How to test for normal distribution with one peak in data?

Question

I would like to test following. Suppose I have a normal distribution with mean 1.5 and sigma 0.5 on interval [0, 3]. Python code:

import numpy as np
import matplotlib.pyplot as plt


mu = 1.5  # mean
sigma = 0.5  # standard deviation
n = 1000  # number of samples

# generate normally distributed random numbers with mean mu and standard deviation sigma
samples = np.random.normal(mu, sigma, n)

# shift and scale the samples to the range 0 to 3
samples = np.clip(samples, 0, 3)

Histogram looks like this:

But now I add one peak around x = 1.96 and my histogram looks like this now:

Python code for generating this peak is:

n = 70  # number of samples

# generate uniformly distributed random numbers between 0 and 1
added_samples = np.random.rand(n)

# scale the samples to the range 1.96 to 1.99
added_samples = 1.96 + added_samples * (1.99 - 1.96)


all_together = np.append(samples, added_samples)

My question is, if there is a statistical test or algorithm that can check for normally distributed data with "one peak"? Some modification of shapiro-wilk test or something.. Thanks a lot.

EDIT: What we know beforehand:

1. Peaks can be only in three points, lets say x = 1.5, x = 1.96 or x = 2.5. We would like to test if there is really a peak in any of these x.
2. We know the mean and standard deviation of the new sample = that is after the peak is present.
3. The additional data are uniformaly distributed around the interval [x, x + 0.03] (that is [1.5,1.53], [1.96,1.99] or [2.5, 2.53])

Answer 1

First, it seems like you should be able to identify the particular alternative hypothesis, i.e., which of the three potential intervals the "additional" data would come from, simply by looking which of the three intervals contains the most excess data over what we would expect under the null hypothesis of a normal distribution.

Next, the probably best way to continue would be a likelihood ratio test between two competing models for the data.

1. A mixture between (a) a mixture of a truncated normal distribution and two point masses at 0 and 3 and (b) a uniform distribution on a prespecified interval.
2. The mixture between the truncated normal and the two point masses, part (a) without (b) above.

Unfortunately, the likelihood functions involved are a bit icky... maybe someone with more time could set these up. It might be best to bootstrap the test itself once we have the likelihoods.

---

A much simpler alternative would be to use a two-sample Kolomogorov-Smirnov test: compare your data against a simulated data sample under the null hypothesis of that mixture between a truncated normal and two point masses with no "additional data".

Here is an example. I will use R, simply because I am more fluent in that.

# a function to simulate data
simulate_obs <- function(n_orig,m_orig,sd_orig,n_add,a_add,b_add) {
    obs <- c(rnorm(n_orig,m_orig,sd_orig),runif(n_add,a_add,b_add))
    pmin(pmax(obs,0),3)
}

# simulation parameters    
n_orig <- 1000
m_orig <- 1.5
sd_orig <- 0.5
n_add <- 70
a_add <- 1.96
b_add <- 1.99

# simulate your "real" data
set.seed(1)
obs_orig <- simulate_obs (n_orig,m_orig,sd_orig,n_add,a_add,b_add)

# simulate under the null hypothesis
null_sample <- simulate_obs(length(obs_orig),mean(obs_orig),sd(obs_orig),0,0,3)
# compare your "real" data to the simulated data:
ks.test(obs_orig,null_sample)

In the present case, this gives $p=0.23$, because the "real" data is quite consistent with coming from a "real" uncontaminated distribution.

Answer 2

1. Compute the probability for the set $S=[1.5,1.53] \cup [1.96,1.99] \cup [2.5, 2.53]$ for a truncated normal with the given parameters, say $p_0$.

2. Run a binomial test for the $H_0:\ p=p_0$ against the alternative $p>p_0$ on data that is 1 for observations in $S$ and 0 for observations not in $S$.

One can probably improve this a bit if you know that there will be only one peak; what I propose here works also (and is maybe in some sense optimal) if you have peaks at two or all three of these locations.