I understand that with $H_0$:
$$b_1 = 0,\ b_2 = 0,\ b_3 = 0$$
the number of restrictions ($q$) is 3.
However, with $H_1$:
$$b_1 + b_2 + b_3 = 0$$
is the number of restrictions now 1?
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2$\begingroup$ Hmm--I count three equality signs in the first example and one in the second. $\endgroup$– whuber ♦Commented Dec 5, 2017 at 15:38
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1 Answer
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In the second, we are testing one linear restriction, so q should be one.
