Re: Normality assumption for ANOVA (was: Effect statistics for non-normality)

[Paul R Swank]

At 08:29 AM 1/19/01 -0500, Bruce Weaver wrote:
>On 17 Jan 2001, Robert J. MacG. Dawson wrote:
snip
>
>Dr. Dawson has touched on something here that I've always found a bit
>puzzling--the oft stated ANOVA assumption that the populations from which
>you sample must be normal. I've always had a bit of trouble seeing why
>that is the case. I'll try to explain why by approaching it gradually.
>
>Everyone agrees, I think, that if you have a population of scores that is
>normally distributed, a z-score calculated as X-Xbar/SD can be referred to
>a table of the standard normal distribution.
>
>If the "population" from which I pull a score is a population of sample
>means (i.e., the sampling distribution of the mean), I simply change the
>formula for z or t to:
>
> X-bar - mu(X-bar)
>z or t = -----------------
> SE(X-bar)
>
>This is the z or t-test for a single sample. Provided that the sampling
>distribution of X-bar is normal (or near enough), I can still refer z to
>the standard normal, or t to the appropriate t-distribution. Now this is
>where the CLT comes into play. It gives the conditions under which the
>sampling distribution of X-bar is normal (or near enough):
>
>
end snip

An interesting example I show my students is to take progressively larger samples from a skewed distribution and observe an approximation of the sampling distribution of X bar, which becomes downright beutiful at relatively reasonable sample sizes and then show them the approximate sampling distribution of S, the sample standard deviation, Which requires a much larger sample size to straighten out. As William Ware said, it's the denominator.

Your formula above needs to indicate that it is the estimate of the standard error in the denominator.
------------------------------------
Paul R. Swank, PhD.
Professor & Advanced Quantitative Methodologist
UT-Houston School of Nursing
Center for Nursing Research
Phone (713)500-2031
Fax (713) 500-2033

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