One of the assumptions for t-tests is that the data must follow a normal distribution.

However, due to the Central Limit Theorem (and this thread): "if the sample is large enough you can use t-test (with unequal variances)". I'm trying to sort out what this means for my case. I think my sample should be large-enough, but how to confirm it?

A Levene's test showed that the two samples don't have an equal variance, hence I plan to use Welch's test (the unequal variance version of the t-test). I've also ran the Shapiro-Wilk test to confirm that one of my two samples doesn't, in fact, follow a normal distribution.

Additional information

I need to run the tests for a few different cases, but to keep things short I'm detailing only two of them.

Sample sizes are 19 and 15, respectively for group1 and group2 (this happens on both the examples: Case1 and Case2).

Results of Shapiro-Wilk's test for normality

sample | p_value   | w     | Result
group1 | 0.104     | 0.918 | Normal
group2 | 0.027     | 0.863 | Not Normal (p<0.05)

sample | p_value   | w     | Result
group1 | 2.054e-05 | 0.663 | Not Normal (p<0.05)
group2 | 0.006     | 0.814 | Not Normal (p<0.05)

Results of Levene's test for equality of variances

p_value | w    | Result
0.154   |2.128 | Equal Variance

p_value | w    | Result
0.0251  |5.521 | Unequal Variance (p<0.05)

Result of the one-tailed (Welch) t-test (H1: group1>group2)

t_statistic | p_value | Result
3.073       | 0.002   | Significant (p<0.05)

t_statistic | p_value | Result
2.475       | 0.012   | Significant (p<0.05)
  • $\begingroup$ I'd recommend the Wilcoxon rank sum test. $\endgroup$ Commented Jan 8, 2014 at 15:38
  • $\begingroup$ Doesn't the Wilcoxon/Mann-Whitney-U test require an equal variance? I was ruling it out for that reason. $\endgroup$ Commented Jan 8, 2014 at 15:41
  • $\begingroup$ How different are the variances? $\endgroup$ Commented Jan 8, 2014 at 15:50
  • 3
    $\begingroup$ How does WMW know if the variances are the same or different? It ignores variances. A deeper problem is that you say absolutely nothing about your sample sizes and little about the character or strength of non-normality in your data, so we don't have enough information to advise well. Tests being significant (at conventional levels) is not much information. Unimportant deviations from normality (e.g.) will often show up on (e.g.) a Shapiro-Wilk test whenever the sample size is large enough for them not to be a problem. But using Welch sounds a good idea. $\endgroup$
    – Nick Cox
    Commented Jan 8, 2014 at 15:51
  • $\begingroup$ How large is the sample size? Both Levene's test and Shapiro-Wilk are based on NHST, and so with large samples, even small deviations result in rejecting the null. Graphics such as a QQ plot could be useful that case. See here for more info. Other than Welch, Using Bayesian estimation with a t-distribution could be useful, more information can be found here $\endgroup$
    – dmartin
    Commented Jan 8, 2014 at 15:53

1 Answer 1


One simple way to convince yourself that the CLT applies or does not apply is with some simulations.

Here is some R code:

testfun <- function(n1=19, n2=15) {
    x <- rexp(n1, 1/3)
    y <- rt(n1, 5) + 3

out <- replicate(10000, testfun(n1=19, n2=15))
abline(v=0.05, col='red')
mean( out <= 0.05 )

This code defines a function (testfun) that generates data from 2 different distributions (t with 5 df and exponential ) that have the same mean (3 in this case) and runs the built in t.test function and returns the p-value.

The replicate then runs this 10,000 times and we look at the results. The histogram should be close to uniform, but in this case we see an excess of values close to 0. The mean function calculates the type I error rate (since the null is true in the simulations), for my run this was a little of 7% when it should be 5%. Is that far enough to cause you concern? or are you happy with that as a "close enough" approximation?

Of course you should probably run this generating data from distributions that are more reasonable for your study, it may be that for something less skewed than the exponential that the differences would be small enough to not worry about.


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