You are getting into a logical trap.

Either you have the assumption that the samples are paired, then the analysis of the two samples as if they were independent does not make sense. Or you don't have this assumption, then the question for paired-sample test makes no sense.

It is irrelevant for the paired-sample t-test how the distribution of the values in the individual groups look like. It is only relevant how the differences look like, and this is one group of differences. These differences will have some variance - there is nothing to what this could be compared.

To make it clearer: the distribution of the values in the groups is completely irrelevant for the paired-sampel t-test. Their distribution can be arbitrary, non-normal to any extend. The only assumption the paired test relies on is that the differences are normal distributed.

And again, very generally: hypothesis tests or significance tests about assumptions are nonsense, anyway.

   This is a fascinating question.  At its core is the robustness of the statistical test you are running, which is not a mathematical concept but a data analytic one.  That issue seems to be the driving difference between your two textbooks - Lehman is mathematical, MacDonald is about analyzing data. 

     The main issue is whether, by violating the assumption, you have substantially impacted your chances of a Type I or a Type II error.

    For the most part, t-test (both paired and independent samples) tends to be robust to violations of variance assumptions, particularly if you have large samples.  The smaller the sample, the more the assumptions matter. 

    If the statistical assessment of probability isn't clear, you can bolster your point by looking at the effect size.  A large difference is a large difference, and a small difference is a small difference, regardless of the likelihood of observing that difference by chance.  Increasingly we are being asked to discuss effect size rather than simply reporting on significance.

A paired-sample t-test is a one-sample t-test on the differences.

The test (the one-sample t-test) cannot "see" that the values are differences, it cannot "see" how these values are calculated at all. The only information the test gets is a single sample of values. The only thing that matters is the fact that the test is based on the normal probability model for the values.

So I agree with McDonald.

However, if the groups don't have similar variances it is likely (but not necessary!) that the differences will not be normal distributed.

Yes, it is important assumption. Though the test is robust. Julia's view is more important. 

Thanks.

I agree with Julia, and  Jochen. I Also supports McDonald. OK Rajiv ! just to think deeply, If it is an assumption, how is it tested using SPSS, and What is the formula to test it? Is the Levene's test (for independent variances)  appropriate for correlated variances?

a) Assumptions are not calculated. Assumptions are given as foundations to build the analysis on it.

b) Testing assumptions is logically non-sense.*

c) What are "correlatedd variances"?

---

* For instance, there is nothing like "normal distributed data" in reality, the assumption is a mental construct, an abstraction. It is illogical to test this formally to "reject the null" (that the distribution is normal). It is known a priori that the null is false, non-rejections are known false-negatives. The important question is if the deviation of the empirical findings from the theoretical concept is *relevant* in the context and the aims of the analysis.

Yes I agree, and I modified it, assumption is tested not calculated, My question is: Can we use the Levene's test to test the assumption of homogeneity of variance in paired samples t-test ? Some argue that Levene's test is only appropriate for independent samples.

You are getting into a logical trap.

Either you have the assumption that the samples are paired, then the analysis of the two samples as if they were independent does not make sense. Or you don't have this assumption, then the question for paired-sample test makes no sense.

It is irrelevant for the paired-sample t-test how the distribution of the values in the individual groups look like. It is only relevant how the differences look like, and this is one group of differences. These differences will have some variance - there is nothing to what this could be compared.

To make it clearer: the distribution of the values in the groups is completely irrelevant for the paired-sampel t-test. Their distribution can be arbitrary, non-normal to any extend. The only assumption the paired test relies on is that the differences are normal distributed.

And again, very generally: hypothesis tests or significance tests about assumptions are nonsense, anyway.

First, let me say very clearly that I agree with Jochen's comments on testing the assumptions of another test. 

Having said that, see Karl Wuensch's document on Pitman's T (on the page linked below) for a test of homogeneity of variance of paired samples. 

HTH.

http://core.ecu.edu/psyc/wuenschk/StatHelp/StatHelp.htm

Hello Ahmad! Regarding your question, yes, the homogeneity of variance is an assumption when conducting a t-test. However, the testing of this assuption is an ongoing controversy. Some basic statistics textbooks addresses this assumption when conducting a t-test while other textbooks do not mention it. But it is an assumption when conducting the t-test. There is research done, however, regarding to whether testing the homogeneity of variance is needed. I did not find a way for attaching here a paper of a study we conducted for testing this assumption. Besides the results of our study, you might find helpful the resources cited. If you are interested, you could access the paper from my profile (sorry that I could not attach it here). The name of the paper is "Parametric Test for two population Means: An empirical comparison of Type I error and Statistical Power. I hope it is of help. Cheer!

 Dear Patricia,  I enjoyed reading your article,  but please keep in mind that I am asking about Paired t-test, while your article is about Independent Samples t-test.

Regards...

I'd also respond as Jochen : the paired-sample t-test is in fact a test on the difference D=X1-X2, with (most often) the null hypothesis that the true mean of the difference is null. Exactly the same test as when comparing the observed mean of a sample to some given norm. 

Then the assumption needed for using the Student probability law is that the sampled variable (D in this case) has a normal distribution. So I dont think that homogeneity is a requirement and I dont see how it could be : imagine that X2 is a multiple of X1, X2=k*X1. Then clearly the variances would be unequal, yet the difference should be normally distributed.

Yes, and this does not only apply for a multiplication but also for a transition, since any linear transform of a normal distributed variable is also normal distributed.

The fact that differences (and sums) of independent variables with normal distributions has a normal distribution is well known and follows from the central limit theorem. Interestingly, also the sums/differences of correlated (non-independent) normal distributed variables are also normal distributed, see

https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables#Correlated_random_variables

http://www.srabbani.com/bivariate.pdf

Hence, if the values of X1 and X2 are normal distributed, then the difference D = X2-X1 is normal distributed, too, no matter whether or not X1 and X2 have similar means and/or variances, and no matter whether or not they are correlated.

The intersting question is if the assumption of the normal distribution of D is resonable when X1 and/or X2 are not normal distributed. This clearly depends on the distributions of X1 and X2 and kind and strength of their correlation. In any case the distribution of D will be a better approximation of the normal distribution than that of X1 and X2 (if it is "good enough" is a different question!).

The t-test for paired samples does not assume that both variables have the same variance. Indeed, the model behind this test is that the difference of the two variables is Gaussian, hence only the variance of the difference is taken into account. See, for instance, the following textbook: Probabilités, Analyse des données et Statistique, Gilbert Saporta, 2ème édition, Editions Technip, Paris, 2006 (page 349).

@ Patricia: Thank you for that piece of information.

The theory is very simple. As the samples (X and Y) are paired you may define U = X - Y and you have one U-sample with a sample mean and a sample variance as good estimates of the mean X - Y difference and variance.

The most powerful test if U comes from a normal population is the one sample t-test for U values. Don't worry about X and Y variances (and covariance), the theory takes them into account in this case. Listen to Jochen. Work with U. even If U is not normal, the one sample t-test is very robust unless U has a distribution with heavy tails. 

I fully agree with the suggestion to focus on the difference D=X1-X2 and, based on the central limit theorem, D will converge to the normal distribution.

Please note, that the convergence to the normal distribution is faster for symmetric variables. Therefore, at least a visual inspection of the X1 and X2 distribution could help.

Focus only on the distribution of the difference. A paired sample t-test requires that the  difference column must be normally distributed. Good luck. 

I disagree with Elie Maza who said that the t-test for paired samples does not assume that both variables have the same variance. When anyone tries to analyze the data using SAS the output included test of Equality of Variances. According to this test you can identify which p value you must select. Moreover, another assumption is needed to make t-test for paired a valid test,  the variables normally distributed or not. In case the variables not normally distributed the alternative test is Mann-Whitney test. to illustrating this subject, I will give you this code in SAS

DATA Z;

INPUT Group TIME;

CARDS;

1          80

1          93

1          83

1          89

1          98

2          100

2          103

2          104

2          99

2          102

;

proc ttest data=z;

class group;

var time;

run;

Testing the normal distribution of a variable

data z;

input w;

cards;

50

34

44

53

48

46

44

39

PROC UNIVARIATE DATA = z normal plot;

HISTOGRAM w / NORMAL CFILL = ltGRE;

INSET N = 'Number of students' MEDIAN (8.2) MEAN (8.2)  STD (8.3) STDMEAN (8.2)

/ POSITION = ne;

RUN;

The paired t–test does not assume that observations within each group are normal, only that the differences are normal.  we can test the differences by zero .(one sample t-test). good luck.

To continue with what everyone has said and try to draw things together I think you are trying to read too much into the statements as they do not conflict - you posited three statements:

1. indicates explicitly that homogeneity of variance is an assumption for paired samples t-test

2. The paired t–test does not assume that observations within each group are normal, only that the differences are normal.

3. it does not assume that the groups are homoscedastic

Comment: The homogeneity of variance is an assumption that the variables being tested/compared are from same population and/or have equal variances. Inherently if you design an experiment  for paired samples t-test you would assume/should know whether this is the case. The homogeneity of variance or equality of variance is actually more important than the assumption of normality. For a group this is not important (as any competent researcher should check the scale variables for normality - not groups within the variable). In other words, are the standard deviations of the two groups pretty close to equal? Most statistical software packages provide a test of equality of variances  along with the results of the t-test and the most common being Levene's test of homogeneity of variance (but not usually for paired t-tests (because of the assumption - usually for independent t-tests). Levene's test works by testing the null hypothesis that the variances of the group are the same. The output probability is the probability that at least one of the samples in the test has a significantly different variance. If this is greater than a selected percentage (usually 5%) then it is considered too great to be able to usefully apply parametric tests. Don't forget this test is concerned with mean differences

Homoscedasticity is mainly considered in regressions as noisy data (heterodescasticity) has implications for forecasting. One of the assumptions of parametric tests is that the within-group standard deviations of the groups are all the same (exhibit homoscedasticity). If the standard deviations are different from each other (exhibit heteroscedasticity), the probability of obtaining a false positive result even though the null hypothesis is true may be greater than the desired alpha level.

So simply checking if the variables are normal (standard practice), the standard deviations are the same/almost same (homoscedastic), and you are making an assumption that the grouped variables are from the same population the paired t-test is preferred choice

The resultant differences become one series and one sample t test is applicable with H0: mean of differences is zero.

On the question of homoscedasticity, I found this on the web:

In conclusion, then, it does not behoove you to permit extreme trepidation and apprehension to emanate from your amygdala when confronted with an egregious predilection for prolix exposition and inveterate hippopotomonstrosesesquipedalianism. (Translation: Don’t let big words scare you.)

Lighten up people:

http://blog.minitab.com/blog/statistics-and-quality-data-analysis/dont-be-a-victim-of-statistical-hippopotomonstrosesquipedaliophobia

http://blog.minitab.com/blog/statistics-and-quality-data-analysis/dont-be-a-victim-of-statistical-hippopotomonstrosesquipedaliophobia

I think that even though is an assumption, you have to test it, as everything that is assumed in research when we applied any statistical model. Frecuently, we supposed that assumptions are satisfied and we did not probe them. Then we have problems with high variability or other kind of problem in our data, and when we explore our dataset we confirmed that some assumpions are not satisfied. So the best thing to do is to test for the assumptions to make sure you are not violating any of them.

I completely agree with the school of thought that says "You do not need homogeneity of variance assumption for paired sample t-test" but the difference between the pre/post should be normally distributed. Statistics instructors need to screen some textbooks very well before recommending it to their students.

The resultant differences become one series and one sample t-test is applicable with null hypothesis: mean of differences is zero .The only assumption  that the paired test relies on is that the" differences are normal distributed" .Therefor  it  dos' not need homogeneity  o f variance assumption for paired sample t-test" but the difference between the pre/post should be normally distributed.

I agree that the only assumption of the paired test is that the differences are normal distributed or at least, a quite symmetric distribution.

Leonel: I think Roberto and Mehdi have answered the question very well: what matters is the set of differences.

I concur with those who say that

1) there is only one distribution in play in the context of the paired t-test, and that's the distribution of differences in sample observations

2) the distribution of the differences of the population data from which the sample observations are drawn needs to be normally distributed for the mean of the differences in sample observations to be t-distributed.

But I have a related but distinct question:

Do we need the populations from which two independent samples are drawn to be homoskedastic for an independent samples t-test? If not, why not? If we do, where does homoskedasticity play a role in the test?

The t-test is based on the t-value, which is a ratio of the sample mean difference and the standard error of this mean difference. The standard error is a function of the variance of the data (the variance "within group"). If the within-group variances cannot be assumed equal in both groups, it is not clear how the standard error should be calculated (this is known as the Beherns-Fisher problem). One may use the average variance (as one would also do if the assumption of equal variances makes sense, to get a better estimate), but one has to account for the additional uncertainty of the standard error. This is done by the Welch–Satterthwaite adjustment of the degrees of freedom (-> Welch t-test).

Practically, very often when we observe heteroscedasticity, the test of a mean difference is not appropriate anyway. Often, the variable is strictly non-negative (bounded at 0, like a mass or a concentration), variance increases with the mean. Instead one should better test proportional changes. A very simple way doing that is to analyze the mean difference of the logarithms of the data (more sophisticated analyses model the log expectation without needing to transform the values).