Reference no: EM13881455
In studying the purchase of durable goods Y (Y = 1 if purchased, Y = 0 if no purchase) as a function of several variables for a total of 762 house- holds, Janet A. Fisher* obtained the following LPM results:
|
Explanatory variable
|
Coef?cient
|
Standard error
|
|
Constant
|
0.1411
|
-
|
|
1957 disposable income, X1
|
0.0251
|
0.0118
|
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(Disposable income = X1)2, X2
|
-0.0004
|
0.0004
|
|
Checking accounts, X3
|
-0.0051
|
0.0108
|
|
Savings accounts, X4
|
0.0013
|
0.0047
|
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U.S. Savings Bonds, X5
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-0.0079
|
0.0067
|
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Housing status: rent, X6
|
-0.0469
|
0.0937
|
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Housing status: own, X7
|
0.0136
|
0.0712
|
|
Monthly rent, X8
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-0.7540
|
1.0983
|
|
Monthly mortgage payments, X9
|
-0.9809
|
0.5162
|
|
Personal noninstallment debt, X10
|
-0.0367
|
0.0326
|
|
Age, X11
|
0.0046
|
0.0084
|
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Age squared, X12
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-0.0001
|
0.0001
|
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Marital status, X13 (1 = married)
|
0.1760
|
0.0501
|
|
Number of children, X14
|
0.0398
|
0.0358
|
|
(Number of children = X14)2, X15
|
-0.0036
|
0.0072
|
|
Purchase plans, X16 (1 = planned; 0 otherwise)
|
0.1760
|
0.0384
|
R 2 = 0.1336
a. Comment generally on the ?t of the equation.
b. How would you interpret the coef?cient of -0.0051 attached to checking account variable? How would you rationalize the negative sign for this variable?
c. What is the rationale behind introducing the age-squared and number of children-squared variables? Why is the sign negative in both cases?
d. Assuming values of zero for all but the income variable, ?nd out the conditional probability of a household whose income is $20,000 pur- chasing a durable good.
e. Estimate the conditional probability of owning durable good(s), given: X1 = $15,000, X3 = $3000, X4 = $5000, X6 = 0, X7 = 1, X8 = $500, X9 = $300, X10 = 0, X11 = 35, X13 = 1, X14 = 2, X16 = 0.