[R] Assumptions for ANOVA: the right way to check the normality
Mike Marchywka
marchywka at hotmail.com
Mon Jan 10 20:51:11 CET 2011
I can't get hotmail to indicate the original text so I'm going to top
post. There seems to be a lot of back and forth here, let me see if these
comment help guide discussion a bit.
I tried to run some histograms of your experiment (prior to a bunch of other things )
and IIRC in many cases
you have counts under 10. At minimum, anything you do or any test you
run you want to do some senistivity analyses and perturb your data a bit.
Your objective of course is important- say you want to calibrate your
response data and try to validate your assumption that your survey
question relfect some continuous variable ( but a respondent can only
round his response to an int as in teh case of taking a temperature for example, otherwise
all you can really say is that these things are like ranks, 7>6>5 etc ).
Personally I always avoid non-parametrics
( just personal bias) but with small samples and a response that is closer
to a rank than a continuous variable with some meaning, it may make sense.
If you plot hitograms of responses versus A and AH, visually they look
different, you could try fitting the histos to various pdf's and see what you
get etc. This is all retro/post-hoc so you may as well explore away.
From: Greg.Snow at imail.org
To: frodo.jedi at yahoo.com
Date: Mon, 10 Jan 2011 11:26:05 -0700
CC: r-help at r-project.org
Subject: Re: [R] Assumptions for ANOVA: the right way to check the normality
What is the question you are really trying to find the answer for? Knowing that may help us give more meaningful answers.
You keep wanting to test the residuals for normality, but it looks like you are doing it because some outdate recipe suggests it rather than that you understand why.
It is fairly easy to create a distribution that is definitely not normal, that gives the wrong answer most of the time if normality is assumed, yet will pass most normality tests most of the time (well except for SnowsPenualtimateNormalityTest, but that one has an unfair advantage in this situation). So just because the residuals look normal (or close enough) does not mean that the theory holds.
R. A. Fisher is said to have said that the quality of a statistician can be judged by the amount of rat droppings under his finger nails. Now if we take that literally, then I must not be very good. But more what he meant is that a statistician must understand the source of the data, not just get a file and put it through some canned routines. So these questions are really for you or the source of your data.
Also remember that the normality of the data/residuals/etc. is not as important as the CLT for your sample size. The main things that make the CLT not work (for samples that are not large enough) are outliers and strong skewness, since your outcome is limited to the numbers 1-7, I don't see outliers or skewness being a real problem. So you are probably fine for fixed effects style models (though checking with experts in your area or doing simulations can support/counter this). But when you add in random effects then there is a lot of uncertainty about if the normal theory still holds, the latest lme code uses mcmc sampling rather than depending on normal theory and is still being developed.
This now comes back to my first question: what are you trying to find out?
You may not need to do anova or that type of model. Some simple hypotheses may be answered using McNemars test on your data. If you want to do predictions then linear models will be meaningless (what would a prediction of -3.2, 4.493, or 8.1 mean on a 7 point likert scale?) and something like proportional odds logistic regression will be much more meaningful. Between those are bootstrap and permutation methods that may answer you question without any normality assumptions.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.snow at imail.org
801.408.8111
From: Frodo Jedi [mailto:frodo.jedi at yahoo.com]
Sent: Saturday, January 08, 2011 3:20 AM
To: Greg Snow
Cc: r-help at r-project.org
Subject: Re: [R] Assumptions for ANOVA: the right way to check the normality
Dear Greg,
many thanks for your answer. Now I have a problem then in understanding how to check
normality in case of ANOVA with repeated measures.
I would need an help with a numeric example, as I haven´tu fully understood how it works with the
proj() command as it as suggested by another R user in this mailing list.
For example, in attachment you find a .csv table resulting from an experiment, you can access it by means of this command:
> scrd<- read.csv(file='/Users/....../tables_for_R/table_quality_wood.csv',sep=',',header=T)
The data are from an experiment where participants had to evaluate on a seven point likert scale
the realism of some stimuli, which are presented both in condition "A" and in condition "AH".
I need to perform the ANOVA by means of this command:
> aov1 = aov(response ~ stimulus*condition + Error(subject/(stimulus*condition)), data=scrd)
but the problem is that I cannot plot as usually do the qqnorm on the residuals of the fit because
lm does not support the Error term present in aov.
I normally check normality through a plot (or a shapiro.test function). Now could you please
illustrate me how will you be able to undestand from my data if they are normally distributed?
Please enlighten me
Best regards
________________________________
From: Greg Snow
To: Ben Ward ; "r-help at r-project.org"
Sent: Fri, January 7, 2011 7:34:05 PM
Subject: Re: [R] Assumptions for ANOVA: the right way to check the normality
A lot of this depends on what question you are really trying to answer. For one way anova replacing y-values with their ranks essentially transforms the distribution to uniform (under the null) and the Central Limit Theorem kicks in for the uniform with samples larger than about 5, so the normal approximations are pretty good and the theory works, but what are you actually testing? The most meaningful null that is being tested is that all data come from the exact same distribution. So what does it mean when you reject that null? It means that all the groups are not representing the same distribution, but is that because the means differ? Or the variances? Or the shapes? It can be any of those. Some point out that if you make certain assumptions such as symmetry or shifts of the same distributions, then you can talk about differences in means or medians, but usually if I am using non-parametrics it is because I don't believe that things are symmetric and the shift idea doesn't fit in my mind.
Some alternatives include bootstrapping or permutation tests, or just transforming the data to get something closer to normal.
Now what does replacing by ranks do in 2-way anova where we want to test the difference in one factor without making assumptions about whether the other factor has an effect or not? I'm not sure on this one.
I have seen regression on ranks, it basically tests for some level of relationship, but regression is usually used for some type of prediction and predicting from a rank-rank regression does not seem meaningful to me.
Fitting the regression model does not require normality, it is the tests on the coefficients and confidence and prediction intervals that assume normality (again the CLT helps for large samples (but not for prediction intervals)). Bootstrapping is an option for regression without assuming normality, transformations can also help.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.snow at imail.org
801.408.8111
> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
> project.org] On Behalf Of Ben Ward
> Sent: Thursday, January 06, 2011 2:00 PM
> To: r-help at r-project.org
> Subject: Re: [R] Assumptions for ANOVA: the right way to check the
> normality
>
> On 06/01/2011 20:29, Greg Snow wrote:
> > Some would argue to always use the kruskal wallis test since we never
> know for sure if we have normality. Personally I am not sure that I
> understand what exactly that test is really testing. Plus in your case
> you are doing a two-way anova and kruskal.test does one-way, so it will
> not work for your case. There are other non-parametric options.
> Just read this and had queries of my own and comments on this subject:
> Would one of these options be to rank the data before doing whatever
> model or test you want to do? As I understand it makes the place of the
> data the same, but pulls extreme cases closer to the rest. Not an
> expert
> though.
> I've been doing lm() for my work, and I don't know if that makes an
> assumption of normality (may data is not normal). And I'm unsure of any
> other assumptions as my texts don't really discuss them. Although I can
> comfortably evaluate a model say using residual vs fitted, and F values
> turned to P, resampling and confidence intervals, and looking at sums
> of
> squares terms add to explanation of the model. I've tried the plot()
> function to help graphically evaluate a model, and I want to make sure
> I
> understand what it's showing me. I think the first, is showing me the
> models fitted values vs the residuals, and ideally, I think the closer
> the points are to the red line the better. The next plot is a Q-Q plot,
> the closer the points to the line, the more normal the model
> coefficients (or perhaps the data). I'm not sure what the next two
> plots
> are, but it is titled Scale-Location. And it looks to have the square
> root of standardized residuals on y, and fitted model values on x.
> Might
> this be similar to the first plot? The final one is titled Residuals vs
> Leverage, which has standardized residuals on y and leverage on x, and
> something called Cooks Distance is plotted as well.
>
> Thanks,
> Ben. W
> > Whether to use anova and other normality based tests is really a
> matter of what assumptions you are willing to live with and what level
> of "close enough" you are comfortable with. Consulting with a local
> consultant with experience in these areas is useful if you don't have
> enough experience to decide what you are comfortable with.
> >
> > For your description, I would try the proportional odds logistic
> regression, but again, you should probably consult with someone who has
> experience rather than trying that on your own until you have more
> training and experience.
> > --
> > Gregory (Greg) L. Snow Ph.D.
> > Statistical Data Center
> > Intermountain Healthcare
> > greg.snow at imail.org
> > 801.408.8111
> >
> > From: Frodo Jedi [mailto:frodo.jedi at yahoo.com]
> > Sent: Thursday, January 06, 2011 12:57 PM
> > To: Greg Snow; r-help at r-project.org
> > Subject: Re: [R] Assumptions for ANOVA: the right way to check the
> normality
> >
> >
> > Ok,
> > I see ;-)
> >
> > Let´s put in this way then. When do I have to use the kruskal wallis
> test? I mean, when I am very sure that I have
> > to use it instead of ANOVA?
> >
> > Thanks
> >
> >
> > Best regards
> >
> > P.S. In addition, which is the non parametric methods corresponding
> to a 2 ways anova?..or have I to
> > repeat many times the kruskal wallis test?
> > ________________________________
> > From: Greg Snow>
> > To: Frodo Jedi>; Robert Baer>;
> "r-help at r-project.org">
> > Sent: Thu, January 6, 2011 7:07:17 PM
> > Subject: RE: [R] Assumptions for ANOVA: the right way to check the
> normality
> >
> > Remember that an non-significant result (especially one that is still
> near alpha like yours) does not give evidence that the null is true.
> The reason that the 1st 2 tests below don't show significance is more
> due to lack of power than some of the residuals being normal. The only
> test that I would trust for this is SnowsPenultimateNormalityTest
> (TeachingDemos package, the help page is more useful than the function
> itself).
> >
> > But I think that you are mixing up 2 different concepts (a very
> common misunderstanding). What is important if we want to do normal
> theory inference is that the coefficients/effects/estimates are
> normally distributed. Now since these coefficients can be shown to be
> linear combinations of the error terms, if the errors are iid normal
> then the coefficients are also normally distributed. So many people
> want to show that the residuals come from a perfectly normal
> distribution. But it is the theoretical errors, not the observed
> residuals that are important (the observed residuals are not iid). You
> need to think about the source of your data to see if this is a
> reasonable assumption. Now I cannot fathom any universe (theoretical
> or real) in which normally distributed errors added to means that they
> are independent of will result in a finite set of integers, so an
> assumption of exact normality is not reasonable (some may want to argue
> this, but convincing me will be very difficult). But looking for exact
> normality is a bit of a red herring because, we also have the Central
> Limit Theorem that says that if the errors are not normal (but still
> iid) then the distribution of the coefficients will approach normality
> as the sample size increases. This is what make statistics doable
> (because no real dataset entered into the computer is exactly normal).
> The more important question is are the residuals "normal enough"? for
> which there is not a definitive test (experience and plots help).
> >
> > But this all depends on another assumption that I don't think that
> you have even considered. Yes we can use normal theory even when the
> random part of the data is not normally distributed, but this still
> assumes that the data is at least interval data, i.e. that we firmly
> believe that the difference between a response of 1 and a response of 2
> is exactly the same as a difference between a 6 and a 7 and that the
> difference from 4 to 6 is exactly twice that of 1 vs. 2. From your
> data and other descriptions, I don't think that that is a reasonable
> assumption. If you are not willing to make that assumption (like me)
> then means and normal theory tests are meaningless and you should use
> other approaches. One possibility is to use non-parametric methods
> (which I believe Frank has already suggested you use), another is to
> use proportional odds logistic regression.
> >
> >
> >
> > --
> > Gregory (Greg) L. Snow Ph.D.
> > Statistical Data Center
> > Intermountain Healthcare
> > greg.snow at imail.org>
> > 801.408.8111
> >
> >
> >> -----Original Message-----
> >> From: r-help-bounces at r-project.org
> project.org> [mailto:r-help-bounces at r-
> >> project.org] On Behalf Of Frodo Jedi
> >> Sent: Wednesday, January 05, 2011 3:22 PM
> >> To: Robert Baer; r-help at r-project.org>
> >> Subject: Re: [R] Assumptions for ANOVA: the right way to check the
> >> normality
> >>
> >> Dear Robert,
[[elided Hotmail spam]]
> >> So you also think that I have to check only the residuals and not
> the
> >> data
> >> directly.
> >> Now just for curiosity I did the the shapiro test on the residuals.
> The
> >> problem
> >> is that on fit3 I don´t get from the test
> >> that the data are normally distribuited. Why? Here the data:
> >>
> >>> shapiro.test(residuals(fit1))
> >> Shapiro-Wilk normality test
> >>
> >> data: residuals(fit1)
> >> W = 0.9848, p-value = 0.05693
> >>
> >> #Here the test is ok: the test says that the data are distributed
> >> normally
> >> (p-value greather than 0.05)
> >>
> >>
> >>
> >>> shapiro.test(residuals(fit2))
> >> Shapiro-Wilk normality test
> >>
> >> data: residuals(fit2)
> >> W = 0.9853, p-value = 0.06525
> >>
> >> #Here the test is ok: the test says that the data are distributed
> >> normally
> >> (p-value greather than 0.05)
> >>
> >>
> >>
> >>> shapiro.test(residuals(fit3))
> >> Shapiro-Wilk normality test
> >>
> >> data: residuals(fit3)
> >> W = 0.9621, p-value = 0.0001206
> >>
> >>
> >>
> >> Now the test reveals p-value lower than 0.05: so the residuals for
> fit3
> >> are not
> >> distributed normally....
> >> Why I get this beheaviour? Indeed in the histogram and Q-Q plot for
> >> fit3
> >> residuals I get a normal distribution.
> >>
> >>
> >>
> >>
> >>
> >>
[ junk deleted, after hotmrail graciously allowed my ui to respond ]
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