Variation

Sample Variance

Measure how spread out the data is. Compute the mean, then average the squared deviations from it — dividing by n−1 (Bessel's correction) to get an unbiased estimate of the population variance. By hand, two loops: one for the mean, one for the squared deviations. With the library, statistics.variance uses the same n−1 denominator.

By hand

With the library

statistics.variance uses n−1 and matches the naive result exactly. np.var defaults to ddof=0 (population variance, divides by n) — pass ddof=1 to get the sample variance. The snapshot shows all three side by side to make the ddof difference concrete.

naive.py 
values = [4, 8, 6, 13, 10, 6, 9]
n = len(values)
total = 0.0
for v in values:
    total = total + v
mean = total / n
sq_diff = 0.0
for v in values:
    sq_diff = sq_diff + (v - mean) ** 2
variance = sq_diff / (n - 1)
print('RESULT:', round(variance, 10))
library.py 
import statistics
import numpy as np
from dalib.display import set_display
set_display()

values = [4, 8, 6, 13, 10, 6, 9]
var_stdlib = float(statistics.variance(values))
var_np_pop = float(np.var(values))
var_np_samp = float(np.var(values, ddof=1))
print('statistics.variance:', var_stdlib)
print('np.var (ddof=0):    ', round(var_np_pop, 4))
print('np.var (ddof=1):    ', var_np_samp)
print('RESULT:', round(var_stdlib, 10))

statistics.variance: 9.0
np.var (ddof=0):     7.7143
np.var (ddof=1):     9.0
RESULT: 9.0

Implementation notes

  • Dividing by n−1 instead of n is Bessel's correction: the sample mean slightly underestimates deviations from the true population mean, so shrinking the denominator compensates. Dividing by n gives the population variance, which is correct only when you have the full population.
  • np.var default is ddof=0 (population). Always pass ddof=1 when you want sample variance from numpy.
  • statistics.variance always uses n−1; there is no ddof parameter.