Testing percentiles¶

Here’s a question from the Reddit statistics forum.

I have two different samples (about 100 observations per sample) drawn from the same population (or that’s what I hypothesize; the populations may in fact be different). The samples and population are approximately normal in distribution.

I want to estimate the 85th percentile value for both samples, and then see if there is a statistically significant difference between these two values. I cannot use a normal z- or t-test for this, can I? It’s my current understanding that those tests would only work if I were comparing the means of the samples.

As an extension of this, say I wanted to compare one of these 85th percentile values to a fixed value; again, if I was looking at the mean, I would just construct a confidence interval and see if the fixed value fell within it…but the percentile stuff is throwing me for a loop.

This is […] related to a research project I’m working on (in my job).

There are two questions here. The first is about testing a difference in percentiles between two groups. The second is about the difference between a percentile from an observed sample and an expected value.

We’ll answer the first question with a permutation test, and we’ll answer the second in two ways: bootstrap resampling and a Gaussian model.

Data¶

Since OP didn’t provide a dataset, we have to generate one. I’ll draw two samples from Gaussian distributions with the same standard deviation and different means.

Here’s what the distributions of the groups look like.

If we compute the 85th percentile in both groups, we see a difference, as expected.

Now let’s see if a difference of that size would be likely if the two samples were actually drawn from the same distribution.

Testing the Difference¶

When we test a difference between two groups, the usual model of the null hypothesis is that the groups are actually identical. If that’s true, the two samples came from the same distribution, so we can combine them into a single sample.

Now we can simulate the null hypothesis by permutation — that is, by shuffling the pooled data and splitting it into two groups with the same sizes as the originals. The following function generates two samples under this assumption, and returns the difference in their 85th percentiles.

If we call it many times, the result is a sample from the distribution of differences under the null hypothesis.

Here’s what it looks like, with a vertical line at the observed difference. I’m plotting it with cut=0 so the estimated density doesn’t extend past the minimum and maximum of the data.

The distribution is multimodal, which is the result of selecting a moderately high percentile from a moderately small dataset — the diversity of the results is limited. However, in this example we are interested in the tails of the distribution, so multimodality is not a problem.

To estimate a one-sided p-value, we can compute the fraction of the sample that exceeds the actual difference.

Or, for a two-sided p-value, we can compute the fraction of the sample that exceeds the actual difference in absolute value.

In this example, the result of the one-sided test would be considered significant at the 5% significance level, but the two-sided test would not. So which is it?

I think it’s not worth worrying about. My interpretation of the results is the same either way: they are inconclusive. Under the null hypothesis, a difference as big as the one we saw would be unlikely, but we can’t rule out the possibility that the groups are identical — or nearly so — and the apparent difference is due to random variation.

Testing a fixed value¶

Now let’s turn to the second question. Suppose we have reason to think that the actual value of the 85th percentile is 12.3, and we would like to know whether the data contradict this hypothesis.

We’ll test group1 first. Here’s the 85th percentile of group1 and its difference from the expected value.

Let’s see if a difference of this magnitude is likely to happen under the null hypothesis. One way to model the null hypothesis is to create a dataset that is similar to the observed data, but where the 85th percentile is exactly as expected. We can do that by shifting the observed data by the observed difference.

The 85th percentile of the shifted data is the expected value, exactly.

Now, to generate samples under the null hypothesis, we can use the following function, which takes a sample, shifts it to have the expected value of the 85th percentile, generates a bootstrap resample of the shifted values, and returns the difference between the 85th percentile of the sample and the expected value.

If we call this function many times, we get a sample of the differences we expect under the null hypothesis.

The following function shows the distribution of the sample with a vertical line at the observed value.

Without computing a p-value, we can see that a difference as big as actual_diff1 is entirely plausible under the null hypothesis. We can confirm that by computing a one-sided p-value.

So the observed difference in the first group is not statistically significant. Now let’s do the same thing for the second group.

Here’s what the distribution of differences looks like under the null hypothesis, with a vertical line at the observed value.

There are no differences in the sample that exceed the observed value.

We can conclude that a difference as big as that is very unlikely under the null hypothesis. There’s not much point in computing a p-value more precisely than that, but if it’s required, we can estimate it if we assume that the tail of the sampling distribution is roughly Gaussian. In that case, we can fit a KDE to the sampling distribution like this.

And use a Pmf object to approximate the estimated density.

Then we can use the corresponding CDF to compute the probability of a value that exceeds the observed difference.

So the p-value is quite small.

Model-based resampling¶

The bootstrap method in the previous section is a good choice if we are unsure about the distribution of the data, or if there are outliers. But multiple modes in the sampling distribution suggest that there might not be enough diversity in the data for bootstrapping to be reliable.

Fortunately, there is another way we might model the null hypothesis: using a Gaussian distribution. If we generate data from a continuous distribution, we expect the sampling distribution to be unimodal.

But there is a problem we have to solve first — we have to make an assumption about the standard deviation of the hypothetical Gaussian distribution. One option is to use the standard deviation of the data.

Now we need to find a Gaussian distribution with a given standard deviation that has the expected 85th percentile. We can do that by starting with a distribution centered at 0, computing it’s 85th percentile and then shifting it.

ppf stands for “percentile point function”, which is another name for the quantile function, which is the inverse of the CDF — it takes a cumulative probability and returns the corresponding quantity.

The following function takes one of the groups, fits a hypothetical model to it, generates a sample from the model, and returns the difference between the 85th percentile of the sample and the expected value.

If we call this function many times, we get the sampling distribution of the test statistic under the null hypothesis.

Here’s what the distribution looks like, compared to the corresponding distribution from the bootstrapped model.

The shapes of the distributions are different, but their ranges are comparable. And the conclusion is the same: a difference as big as actual_diff1 is entirely plausible under the null hypothesis.

Now let’s try the same test with Group 2.

Here’s the result, along with the result from the bootstrap model.

Again, the shapes of the distributions are different, but the conclusion is the same. A difference as big as actual_diff2 is unlikely under the null hypothesis.

As usual, the two-sided p-value is bigger by a factor of two, roughly, but the difference never matters in practice.

Under this model of the null hypothesis, the probability is small that the 85th percentile of the data would exceed the expected value by so much.

Discussion¶

This example demonstrates a kind of inconsistency in hypothesis testing. We found that Group 1 is not significantly different from the expected value — in the technical sense of significantly — but Group 2 is. So that suggests that Group 1 and Group 2 are different from each other, but when we test that hypothesis, the difference is not statistically significant.

People who are new to hypothesis testing find results like this surprising, but they are not rare. Generally, they are a consequence of the logic of null hypothesis testing and the arbitrariness of the significance threshold.

I think it helps to interpret p-values qualitatively.

• A p-value greater than 10% means that the observed effect is plausible under the null hypothesis, and could happen by chance.

• A p-value less than 1% means that an observed effect is unlikely under the null hypothesis — so it is unlikely to have happened by chance.

• Anything in between is inconclusive.

There is nothing special about 5%.

Data Q&A: Answering the real questions with Python

Copyright 2024 Allen B. Downey

License: Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International

Bootstrapping percentiles¶

Here’s a question from the Reddit statistics forum.

I’m trying to figure out how to determine the confidence interval for the .2 percentile temperature for specific set of observed temperatures (all hourly temperatures during January, February, and December since 2000). I have recordings for 53128 of the 53424 possible hourly recordings.

How would I go about saying that I am X% sure that the actual .2 percentile value is between two numbers? Could anyone provide any insight on how to accomplish this. Thank you.

OP provided a link to the data, so this is a question we can answer! For computing confidence intervals, my first choice is bootstrap resampling, but as it turns out, it does not work well for this problem. I’ll show what goes wrong and how to fix it. Then we’ll answer a follow-up question about quantifying the effect of missing data.

Data¶

I downloaded the data as a CSV file, which we can read into a Pandas DataFrame.

There are 53424 measurements, of which 306 are missing.

The range of temperatures is from -9.9 degF to 89 degF.

And the 0.2 percentile is 1 degF.

Basic bootstrap¶

The following function takes the cleaned data, resamples it, and computes the 0.2 percentile of the bootstrapped sample.

If we call this function 1001 times, we get a sample from the sampling distribution of the percentile.

Here’s what that sample looks like.

Immediately we can see that something has gone wrong. The resampling process produces only 8 unique values.

If we try to compute a CI by pulling percentiles from the sample, the results are not credible.

This example demonstrates a limitation of bootstrap resampling — it does not work well when there are a small number of unique values.

However, because the data are temperature measurements, they are actually continuous quantities. So one option is to replace bootstrapping with a model that generates continuous quantities. We’ll try that with a normal model, see that it does not work, and they try again with KDE.

Resampling from a normal model¶

If we look at the CDF of the data, it resembles the characteristic sigma of the normal distribution.

So let’s see how it compares to a normal model. I’ll estimate the parameters by computing the mean and standard deviation of the data.

And compute the normal CDF within 4 standard deviations of the mean.

Here’s what the model looks like compared to the data.

It looks pretty good, but there are places where the data clearly deviate from the model. That’s enough to make me worry, but let’s proceed and see how it goes.

The following function takes the cleaned data, generates a random sample from the normal model, and returns the 0.2 percentile of the sample.

If we call it 1001 times, we hope the result is a sample from the sampling distribution of the percentile.

And at first glance it looks good.

But notice that range of the sampling distribution does not include the 0.2 percentile of the data, which is 1. We can compute a 90% CI, but again, it is not credible.

To see what went wrong, let’s look at the normal model and the data again, this time with the y axis on a log scale. The log scale is like a microscope that lets us see more clearly what is happening in the tail of the distribution.

On a linear scale, it seemed like the normal model might be good enough; on a log scale, it is clear that the data deviate from the model in the left tail.

In retrospect, it is not a surprise if a simple two-parameter model fails to capture every detail of the distribution — the world is a complicated place. So let’s try a nonparametric approach.

Resampling with KDE¶

We can use kernel density estimation (KDE) to model the distribution of the data, then use the model to resample. Here’s how we estimate the distribution.

To see what the result looks like, we can approximate the density of the model with a discrete PMF.

And then compare the CDF of the model with the CDF of the data.

The result shows that KDE is doing what it is meant to do — fitting a continuous distribution to the data with minimal assumptions.

The following function takes the cleaned data, uses the KDE model to generate a random sample, and returns the 0.2 percentile of the sample.

If we call it 1001 times, we hope once again that the result is a sample from the sampling distribution of the percentile.

And this time we get a better result. The sampling distribution looks good, and it contains the actual percentile of the data.

And the width of the 90% CI is plausible.

So with a couple of false starts, we have answered the original question. But it turns out there’s more.

Fill missing values¶

In a follow-up message, OP wrote:

Just in case it helps any, here’s what I’m ultimately trying to accomplish with this endeavor… I am trying to come up with a plausible way of demonstrating that the .2 percentile value (1 degF) that is derived from this data set is sufficiently representative of what the value would be if there were no missing data points (hourly readings) from the dataset.

OK, that’s a different question! However, the resampling framework can be extended naturally to estimate the effect of missing data. Here’s a function that takes the original data — including NaNs — and fills the missing values with a random selection of valid values. For historical reasons, this way of filling missing values is called “hot deck imputation”.

To test it, we can check that the result has no NaNs.

Now we can include fill_missing as part of the resampling pipeline. The following function takes the original data, fills missing values, generates a sample from a KDE model, and returns the 0.2 percentile of the sample.

If we call it many times, we get a sample from a distribution that represents the uncertainty of the estimate due to a combination of missing data and random sampling.

Here’s what the result looks like, compared to the sampling distribution from the previous section, which represents only uncertainty due to random sampling.