Re: F-test questions (freshman level)

On 10 Mar 2006 10:09:37 -0800, "Frank Palmer" <F_L_Palmer@xxxxxxxxx>
wrote:

I recently did some ANOVA work on a set of data with four columns of 6
elements each. Even I could figure out that the degrees of freedom were
3 and 20. Then I looked up the table and chose n1 as 20 and n2 as 3. I
was shocked by the answer, which suggested that the differences were
coincidental. (The table sure _looked_ like the columns were different
-- only one cell of the first three columns had a value as great as the
mean of the fourth.)

Then, I looked again and decided that I had used the wrong numbers.
With n1 = 3
and n2 = 20, the .01 confidence value would be 4.9. (I'm not sure that
"n1" and "n2" are universally differentitated the way the book does.)

Two questions. The first, of practical importance, is whether this is
the right way of doing it.

The second, merely curiosity, is _how_ a value of F with n1 = 20 and n2
= 3 could be used. If you have twenty-one columns, and only two values
in each column, then you'd have n2 = 21 -- wouldn't you?

Another poster has pointed out that you don't need to
have the same N in each group.

In addition - The "F-statistic" is not wedded to the
univariate ANOVA table. "Fisher's variance ratio test"
can arise in a number of different circumstances.

Example.
- The t-test procedure in SPSS gives a test for equal
variances between the two groups, which can serve as
a warning that you might think about which of the two
forms of t-test you should use. Years ago, before a
more robust algorithm was implemented, SPSS tested that
as the ratio of "larger variance over smaller variance",
which is an F with the corresponding degrees of
freedom: F(Numerator, Denominator), as you should
learn, is the universal convention. About half the time,
that test would have the larger d.f. in the numerator
(which seldom happens for your ANOVAs).

--
Rich Ulrich, wpilib@xxxxxxxx
http://www.pitt.edu/~wpilib/index.html
.

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