How to use one-sample t-test on skewed distributions?

Question

I have been reading that people use the one-sample t-test also for skewed underlying distributions, saying that for a high enough number of datapoints (I read for example N=30 and N=100 in some places) the central limit theorem means that you can apply the t-test even to non-normal data. I struggle to understand the argument.

In case of the one-sample t-test I only have one sample, whose distribution is not normal. I don't have multiple samples with multiple means, that would follow a normal distribution according to the CLT. In my understanding of the CLT it does not apply to a one sample case? Why do people say that with enough datapoints in my one sample these two cases are equivalent? Is this just obvious and I am not seeing it? Where do people get the N=30 from, and how do I find out which N is appropriate for my case?

Answer 1

I am not sure where you read that, or what the exact argument was. I think maybe you got it from some site comparing t test to z test. Googling on "one sample t test large sample" found some references to this. I don't think it's relevant here. Here, neither the t nor the z is appropriate.

Intuitively, the one sample t test is a test of whether the mean is equal to 0 (you can adjust this to another value). The mean is generally not recommended as a measure of central tendency for skewed distributions (although there are exceptions) so we usually shouldn't use the t-test on such distributions.

The assumptions of the t-test include one sample normality. I've not seen any exception for large samples. Laerd statistics lists four assumptions. Briefly, they are:

1. DV is measured at interval or ratio level.
2. Data are independent.
3. No significant outliers.
4. DV is approximately normally distributed.

(See the link for details).

So, I suggest using an alternative test. Which one would depend on your exact question and the shape of the distribution (e.g. are there outliers?)

Answer 2

On the one hand, Peter Flom makes a good point that the mean may not be the "best" way to describe a skewed distribution.

Then again, it may well be. (I do retail forecasting. Retail sales are usually skewed, and we are still interested in expected sales, e.g., for promotion planning.)

And then, it turns out that even for highly skewed original data, the mean is asymptotically normally distributed. $N=30$ is empirically a not too unreal sample size for this to be "true enough".

As an example, let's simulate a population that is Beta(2,0.5) distributed. This is quite skewed indeed:

However, let's now draw a sample of size $N=30$ and calculate the mean. Let's do this many times to get an idea of the distribution of the mean. Here is a histogram:

We see that this already looks "much more normal" than the population. Of course, it is not normal, since the means all will be between zero and one. But it is close enough for a one-sample t test to have close to the correct size and power.

In case of doubt, you could always bootstrap the mean. This is also a little questionable for small sample sizes, but it at least gives you an additional data point.

This is related, for the two-sample case: t-test when the data population is not normally distributed

R code:

nn <- 1e5
population <- rbeta(nn,2,0.5)
hist(population)

means <- replicate(1e4,mean(sample(population,30,replace=FALSE)))
hist(means)