http://spssx-discussion.165.s1.nabble.com/Question-About-Tests-of-Normality-and-Choice-of-Statistical-Analysis-tp5728302p5728329.html
Diane,
Haven't we covered this before?
There's a good point or two behind your "rant", and I will point out those, plus,
where I disagree.
1. Single Likert items are generally not a good choice for major analyses.
Use composite scores. But if the score is not arguably "interval", it does not
qualify for the description "Likert". The same goes for "skewed"; but, even there,
it is hard to discern major disadvantages of using ANOVA instead of a logistic approach.
On the other hand, it is *easy* to point to means where the meaning is inherent.
2. When reporting on sets of Likert items, you might get away with showing
logistic results -- which I would generally consider as technically superior.
3. If you do focus on a single Likert item, I expect that you will strongly disappoint
your reviewers, editors and readers if you do not provide them with the simple MEANs,
despite your animadversion. (We flat-earthers live in small cities where it is safe to
ignore the curvature owing to advantages of scale.)
4. I am pleased that you do not like the use of rank-transformed scores for Likert; my
main purpose was to deprecate those, which was on over-interpreted recommendation
from the 1950s which keeps recurring. I thought Brian was addressing that.
5. And then there is this confusing statement, which I will address with an example:
"Both normal based [t-tests] or rank based [mann-whitney]
inferential tests
on means or rank are meaningless, as they assume
metric data, i.e. difference
between agree and strongly agree is same as
between neutral and agree.
This is a nonsensically improbable
hypothesis."
Every 1-d.f. test will use "a metric". The problem with the rank-transform is that, in the
cases that it is apt to make a difference, it is liable to use a palpably BAD choice of metric.
You can confirm this when you look at the transformed values.
(Plus, as it happens, the tests tend to be inaccurate when they are based on
"variance estimates adjusted for ties." Conover showed that doing the rank-transform
followed by ordinary ANOVA seemed to be generally more accurate.)
The logistic uses a metric which is based on the assumption that distances should be
(or can usefully be) defined by the Ns of the sample subsets, using cumulative ranks,
and the "logistic" ought to describe the outcome. (By the way, "probit" starts the
same way with ranks, but uses "normal" instead of "logistic" as the basis for its
distances.)
One reason that the ANOVA and the logistic agree better is that the logistic makes
rather less change to the metric, where the rank-transform (an intermediate step on
the way to the logistic) gives results that, well, *I* do not like. But the logistic
*definitely* does use a varied metric; that is why it gives (a slight) variation in results.
Here is what Rank-transform does to unequal N, for one skewed example; skewing is where
the differences mainly appear. The implicit "metric" is in the final column. The distance
between the 0-1 responses is used to rescale the responses for scores 2 and 3. This gives
a direct comparison to the original 0-3 metric.
Score N range aveRank 0|1|what?
0 55 1-55 28 0
1 31 56-86 72 1
2 9 87-95 92 1.42
3 5 96-100 98 1.48
Instead of analyzing (0,1,2,3), the rank-transform uses the metric which works
exactly like (0, 1, 1.42, 1.48), since linear transforms will be transparent .
On the other hand, even for this amount of skew, the logistic gives the implicit metric
as (0, 1, 1.75, 2.5). There is still a decreased gap between the higher numbers, but the
logistic "undoes" most of the compression of the range. The logistic would be computed
from the percentile of the average rank, log(P/(1-P) ), which is too messy to go into.
--
Rich Ulrich
Date: Thu, 1 Jan 2015 16:32:20 +0000
From:
[hidden email]Subject: Re: Question About Tests of Normality and Choice of Statistical Analysis for LIkert items
To:
[hidden email]Contemplating an inferential t-test for a Likert ITEM suggests that you are using MEAN of an ordinal variable as a descriptive measure.
This is the statistical equivalent in believing in a flat earth. Move on
Likert items are ORDINAL. Consequently the only appropriate descriptive measure are raw probabilities of items 1 -n or cumulative probabilities of [<=1, <=2,…<= n-1]
If you are using raw probabilities then chi=square contingency test [pearson or log likelihood] is good for single predictor variable, and multinomial for > 1 predictor.
If you are using cumulative probabilities then ordinal regression [either logit or probit] is good
SPSS REGRESSION [ordinal or binary logit] is good fro between grip designs
SPSS MIXED is good for designs that include repeated measure, or GENERALIZED linear models with generalised estimating equation
Its pointless to ask whether ordinal measure are normally distributed . The answer is ALWAYS NO, higher mean implies negative skew, lower mean positive skew
Both normal based [t-tests] or rank based [mann-whitney] inferential tests on means or rank are meaningless, as they assume metric data, i.e. difference between agree and strongly agree is same as between neutral and agree. This is a nonsensically improbable hypothesis.
There is no excuse for any SPSS user, or anyone else as there always R, doing inappropriate t-tests on Likert items. The right tests are easily available.
A user who is unfamiliar with these kinds of tests should consult someone with statistical expertise and ask the RIRHT question, which is:
‘how do I analyse this Likert item data’
not the wrong question,
‘how do i test if this is data is normally distributed’
when consulting it is always a good idea to give expert FULL picture of problem to be solved and data collected [or better to be collected]. assuming you know the right test and skiing how to do that test is a recipe for disaster. if you are unlucky expert will tell you how to do the test without probing whether it is the right test
End of RANT, Happy New Year
NB if you have Likert SCALE rather than a Likert item, then assumption of metric properties may be appropriate.
best
Diana
Bruce,
That is very nice.
But you never even mentioned the assumptions of the relevant non-parametric
tests that are based on the rank-transformation: continuous data of similar-
shape distributions in both samples, and few ties. Some of your examples
("Normal versus skewed") would not be appropriate for testing by ranks.
Likert-type items deserve normal testing for various reasons, including the
occasional weird scoring that you can observe as resulting from rank-transforms.
Continuous items with similar skew, etc., usually should be transformed by taking
logs or reciprocal (whatever is appropriate) to "normalize"
- That improves both the metric and the test. I can regard rank-testing as a sloppy,
time-saving expedient, compared to doing a transformation that is apparent.
- If there is not a transformation available, then there is big doubt about whether
these data fit the non-par assumption.
--
Rich Ulrich
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