Look in your favorite ANOVA text for the chapter on post hoc comparisons,
and in particular the method of Scheffe' (called in some texts the
"S-method"). The linear combination you describe is properly called a
"contrast" in this context, a contrast being merely a linear combination
of means with the constraint that the coefficients of the means sum to 0
(as your +0.5, +0.5, -1.0 do).
Briefly, one calculates the standard t:
(contrast value) - (hypothesized value)
---------------------------------------
(standard error of contrast)
where "contrast value" is the observed value using the sample means,
and the std. error is the square root of the sampling variance,
(error MS)*SUM(c^2/n)
where "c" is the coefficient (0.5 or -1.0) and "n" is the number of
observations in the corresponding group whose mean is of interest, and
"error MS" is the error mean square from the ANOVA.
If the contrast represents a "planned comparison", the value of t is
compared to the t-table with df = the number of degrees of freedom for
error in the ANOVA, and if there are k such planned comparisons of
interest, the critical value of t at alpha/k should be invoked for
deciding on significance at alpha.
If the contrast is a post hoc comparison, arising mainly from the nature
of the results rather than from prior theoretical argument (so that, in
principle, one might have been willing to consider ANY possible contrast)
the proper critical value is the square root of m*F(m-1,df;1-alpha) --
that is, the (1-alpha) cumulative distribution point of F with m-1 and df
degrees of freedom, multiplied by m, where m = the number of groups in the
ANOVA.
Although your expert is interested in a contrast involving 3 means, the
ANOVA may have involved m > 3 groups or samples. That's OK -- the c's
for the other, unmentioned, groups are equal to 0, so they don't
contribute to the contrast nor to its standard error. But you have the
advantage of the larger number of degrees of freedom for error using all
m groups.
Hope this helps...
-- DFB.
-----------------------------------------------------------------------
Donald F. Burrill [EMAIL PROTECTED]
184 Nashua Road, Bedford, NH 03110 (603) 471-7128
On Thu, 27 Jun 2002, Paige Miller wrote:
> Suppose I have three (or more) samples, from three (or more) different
> populations. According to my subject matter expert, he wants to estimate
> a linear combination of the means, say for example
>
> 0.5*mu1 + 0.5*mu2 - mu3
>
> where mu1, mu2 and mu3 are the population means. I know how to compute
> this estimate, it is done by simply replacing the population means with
> the sample means. If I assume the original populations are normal and
> that the population variances are equal, I can compute the variance of
> this linear combination. Pretty straightforward stuff.
>
> However, I want to create a t-test to test the null hypothesis that this
> linear combination of means is equal to zero, using an estimate of
> variance derived from the data, rather than a population variance, which
> is unknown. In doing so, I run into the mathematical difficulty that I
> do not know the proper degrees of freedom for this test. (And yes, I
> know that for the special case of estimating mu1 - mu2 there are
> textbook formulas for this t-test, however I really am interested in
> linear combinations of more than two means).
>
> I feel like I am missing something very obvious; however, if someone
> knows, or can point me to the proper formula for a t-test on a linear
> combination of means, it would be greatly appreciated.
The general rule is, the number of degrees of freedom for a t-test is the
number of d.f. associated with the variance estimate one is using. In the
context of a one-way ANOVA, that is the within-group MS, aka the error MS.
In a more complex ANOVA, one that has more than one possible error MS,
use the error MS corresponding to the source of variation represented by
your mu1, mu2, mu3 (and possibly others, if m > 3). (Not the MS for that
source, but the denominator MS used in testing the null hypothesis that
that source shows no differences among all its levels.)
> Paige Miller
> [EMAIL PROTECTED]
> http://www.kodak.com
.
.
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