Dana, another slightly different spin of what Clyde is saying, is to think of the hypothesis as a logical construct.
When you write b1=b2=b3, you very clearly view this as the logical construct (b1=b2) AND (b2=b3) AND (b1=b3). However one of these three elements in the logical construct is automatically implied by the other two. So if you prefer, you can view the issue not as a hypothesis testing or linear algebra issue, but as a logical issue. For example
IF (b1=b2) AND (b1=b3) THEN (b2=b3). The third one is unavoidably logically implied by the combination of the first two.
Therefore if I try to test such implied constraint, Stata automatically drops it, and correctly states in the numerator degrees of freedom of the F test that in fact I am testing only two constraints:
When you write b1=b2=b3, you very clearly view this as the logical construct (b1=b2) AND (b2=b3) AND (b1=b3). However one of these three elements in the logical construct is automatically implied by the other two. So if you prefer, you can view the issue not as a hypothesis testing or linear algebra issue, but as a logical issue. For example
IF (b1=b2) AND (b1=b3) THEN (b2=b3). The third one is unavoidably logically implied by the combination of the first two.
Therefore if I try to test such implied constraint, Stata automatically drops it, and correctly states in the numerator degrees of freedom of the F test that in fact I am testing only two constraints:
Code:
. sysuse auto, clear (1978 Automobile Data) . reg price mpg headroom weight, noheader ------------------------------------------------------------------------------ price | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- mpg | -56.19416 85.07654 -0.66 0.511 -225.874 113.4856 headroom | -675.5962 392.3504 -1.72 0.090 -1458.115 106.922 weight | 2.061945 .6586383 3.13 0.003 .748332 3.375557 _cons | 3158.306 3617.449 0.87 0.386 -4056.468 10373.08 ------------------------------------------------------------------------------ . test (mpg=headroom) (headroom=weight) (mpg=weight) ( 1) mpg - headroom = 0 ( 2) headroom - weight = 0 ( 3) mpg - weight = 0 Constraint 2 dropped F( 2, 70) = 1.68 Prob > F = 0.1946
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