P(k=0) = [3!/0!3!] (0.21)P(k=0) = [3!/0!3!] (0.21)00(0.79)(0.79)33 = 0.493 = 0.493 P(k=1) = [3!/1!2!] (0.21)P(k=1) = [3!/1!2!] (0.21)11(0.79)(0.79)22 = 0.393 = 0.393 P(k=2) = [3!/2!1!] (0.21)P(k=2) = [3!/2!1!] (0.21)22(0.79)(0.79)11 = 0.105 = 0.105 P(k=3) = [3!/3!0!] (0.21)P(k=3) = [3!/3!0!] (0.21)33(0.79)(0.79)00 = 0.009 = 0.009

2323

Assuming the binomial

The probability of zero men on base is The probability of zero men on base is 0.4930.493

the total number of observations is 765the total number of observations is 765 so the expected number of observations for so the expected number of observations for

zero men on base is 0.493*765=377.1zero men on base is 0.493*765=377.1

2424

Goodness of Fit# 0 1 2 3 Sum

Observed 421 227 96 21 765

binomial 377.1 300.6 80.3 6.9 764.4

(Oj – Ej) 43.9 -73.6 15.7 14.1

(Oj–Ej)2/Ej 5.1 18.0 2.6 28.8 54.5

0.00

0.05

0.10

0.15

0.20

0.25

0 5 10 15 20

CHI

DE

NS

ITY

Chi Square, 3 degrees of freedomChi Square, 3 degrees of freedom

5%5%

7.817.81

2626

Conjecture: Poisson where np = 0.63

P(k=3) = 1- P(k=2)-P(k=1)-P(k=0)P(k=3) = 1- P(k=2)-P(k=1)-P(k=0) P(k=0) = eP(k=0) = e--k k /k! = e/k! = e-0.63 -0.63 (0.63)(0.63)00/0! = 0.5326/0! = 0.5326 P(k=1) = eP(k=1) = e--k k /k! = e/k! = e-0.63 -0.63 (0.63)(0.63)11/1! = 0.3355/1! = 0.3355 P(k=2) = eP(k=2) = e--k k /k! = e/k! = e-0.63 -0.63 (0.63)(0.63)22/2! = 0.1057/2! = 0.1057

2727

Average # of men on base# 0 1 2 3

fraction 0550 0.298 0.125 0.027

product 0 0.298 0.250 0.081

Sum of products = n*p = 0.298+0.250+0.081 = 0.63Sum of products = n*p = 0.298+0.250+0.081 = 0.63

21.03/63.0/ˆˆ npnp

2828

Conjecture: Poisson where np = 0.63

P(k=3) = 1- P(k=2)-P(k=1)-P(k=0)P(k=3) = 1- P(k=2)-P(k=1)-P(k=0) P(k=0) = eP(k=0) = e--k k /k! = e/k! = e-0.63 -0.63 (0.63)(0.63)00/0! = 0.5326/0! = 0.5326 P(k=1) = eP(k=1) = e--k k /k! = e/k! = e-0.63 -0.63 (0.63)(0.63)11/1! = 0.3355/1! = 0.3355 P(k=2) = eP(k=2) = e--k k /k! = e/k! = e-0.63 -0.63 (0.63)(0.63)22/2! = 0.1057/2! = 0.1057

2929

Goodness of Fit# 0 1 2 3 Sum

Observed 421 227 96 21 765

Poisson 407.4 256.7 80.9 20.0 765

(Oj–Ej)2/Ej 0.454 3.44 2.82 0.05 6.76

0.00

0.05

0.10

0.15

0.20

0.25

0 5 10 15 20

CHI

DE

NS

ITY

Chi Square, 3 degrees of freedomChi Square, 3 degrees of freedom

5%5%

7.817.81

3131

Likelihood Functions

Review OLS LikelihoodReview OLS Likelihood Proceed in a similar fashion for the probitProceed in a similar fashion for the probit

3232

Likelihood function The joint density of the estimated residuals The joint density of the estimated residuals

can be written as:can be written as:

If the sample of observations on the If the sample of observations on the dependent variable, y, and the independent dependent variable, y, and the independent variable, x, is random, then the observations variable, x, is random, then the observations are independent of one another. If the errors are independent of one another. If the errors are also identically distributed, f, i.e. i.i.d, are also identically distributed, f, i.e. i.i.d, thenthen

)ˆ.....ˆˆˆ( 1210 neeeeg

3333

Likelihood function Continued: If i.i.d., thenContinued: If i.i.d., then

If the residuals are normally distributed:If the residuals are normally distributed:

This is one of the assumptions of linear This is one of the assumptions of linear regression: errors are i.i.d normalregression: errors are i.i.d normal

then the joint distribution or likelihood then the joint distribution or likelihood function, L, can be written as:function, L, can be written as:

)ˆ()...ˆ(*)ˆ()ˆ...ˆˆ( 110110 nn efefefeeeg

2]/)0ˆ[(2/12 )2/1(),0(~)ˆ( iei eNef

3434

Likelihood function

and taking natural logarithms of both sides, where and taking natural logarithms of both sides, where the logarithm is a monotonically increasing the logarithm is a monotonically increasing function so that if lnL is maximized, so is L:function so that if lnL is maximized, so is L:

1

0

22

2

]ˆ[)2/1(2/2/2

]/)0ˆ[(2/11

0110

*)2/1(*)/1(

)2/1()ˆ...ˆˆ(

n

ii

i

enn

en

in

eL

eeeegL

3535

Log-Likelihood

Taking the derivative of lnL with respect to Taking the derivative of lnL with respect to either a-hat or b-hat yields the same either a-hat or b-hat yields the same estimators for the parameters a and b as with estimators for the parameters a and b as with ordinary least squares, except now we know ordinary least squares, except now we know the errors are normally distributed.the errors are normally distributed.

21

0

22

1

0

222

]*ˆˆ[)2/1()2ln(*)2/(]ln[*)2/(ln

ˆ)2/1()2ln(*)2/(]ln[*)2/(ln

i

n

ii

n

ii

xbaynnL

ennL

3636

Probit Example: expenditures on lottery as a % of household Example: expenditures on lottery as a % of household

incomeincome lotterylotteryii = a + b*income = a + b*incomei i + e + eii

if lotteryif lotteryi i >0, i.e. a + b*income>0, i.e. a + b*incomei i + e + ei i >0, then Bern >0, then Bernii , ,

the yes-no indicator variable is equal to one and ethe yes-no indicator variable is equal to one and e i i >- a >- a

- b*income- b*incomeii

this determines a threshold for observation i in the this determines a threshold for observation i in the distribution of the error edistribution of the error eii

assume assume

),0(~ 2Nei

Density Function for the Standardized Normal Variate

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-5 -4 -3 -2 -1 0 1 2 3 4 5

Standard Deviations

Den

sity

2]1/)0[(2/1*]2/1[)( zezf

ii

/)0(/)0*(: ii eincomebathreshold


0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-5 -4 -3 -2 -1 0 1 2 3 4 5

Standard Deviations

Den

sity

2]1/)0[(2/1*]2/1[)( zezf

ii

/)0*(: iincomebathreshold

Area above the Area above the thresholdthresholdis the probability of is the probability of playing the lottery forplaying the lottery forobservation i, Pobservation i, Pyesyes


0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-5 -4 -3 -2 -1 0 1 2 3 4 5

Standard Deviations

Den

sity

2]1/)0[(2/1*]2/1[)( zezf

ii



PPno no for for

observation iobservation i

4040

Probit

Likelihood function for the observed Likelihood function for the observed samplesample

Log likelihood:Log likelihood:

n

i

Bernyes

Bernnonoyes

Bern Bernyesnonoyes

ii iPiPnnnLIK

PPnnnLIK

1

)1(

0 1

)(*)(*)!!/(!

*)!!/(!

n

iiyesiinoinoyes PBernPBernnnnLIK

1

lnln)1()]!!/(!ln[ln

4141

incomeba

inoP*

2

2

)/]0)([2/1(

*

)/]0)([2/1(*

*]2/1[

*2/1

i

i

ii

e

incomebaiyes

eincomeba

ino

eP

eP


0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-5 -4 -3 -2 -1 0 1 2 3 4 5

Standard Deviations

Den

sity

2]1/)0[(2/1*]2/1[)( zezf

ii



PPno no for for

observation iobservation i

4343

Probit

Substituting these expressions for PSubstituting these expressions for Pno no and and

PPyes yes in the ln Likelihood function gives the in the ln Likelihood function gives the

complete expression.complete expression.

4444

Probit

Likelihood function for the observed Likelihood function for the observed samplesample

Log likelihood:Log likelihood:

n

i

Bernyes

Bernnonoyes

Bern Bernyesnonoyes

ii iPiPnnnLIK

PPnnnLIK

1

)1(

0 1

)(*)(*)!!/(!

*)!!/(!

n

iiyesiinoinoyes PBernPBernnnnLIK

1

lnln)1()]!!/(!ln[ln

4646

Outline

I. ProjectsI. Projects II. Goodness of Fit & Chi SquareII. Goodness of Fit & Chi Square III.Contingency TablesIII.Contingency Tables

4747

Part I: Projects

TeamsTeams AssignmentsAssignments PresentationsPresentations Data SourcesData Sources GradesGrades

4848

Team One

: Project choice: Project choice : Data Retrieval: Data Retrieval : Statistical Analysis: Statistical Analysis : PowerPoint Presentation: PowerPoint Presentation : Executive Summary: Executive Summary : Technical Appendix: Technical Appendix : Graphics (Excel, Eviews, other): Graphics (Excel, Eviews, other)

4949

Assignments

1. Project choice: Markus Ansmann1. Project choice: Markus Ansmann 2. Data Retrieval: Theodore Ehlert2. Data Retrieval: Theodore Ehlert 3. Statistical Analysis: David Sheehan3. Statistical Analysis: David Sheehan 4. PowerPoint Presentation: Qun Luo4. PowerPoint Presentation: Qun Luo 5. Executive Summary: Steven Comstock5. Executive Summary: Steven Comstock 6. Technical Appendix: Alan Weinberg6. Technical Appendix: Alan Weinberg 7. Graphics: Gregory Adams7. Graphics: Gregory Adams

5050

PowerPoint Presentations: Member 4 1. Introduction: Members 1 ,2 , 31. Introduction: Members 1 ,2 , 3

– WhatWhat– WhyWhy– HowHow

2. Executive Summary: Member 52. Executive Summary: Member 5 3. Exploratory Data Analysis: Members 3, 73. Exploratory Data Analysis: Members 3, 7 4. Descriptive Statistics: Member 3, 74. Descriptive Statistics: Member 3, 7 5. Statistical Analysis: Member 35. Statistical Analysis: Member 3 6. Conclusions: Members 3 & 56. Conclusions: Members 3 & 5 7. Technical Appendix: Table of Contents, 7. Technical Appendix: Table of Contents,

Member 6Member 6

5151

Executive Summary and Technical Appendix

5252

I. Your report should have an executive summary of one to one

and a half pages that summarizes your findings in words for a non-

technical reader. It should explain the problem being examined

from an economic perspective, i.e. it should motivate interest in the

issue on the part of the reader. Your report should explain how you

are investigating the issue, in simple language. It should explain

why you are approaching the problem in this particular fashion.

Your executive report should explain the economic importance of

your findings.

The technical details of your findings you can attach as an

appendix.

5353

GradesComponent A B C Introduction Exec. Summy Explor. Descriptive Stat. Anal. Conclusions Tech. Appen. Graphics Overall Proj.

5454

Data Sources FRED: Federal Reserve Bank of St. Louis, FRED: Federal Reserve Bank of St. Louis, http://http://

research.stlouisfed.org/fredresearch.stlouisfed.org/fred//– Business/FiscalBusiness/Fiscal

Index of Consumer Sentiment, Monthly (1952:11)Index of Consumer Sentiment, Monthly (1952:11) Light Weight Vehicle Sales, Auto and Light Truck, Monthly Light Weight Vehicle Sales, Auto and Light Truck, Monthly

(1976.01)(1976.01)

Economagic, Economagic, http://http://www.economagic.comwww.economagic.com// U S Dept. of Commerce, U S Dept. of Commerce, http://http://

www.commerce.govwww.commerce.gov//– PopulationPopulation– Economic Analysis, Economic Analysis, http://http://www.bea.govwww.bea.gov//

5555

Data Sources (Cont. ) Bureau of Labor Statistics, Bureau of Labor Statistics, http://http://

stats.bls.govstats.bls.gov// California Dept of Finance, California Dept of Finance, http://http://

www.dof.ca.govwww.dof.ca.gov//

1 power 14 goodness of fit & contingency tables. 2 ii. goodness of fit & chi square u...

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