Asymptotic Methods in Statistical Inference - Statistics Centre

59
Put = − and test for treatment differences.
Assume that − = 0 ∼ for a
‘shift’ parameter , with 0 indicating that
treatment A is superior to treatment B. Then
( ≤ ) = ( − ≤ )
= ( 0 + − ≤ ) = ( − )
where is the d.f. of 0 − .Since 0 − and
− 0 are distributed in the same way (by virtue
of the random assignments within pairs) we have
that is symmetric: (−) =1− (). Here we
have assumed that is continuous, and will also
assume it to be differentiable at 0 with derivative
(0) 0.
• Consider the sign test. Put + =numberof
positive and = + . Forany we have
+ ∼ ( (0) = 1 − (−) = ())
Under : = = ∆ √ + ( −12 ) the
test statistic has mean ( ) and variance
( )(1− ( )) .