Re: Overlapping CIs vs t-tests?

On Jun 3, 3:56 am, John Uebersax <jsueber...@xxxxxxxxx> wrote:
Suppose a medical researcher wants to compare two groups on a
continuous variable. Instead of performing a t-test, the researcher
proposes to compute the 95% CI within each group, and to consider
whether these confidence intervals overlap.

Non-overlapping CI's would then be taken as a surrogate indicator of a
statistically significant mean difference between groups.

Clearly this is not ideal, but my question is whether it is acceptable
as an expedient device -- would it give results in the "general
ballpark" of results gotten by a t-test? Would the only difference
between the two approaches depend on the difference between the t and
z distributions, for instance?

Thanks in advance.

John Uebersax PhD

I'm quite sure that if the confidence intervals do not overlap, the
means are significantly different. I am also quite sure that if the
confidence intervals do overlap, then you can't tell what the properly
computed t-test would show. Very simple algebra confirms this. I don't
think it has anything to do with the t versus z distribution, it has
to do with the fact that the t-test uses pooled variances and pooled
degrees of freedom.

But the bottom line is the computations involved in doing t-tests are
not difficult, and so I would say that this is *not* acceptable
procedure "as an expedient device".

--
Paige Miller
paige\dot\miller \at\ kodak\dot\com
.