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Lecture Twelve

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Outline

• Projects

• Failure Time Analysis

• Linear Probability Model

• Poisson Approximation

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Part I: Projects

• Teams

• Assignments

• Presentations

• Data Sources

• Grades

4

Team One

• Hongtao Xu: Project choice

• Sasha Hochstadt: Data Retrieval

• Logan McCleod: Statistical Analysis

• Heather Samoville: PowerPoint Presentation

• Christian Helland: Executive Summary

• Meng Yu: Technical Appendix

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Assignments

• 1. Project choice

• 2. Data Retrieval

• 3. Statistical Analysis

• 4. PowerPoint Presentation

• 5. Executive Summary

• 6. Technical Appendix

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

– What– Why– How

• 2. Executive Summary: Member 5

• 3. Exploratory Data Analysis: Member 3

• 4. Descriptive Statistics: Member 3

• 5. Statistical Analysis: Member 3

• 6. Conclusions: Members 3 & 5

• 7. Technical Appendix: Table of Contents, Member 6

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Executive Summary and Technical Appendix

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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.

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Grades

Component A B CIntroductionExec. SummyExplor.DescriptiveStat. Anal.ConclusionsTech. Appen.Overall Proj.

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

http://research.stlouisfed.org/fred/– Business/Fiscal

• Index of Consumer Sentiment, Monthly (1952:11)

• Light Weight Vehicle Sales, Auto and Light Truck, Monthly (1976.01)

• Economagic, http://www.economagic.com/

• U S Dept. of Commerce, http://www.commerce.gov/– Population– Economic Analysis, http://www.bea.gov/

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Data Sources (Cont. )

• Bureau of Labor Statistics, http://stats.bls.gov/

• California Dept of Finance, http://www.dof.ca.gov/

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Part II: Failure Time Analysis

• Exponential– survival function– hazard rate

• Weibull

• Exploratory Data Analysis, Lab Seven

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Duration of Post-War Economic Expansions in Months

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Trough Peak DurationOct. 1945 Nov. 1948 37Oct. 1949 July 1953 45May 1954 August 1957 39April 1958 April 1960 24Feb. 1961 Dec. 1969 106Nov. 1970 Nov. 1973 36March 1975 January 1980 58July 1980 July 1981 12Nov. 1982 July 1990 92March 1991 March 2000 120

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Estimated Survivor Function for Ten Post-War Expansions

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Duration # Ending # At Risk F(t) Survivor0 0 10 0 112 1 10 0.1 0.924 1 9 0.2 0.836 1 8 0.3 0.737 1 7 0.4 0.639 1 6 0.5 0.545 1 5 0.6 0.458 1 4 0.7 0.392 1 3 0.8 0.2106 1 2 0.9 0.1120 1 1 1 0

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Figure 2: Estimated Survivor Function for Post-War Expansions

0

0.2

0.4

0.6

0.8

1

1.2

0 20 40 60 80 100 120 140

Duration in Months

Su

rviv

or

Fu

nct

ion

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Figure 3: Exponential Trendline Fitted to Estimated Survivor Function

y = 1.1972e-0.0217x

R2 = 0.9533

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 20 40 60 80 100 120

Duration in Months

Su

rviv

or

Fu

nct

ion

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Figure 4: Constrained Expontial trendline, Fitted to Estimated Survivor Function

y = e-0.019x

R2 = 0.9313

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 20 40 60 80 100 120

Duration in Months

Su

rviv

or

Fu

nc

tio

n

Exponential Distribution

• Hazard rate: ratio of density function to the survivor function:

• h(t) = f(t)/S(t)

• measure of probability of failure at time t given that you have survived that long

• for the exponential it is a constant:

• h(t) = )exp(/)exp( tt

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Duration # Ending # At Risk Inter. Haz.0 0 10 012 1 10 0.100024 1 9 0.111136 1 8 0.125037 1 7 0.142939 1 6 0.166745 1 5 0.200058 1 4 0.250092 1 3 0.3333106 1 2 0.5000120 1 1 1.0000

Cumulative Hazard Function

• In general:

• For the exponential,

t

duuhtH0

)()(

t

tdutH0

)(

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Duration # Ending # At Risk Inter. Haz.Cum. Hazard0 0 10 0 012 1 10 0.1000 0.100024 1 9 0.1111 0.211136 1 8 0.1250 0.336137 1 7 0.1429 0.479039 1 6 0.1667 0.645645 1 5 0.2000 0.845658 1 4 0.2500 1.095692 1 3 0.3333 1.4290106 1 2 0.5000 1.929120 1 1 1.0000 2.929

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Cumulative Hazard Function: Postwar Expansions

y = 0.0223x - 0.2422R2 = 0.9288

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

0 20 40 60 80 100 120 140

Duration in Months

Cu

mu

lati

ve H

azar

d

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Cumulative Hazard Function, Postwar Expansions

y = 0.0192xR2 = 0.9015

0

0.5

1

1.5

2

2.5

3

3.5

0 20 40 60 80 100 120 140

Duration in Months

Cu

mu

lati

ve H

azrd

Weibull Distribution• F(t) = 1 - exp[

• S(t) =

• ln S(t) = - (t/

• h(t) = f(t)/S(t)

• f(t) = dF(t)/dt = - exp[-(t/t/

• h(t) = (t/

• if h(t) = constant

• if h(t) is increasing function

• if h(t) is a decreasing function

])/( t

])/(exp[ t

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Source:Wayne Nelson, Applied Life data Analysis(1982) John WileyDiesel Generators, hours to fan failure, (+ indicates running time, i.e. still running whenlast observed)

Hours # Ending # At Risk Interval Interval Hazard Rate Cumulative Hazard Rate450

460+11501150

1560+1600

1660+1850+1850+1850+1850+1850+2030+2030+2030+

207020702080

2200+

Lab Seven

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Source:Wayne Nelson, Applied Life data Analysis(1982) John WileyDiesel Generators, hours to fan failure, (+ indicates running time, i.e. still running when last observed)

Hours # Ending # At Risk Interval Interval Hazard Rate Cumulative Hazard Rate450 1 70 450

460+ 681150 2 68 7001150

1560+ 651600 1 65 450

1660+ 631850+ 621850+ 611850+ 601850+ 591850+ 582030+ 572030+ 562030+ 55

2070 2 55 47020702080 1 53 10

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Source:Wayne Nelson, Applied Life data Analysis(1982) John WileyDiesel Generators, hours to fan failure, (+ indicates running time, i.e. still running whenlast observed)

Hours # Ending # At Risk Interval Interval Hazard Rate Cumulative Hazard Rate450 1 70 450 0.0143 0.0143

460+ 681150 2 68 700 0.0294 0.04371150

1560+ 651600 1 65 450 0.0154 0.0591

1660+ 631850+ 621850+ 611850+ 601850+ 591850+ 582030+ 572030+ 562030+ 55

2070 2 55 470 0.0364 0.095520702080 1 53 10 0.0189 0.1143

2200+ 51

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Cumulative Hazard Rate for Fan Failure

y = 4E-05x + 0.0089

R2 = 0.9816

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

Duration in Hours

Cu

mu

lati

ve H

azar

d

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Part III: Linear Probability Model

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LOTTERY AGE CHILDREN EDUCATION INCOME 5.000000 50.00000 2.000000 15.00000 41.00000 7.000000 26.00000 0.000000 10.00000 22.00000 0.000000 40.00000 3.000000 13.00000 24.00000 10.00000 46.00000 2.000000 9.000000 20.00000 5.000000 40.00000 3.000000 14.00000 32.00000 5.000000 39.00000 2.000000 15.00000 42.00000 3.000000 36.00000 3.000000 8.000000 18.00000 0.000000 44.00000 1.000000 16.00000 47.00000 0.000000 47.00000 4.000000 20.00000 85.00000 6.000000 52.00000 1.000000 10.00000 23.00000 0.000000 51.00000 2.000000 18.00000 61.00000 0.000000 41.00000 2.000000 17.00000 70.00000 12.00000 42.00000 2.000000 9.000000 22.00000 7.000000 53.00000 1.000000 12.00000 27.00000 11.00000 72.00000 1.000000 9.000000 25.00000

0

5

10

15

0 20 40 60 80 100

INCOME

LOT

TE

RY

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Bernoulli Variable: Bern

• Bern = 0*(lottery=1) + 1*(lottery>0)

• Linear Probability Model: dummy dependent variable

• Bern(i) = c + a*income + b*age +d*children + f*education + e(i)

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BERN LOTTERY1 51 70 01 101 51 51 30 00 01 60 00 01 121 71 11

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Bern age children education income

0 40.49 1.78 15.57 47.565

1 44.19 1.78 11.94 28.545

Averages for Players and Non-Players

Part IV. Poisson Approximation to Binomial

• Conditions:

• f(x) = {exp[-] x }/x!

• Assumptions:– the number of events occurring in non-

overlapping intervals are independent– the probability of a single event occurring in a

small interval is approximately proportional to the interval

– the probability of more than one event in an interval is negligible

50,1)1(,0 npp

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Example

• Ten % of tools produced in a manufacturing process are defective. What is the probability of finding exactly two defectives in a random sample of 10?

• Binomial: p(k=2) = 10!/(8!2!)(0.1)2(0.9)8 = 0.194

• Poisson , where the mean of the Poisson, equals n*p = 0.1 p(k=2) = {exp[-1] 12 }/2! = 0.184