Normality test after rounding

Question

Suppose we have a random variable $Y$ that generates data $(Y_1,...,Y_N)$ for $N \in \mathbb{N}$ big, i.e. $N \geq 10'000$. The mean of the random variable $Y$ is $\mu = 0.52$ and the standard deviation is $\sigma = 0.002$.

We do not have access to the data $(Y_1,...,Y_N)$ directly, rather a device measures the outcomes of $Y$ and give us the data $(\tilde{Y_1},...,\tilde{Y_N})$ such that $\tilde{Y_i}$ is rounded at the third decimal for $i = 1,...,N$. As an example, suppose that $Y_i = 0.5236$, then $\tilde{Y_i} = 0.524$.

The goal is to know if the random variable $Y$ is normally distributed, by applying a normality test for example. The problem is that we can only do the normality test on the data
$(\tilde{Y_1},...,\tilde{Y_N})$ that are not normally distributed anymore because they were transformed and the rounding is 'too strong' in regard with the standard deviation implying a strong deviation from normality.

I have found an article of minitab explaining that the Ryan-Joiner test can be an alternative and works good in case of rounding https://blog.minitab.com/en/the-statistical-mentor/normality-tests-and-rounding.

However, it does not always work in my case. I have also made some simulations for $N = 30'000$ (which corresponds to my practical case) to illustrate what happen.

In red is $(Y_1,...,Y_N)$ if $Y$ is a normal distribution and in black is $\tilde{Y}$. As we can see, the distribution of $\tilde{Y}$ has a form of a bell shape but with a big peak.

So, I tried to take a subsample (of size 50 and 100) of the $\tilde{Y}$ and do a normality test on it, it works for some cases. I was wondering if it is a good approach because in any case, a normality test on a sample of size 30'000 can rarely be passed.

Has anyone any idea how to overcome this problem?

Answer 1

Here is an informal test you could run: collect your rounded observations $(\tilde{Y_1},...,\tilde{Y_N})$ and plot the "peaky" histogram as you did. Derive the mean and standard deviation from the rounded data.

Then: simulate normally distributed data using the estimated mean and SD. Round the simulated data. Plot the corresponding histogram.

Do the simulation 19 times, ending up with 19 simulated histograms. Lay these out in a $5\times 4$ grid, and insert the "true" histogram at a random spot.

Then go and ask people whether they can identify the "off" histogram. If they can't, this is at least some evidence that the deviation between the simulated data and the actual observations can't be "too strong".

This has been called an "inter-ocular trauma test for significance".

Below is an example. Can you spot which distribution comes from gamma distributed original data (which I intentionally generated to have a mean and standard deviation close to what you report for your data?)? If so, is the deviation strong enough to pose any problems in the subsequent analysis you need this information for?

R code (with the solution):

nn <- 30000
set.seed(1)
true_data <- round(rgamma(nn,40000,75000),3)
simulated_data <- replicate(19,round(rnorm(nn,mean(true_data),sd(true_data)),3))
all_data <- cbind(simulated_data[,1:13],true_data,simulated_data[,14:19])
opar <- par(mfrow=c(5,4),mai=c(.5,.1,.1,.1))
    for ( ii in 1:20 ) {
        hist(all_data,breaks=seq(min(all_data)-0.0005,max(all_data)+0.0005,by=0.0001),
            yaxt="n",main="",xlab="",ylab="")
    }
par(opar)

You could of course turn something like the "test" above into a formal parametric bootstrap: use a "reasonable" test (e.g., the Shapiro-Wilk test for normality), calculate the test statistic (and discard the p value, it's useless). Then simulate many normally distributed datasets with the same mean and standard deviations, round them, and again calculate the Shapiro-Wilk test statistic. Finally, derive a p statistic, by taking the empirical cumulative distribution function (ECDF) of the bootstrapped test statistics and evaluating this at the test statistic derived from the true data.

In the example above, this gives you result of $p=0.11$. Any significant result here would, again, be a result of your large sample size, and only give you statistical significance, not clinical significance, which you can only assess by placing the entire exercise in the larger context.

Here is a histogram of the Shapiro-Wilk test statistics from 1000 bootstraps, together with the test statistic from the "true" data (from an underlying gamma distribution) inserted as a red line.

More R code (where I took a sample of size 5000 from the original data of size 30000 because R's shapiro.test can't handle more data):

shapiro_true <- shapiro.test(sample(true_data,5000))$statistic
shapiro_sim <- replicate(1000,
 shapiro.test(round(rnorm(5000,mean(true_data),sd(true_data)),3))$statistic)
hist(shapiro_sim,xlim=range(c(shapiro_sim,shapiro_true)))
abline(v=shapiro_true,col="red",lwd=2)
ecdf(shapiro_sim)(shapiro_true)

Answer 2

You could run a chi-squared test. A chi-squared test would partition data into categories. You can use as categories a fixed number of rounded values that you expect with large enough probability, say $k$, plus "anything larger" and "anything smaller", so you have $k+2$ categories. Estimating mean and variance, you can compute the expected probabilities for all these categories, and then run a standard chi-squared test with k-1 degrees of freedom (number of categories $k+2$ minus one is standard and you have to subtract 2 degrees of freedom for having estimated two parameters). The good thing about this is that it is based on rounding by construction, so the very fact that you have rounded data will not affect this.

Note however that nothing in real life is ever normally distributed, so there is no way to "know" that anything is normally distributed. Neither do you ever "need" normality, as most things that are based on normality work approximately on non-normal data (and those that don't shouldn't be touched with a bargepole). It is only ever necessary (and possible) to detect problematic deviations from normality that will mislead your analysis, and this always depends on the analysis you want to do and how you want to interpret it.

The chi-squared (or any) test of normality will not by any means prove normality; it will rather only ask whether data cannot be statistically distinguished from normality, according to what the test statistic measures (which in this case does not measure distortion by rounding). Note in particular that much inference that supposedly "assumes" normality will still work well with many non-normal distributions, so that in your situation normality may be technically rejected but inference may still be fine (although occasionally it may not; but this is a different issue than what a normality test addresses).