Re: 95% significance test

Posted by David Marso on
URL: http://spssx-discussion.165.s1.nabble.com/95-significance-test-tp5717562p5717585.html

Actually,  The divisor (according to the Algorithms is K *(K-1)/2) K=4: 12/2 = 6 so .05/6=.00833...
However I am at a loss to the true 'correct' applicability of statistical testing within CTABLES and stand by my assertion in another thread from earlier today.
@ Paul: IIRC from my school days Sometimes Scheffe is more powerful/less conservative than Bonferroni.
That is when there are MANY comparisons considered.
See bottom of http://www.itl.nist.gov/div898/handbook/prc/section4/prc473.htm for example.
I guess one could build out some simulations and graph the boundary cases.
Maybe another day, or if this discussion gets too interesting ;-)
--

David Marso wrote

Indeed!  
In fact Bonferroni is very (the most) conservative approach to multiple comparisons.
In your case where k=4 the effective level of significance for each comparison would be .05/4 = .0125 .

mils wrote

Wow, thanks David, I wasn't aware that the Bonferroni adjust will make such a difference.

Thanks,

mils