Normal Distribution - Science topic

There is no assumption about normality of predictor or outcome variables in linear regression (only about the normality of residuals/errors). Also, linear regression is designed to handle correlated predictors, so there should be no problem unless the correlations are extremely high.

If you look up the assumptions for an OLS general linear model, you'll see that there is no assumption that the dependent variable is normally distributed.

Usually the assumption is written as the errors from the model being normally distributed. These are approximated by the residuals.

Pretty much any textbook on the design and analysis of experiments should be a reference for this. Slides in the following presentation are taken from Montgomery, Design and Analysis of Experiments. See slide 8, and then slides 19 - 22.

Daniel Wright , no, it just results in the default being the Elfving method. I don't think I had any great reason to choose one method over another for the default.

You state "I want to compare two groups with each other." What aspect of the two groups do you want to compare?

What the model requires is that the errors are mutually independent and identically distributed in a way that is approximating a Normal distribution with mean equal to 0 and variance equal to σ ².

Also, I remind you that linear regression is a mathematical function based on the equation of the geometric straight line.

The rest comes by itself.

My favorite reference is attached here.. David Booth

Joel Coll Ferrer , your data is clearly not normal distributed, and assuming that it would be a sample from a normal distribution is very clearly wrong. This is independent of the sample size.

However, you want to make inference abot the mean of that (unknown, clearly not normal) distribution. For this inference the distribution of the statistic (the sample mean) is relevant.

If -hypothetcally!- the data were taken from a normal distributuon, then we would know that the sample mean could be interpreted as a value taken from a normal distribution, too. And if this would be the case, then we could make inference based on the normal distribution model. This is what linear models (t-test, ANOVAs, liner regressions) do. If the assumption that the data are approximately normal distributed is quite reasonabble, then the inference made are also quite reasonable or approximately correct. This is again (quite) independent of the sample size.

You data is very clearly not from a normal distribution. Assuming that it is from a distribution that would be at least approximately normal is clearly not reasonable. Hence, inference based on such assumptions may be considerably wrong. To be able to make some reasonably correct inference would require to know the distribution of the sample mean. It might be possible to derive this, but this might be difficult and require the help of a mathematical statistician.

The distribution of the sample mean depends on the distribution of the data. The CLT tells us that for large sample sizes it will approximate the normal distribution reasonably well. Hence, large samples allow you to do reasonable inference about the mean using the standard linear models although we know that the distribution of the data is not normal.

Calculate with the help of this neat Normal Distribution Applet/Calculator:

Beyond the lack of clarity of your exposition, is it possible that the university you attend doesn't have adequately trained statisticians?

In small samples the distribution of the residuals of a model need to be normally distributed for hypothesis tests on the coefficients to follow the standard distributions. In large samples the Central Limit Theorem applies and the coefficients follow conventional distributions despite the residuals not following a normal distribution. However, the tests for a long run levels relationship in an ARDL model are nonstandard. Nevertheless, my understanding is that in large samples the ARDL model can have nonnormally distributed residuals and the simulated nonstandard critical values will be valid. However, this will not be the case in small samples. Others may have more information on whether the nonstandard critical values are valid in large samples when the ARDL model's residuals are non-normal.

Zoiya Naz ANOVA requires the dependent variable be measured on a continuous scale. I recommend binary logistic regression.

Non-normality mostly affects the test statistics (chi-square test of model fit) and parameter standard errors when using maximum likelihood estimation in SEM. To obtain adjusted ("robust") test statistics, standard errors, and p-values with non-normal data, you can use the Bollen-Stine bootstrap option in AMOS.

It depends on the outliers but always remember that normally distributed means normally distributed and if there are outliers your distribution at best is a contaminated normal. So how much contamination is really important? Investigate that if you can but please remember that there are robust methods to help you. See the attachment for more details. Best wishes David Booth

You have to decide from which direction you want to look at the relationships. When you have a continuous and a dichotomous variable, you may ask how the expectation of the continuous response variable depends on the value of the dichotomous predictor or how the expectation of the dichotomous response variable depends on the value of the continuous predictor variable. In either way, the question can be addressed by some form of regression model that depends on the characteristic of the response variable.

When the response is dicotomous, a logistic regression model is used, because the disitribution of the dichotomous variable must be Bernoulli. When the response is continuous, the form of the regression model depends on the assumed distiribution model of the response variable, which without specification can be any continuous distribution model.

Such models can include and kind and any muber of predictor variables. Nominal predictor variables in such regression models are typically 0/1 coded. If the predictor is dichotomous, one of the model parameters then reflects the expected difference in the response between the "groups".

For continuous predictors, the form of the relationship with the response must be specified, which may be simply linear but may also follow some arbitrary nonlinear relationship.

One can test hypotheses about parameter values in such models, typically in form of likelihood-ratio tests (the test statistic is the ratio of the likelihood under the unrestricted model and the likelihood under the model restricted at the hypothesized parameter values). In most cases, the sampling distribution of the likelihood ratio is not known, but Wilk's theorem sais that the distribution of the log likelihood ratio is at least approximately Chi². So one can do Chi² tests, which are usually approximate likelihood-ratio test. Tests about a single parameter can be equivalently formulated as z-tests (when Z is z-distributed, then Z² is chi²-distributed).

If the distribution of the response is normal, then the sampling distribution of the log likelihood ratio is exactly Chi², and z-tests about individual parameters are also exact tests.* However, the normal distribution has an independent parameter for the variance, which must be known/given. If this must be estimated from the data, too, the additional uncertainty can be accounted for with F- and t-tests, respectively. These tests are exact for normal distributed response variables with unknown variance. For response variables with other disributions that have an independent variance parameter (e.g. Gamma) the F/t test are superior to the Chi²/z test but not exact as for the normal distributed response.

* "exact" means that the derivation is mathematically exact and not approximate. It does not mean that the actual results are "exact". This "exact"-thing is not very relevant in practice. More relevant is that we assume what a reasonable distribution model for the response might be, and that the assumed functional forms of the relationships are reasonable, too.

Mann–Whitney U test

It may be useful for you take a look at the following article:

(Multinominal) logistic regression. Possibly after a transformation of the IV to linearize the relationship between the IV and the log odds ratios. I suggest you contanct a local statistician.

-There should be homogeneity of variances

The variances of the populations from which the groups are sampled are all assumed to be identical. This assumption is required to make and use a single pooled variance estimate from all the observed data.

-Dependent variable should be normally distributed for each category of the independent variable

The distribution of the dependent variable is assumed to be normal, independent if the sampled population. This assumption is required to derive the sampling distribution of the test statistics from which the p-values are calculated.

-what does 95% confidence interval in one-way ANOVA mean? Does this mean F statistic is 95% likely to be true compared to population mean?

No. Your idea makes no sense. A 95% confidence interval is (very typically) the interval of all hypotheses about the value of a parameter that can not be rejected at a 5% level of significance.

-For significant F statistic, I should use dunnett's post hoc test to compare means of groups to control mean. Tukey post hoc would be if I want to compare means of each group to each other.

Both procedures, Dunnett's and Tukey's, are based on series of t-tests performed using the pooled variance estimate to get the standard errors of the mean differences. Dunnett's procedure controls the family-wise error-rate of "multiple-to-one" comparisons, Tukey's procedure controls the family-wise error-rate for "all-pairwise" comparisons. There is no ANOVA required, and there is no need that the ANOVA is significant. Fisher's "least significant difference" method for exactly 3 tests between 3 groups is an ANOVA-protected procedure of controlling the family-wise error-rate of these 3 tests (only when the ANOVA is significant at the same level).

- In simple terms, What is Bon Ferroni post hoc and Sidak post hoc for?

Bonferroni 's procedure is a particularily simple procedure to control the family-wise error-rate across any family of tests. You calculate the least-significant difference tests (t-tests using the pooled variance estimate), and then you simply compare the p-values from these tests to the level of significance divided by the number of tests in order to control the family-wise error rate (this is called the Bonferroni-correction). Sidak's procedure is extremely similar, using a slightly more difficult to calculate correction factor. The differences seem mostly of academic (not practical) interest to me (for instance, see Journal of Modern Applied Statistical Methods 14(1), 12-23).

Both correction methods can be applied any series of p-values, which may come from entirely unrelated tests.

There are improvements of these procedures that slightly increase the power, particularily when the number of tests is large. The most frequently used improvement is that by Holm (aka Bonferroni-Holm or Sidak-Holm), which starts with sorting the p-values and then adjusting the stringency of the tests while "stepping down" the list of sorted p-values.

Note: "post-hoc tests" are tests about subgroups of the data that can be performed only after the entire data of all groups is lying on the table. The typical post-hoc tests are t-tests, that are performed using a pooled variance estimate to calculate the standard errors. This pooled variance estimate can be calculated only when all the data are available: you can compare group A and B only after you also have the data for the groups C, D, E, .... This logically is unrelated to omnibus procedures like ANOVA.

The sums of ranks aren't always very helpful for the reader because they depend on the sample size of the data analyzed, and aren't in the same units as the original data.

One simple way to express the change in values is to present the median of the differences in scores. Or a five-number summary of these differences. Or a histogram of these differences:

A helpful plot is a bivariate plot of Before and After, with a one-to-one line superimposed. Here, points above and to the left of the line indicate an observation where the value increased.

You could also calculate the number of observations that increased, decreased, or were tied, and express this as a proportion of the total observations.

Another effect size statistic is the matched-pairs rank biserial correlation coefficient. This is a standardized effect size measurement and matches the signed rank test. See: King, B.M., P.J. Rosopa, E.W. and Minium. 2000. Statistical Reasoning in the Behavioral Sciences, 6th. Wiley.

Your model may be saturated or "just identified" (df = 0). This would be the case if you estimated all possible paths between variables. You should still be able to look at the parameter estimates for a saturated/just-identified model. You just don't get an overall chi-square test of model fit (because a saturated model by definition reproduces the observed data perfectly).

When I hear things like "...patients distributed in three groups (Ctrl, mild and severe groups)..." my alarm starts ringing. Very often (but not always), something quantitative is measured (e.g. beta-amyloid, reaction time, number of correctly perfomred tasks, etc) to get a "diagnosis" which is then broken down by often arbitraryly defined cut-offs into mild/moderate/severs and whatever. This might be reasonable for the clinical physisian but definitively not for research work. But that is not the point of your question...

But there is another point, before I come to your question: Quantification by WB densitometry is very often lacking appropriate controls. You should make sure that changes in the protein concentration (in the relevant range) can be and will be detected. You need an external standard for this. This is certainly laborious, but not doing this, the whole procedure is pretty much based on hopes.

Ok, now to your question: protein concentration is gamma- or log-normal distributed. So the log-treansformed values should be approximately normal distributed - and this is exactly what you see. Testing hypotheses about differences in log concentrations (-> ANOVA, t-test, regression) is ok.

Jos Feys , normalizig the band intensity values of the protein of interest from a WB to that of a loading control is common practice in the field. Each protein preparation has a different yield. Measureing a protein that is assumed to have a constant concentration in all samples (in the living cells) allows to eliminate most of all these differences between samples by normalization. Additionally there are differences between every blot/membrane (it very sensitively depends on many things), and these effects may also be protein/antibody specific. Hence, comparing data from different membranes requires additional normalization to a standard (that is measured on each membrane).

Linear regression only makes a normality assumption with regard to the residuals (errors). I would take a look at residual distribution plots (e.g., histograms) and residual scatter plots and see what they look like. Unless the residual plots reveal strong non-normality or other issues (e.g., non-linearity), you may be fine simply running conventional OLS linear regression and computing the corresponding R-squared.

If there is a need to transform the data to a normal distribution, then any transformation that does that would be acceptable. The one I proposed is very general, but certainly not the only option.

Hello Mariyana,

I'm not quite clear on what you have. Is it: (a) two groups, one of which received an intervention and one of which did not, and each group was subsequently measured on some relevant variable; or (b) two groups, each measured pre-intervention and post-intervention on some relevant measure? (or something else?)

If (a), then assuming groups were in fact comparable prior to the intervention, the fact that the intervention group has a slightly higher SD than the non-intervention group simply means the impact of the intervention wasn't equal for all participants in the intervention condition. Of course, if groups weren't comparable at the outset, then you simply can't be sure this wasn't merely an a priori difference between groups/batches.

If (b), pretty much the same thing: post-intervention scores are more variable than pre-intervention scores, which implies some persons benefited more from the intervention than did others.

It's not apparent why this might be a study limitation, unless you're worried about a homogeneity of variance assumption for a significance test (as in option "a", above). If that's the case, then you can:

1. Opt for bootstrap/resampling methods to estimate group differences (and the CI for same);

2. Apply an exact/permutation test;

3. Use robust regression model;

4. If it makes sense to do so, try transforming scores on the measure of interest so as to stabilize variances;

5. Use one of the common methods for adjusting/penalizing t-test/anova models for violation of homogeneity, such as: Welch method; Brown-Forsythe method.

Good luck with your work.

The paired t test makes no assumption about the distribution of the two samples, just (as you suspect) their differences. These are assumed to be sampled from a normal distribution. (This is also equivalent to assuming normality of residuals/errors in a regression model).

In any case I see little value in a Shapiro-Wilk test (or similar) for normality in testing assumptions of tests. The issue is statistical power. In large samples the tests detect minor violations of assumptions that are not likely to impact the interpretation. In small samples they fail to detect major violations. Just use graphical methods or descriptive statistics.

If you have trouble interpreting these then simulation can be useful learning tool. For a variable with mean m and SD s simulate draws from a perfection normal distribution and plot the shape (or calculate descriptives) with the same n. Repeat it a few times. This can give you a feel for how "normal" data should look if it meets the assumptions.

I don't see why not. If you're meeting the assumptions, then I would expect that your asymptotic/theoretical ML standard errors are similar to the robust/bootstrap standard errors (you should verify this for the unstandardized solution--simply compare the ML and bootstrap standard errors and p values).

It is unlikely that you are "perfectly" meeting the multivariate normality assumption anyway. The fact that the normality hypothesis is not rejected in your sample does not mean it is true. It could just be a sign of "insufficient" power.

I want to emphasize Daniel Wright's comment about assumptions applying to the populations from which you sampled. Textbooks often present the F-test for one-way ANOVA as if it is an exact test. But in order for it to be an exact test, you would need to have random samples from k populations that are perfectly normally distributed with exactly equal variances. (In addition to that, each observation would have to be perfectly independent of all other observations.) Even if it was possible to meet those conditions (which it is not if you are working with real data), the samples would not be perfectly normal, and would not have exactly equal sample variances.

Because it is not possible to meet the conditions described above (at least when you are working with real data, not simulated data), the F-test for ANOVA is really an approximate test. And when you are using an approximate test, the real question is whether the approximation is good enough to be useful.* That's how I see it. YMMV. ;-)

* Yes, I am borrowing "useful" from George Box's famous statement(s) about all models being wrong, but some being "useful". Several variations on that statement can be found here:

https://en.wikiquote.org/wiki/George_E._P._Box

It depends on what statistical analyses/estimators you are planning to use. For example, for conventional CFA/SEM with maximum likelihood (ML) estimation (assuming continuous variables), extreme skew could be a problem because ML estimation rests on the assumption of multivariate normality of the variables. Although robust ML estimators are available, skewness could still be a problem because the inter-item covariances/correlations can be affected/biased by skewness (and, as a consequence, also the factor structure). In that situation, it may be useful to try out alternative estimation methods (e.g., WLSMV estimation using polychoric correlations vs. ML estimation) and conduct sensitivity analyses (i.e., checking how much the results differ across different estimation methods).

I recommend studying the extensive literature on non-normality and use of categorical (ordinal) variables in CFA/SEM. A Google Scholar search using terms like "ordinal items CFA", "non-normal data CFA" and the like will point you to the relevant papers.

Dewan -

I have not really worked with panels, but in general, the estimated variance of the prediction error is made up of a part for the model (coefficients), and a part for the estimated residuals.  (The former also involves the latter.)  The former becomes smaller with larger sample sizes, and the central limit theorem applies to estimating the coefficients.  The difference between the former and latter parts is a difference between standard errors and a standard deviation.  If the latter is a big part of your estimated variance of the prediction error, then your prediction interval will not be in a t or normal shape, and I suppose rather skewed, which may make it harder to interpret. 

Not having close to normally distributed residuals then impacts your prediction interval.  I think it also may make it not the best of all predictions, but I'm not so familiar with that. 

Cheers - Jim

Yes, it is possible to estimate Spearman's correlation (rank correlation). It gives you the monotone association (rather than the linear association) of the two variables.

But I don't see that septal wall thickness is not approximately normal:

with the group that is not normally distributed, is it close to being normally distributed or far from being normally distributed

I think the dispositive questions are, How different are the distributions of the two groups ? What hypothesis do you want to test ? ... If the distributions of the two groups are quite different, you might ask if comparing the means is the most meaningful approach for your situation.

Not necessarily!

How did you determine the index is normally distributed?

consider the discussion here:

Good luck,

Hamid

Hello again Brian,

Should you opt for transforming one or more variables, I'd suggest reporting summary statistics for both original & transformed values (especially if DV was transformed). Do include correlations (transformed values) as part of summary stats.

However, for interpreting estimated regression coefficients, your discussion needs to make clear that you're talking about predicted score differences in the transformed metric(s), not the original.

Good luck finishing your thesis.

There are several confusions and misconceptions already in the question, so let me start with a small refresher to make these things clear:

The t-test tests a mean difference. It assumes that the values are sampled from a normal distribution. In Student's version, the standard error of the mean difference is estimated by the pooled variance of the observations in both groups. Therefore, there is the additional assumption that the population variances in both groups are identical. The Welch t-test claculates the standard error of the mean difference from the two individual sample standard deviations and uses a t-distriobution modified degrees of freedom. It therefore relaxes the assumption of equal variances. It has been shown that the t-test is quite robust against inhomogeneous variance, particularily when the sample sized are similar. It is also very robust against non-normality, as long as the distributions are more or less symmetric. The robustness increases with increasing sample size (because it's the distribution of the mean difference that must eventually be normal distributed, and the central limit theorem sais that the distribution of the mean difference approximates the normal distribution increasingly well with increasing sample size). Skewed distributions are problematic as the strongly decrease the power of the t-test. The test still faithfully tests the mean difference, but it needs rather large sample sizes to "detect" differences (give small p-values).

The Mann-Whithey U-test tests stochastic equivalence and makes no assumptions on the distributions from which the values are sampled. Note that the tested hypothesis is not the same as that of the t-test. Only when additional assumptions are imposed on the distributions, the Mann-Whithey U-test can effectively test a mean difference, e.g. the distributions differ only by a location shift, what means, particularily, that they have the same variance! This is quite a strong assumption and I have no practical example where this might be justified (happy if someone could point out some examples, though!). Also note that, for instance, two distributions A and B may be constructed so that the mean difference A-B is positive and at the same time Pr(A>B) < 0.5. If the U-test is used as a "substitute" for the t-test, it would lead to the opposite interpretation ("B is kind of larger than A"; although the data suffice to see that the expectation B is smaller than the expectation of B).

So here are the confusions:

- The t-tests are quite robust against violations of their assumptions.

- The robustness depends the actual distributions and on the sample size.

- Violations do not invalidate the test, but limit its power.

- The U-test is generally NOT testing a mean difference. It's therefore generally not a "substitute" for the t-test.

and further:

- The assumptions are about the distributions of the variables (population distributions, if you like). But what you see is only a sample. A sample can be misleading (a sample from a symmetric distribution may be skewed and vice versa), and a sampel may be too small to reveal any reliable information about the distribution if was sampled from. This means that it's hard to justify distributional assumptions by looking at sample data.

Finally,

- No real variable has a distribution that would be described EXACTLY (fully correctly) by a distribution model (be it the normal distribution or some other distribution). Hence, one can say that your variables are never normal distributed. But this is not the point. The point is if the distributions can reasonabley well be described by a normal distribution. And the meaning of "reasonably" depends also on the sample size.

Now whe you write "The results from the 1 group are not normally distributed" implies that you measure two different variables in the two groups, and with this in mind your task ist like comparing apples and peaches. It makes no sense.

Practically: how did you come to the conclusion that one variable is normal distributed and the other is not? It's likely that this procedure is already logically flawed. I presume that the variables are not normal distributed, and that the difference in the sample distributions to a normal distribution is just a bit more obvious in one sample. This is often the case when the variables is strictly positive and accordingly has a skewed distribution with a particular mean-variance relationship and that the mean of one sample is smaller.

The description "not normal" isn't very useful. ANOVA can be somewhat robust against deviations from normality, but if the underlying distributions of the group populations are very skewed or distinctively different, the results from ANOVA may not be very reliable.

I recommend thinking of ANOVA as a form of general linear model. See slide 8 here: https://staff.emu.edu.tr/adhammackieh/Documents/courses/ieng581/lecture-notes/ch03.pdf . You'll see it's the errors from the model that are assumed to be normally distributed. From this perspective, you can fit the model and then examine the residuals with a histogram or quantile-quantile plot.

Also with ANOVA, heteroscedasticity may be more of an issue. I like to plot the residuals vs. the predicted values to examine this. There are ways to address heteroscedasticity. In the one-way case or t-test, you can use Welch's anova or t-test.

I will suggest reading

In general, I would not expect such a large drop in reliability for removing 7 out of 355 observations. So, compare the correlation matrices from the full data set and the reduced data set.

Aligned ranks transformation anova may work for your situation. Maybe also ordinal regression or quantile regression. Or an appropriate generalized linear model.

If I understood correctly, you have two groups, and for each group you have pre and post data. In this case, the distribution of pre- or post-data is irrelevant. Only the distribution of post minus pre is relevant, and here you should also separate the two groups.

PS: I don't know how you came to the conclusion that some scores are normal distributed and others are not. Often, normal distribution tests are used in this context - what is rather stupid. Such test will sometimes allow you to reject the tested hypothesis (the sample is from a normal distributed RV), sometimers it won't. If it does, you don't know if the difference to the normal distribution is of any relevance, and if if does not, you don't know if there are relevant differences but the sample size is too small to recognize this. It's better to make an informed judgement which hypotheses make sense and which are clearly nonsensical.

Laxman Singh Bisht Thank you for the reply!

Since my database included negative numbers, I added the absolute of the most negative number, which is -4.5, to all the values. In that case, my values got 0 and positive. And then I did Box-cox transformation on them. After I checked the normality with Shapiro-Wilk Test, ERS variable got normally distributed.

However, I am not sure whether it is valid to play with the data like that since we want to publish an article with ERS data??

Jochen Wilhelm Thank you for the reply!

Actually, my Professor asked me to find a way to make the current data more normally distributed to do regression analysis in twin data. And, since there are negative values in the data, I believe it is not possible to transform the data. That's why I am looking for a way to apply a suitable transformation.

A few comments. First, it doesn't matter whether each individual mouse's data is normally distributed, but whether the residuals from the entire group (handled, nonhandled) approximate a normal distribution. Second, the issue is not whether the data are normally distributed, but whether the residuals are (roughly) normally distributed. I recommend generating Q-Q plots of the residuals to inspect this assumption if you have not already done so. Third, duration (i.e., time) data are rarely normally distributed, so tests based on a normally-distributed model may not be appropriate. In most cases, time data are best modeled by a gamma distribution. Fourth, before transforming the data, you need to consider whether you're interested in making absolute time comparisons between the groups or only relative comparisons. If you're interested in absolute time differences, do NOT transform the data. If you're only interested in making relative comparisons, transforming the data should not be a problem.

Jim McLean has provided some very good advice, and David Morgan has raised a good point about individual items vs. summated scales... You might also consider a resampling/bootstrapping approach for "back up."

It is widely believed, but unfortunately untrue, that we must check for normality before we do a statistical text. In fact it is the errors that must be normal. So we run the ANOVA to compare means and look at the residuals to see if they are normal. Looking at your data, the residuals may well be normal and homogeneous, even though the response variable (all 6 values) are not.

With only 6 values, the ANOVA will have little power to detect even large differences, where the means for your data differ little. The result "not significant" will be due either to no real difference or to a test with little power to detect a difference. Another way to say this is that you will not be able to reject the null of no difference for either of two reasons -- because there is none real difference or because the sample size was too small to detect even a large difference.

A non-parametric test such as Kruskal Wallis is of little help. It does not test differences in means. It gives a p-value that is not about the difference in means. So you can't use it to reject a null hypothesis of no difference between means.

I would recommend that you consider a larger sample size.

Good luck with your research,

David Schneider

I'm not sure I understand why you transformed Y. Perhaps I'm missing something, but leaving Y untransformed and simply including the quadratic term X^2 should give you what you are looking for (an inverted u-shaped relationship). If the residual non-normality is due to the quadratic effect, then including X^2 should account for this.

Look at the severity of the violation. It may be just negligible given the large sample size.

The assumption should be sensible and resonable. Judging if this is the case requires to understand the characteristic of the variable you are measuring. If you don't understand your variable, you possibly should not test hypotheses at all, as you will not be able to interpret the tested hypotheses anyway. If you have some justifyable opinion that the variable should have an approximate normal distribution but you are not sure if this is sufficiently correct, the assumtion is quite save or conservative. The problem may only be that the data remin unconclusive, what is your personal problem but not a scientific one.

That brings me to using non-parametric tests: usually, these tests are about a different hypothesis, and again you should be clear about which hypothesis you actually want to test. It is not useful to test some weird or misunderstood hypothesis just because one knows how to test it. There should be a scientific rationale behind your choice (a feasability argument is not a scientific rationale). Despite this, your tiny sample size prohibits non-parametric analyses anyway. There is not enough information in the data to reliably estimate the sampling distribution, and non-pareametric tests will have almost no power (again: power to test a hypothesis that may not be the one you actually like to test!).

Please remember that most people that do normality tests get it wrong. For these tests large p-values favor the normal hypothesis not small p-values. See Wikipedia for your test. Best wishes David Booth

"In case of coefficient of variation we measure standard deviation (non zero value) with something what can be zero."

No, this is not true. The CV is not a useful measure in this case and no sensible person would use it in such instances.

The CV is helpful in cases where you measure something positive (length, mass, intensity, concentration, activity etc) with different methods/principles so that it is reported in different units. In such cases, the CV serves as a measure for the relative precision that can be compared between the methods/principles.

Try Aligned Rank Transform (ART) method via the R

package ARTool proposed by Wobbrock et al. (2011) to prepare data for a non-parametric

ANOVA. There is also a non-R based version as well.

Wobbrock, J. O., Findlater, L., Gergle, D., & Higgins, J. J. (2011, May). The aligned rank

transform for nonparametric factorial analyses using only anova procedures.

In Proceedings of the SIGCHI conference on human factors in computing systems (pp.

143-146).

The first thing you save to decide is if you are okay with treating the Perception scale responses as a continuous variable or would rather treat it as an ordinal variable.

a) If you have values from 3 to 15 (13 unique values), it might be okay to treat the response as a quasi-continuous variable. In that case, you might try a regular Ordinary Least Squares (OLS) general linear model. This is the same as anova or multiple regression, except that you can include whatever terms and interactions you want on the right hand side. ... This model has assumptions, that you should understand. In particular for the normality assumption, understand that the model assumes a response variable that represents a population that is conditionally normal, not that the data itself is normal. One way to assess this is to fit the model and then look at the residuals from the model.

a2) A non-parametric approach like aligned ranks transformation anova may work for what you need.

b) If you want to treat your response variable as ordinal in nature, ordinal regression will likely work. This is a good idea especially if you have maybe 10 or fewer unique response values. (Like if all responses were, say, 9, 10, 11, 12, 13, 14, or 15).

Beyond the scarcity of information, are you sure of the relationship between variables?

In general, the t-Test (and equivalently ANOVA) are relatively robust with regard to non-normality, especially with large sample sizes. So, if you have a small sample, I would use non-parametric tests, but otherwise the data as linear.

Games-Howell test is the sound choice since it is designed for situations in which the variances are unequal. Concerning the effect size calculation, you could consider free online effect size calculators such as the one referenced below.

Uanhoro, J. O. (2017). Effect Size Calculators. https://effect-size-calculator.herokuapp.com/

Good luck,

Looking for the association among three categorical variables is generally done with log-linear modeling, though you seem to want to predict species from the other two variables. There are methods around for plotting these associations, but one of the more straight-forward would be to combine what you are calling your explanatory variables into one with 12 categories and do a biplot from a correspondence analysis. Details will be covered in most categorical data analysis books and some general intro to stats books.

With almost certainty: it is not normal (unless it is a synthetic variable, and even then it ay be doubtful, depending on the RNG used).

Note 1: the test cannot give any indication or evidence for the assumption that the tested distribution is normal. It only checks if the given sample size is already sufficient to reject this hypothesis. Failing to reject only means that the sample size is too small.

Note 2: I bet that answering the question "are my data sampled from a normal distribution?" is not what you need. I guess that the correctly formulated question is rather: "are my data sampled from a distribution that is sufficiently similar to a normal distribution to warrant conclusions from analysis methods assuming that the data are samples from a normal distribution?". This is a considerably different question, and this cannot be answered by "Normality test" like Shapiro-Wilk etc.

For highest returns, I would recommend the General Extreme Value distribution. Our analysis in the Turkish Stock Exchange by using this distribution showed, however, that highest returns follow the Frechet distribution.

Hello Wim,

I think your scenario doesn't match well to the situation for which Fisher initially developed the z-transformation (r-to-z). The issue with the distribution of the Pearson correlation (r) is that it does not have a constant variance and the r values don't behave linearly very well (unlike r-squared). So, the z-transform is principally a variance-stabilizing measure, which does tend to yield a z-variate that conforms better to the normal distribution than does r.

But, you're presuming that the variate to be transformed has an underlying normal distribution; that's unlike the r-to-z situation. As well, I'm guessing that your intention is to convert some proportion of area (of non-overlap) to a z-score or some other standard score (not the z-transformation, above), which is also presuming a normal distribution for whatever attribute is under investigation, which may not be the case (and frequently is not in practice). Why not stick with proportions as your explanatory ES, as in:

1. CLES of McGraw & Wong: Proportion of times randomly selected case from batch A has score that exceeds that of randomly selected case from batch B?

2. Cliff's dominance statistic: CLES - proportion of times that B cases exceed A cases?

3. Vargha & Delaney's A statistric (similar to Cliff's d, though it also splits tied cases and assigns half to one batch and half to the other).

Each of these is exact. There is a pretty good closed form expression for the variance of d, if this is important.

Good luck with your work.

Thank you!

Hi, Chukwuma Ofor ,

Please check this article on factors versus composite model estimation:

Here is a quick overview:

And here is a results comparison:

To follow on from comment by Salvatore S. Mangiafico .you might think about what difference this step makes in the eventual results.

In this small part of the procedure, you have a test statistic consisting of a ratio, where the divisor is an estimated standard deviation, and the two versions you mention either do or do not take into account the fact that you have an estimated standard deviation.

(i) there is a possible variant where you replace the estimated standard deviation (estimated locally in the sequence) by something obtained more broadly, and so less subject to error, and for which the normal approximation is better.

(ii) the cases where the test statistic might be most misleading arise when the estimated standard deviation is either unusually small or unusually large. So you might want to look at whether the actual pattern of data in such cases really do justify being counted as turning points. Perhaps the short section of the series is too smooth or too rough to support a conclusion.

(iii) you might consider an alternative way of assessing this, or any other test statistic, (getting a "p-value") which could be via a permutation test of some sort.

But, for the two versions you mention, if the later arguments of the overall procedure really only choose to look at smallest apparent p-values without using a formal threshold (such as 0.05), then there may be no contradiction between the lists of points selected as turning points (except perhaps in the numbers in the lists).

You can look at the conditional distribution of the data. That is, run the model, and then look at the residuals from the analysis. ... There may be other appropriate approaches like generalized linear model or a nonparametric approach like aligned ranks anova.

Dr. Wilhelm,

Thanks for your quick reply.

I am not familiar with math or statistics, so your comments are very helpful. Thank you very much for your valuable information.

Just go for normal. The means and standard errors will be close.

Normal distribution means that the collected data is distributed like a bell shap. Begin with the low measured values and the number increased till it reach a peak value and then reduced. The normal distribution curve is a symmetrical curve. The data can be tested using Shapero test. If the test is not significant p>0.05 then the data is normally distributed, while if p<0.05 or the test is significant, the distribution is abnormal.

Christian and Ali, thanks for your posts. Appreciated. I'll follow up on both of them.

Robert