Simultaneous Hypothesis Tests

#1
Hi, this is my first post here, so I hope this is the right place to write it.

I was wondering how can I test these hypothesis simultaneously:

Ho: x1<>0 ; x2=0 ; x3=0
Ha: x1=0 ; x2<>0 ; x3<>0

I do not want to do sequential tests, like test for x1 first, then x2, and so on.

I know I can use the General Linear Hypothesis to test things like:

Ho: x1=0, x2=0, x3=0
Ha: x1<>0, x2<>0, x3<>0

But in my case, I have a mix of = and <>.

Any help will be appreciated. Thanks! :)
 

JohnM

TS Contributor
#4
Actually, something similar to this comes up in pharmaceutical "equivalence" testing, or in any situation where you need to demonstrate or "prove" equivalence rather than fail to demonstrate a difference.

Say you need to show that two processes yield statistically similar results. The traditional method is-

Ho: mu1 = mu2
Ha: mu1 <> mu2

Then run a t-test, and if it's not significant, then you fail to reject Ho and therefore the two processes are "equivalent."

The equivalence testing method goes something like this:

select a "delta" between the two processes, within which you can say that two processes are equivalent

Ho: mu1-mu2 > delta
Ha: mu1-mu2 < delta

If the confidence interval around the difference falls within delta, reject Ho and therefore "demonstrate" equivalence

This probably doesn't apply to sequential testing per se, but the links on this page may help shed some light on it:

http://www.graphpad.com/index.cfm?cmd=library.page&pageID=36&categoryID=4
 
#5
I think I get the idea of equivalence testing, however, I don't quite see how it is related to my problem. Could you please help me see it? :p

Thanks a lot!


PS.

Shouldn't it be:

Ho: mu1-mu2 < delta
Ha: mu1-mu2 > delta

instead of:

Ho: mu1-mu2 > delta
Ha: mu1-mu2 < delta ?

so mu1-mu2 < delta being true indicates their difference is quite small, so mu1 is almost equal to mu2?
 
#6
Oh, I think I see what you mean, tell me if I'm right please:


Ho: mu1-mu2 < delta <---> Ho: mu1<>mu2
Ha: mu1-mu2 > delta <---> Ha: mu1=mu2

while

Ho: mu1-mu2 > delta <---> Ho: mu1=mu2
Ha: mu1-mu2 < delta <---> Ho: mu1<>mu2 ?



So, in my case I will say:

Ho: x1<>0 ; x2=0 ; x3=0 <---> Ho: x1-0 < delta ; x2=0 ; x3=0
Ha: x1=0 ; x2<>0 ; x3<>0 <---> Ho: x1-0 > delta ; x2=0 ; x3=0

something like that?



Thanks a lot!!! :)
 

JohnM

TS Contributor
#7
Escualida,

Equivalence testing sort of "reverses" Ho and Ha.

Ho: mu1 - mu2 > delta --> implies mu1 <> mu2

Ha: mu1 - mu2 <= delta --> implies mu1 = mu2, or mu1 is "close enough" to mu2 to establish equivalence, by our "definition" of equivalence

  • If we find that the confidence interval around (mu1-mu2) is within +/- delta, then we reject Ho in favor of Ha.
  • If we find that the confidence interval around (mu1-mu2) overlaps delta, then we say that equivalence is "uncertain."
  • If we find that the confidence interval around (mu1-mu2) is outside +/- delta, then we fail to reject Ho.
The page of links I mentioned previously will have more details behind the logic.

JohnM
 
#8
Thanks. Everything's clear now.

However, there's not known way to test:

Ho: x1<>0 ; x2=0 ; x3=0 <---> Ho: x1-0 > delta ; x2=0 ; x3=0
Ha: x1=0 ; x2<>0 ; x3<>0 <---> Ho: x1-0 < delta ; x2=0 ; x3=0

right?



What if I test:
Ho: x1=x2=x3=0 (simultaenously)
Ha: at least one of them is different from zero

I would be expecting to reject Ho. Then, I just need to see which one(s) is different from zero (hopefully it would be x1). How can I know which X is different from zero?

Will that reasoning be valid to solve my original problem?



I know how to handle this case ("zero" is not there):
Ho: x1=x2=x3 (simultaenously)
Ha: at least one of them is different

If I reject Ho, then I have to use the Tukey-Kramer procedure (to see which ones are different). This procedute from doing a sequence of tests, because it calculates confidence intervals and looks at wheter of zero is there or not (so they are not significantly different).



Thanks again!!!
 
Last edited: