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Chapter 61 · Solutions

Why We Sample — Worked Solutions ✅

Five challenges, each verified in code. Try them yourself first.

⚙️ Setup

import numpy as np
rng = np.random.default_rng(610)

CHALLENGE 1

The sampling-error law

Draw 2,000 samples of n=50 from Normal(100, 20) and check the observed SE against sigma/sqrt(n).

means = np.array([rng.normal(100,20,50).mean() for _ in range(2000)])
print(f"observed SE  = {means.std():.3f}")
print(f"sigma/sqrt(n)= {20/np.sqrt(50):.3f}")
print("match: the standard error of the mean is sigma/sqrt(n)")

observed SE  = 2.796
sigma/sqrt(n)= 2.828
match: the standard error of the mean is sigma/sqrt(n)

CHALLENGE 2

The fraction myth

Draw n=200 from a population of 2,000 and from one of 200,000. Show the standard error is essentially the same.

def se(N, n=200, reps=1500):
    pop = rng.normal(100, 20, N)
    return np.array([rng.choice(pop, n, replace=False).mean() for _ in range(reps)]).std()
print(f"N=2,000   (10%  sampled): SE = {se(2_000):.3f}")
print(f"N=200,000 (0.1% sampled): SE = {se(200_000):.3f}")
print("same precision: SE rides on n, not on the fraction n/N")

N=2,000   (10%  sampled): SE = 1.322
N=200,000 (0.1% sampled): SE = 1.459
same precision: SE rides on n, not on the fraction n/N

CHALLENGE 3

Bias cannot be outvoted

A convenience method reaches only group A (mean 60); the population is half A, half B (mean 40, true mean 50). Show the estimate stays wrong as n grows.

A = rng.normal(60, 8, 500_000); B = rng.normal(40, 8, 500_000)
pop = np.concatenate([A, B]); print(f"true population mean = {pop.mean():.1f}")
for n in [100, 1000, 50000]:
    conv = rng.choice(A, n, replace=False)        # only group A is reachable
    print(f"convenience sample n={n:>6}: estimate = {conv.mean():.1f}  (bias +{conv.mean()-pop.mean():.0f})")
print("more data sharpens the WRONG answer: bias is systematic, not random")

true population mean = 50.0
convenience sample n=   100: estimate = 60.2  (bias +10)
convenience sample n=  1000: estimate = 60.0  (bias +10)
convenience sample n= 50000: estimate = 60.0  (bias +10)
more data sharpens the WRONG answer: bias is systematic, not random

CHALLENGE 4

Quadruple for half

Show the standard error at n=400 is half the standard error at n=100.

pop = rng.normal(100, 20, 1_000_000)
se100 = np.array([rng.choice(pop,100,replace=False).mean() for _ in range(1500)]).std()
se400 = np.array([rng.choice(pop,400,replace=False).mean() for _ in range(1500)]).std()
print(f"SE at n=100: {se100:.3f}")
print(f"SE at n=400: {se400:.3f}")
print(f"ratio = {se400/se100:.2f}  (~0.5: 4x the data halves the error)")

SE at n=100: 1.935
SE at n=400: 0.991
ratio = 0.51  (~0.5: 4x the data halves the error)

CHALLENGE 5

Precision per dollar

If measuring one unit costs $1, compare the precision (1/SE) of a census of N=1,000,000 to a sample of n=2,000.

pop = rng.normal(100, 20, 1_000_000)
se_sample = np.array([rng.choice(pop,2000,replace=False).mean() for _ in range(1000)]).std()
prec_sample = 1/se_sample
prec_census = 1/ (20/np.sqrt(1_000_000))   # SE of the full census
print(f"sample n=2,000 : cost $2,000,   precision = {prec_sample:.2f}")
print(f"census N=1e6   : cost $1,000,000, precision = {prec_census:.2f}")
print(f"the census costs 500x more for only {prec_census/prec_sample:.1f}x the precision")

sample n=2,000 : cost $2,000,   precision = 2.22
census N=1e6   : cost $1,000,000, precision = 50.00
the census costs 500x more for only 22.5x the precision

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Statistics, Data Science and AI: A Visual Handbook · © 2026 John Fisher

←Back to the bookChapter 61 · Why We Sample