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* CHAPTER 4 - STATISTICS FOR BUSINESS AND ECONOMICS, 5th Edition            *
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* Example 4.1 , p. 81
*
* Read this example carefully.  Be sure you understand the material.
*
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* Example 4.2 , p. 81
*
* Read this example carefully.  Be sure you understand the material.
*
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* Example 4.3 , p. 84
*
* Read this example carefully.  Be sure you understand the material.
*
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* Example 4.4 , p. 84
*
* Read this example carefully.  Be sure you understand the material.
*
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* Example 4.5 , p. 86
*
* Read this example carefully.  Be sure you understand the material.
*
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* Example 4.6 , p. 87
*
* Read this example carefully.  Be sure you understand the material.
*
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* Example 4.7, p. 87
*
* Read this example carefully.  Be sure you understand the material.
*
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* Example 4.8 , p. 89
*
* Read this example carefully.  Be sure you understand the material.
*
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* Example 4.9 , p. 90
*
* Read this example carefully.  Be sure you understand the material.
*
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* Example 4.10, p. 91
*
* The GEN1 command is used to generate the constant.  Where:
*
*    NA = Number of incomes above $50,000
*    N  = Number of households inthe Dakota County
*    PA = Probability of Event A
*
* Data from Census
*
GEN1 NA=31496
GEN1 N=54345
*
* Probability of Event a given that the Family income is greater than
* $50,000.
*
GEN1 PA=NA/N
PRINT PA
*
* Data from Sales Management magazine
*
GEN1 NA=32047
GEN1 N=55100
*
* Probability of Event A from the Sales Management magazine source
*
GEN1 PA=NA/N
PRINT PA
*
DELETE / ALL
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* Example 4.11, p. 94
*
* Where:
*
*    TICKETS = Total Number of Tickets
*    MAJORP  = Win Major Prize
*    MINORP  = Win Minor Prize
*    PA      = Probability that Selected ticket wins a major prize
*    PB      = Probability that Selected ticket wins a minor prize
*    PAUB    = Probability that Ticket wins some prize
*
GEN1 TICKETS=1000
GEN1 MAJORP=10
GEN1 MINORP=100
GEN1 PA=MAJORP/TICKETS
GEN1 PB=MINORP/TICKETS
PRINT PA PB
GEN1 PAUB=PA+PB
PRINT PAUB
*
DELETE / ALL
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* Example 4.12, p. 95
*
* Read this example carefully.  Be sure you understand the methodology.
*
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* Example 4.13, p. 95
*    
* Where:
*    PA   = Reserves exceed 2 billion barrels
*    PB   = Reserves between 1 and 2 billion barrels
*    PAUB = Reserves exceed 1 billion barrels
*   
GEN1 PA=0.1
GEN1 PAUB=0.5
GEN1 PB=PAUB-PA
PRINT PB
*
DELETE / ALL
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* Example 4.14, p. 97
*
* Read this example carefully.  Be sure you understand the methodology.
*
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* Example 4.15, p. 97
*
* Where:
*    PA   = Customers uses mustard
*    PB   = Customers uses ketchup
*    PAIB = Customers uses both
*    PAUB = Probability Customers uses at least one of these
*
GEN1 PA=0.75
GEN1 PB=0.80
GEN1 PAIB=0.65
GEN1 PAUB=PA+PB-PAIB
PRINT PAUB
*
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* Example 4.16, p. 99
*
* Where:
*    PACB = Conditional Probability of Event A given Event B
*    PBCA = Conditional Proabbility of Event B given Event A
*
GEN1 PACB=PAIB/PB
GEN1 PBCA=PAIB/PA
PRINT PACB PBCA
*
DELETE / ALL
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* Example 4.17, p. 100
*
* Where:
*    A    = Respondent answered Yes
*    E1   = Respondent answered question B
*    E2   = Respondent answered question B
*    PAE1 = Probability of a "Yes" answer given that Question A was
*           answered
*    PAE2 = Probability of a "Yes" response given Question B was answered
*
GEN1 A=0.37
GEN1 E1=0.50
GEN1 E2=0.50
GEN1 PAE1=0.50
GEN1 PE1A=(PAE1)*E1
GEN1 PE2A=A-(PE1A)
GEN1 PAE2=(PE2A)/E2
PRINT PE1A PE2A PAE2
*
DELETE / ALL
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* Example 4.18, p. 101
*
* Where:
*    A    = Bachelor degree holder is a woman
*    B    = Bachelor degree is in business
*    PACB = Probability of the Business degree goes to a woman
*
GEN1 PA=0.48
GEN1 PB=0.175
GEN1 PAIB=0.06
GEN1 PAPB=PA*PB
GEN1 PACB=PAIB/PB
PRINT PAPB PACB
*
DELETE / ALL
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* Example 4.19, p. 102
*
* Where:
*    E1 = Computer 1 works for one year without repair
*    E2 = Computer 2 works for one year without repair
*    E3 = Computer 3 works for one year without repair
*
GEN1 E1=0.90
GEN1 E2=0.90
GEN1 E3=0.90
GEN1 PE1E2E3=E1*E2*E3
PRINT PE1E2E3
*
DELETE / ALL
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* Example 4.20 , p. 102
*
* Read this example carefully.  Be sure you understand the methodology.
*
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* Example 4.21, p. 103
*
* Read this example carefully.  Be sure you understand the methodology.
*
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* Example 4.22, p. 113
*
* Where:
*    A1 = Driver had been drinking
*    A2 = Driver had not been drinking
*    C1 = Driver was involved in a crash
*    C2 = Driver was not involved in a crash
*    PC = Probability of a crash
*
GEN1 PA1C1=0.324
GEN1 PA1C2=0.103
GEN1 PC=PA1C1/PA1C2
PRINT PC
*
DELETE / ALL
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* Example 4.23, p. 117
*
* Where:
*    H1 = The person is HIV positive
*    H2 = The person is not HIV positive
*    T1 = Test says that the person is HIV positive
*    T2 = Test says that the person is not HIV positive
*
GEN1 H1=0.10
GEN1 T1H1=0.90
GEN1 T1H2=0.10
GEN1 H2=0.90
GEN1 T2H1=0.10
GEN1 T2H2=0.90
*
* Joint Probabilities
*
GEN1 H1IT1=T1H1*H1
GEN1 H1IT2=T2H1*H1
GEN1 H2IT1=T1H2*H2
GEN1 H2IT2=T2H2*H2
GEN1 T1=H1IT1+H2IT1
GEN1 T2=H1IT2+H2IT2
PRINT H1IT1 H1IT2 H2IT1 H2IT2
PRINT T1 T2
*
* Conditional Probability of HIV positive
*
GEN1 H1CT1=H1IT1/T1
PRINT H1CT1
*
* Conditional Probability of not HIV positive
*
GEN1 H2CT2=H2IT2/T2
PRINT H2CT2
*
DELETE / ALL
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* Example 4.24, p. 119
*
* Where:
*    A = Error in account balance
*    B = Unusual value based on historical figures
*
GEN1 A=0.15
GEN1 B=0.20
GEN1 BA=0.60
GEN1 AB=(BA*A)/B
PRINT AB
*
DELETE / ALL
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* Example 4.25, p. 120
*
* Where:
*    D1 = Customer has dinner with the salesperson
*    D2 = Customer does not have dinner with the salesperson
*    P1 = Customer purchases a car
*    P2 = Customer does not purchase a car
*
GEN1 P1=0.10
GEN1 D1P1=0.40
GEN1 D1P2=0.10
GEN1 P2=0.90
GEN1 D2P1=0.60
GEN1 D2P2=0.90
*
* Probability of purchase given the dinner with the salesperson.
*
GEN1 P1D1=(D1P1*P1)/((D1P1*P1)+(D1P2*P2))
PRINT P1D1
*
* Probability of purchase given the customer does not have dinner with
* the salesperson.
*
GEN1 P1D2=(D2P1*P1)/((D2P1*P1)+(D2P2*P2))
PRINT P1D2
*
DELETE / ALL
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* Example 4.26, p. 127
*
* Read this example carefully.  Be sure you understand the methodology.
*
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* Example 4.27, p. 128
*
* Read this example carefully.  Be sure you understand the methodology.
*
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*
STOP