A Note on Screening Regression Equations - Common Errors in Statistics - page 169

() - ()
-

g l

a

1 g lr

ar

= ˙ 40 ..

1

12 in the
numerator and 100 - 12 = 88 in the denominator. (However, q n ,a is still
quite variable, its standard deviation being about 3.) On this basis, a P
value on the order of 10 -4 may be anticipated.
What about the t tests? Take l¢>l, corresponding to level a¢<a. The
nominal level for the test is a¢, but the real level is

The number of degrees of freedom should be around ar n

1 Ó 1 -

- ()

ar
lr

˛ .

PZ >¢ l

a 1 g

Since g (l) > a, it follows that 1 - ar > 1 - g (l)r. Keep a = 0.25, so
l 1.15; take a¢=5 percent, so l¢=1.96; keep r = 2 . Now

1 -

- () =

ar
lr

l ¢ 23 .

1 g

and the real level is 9 percent. This concludes the example.
Turn now to the proofs. Without loss of generality, suppose the i th
column of X has a 1 in the i th position and 0's everywhere else. Then

ˆ

b i = Y for i = 1 ,..., p ,

i

and the sum of squares for error in the first-pass regression corresponding
to the model (1) is

n

Â

Y i

2

.

ip =+ 1

Thus

p n

 YY

2 2 2

R n =

i i

i = 1 i = 1

and

p

1 1

n

 Â

F = Y

2

Y .

2

n i i

p n - p

i = 1 ip = + 1

Now (3) follows from the weak law of large numbers. Of course, E ( R n )
and var R n are known: see Kendall and Stuart (1969).