Date: Tue, 8 Mar 2011 04:30:23 -0800
Reply-To: Bruce Weaver <firstname.lastname@example.org>
Sender: "SPSSX(r) Discussion" <SPSSX-L@LISTSERV.UGA.EDU>
From: Bruce Weaver <email@example.com>
Subject: Re: T-test or nonparametric test? Confused.
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Several points here.
1. All t-tests have the form t = (statistic - parameter|H0) / SE(statistic).
The test will be pretty good if the sampling distribution of the statistic
in the numerator is approximately normal. As the sample size increases, the
sampling distribution of the statistic converges on a normal distribution
(almost regardless of population shape).
2. I said "approximately" in point 1, because as George Box said, normal
distributions (and straight lines) do not exist in nature.
3. The t-test would be an exact test only if the two populations were
perfectly normal, and the two population variances exactly equal. Since
neither of those conditions will ever hold, the t-test on real data is an
approximate test. Therefore, the real question becomes whether the
approximation is good enough to be "useful". (I refer to another George Box
quote here: "All models are wrong. Some are useful.")
4. Normality (which never holds) applies to the POPULATIONS from which you
5. If you are going to assess normality via plots, you can't do it by
looking at one plot--you need two plots, one for each group. (Suppose the
two populations were perfectly normal, but with different means. In this
case, you would most likely see a bimodal distribution if you made one
6. The t-test is quite robust to non-normality of populations. If your
library has it, take a look at Figure 12.2 in Statistical Methods in
Education and Psychology , 3rd Edition (by Glass & Hopkins). It shows that
the t-test performs quite well under the following circumstances:
R/R – both populations rectangular; n1 = n2 = 5
S/S – both populations skewed; n1 = n2 = 15
N/S – one population normal, one skewed; n1 = n2 = 15
R/S – one population rectangular, one skewed; n1 = n2 = 15
L/L – both populations leptokurtic (i.e., tails thicker than the normal
distribution); n1 = 5, n2 = 15
ES/ES – both populations extremely skewed in same direction; n1 = 5, n2 = 15
M/M – both populations multimodal; n1 = 5, n2 = 15
SP/SP – both populations spiked; n1 = 5, n2 = 15
T/T – both populations triangular; n1 = 5, n2 = 15
And for dichotomous populations with:
P = .5, Q = .5, n = 11
P = .6, Q = .4, n = 11
P = .75, Q = .25, n = 11
Bridgette Portman wrote:
> Hi everyone,
> This is a rather elementary statistics question and I feel kind of stupid
> asking it. But I've managed to thoroughly confuse myself. I hope someone
> can help me out.
> I've collected survey data from 260 respondents. As I'm analyzing
> demographic information, I have noticed that the distribution of ages in
> my sample is not normal. In fact, it is bimodal, with peaks around 20 and
> 60, and a trough around 40. This was due to my sampling method, not to any
> intrinsic pattern in the population I was sampling from. I want to be able
> to compare ages between various groups, such as men and women, in my
> sample. But can I use a t-test, given the abnormal distribution? Should I
> use a nonparametric test like Mann-Whitney U instead?
> The reason I'm confused is that the bimodality is in my sample alone, do
> to my sampling technique. The ages in the population as a whole, I'm sure,
> has an underlying normal distribution. I am studying political activists,
> and in order to get at them, I sampled from a) college student political
> clubs, and b) actual political parties. The college kids tended to be
> around 20, while the party people were 50+. I know one alternative would
> be to just recode age into something like "below 40" and "above 40" but
> I'd rather avoid doing that if I can.
> Can anyone offer advice?
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