s 1 2 s 1 2 s 1 2 2 n p 1 p s 2 s 1 2 s 1 2 2 n p 1 p c Show that for very

# S 1 2 s 1 2 s 1 2 2 n p 1 p s 2 s 1 2 s 1 2 2 n p 1 p

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s 1 2 - s 1 2 s 1 2 < 2 n ( p 1 p 0 ) s 0 2 s 1 2 s 1 2 < 2 n ( p 1 p 0 ) (**) (c) Show that for very large values of n the condition in (b) is approximately equal to e R ' e R e U ' e U e U ' e U < 2 n ( p 1 p 0 ) where e R is the vector of residuals for the restricted model with p 0 parameters and e U the vector of residuals for the full unrestricted model with p 1 parameters. Answer: Since e R is the vector of residuals for the restricted model with p 0 parameters and e U the vector of residuals for the full unrestricted model with p 1 parameters, we have: s 0 2 = e R ' e R and s 1 2 = e U ' e U Which also means: (**) = e R ' e R e U ' e U e U ' e U < 2 n ( p 1 p 0 ) (***) (d) Finally, show that the inequality from (c) is approximately equivalent to an F-test with critical value 2, for large sample sizes. We have: F = e ( ¿¿ 0 ' e 0 e 1 ' e 1 )/ g e 1 ' e 1 /( n k ) ¿ g = number of explanatory factors to be removed from the unrestricted model = p 1 – p 0 For large sample sizes, n is very large, which means: n – k n The critical value 2, so: F < 2 e ( ¿¿ 0 ' e 0 e 1 ' e 1 )/ g e 1 ' e 1 /( n k ) ¿ < 2 e ( ¿¿ 0 ' e 0 e 1 ' e 1 )/( p 1 p 0 ) e 1 ' e 1 / n ¿ < 2 e R ' e R e U ' e U e U ' e U x n p 1 p 0 < 2 e R ' e R e U ' e U e U ' e U < 2 n ( p 1 p 0 ) Which proved that F-test = (***) #### You've reached the end of your free preview.

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