Chi-Square Confidence Intervals to Bootstrap Intervals

 2026-05-25

#statistics #chi-square #bootstrap #confidence-intervals

From confidence interval derived from $\chi^2$ to boostrap confidence interval?¶

In [1]:

from math import sqrt
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as stats
import seaborn as sns

$\chi^2$¶

Let's start with creating a normally distributed data $\mu = 25000$ and $\sigma = 500$. This will be our population, and we are going to estimate variance using the chi-square CI formula.

In [2]:

data = np.random.normal(loc=50, scale=20, size=100_000)

In [3]:

fig, ax = plt.subplots()
sns.kdeplot(data)
plt.show()

No description has been provided for this image

Yes, our data looks somewhat normal. Let's continue.

In [4]:

sample_size = 11
sample = np.random.choice(data, sample_size)
sample_variance = np.var(sample, ddof=1)
sample_variance, sqrt(sample_variance)

Out[4]:

(np.float64(447.92889896496445), 21.164330817792575)

We are going to set our sample size as 11, so the $df$ will be 10.

So for this specific sample/sample size, we want %95 confidence. We are going to use the below formula:

$$ \frac{(n-1)s^2}{\chi_{a/2, k}^2} \lt \sigma^2 \lt \frac{(n-1)s^2}{\chi_{1-a/2, k}^2} \\ $$

$$ \frac{10s^2}{\chi_{0.025, k=10}^2} \lt \sigma^2 \lt \frac{10s^2}{\chi_{0.975, k=10}^2} $$

In [5]:

a = 0.05
chi2_lower = stats.chi2.ppf(1 - (a / 2), sample_size - 1)
chi2_upper = stats.chi2.ppf(a / 2, sample_size - 1)

We don't directly use the chi-squared distribution, but the values we used for $\chi_{0.975}$ and $\chi_{0.025}$ comes from the two ends of this distribution.

In [6]:

chi_max = stats.chi2.ppf(0.999, sample_size - 1)
x_ticks = np.linspace(0, chi_max, 5000)
chi = stats.chi2.pdf(x_ticks, sample_size - 1)

In [7]:

x_shade = np.linspace(chi2_lower, chi2_upper, 500)
y_shade = stats.chi2.pdf(x_shade, sample_size - 1)

In [8]:

fig, ax = plt.subplots()
ax.plot(x_ticks, chi)
ax.fill_between(x_shade, y_shade, alpha=0.3)
ax.set_ylim(0)
ax.set_xlim(0)
plt.show()

Result¶

In [9]:

highest_expected_chaos = (sample_size - 1) * sample_variance / chi2_upper
lowest_expected_chaos = (sample_size - 1) * sample_variance / chi2_lower
lowest_expected_chaos, highest_expected_chaos

Out[9]:

(np.float64(218.68135557948892), np.float64(1379.5277302333639))

Now, we can interpret this result as "Variance is in these intervals with %95 confidence", but this is not true. It actually means that, "In all the '%95 samples', %95 of the confidence intervals would contain the population variance."

Bootstrap Confidence Interval¶

Chi-sqarued distribution is very useful for finding a CI for variance, when the underlying dataset is normally distributed. However, real-world data may not follow a perfectly normal distribution. In a situation like that, we can use bootstrap confidence interval.

In [10]:

bootstrap_variance = []
m = np.mean(data)
s = np.std(data)
sample = np.random.choice(data, 1000, replace=False)

for _ in range(5000):
    bootstrap_variance.append(
        np.var(np.random.choice(sample, len(sample), replace=True), ddof=1)
    )

Keep in mind that, when dealing with a sample, we should always use $\delta df=1$, even if we are bootstrapping (using the whole sample again).

In [11]:

fig, ax = plt.subplots()
sns.kdeplot(bootstrap_variance, ax=ax)
plt.show()

In [12]:

print(f"CI for variance is: {np.quantile(bootstrap_variance, [0.025, 0.975])}")
print(f"Population variance is: {np.var(data)}")
print(
    f"Mean of the bootstrap distribution of the variance: {np.mean(bootstrap_variance)}"
)

CI for variance is: [336.17646994 406.14348428]
Population variance is: 402.1095537256238
Mean of the bootstrap distribution of the variance: 370.3756709701916

 Back to Index