One Way Anova Hypothesis Test My professor asked this question 
Why is the hypothesis test correct 


*

*$H_0: \mu_1=\mu_2=\mu_3$
$H_a:$ at least one different 
And these hypothesis wrong

*$H_0: \mu_1=\mu_2$ 

*$H_0: \mu_2=\mu_3$,

*$H_0: \mu_1=\mu_3$
So basically why cant we say that if we reject 2, 3, and 4 than why cant we reject 1 or if we reject 2, 3, or 4 than we can reject 1? He basically wants to know why we must use the first hypothesis test and not the last three? 
 A: Say you use a test under which you have a $5\%$ chance of rejecting $\mu_1=\mu_2$ given that that null hypothesis is actually true, and you also have a $5\%$ chance of rejecting $\mu_2=\mu_3$, given that that is true, and you have a $5\%$ chance of rejecting $\mu_2=\mu_3$ given that that is true.  What, then, is the probability that you reject at least one of those, given that all three are true?  That's harder to figure out because of the nature of the dependence among the tests.  I'm not sure what the answer is without working out the details, but I know it's more than $5\%$.  If you want to limit your probability of type I error to $5\%$ when testing $\mu_1=\mu_2=\mu_3$, you can't do it by treating those three hypotheses separately and testing at the $5\%$ level each time.  Possibly that was the professor's concern.  (But I'd have to know more than you've said in your question to be sure of that.)
A: The first case, you are asking whether the data leads to values of $\mu_1,\mu_2,\mu_3$ too far from the line $x=y=z$.  In the second case, you are asking are they too far from the three planes $x=y,y=z$ and $x=z$.  The '95% level' will be different shapes in the two cases, so you might accept the hypothesis in one case and reject it in the other.  If the data is extreme, both will be rejected, but we are interested in borderline cases.
