Sampling and Estimation

Standard Error

The standard error (SE) of the mean measures how much the sample mean varies across repeated samples: SE = s/√n, where s is the sample standard deviation (ddof=1). Two loops: mean, then sample std; divide by √n. Library: scipy.stats.sem(sample) — defaults to ddof=1 (sample std), matching the naive. Same fixed sample as seeded-sample-mean.

By hand

Sample [4, 7, 2, 9, 5, 9], n=6, mean=6.0. Loop 2: squared deviations [4,1,16,9,1,9] sum to 40; s=√(40/5)=√8≈2.828427. SE = √8/√6 = 2/√3 ≈ 1.154701.

naive.py
Replay: real traced execution (multi-file project)
import math
sample = [4, 7, 2, 9, 5, 9]
n = len(sample)
total = 0.0
for v in sample:
    total = total + v
mean = total / n
sq_diff = 0.0
for v in sample:
    sq_diff = sq_diff + (v - mean) ** 2
s = math.sqrt(sq_diff / (n - 1))
se = s / math.sqrt(n)
print('RESULT:', round(se, 6))

  1. import math

    1import math2sample = [4, 7, 2, 9, 5, 9]

  2. sample ← [4, 7, 2, 9, 5, 9]

    1import math2sample = [4, 7, 2, 9, 5, 9]3n = len(sample)
    values this step[4, 7, 2, 9, 5, 9]sample

  3. n ← 6

    2sample = [4, 7, 2, 9, 5, 9]3n = len(sample)4total = 0.0
    values this step6n

  4. total ← 0.0

    3n = len(sample)4total = 0.05for v in sample:
    values this step0.0total

  5. v ← 4, total ← 4.0

    pass 1 of 6
    4total = 0.05for v in sample:6    total = total + v7mean = total / n
    values this step4v0.0 → 4.0total
    All 6 passes — pass 1 is the card above
    passvtotal
    140.0 → 4.0
    24 → 74.0 → 11.0
    37 → 211.0 → 13.0
    42 → 913.0 → 22.0
    59 → 522.0 → 27.0
    65 → 927.0 → 36.0

  6. for v in sample:

    4total = 0.05for v in sample:6    total = total + v

  7. mean ← 6.0

    6    total = total + v7mean = total / n8sq_diff = 0.0
    values this step6.0mean

  8. sq_diff ← 0.0

    7mean = total / n8sq_diff = 0.09for v in sample:
    values this step0.0sq_diff

  9. v ← 4, sq_diff ← 4.0

    pass 1 of 6
    8sq_diff = 0.09for v in sample:10    sq_diff = sq_diff + (v - mean) ** 211s = math.sqrt(sq_diff / (n - 1))
    values this step9 → 4v0.0 → 4.0sq_diff
    All 6 passes — pass 1 is the card above
    passvsq_diff
    19 → 40.0 → 4.0
    24 → 74.0 → 5.0
    37 → 25.0 → 21.0
    42 → 921.0 → 30.0
    59 → 530.0 → 31.0
    65 → 931.0 → 40.0

  10. for v in sample:

    8sq_diff = 0.09for v in sample:10    sq_diff = sq_diff + (v - mean) ** 2

  11. s ← 2.8284271247461903

    10    sq_diff = sq_diff + (v - mean) ** 211s = math.sqrt(sq_diff / (n - 1))12se = s / math.sqrt(n)
    values this step2.8284271247461903s

  12. se ← 1.1547005383792517

    11s = math.sqrt(sq_diff / (n - 1))12se = s / math.sqrt(n)13print('RESULT:', round(se, 6))
    values this step1.1547005383792517se

  13. stdout ← RESULT: 1.154701

    12se = s / math.sqrt(n)13print('RESULT:', round(se, 6))
    values this stepRESULT: 1.154701stdout

With the library

scipy.stats.sem(sample) uses ddof=1 by default — the same denominator as numpy.std(ddof=1). The snapshot shows s separately so the two-step structure is visible.

library.py 
import numpy as np
from scipy.stats import sem
from dalib.display import set_display
set_display()

sample = [4, 7, 2, 9, 5, 9]
n = len(sample)
s = float(np.std(sample, ddof=1))
se = float(sem(sample))
print('n:', n)
print('s (ddof=1):', round(s, 6))
print('RESULT:', round(se, 6))

n: 6
s (ddof=1): 2.828427
RESULT: 1.154701

Honesty

This lesson shows the computation of the standard error exactly, on a tiny pinned sample. The value is arithmetically correct and reproducible, but computing SE from a handful of points demonstrates the formula, not a dependable estimate of precision — with so few observations the SE itself is unstable and rests on assumptions (independent, identically distributed draws). Read it as "how SE = s/√n is computed," not as a trustworthy margin; real precision claims need an adequate sample size and assumption checks.

Implementation notes

  • SE = s/√n: larger n → smaller SE → more precise estimate. Quadrupling the sample size halves the SE (square-root law).
  • ddof=1 trap: scipy.stats.sem defaults to ddof=1; using np.std (ddof=0 default) for s and then dividing by √n would give a slightly different (biased) result. Always match ddof on both sides.
  • Cross-reference: standard-deviation (ch02) for the s computation; seeded-sample-mean (this chapter) for the mean used here.