*       Hausman specification test of error in variables
* REFERENCE:   J. A. Hausman, "Specification Tests in Econometrics",
*              Econometrica, Vol. 46, 1978, pp. 1251-1271.
*
* DESCRIPTION:   A regressor may be measured with error that is not
* independent of the equation error. Under the null hypothesis of
* no misspecification, OLS will be a consistent and efficient estimator.
* But under the alternative of misspecification OLS will be biased and
* inconsistent.
* The instrumental variables (IV) estimator will not be asymptotically 
* efficient under the null hypothesis but will be appropriate in the 
* presence of misspecification. 
* The Hausman specification error test compares the difference 
* between the two estimators.
*
* EXAMPLE:   The data set on investment (I), consumption (C) and 
* income (Y) is from Griffiths, Hill and Judge (1993, Table 14.2, p. 464).
* It is assumed that income is measured with error and an appropriate
* instrumental variable is investment.
SAMPLE 1 20
READ I C Y / LIST
2.0  15.30  17.30
2.0  19.91  21.91
2.2  20.94  22.96
2.2  19.66  21.86
2.4  21.32  23.72
2.4  18.33  20.73
2.6  19.59  22.19
2.6  21.30  23.90
2.8  20.93  23.73
2.8  21.64  24.44
3.0  21.90  24.90
3.0  20.50  23.50
3.2  22.83  26.05
3.2  23.49  26.69
3.4  24.20  27.60
3.4  23.05  26.45
3.6  24.01  27.61
3.6  25.83  29.43
3.8  25.15  28.95
3.8  25.06  28.86
* Estimation using the consistent estimator (IV) under both
* the null and the alternative hypotheses;
* in SHAZAM, the 2SLS command is used for instrumental variable estimation
2SLS C Y (I) / DN COEF=B1 PCOV COV=V1
GEN1 SIGIV=$SIG2
* Estimation using the efficient estimator (OLS) under the null
OLS C Y / DN COEF=B0 COV=V0 NOMULSIGSQ
* Use the error variance estimate obtained from IV estimation.
MATRIX V0=SIGIV*V0
PRINT V0
* Compute the Hausman specification test statistic using the method
* illustrated in Griffiths, Hill and Judge (1993, p. 465) 
SAMPLE 1 2
GENR Q=B1-B0
MATRIX VQ=V1-V0
MATRIX M=(Q(1)**2) / VQ(1,1)
* The statistic M is distributed chi-square with 1 degree of freedom
* under the null hypothesis. The 5% critical value is 3.84.
PRINT M
STOP