用不显著来验证“没有”处理效应总是不大合适，
所以要反过来做。

比如没有”处理效应 is mu<=0 处理效应 is mu>0

The folloing is not a good idea

H0： mu<=0
Ha: mu>0

A better way is:

define a delta that is threshhold for 处理效应

H0:mu>delata
Ha:mu<delta

So when you get significant p-value, you can
accept Ha that there is no 处理效应