Confidence Intervals

A single number such as X̄ = 52 is a point estimate of μ. It is almost certainly not exactly right. A confidence interval gives a range of plausible values together with a measure of how much we trust it.

The idea

A 100(1 − α)% confidence interval for a parameter θ is an interval computed from the sample by a procedure that captures the true θ in 100(1 − α)% of samples. For a 95% interval, α = 0.05.

Interpretation, which GATE likes to test: the parameter is fixed and the interval is random. "95% confidence" means that if we repeated the sampling many times, about 95% of the intervals built this way would contain μ. It does not mean that there is a 95% probability that μ lies in the particular interval we computed: that interval either contains μ or it does not.

Mean with known σ: the z interval

If the population is normal, or n is large (by the CLT), and σ is known:

X̄ ± z_α/2 · σ/√n

where z_α/2 is the value with upper-tail area α/2 under N(0, 1):

The term E = z_α/2 σ/√n is the margin of error; the interval's width is 2E.

Example. n = 36, X̄ = 52, σ = 6. The standard error is 6/6 = 1, so the 95% interval is 52 ± 1.96 × 1 = (50.04, 53.96).

What makes an interval wider?

• higher confidence (larger z): a 99% interval is wider than a 95% one;
• more variability (larger σ);
• smaller samples (the width shrinks like 1/√n).

Mean with unknown σ: the t interval

Usually σ is unknown and is replaced by the sample standard deviation S. That adds uncertainty, so the normal value z is replaced by a value from the t distribution with n − 1 degrees of freedom (for a normal population):

X̄ ± t_{α/2, n−1} · S/√n

Example. n = 10, X̄ = 20, S = 3. For 95% confidence, t_{0.025, 9} = 2.262, so the interval is 20 ± 2.262 × 3/√10 = 20 ± 2.146 = (17.854, 22.146). With z = 1.96 it would have been narrower, 20 ± 1.859, but too optimistic.

For large n, t_{α/2, n−1} is close to z_α/2, and the two intervals agree.

Proportion

To estimate a population proportion p from x successes in n trials, use p̂ = x/n. For large n, p̂ is approximately normal with standard error √(p(1 − p)/n), so an approximate interval is

p̂ ± z_α/2 √(p̂(1 − p̂)/n)

Example. In a survey of 400 people, 240 prefer product A. p̂ = 0.6, and the 95% interval is 0.6 ± 1.96 × √(0.6 × 0.4/400) = 0.6 ± 1.96 × 0.0245 = 0.6 ± 0.048, i.e. about (0.552, 0.648).

Choosing the sample size

To get a margin of error at most E with known σ, solve E = z_α/2 σ/√n for n:

n = (z_α/2 σ / E)², rounded up.

Example. With σ = 10, how many observations give a 95% margin of error of 2? n = (1.96 × 10/2)² = 96.04, so 97 observations.

For a proportion with no prior guess, use p = 0.5 (the worst case): n = z² × 0.25 / E². For E = 0.03 at 95%, n = 1.96² × 0.25/0.0009 ≈ 1067.1, so 1068.

import random
from math import sqrt, ceil
from statistics import NormalDist, mean, stdev

z = NormalDist().inv_cdf(0.975)
print(f"z interval: 52 +/- {z * 6 / sqrt(36):.3f} -> ({52 - z:.2f}, {52 + z:.2f})")
t9 = 2.262
half = t9 * 3 / sqrt(10)
print(f"t interval: 20 +/- {half:.3f} -> ({20 - half:.3f}, {20 + half:.3f})")
p = 240 / 400
e = z * sqrt(p * (1 - p) / 400)
print(f"proportion: {p} +/- {e:.4f} -> ({p - e:.3f}, {p + e:.3f})")
print("sample size for E = 2, sigma = 10:", ceil((1.96 * 10 / 2) ** 2))

# What "95% confidence" means: about 95% of intervals contain the true mean.
random.seed(11)
mu, sigma, n, hits, reps = 50, 8, 25, 0, 10000
for _ in range(reps):
    xbar = mean(random.gauss(mu, sigma) for _ in range(n))
    if abs(xbar - mu) <= z * sigma / sqrt(n):
        hits += 1
print(f"{hits} of {reps} intervals contain the true mean ({hits / reps:.1%})")

z interval: 52 +/- 1.960 -> (50.04, 53.96)
t interval: 20 +/- 2.146 -> (17.854, 22.146)
proportion: 0.6 +/- 0.0480 -> (0.552, 0.648)
sample size for E = 2, sigma = 10: 97
9476 of 10000 intervals contain the true mean (94.8%)

Practice questions