Dear Prakash, dear all,

I see a lot of times a misunderstanding revolving around the "normality of distribution" to say so.

So, please keep reading this, hopefully it will save you time and make you more aware of what you are doing when you perform a regression or an ANOVA.

Generally, when one asks this question, it is because (s)he is applying some kind of regression method (of which the most common is the ANOVA, when the independent variable(s) are nominal or ordinal) to see which independent variables (or interactive combination of these) explain variance of the dependent variable.

The usual mistake is to think that the distribution of the dependent variable per se (i.e., its marginal distribution) ought to be normal (or close enough to a normal distribution). This in NOT the case!

What ought to have a distribution close enough to a normal distribution is the distribution of the dependent variable CONDITIONAL TO EACH LEVEL OF THE INDEPENDENT VARIABLE.This is a very different thing.

Here is a very simple example to make myself understood, a case most people would analyse through an One-Way ANOVA.

Imagine you want to compare reading scores in kids aged 9 who have learned to read with 3 different methods.

Dependent variable : reading score, numeric

Independent variable : reading methods (M1, M2, M3), a nominal variable

Comparing distributions without making any assumption about what kind of distribution one is comparing is virtually impossible. The group model (another name for the ANOVA) circumvents this HUGE problem by making 3 key assumptions (plus a forth shallow one):

ASSUMPTION 1: The distribution of the scores IN EACH GROUP is close enough to the normal distribution so that one can reasonably assume a normal distribution of the scores IN EACH GROUP

ASSUMPTION 2: The variability of the scores around the group mean is not different between the 3 groups, i.e., the variance is "the same" in all groups that are to be compared.

These 2 assumptions are crucial because they change the problem from one of comparing distributions of the type one knows nothing about to that of comparing distributions that are all normal distributions (assumption 1). A normal distribution has 2 parameters, its variance (or the squared root of that, standard deviation) and its mean. Because of the 2nd assumption, the variance is considered the same in all groups, so the distributions of the scores are the same in all respects in all 3 groups (I mean they have the same shape), except for the mean, which is allowed to be different (i.e., not all distributions are placed on top of each other, they may be shifted away one from another).

This is how from a problem consisting in comparing distributions of the type one knows nothing about one gets to a problem consisting in comparing means (of the 3 groups). Please take a moment to digest / to think about this.

ASSUMPTION 3: The scores in each group are independent one from another. This is an assumption that has to be made in order to have a tractable computation of the F statistic that is used in the ANOVA process in order to get the probability of having the data one has if the null hypothesis is true, also known as the p-value. Beware of the violation of this assumption! (more on that bellow)

ASSUMPTION 4: The computation of the F statistic make the assumption that the statistic is computed under the null hypothesis. This is no biggie, since one always starts under the null hypothesis, computes a statistic and the p-value associated with the statistic gives the probability of having the data one has if the null hypothesis is true (i.e., is the p-value is lower than 0.05, it is considered unreasonable to make the assumption the null hypothesis is correct, so the null hypothesis has to be rejected, and so the alternative hypothesis, the one that supposes there is a difference between the means, is the correct one).

If any of the assumptions above are violated, then whatever result the ANOVA gives you is worthless.

In other words, when you run an ANOVA you must make sure the first three assumptions are not violated, otherwise the results the program spits as an output are totally worthless (because you do not have the means to know whether they are real or falsely obtained) and you should not trust them!!!!!!!

So, how you ensure the 3 key assumptions are met:

Test of Assumption 1: A test such as the Shapiro-Wilk is to be performed (on the model corresponding to the alternative hypothesis under R, maybe differently so in SPSS I couldn't tell). IF you are not sure how the program you are using does the things, be sure to test the normality of the distribution of the scores FOR EACH GROUP (re-read above Assumption 1 if you don't remember why).

Test of Assumption 2: The variability of the scores around the group mean should not be different between the 3 groups ; one makes sure of that by comparing the variances in the 3 groups by a test (e.g., Levene's test of equality of variances), and because this condition required that the variances be NOT different, one is happy (i.e., can carry on with the ANOVA) if the test yields a p-value HIGHER that 0.05 (or higher than 0.10 if you want to be really sure). If Levene's test result yields a statistic (i.e., a number) to which is associated a p-value of less than 0.05, whatever result the ANOVA gives you is worthless.

Test of Assumption 3: There is no formal test for this, the idea is that the reading score of each kid must be independent from the reading score of all the other kids. Most notably, this means no mass testing (but independent testing), not including kids from the same class, the same school, or anything that creates a bulk/grouping effect.

OK, now back to something more related to your question.

Imagine the 3 reading methods have all different efficiencies, so that the mean of group M1 < the mean of group M2 < the mean of group M3.

Because of the assumptions we made, this means you have to imagine (or better yet, draw this on a sheet of paper) 3 normal distributions of the same shape (the standard deviation is the same for the 3 groups, and standard deviation is the shape-parameter of the normal law), with distribution of M1-group scores at the left, distribution of M2-group scores shifted a little bit to the right (but overlapping quite a bit with the previous distribution), and finally distribution of M3-group scores shifted even more to to the right (but still overlapping quite a bit with M2-group score distribution, and a bit less with M1-group score distribution). All in all, if you look at all these distributions at the same time (i.e., irrespective of the group) the shape is one of a 3-hump-and-2-depression form, clearly not a normal distribution. This is the shape of the distribution of the scores irrespective of the group they come from, and if one understands the logic of the ANOVA (and that of its two first assumptions), then one understands that there is no assumption to be made on this marginal distribution of the dependent variable. The assumptions concern the distributions (in plural!) conditional on the levels of the independent variable (i.e., here, conditional on the groups).

The same assumptions are to be made in regression (ANOVA being just a special case of regression), i.e, when the independent variable is numeric: for each value of the independent variable the distribution of the dependent variable must be normal.

HTH.

Cheers,

SCM

look up the KS test

this is the best direct way to test how normally your data is distributed. Simply generate the cumulative frequency distribution for a normal distribution and compare that your data.

A simple way to do this is to use the KS Stat calculator at

http://www.physics.csbsju.edu/stats/KS-test.n.plot_form.html

SPSS does this in either the Explore or the Descriptives command (I haven't used it in a while). You have to check the appropriate box to specify it. It actually gives values to indicate skewness and (I think) kurtosis of the distribution of your data.

moment of inertia analysis is not as good of indicator of a normal distribution

unless you have large N. For small N, the KS test is more reliable to check for normality.

The following tests can be used:

Kolmogorov–Smirnov test;

Cramér–von Mises criterion;

Anderson–Darling test;

Shapiro–Wilk test;

Chi Square test;

Akaike information criterion;

Hosmer–Lemeshow test;

Kolmogorov–Smirnov test and Chi Square test are more commonly used and are included in softwares.

Kurtosis is a simple measure for this purpose.

In matlab

chi2gof

Chi-square goodness-of-fit test

h = chi2gof(x)

[h,p] = chi2gof(...)

[h,p,stats] = chi2gof(...)

[...] = chi2gof(X,'Name',value)

Description

h = chi2gof(x) performs a chi-square goodness-of-fit test of the default null hypothesis that the data in vector x are a random sample from a normal distribution with mean and variance estimated from x, against the alternative that the data are not normally distributed with the estimated mean and variance. The result h is 1 if the null hypothesis can be rejected at the 5% significance level. The result h is 0 if the null hypothesis cannot be rejected at the 5% significance level.

One-sample Kolmogorov-Smirnov test in MATLAB

kstest

One-sample Kolmogorov-Smirnov test

Syntax

h = kstest(x)

h = kstest(x,CDF)

h = kstest(x,CDF,alpha)

h = kstest(x,CDF,alpha,type)

[h,p,ksstat,cv] = kstest(...)

Description

h = kstest(x) performs a Kolmogorov-Smirnov test to compare the values in the data vector x to a standard normal distribution. The null hypothesis is that x has a standard normal distribution. The alternative hypothesis is that x does not have that distribution. The result h is 1 if the test rejects the null hypothesis at the 5% significance level, 0 otherwise

It's always worth asking - so what? What difference does it make if your data is or is not normal by some statistical test? How will you know if it is normal enough? If you apply any of the tests given above (I prefer the KS test and the D-value it returns can be useful) and get a number, what will you do?

If you plot it, does it look normal? If it looks normal then you may treat the data as normal in any analysis.

Another way to look at this is to perform the following experiment - draw 25 or 100 or 200 or 1000 samples from a standard normal distribution (so the data is by definition normally distributed) and use any of the tests above to compare it to the normal distribution. As the number of samples in your dataset rises the frequency with which the data passes the normality test will decline. Does this mean that the data, which was drawn from the standard normal distribution, is not normally distributed? No, it just means the tests cannot be applied without thought.

Dear Prakash, dear all,

I see a lot of times a misunderstanding revolving around the "normality of distribution" to say so.

So, please keep reading this, hopefully it will save you time and make you more aware of what you are doing when you perform a regression or an ANOVA.

Generally, when one asks this question, it is because (s)he is applying some kind of regression method (of which the most common is the ANOVA, when the independent variable(s) are nominal or ordinal) to see which independent variables (or interactive combination of these) explain variance of the dependent variable.

The usual mistake is to think that the distribution of the dependent variable per se (i.e., its marginal distribution) ought to be normal (or close enough to a normal distribution). This in NOT the case!

What ought to have a distribution close enough to a normal distribution is the distribution of the dependent variable CONDITIONAL TO EACH LEVEL OF THE INDEPENDENT VARIABLE.This is a very different thing.

Here is a very simple example to make myself understood, a case most people would analyse through an One-Way ANOVA.

Imagine you want to compare reading scores in kids aged 9 who have learned to read with 3 different methods.

Dependent variable : reading score, numeric

Independent variable : reading methods (M1, M2, M3), a nominal variable

Comparing distributions without making any assumption about what kind of distribution one is comparing is virtually impossible. The group model (another name for the ANOVA) circumvents this HUGE problem by making 3 key assumptions (plus a forth shallow one):

ASSUMPTION 1: The distribution of the scores IN EACH GROUP is close enough to the normal distribution so that one can reasonably assume a normal distribution of the scores IN EACH GROUP

ASSUMPTION 2: The variability of the scores around the group mean is not different between the 3 groups, i.e., the variance is "the same" in all groups that are to be compared.

These 2 assumptions are crucial because they change the problem from one of comparing distributions of the type one knows nothing about to that of comparing distributions that are all normal distributions (assumption 1). A normal distribution has 2 parameters, its variance (or the squared root of that, standard deviation) and its mean. Because of the 2nd assumption, the variance is considered the same in all groups, so the distributions of the scores are the same in all respects in all 3 groups (I mean they have the same shape), except for the mean, which is allowed to be different (i.e., not all distributions are placed on top of each other, they may be shifted away one from another).

This is how from a problem consisting in comparing distributions of the type one knows nothing about one gets to a problem consisting in comparing means (of the 3 groups). Please take a moment to digest / to think about this.

ASSUMPTION 3: The scores in each group are independent one from another. This is an assumption that has to be made in order to have a tractable computation of the F statistic that is used in the ANOVA process in order to get the probability of having the data one has if the null hypothesis is true, also known as the p-value. Beware of the violation of this assumption! (more on that bellow)

ASSUMPTION 4: The computation of the F statistic make the assumption that the statistic is computed under the null hypothesis. This is no biggie, since one always starts under the null hypothesis, computes a statistic and the p-value associated with the statistic gives the probability of having the data one has if the null hypothesis is true (i.e., is the p-value is lower than 0.05, it is considered unreasonable to make the assumption the null hypothesis is correct, so the null hypothesis has to be rejected, and so the alternative hypothesis, the one that supposes there is a difference between the means, is the correct one).

If any of the assumptions above are violated, then whatever result the ANOVA gives you is worthless.

In other words, when you run an ANOVA you must make sure the first three assumptions are not violated, otherwise the results the program spits as an output are totally worthless (because you do not have the means to know whether they are real or falsely obtained) and you should not trust them!!!!!!!

So, how you ensure the 3 key assumptions are met:

Test of Assumption 1: A test such as the Shapiro-Wilk is to be performed (on the model corresponding to the alternative hypothesis under R, maybe differently so in SPSS I couldn't tell). IF you are not sure how the program you are using does the things, be sure to test the normality of the distribution of the scores FOR EACH GROUP (re-read above Assumption 1 if you don't remember why).

Test of Assumption 2: The variability of the scores around the group mean should not be different between the 3 groups ; one makes sure of that by comparing the variances in the 3 groups by a test (e.g., Levene's test of equality of variances), and because this condition required that the variances be NOT different, one is happy (i.e., can carry on with the ANOVA) if the test yields a p-value HIGHER that 0.05 (or higher than 0.10 if you want to be really sure). If Levene's test result yields a statistic (i.e., a number) to which is associated a p-value of less than 0.05, whatever result the ANOVA gives you is worthless.

Test of Assumption 3: There is no formal test for this, the idea is that the reading score of each kid must be independent from the reading score of all the other kids. Most notably, this means no mass testing (but independent testing), not including kids from the same class, the same school, or anything that creates a bulk/grouping effect.

OK, now back to something more related to your question.

Imagine the 3 reading methods have all different efficiencies, so that the mean of group M1 < the mean of group M2 < the mean of group M3.

Because of the assumptions we made, this means you have to imagine (or better yet, draw this on a sheet of paper) 3 normal distributions of the same shape (the standard deviation is the same for the 3 groups, and standard deviation is the shape-parameter of the normal law), with distribution of M1-group scores at the left, distribution of M2-group scores shifted a little bit to the right (but overlapping quite a bit with the previous distribution), and finally distribution of M3-group scores shifted even more to to the right (but still overlapping quite a bit with M2-group score distribution, and a bit less with M1-group score distribution). All in all, if you look at all these distributions at the same time (i.e., irrespective of the group) the shape is one of a 3-hump-and-2-depression form, clearly not a normal distribution. This is the shape of the distribution of the scores irrespective of the group they come from, and if one understands the logic of the ANOVA (and that of its two first assumptions), then one understands that there is no assumption to be made on this marginal distribution of the dependent variable. The assumptions concern the distributions (in plural!) conditional on the levels of the independent variable (i.e., here, conditional on the groups).

The same assumptions are to be made in regression (ANOVA being just a special case of regression), i.e, when the independent variable is numeric: for each value of the independent variable the distribution of the dependent variable must be normal.

HTH.

Cheers,

SCM

Thank you Prakash for voting up my answer.

Though I understand that is does not answer the actual question asked by Prakash (one about Matlab code), I think it gives some insight into the question at hand. And while I don't take it personally when my input is voted down, I prefer a sensible argument to a quick and mean downvote. So here is my message to the person who voted down my answer: Could you, dear sir or madam, be so kind and write down here why you don't agree with my answer? If your argument is sensible we may all learn something. And personally, I appreciate being proven wrong IF I am wrong.

Cheers,

SCM

Mauro,

The Shapiro-Wilk ought yo be performed on the residuals, and it is indeed performed on the residuals. However, the point of my (rather lengthy) post was to insist on the fact that the normality (of the residuals !) is supposed (and ought to be tested) on a model that supposes an effect, i.e., CONDITIONALLY on the levels (or values) of the independent variable (I speak to the simplest case, where there is only an independent variable).

HTH.

Cheers,

Serban

use the KS test - make up a model normal distribution and compare

the data to that.  This test is the most direct and powerful for determining p value for deviation from normality

http://homework.uoregon.edu/pub/class/355/ks.html