09-01: Confidence Interval for a Mean¶

A single number estimate is almost certainly wrong. A confidence interval reports a range plus a statement of how often that procedure captures the truth — the first genuinely inferential tool in the course.

Point vs. Interval Estimates¶

Point estimate      A single value:  x̄ = 24.6
                    Best guess, but you know it isn't exactly μ.

Interval estimate   A range:  24.6 ± 1.2  →  (23.4, 25.8)
                    Carries its own uncertainty with it.

x̄ is the point estimate of μ, p̂ of p, and s² of σ². Each is unbiased — right on average across repeated samples.

The Anatomy of Every Confidence Interval¶

Estimate   ±   (critical value)  ×  (standard error)
             └──────────── margin of error, E ────────────┘

Everything in Chapters 09–11 is this one line with different pieces plugged in.

Case 1 — `σ` Known (the z-interval)¶

Use when the population standard deviation is genuinely known, or n is large and σ is well established.

                    σ
x̄  ±  z_(α/2) · ───────
                   √n

Conditions:  random sample AND (population normal OR n ≥ 30)

Critical z-values¶

Confidence	`α`	`α/2`	`z_(α/2)`
90%	0.10	0.050	1.645
95%	0.05	0.025	1.960
98%	0.02	0.010	2.326
99%	0.01	0.005	2.576

Case 2 — `σ` Unknown (the t-interval) — the usual case¶

When you must estimate σ with s, the extra uncertainty widens the interval. That is what the Student t-distribution encodes.

                    s
x̄  ±  t_(α/2, df) · ───────           df = n − 1
                   √n

Conditions:  random sample AND (population approximately normal OR n ≥ 30)

The t-distribution¶

Bell-shaped and symmetric about 0, like z, but with heavier tails.
Its shape depends on the degrees of freedom, df = n − 1.
Smaller df → fatter tails → larger critical value → wider interval.
As df → ∞, t → z. At df = 30, t = 2.042 vs. z = 1.960; at df = 100, t = 1.984.

Tip

Which one? If you computed s from your own data — and you almost always did — use t. Reserve z for when σ is given in the problem statement.

Interpreting a Confidence Interval Correctly¶

For a 95% CI of (23.4, 25.8):

✅ Correct. "We are 95% confident that the true population mean lies between 23.4 and 25.8." Or, more precisely: "If we repeated this sampling procedure many times, 95% of the intervals produced would contain μ."

❌ Wrong. "There is a 95% probability that μ is between 23.4 and 25.8." μ is a fixed constant, not a random variable — it either is or is not in this particular interval. The randomness lives in the interval, not in μ.

❌ Wrong. "95% of the data lies between 23.4 and 25.8." That would be a prediction interval, and it is much wider.

❌ Wrong. "95% of sample means fall in this interval."

What Changes the Width¶

Width  =  2 × E  =  2 × t · s/√n

Change	Effect on width	Why
Higher confidence (95% → 99%)	Wider	Larger critical value
Larger `n`	Narrower (by `√n`)	Smaller standard error
Larger `s`	Wider	More variable data

You cannot have high confidence and a narrow interval and a small sample. Pick two.

Sample Size for a Desired Margin of Error¶

Solve E = z·σ/√n for n:

          ( z_(α/2) · σ )²
n  =   ( ───────────────── )              ALWAYS round UP
          (       E       )

If σ is unknown, estimate it from a pilot study, from prior work, or with the range rule of thumb σ ≈ range/4.

Example. 95% confidence, E = 2, σ = 10:

n = (1.96 × 10 / 2)² = (9.8)² = 96.04  →  n = 97

To halve E to 1: n = (19.6)² = 384.2 → 385. Four times the sample for twice the precision.

Worked Examples¶

(a) `σ` known — z-interval¶

A sample of n = 50 bulbs has x̄ = 1,250 hours; the manufacturer's process has σ = 85 hours. Build a 95% CI for μ.

SE = 85 / √50 = 12.021
z  = 1.96
E  = 1.96 × 12.021 = 23.56

CI = 1250 ± 23.56  =  (1226.4, 1273.6) hours

(b) `σ` unknown — t-interval¶

A sample of n = 15 commute times gives x̄ = 27.4 minutes and s = 6.2 minutes. Build a 95% CI.

df = 14,   t(0.025, 14) = 2.145

SE = 6.2 / √15 = 1.601
E  = 2.145 × 1.601 = 3.434

CI = 27.4 ± 3.43  =  (23.97, 30.83) minutes

Note how much wider this is than the z-interval would be: using z = 1.96 would have given ± 3.14. The extra 0.29 minutes is the price of not knowing σ.

Reporting it: "The mean commute time is 27.4 minutes (95% CI: 24.0 to 30.8 minutes)."

Confidence Interval for a Variance or Standard Deviation¶

Uses the chi-square distribution, which is right-skewed — so the interval is not symmetric about s².

    (n − 1)s²                (n − 1)s²
   ───────────   <  σ²  <   ───────────           df = n − 1
    χ²_(α/2)                χ²_(1−α/2)

Take square roots for σ.   Requires the population to be NORMAL — this
procedure is not robust to departures from normality.

Excel¶

' ── z-interval  (σ known) ───────────────────────────────────────────
' B1 = x̄ = 1250, B2 = σ = 85, B3 = n = 50, B4 = confidence = 0.95
=CONFIDENCE.NORM(1-B4, B2, B3)      ' margin of error E   -> 23.5604
=B1-CONFIDENCE.NORM(1-B4,B2,B3)     ' lower bound         -> 1226.44
=B1+CONFIDENCE.NORM(1-B4,B2,B3)     ' upper bound         -> 1273.56

' Built by hand
=NORM.S.INV(1-(1-B4)/2)             ' critical z          -> 1.95996
=B2/SQRT(B3)                        ' standard error      -> 12.0208
=NORM.S.INV(0.975)*B2/SQRT(B3)      ' E                   -> 23.5604

' ── t-interval  (σ unknown — the usual case) ────────────────────────
' D1 = x̄ = 27.4, D2 = s = 6.2, D3 = n = 15, D4 = 0.95
=CONFIDENCE.T(1-D4, D2, D3)         ' margin of error E   -> 3.4335
=D1-CONFIDENCE.T(1-D4,D2,D3)        ' lower bound         -> 23.9665
=D1+CONFIDENCE.T(1-D4,D2,D3)        ' upper bound         -> 30.8335

' Built by hand
=T.INV.2T(1-D4, D3-1)               ' two-tailed critical t -> 2.14479
=T.INV(1-(1-D4)/2, D3-1)            ' same value, one-tailed form
=T.INV.2T(0.05,14)*D2/SQRT(D3)      ' E                     -> 3.4335

' ── Straight from raw data ──────────────────────────────────────────
=AVERAGE(A2:A16)
=STDEV.S(A2:A16)/SQRT(COUNT(A2:A16))                       ' standard error
=AVERAGE(A2:A16)-T.INV.2T(0.05,COUNT(A2:A16)-1)*STDEV.S(A2:A16)/SQRT(COUNT(A2:A16))
=AVERAGE(A2:A16)+T.INV.2T(0.05,COUNT(A2:A16)-1)*STDEV.S(A2:A16)/SQRT(COUNT(A2:A16))

' Analysis ToolPak ▸ Descriptive Statistics ▸ tick
'   "Confidence Level for Mean: 95%"  →  reports the margin of error E

' ── Sample size ─────────────────────────────────────────────────────
=ROUNDUP((NORM.S.INV(0.975)*10/2)^2, 0)     ' -> 97
=ROUNDUP((NORM.S.INV(0.975)*10/1)^2, 0)     ' -> 385

' ── CI for a variance / standard deviation ──────────────────────────
=(D3-1)*D2^2/CHISQ.INV.RT(0.025, D3-1)      ' lower bound for σ²
=(D3-1)*D2^2/CHISQ.INV.RT(0.975, D3-1)      ' upper bound for σ²
=SQRT((D3-1)*D2^2/CHISQ.INV.RT(0.025,D3-1)) ' lower bound for σ

R¶

# ── From summary statistics ────────────────────────────────────────
ci_z <- function(xbar, sigma, n, conf = 0.95) {
  z <- qnorm(1 - (1 - conf)/2)
  e <- z * sigma / sqrt(n)
  c(estimate = xbar, margin = e, lower = xbar - e, upper = xbar + e)
}
round(ci_z(1250, 85, 50), 3)
# estimate   margin    lower    upper
# 1250.000   23.560 1226.440 1273.560

ci_t <- function(xbar, s, n, conf = 0.95) {
  t <- qt(1 - (1 - conf)/2, df = n - 1)
  e <- t * s / sqrt(n)
  c(estimate = xbar, margin = e, lower = xbar - e, upper = xbar + e)
}
round(ci_t(27.4, 6.2, 15), 3)
# estimate   margin    lower    upper
#   27.400    3.434   23.966   30.834

qt(0.975, df = 14)      # critical t -> 2.144787
qnorm(0.975)            # critical z -> 1.959964

# ── From raw data — t.test() gives the CI directly ─────────────────
# (a different 15-commute sample, to show the raw-data route)
commute <- c(22, 31, 28, 19, 35, 27, 24, 33, 29, 21, 30, 26, 32, 25, 29)
t.test(commute)                       # default: 95% CI for the mean
t.test(commute, conf.level = 0.99)$conf.int
t.test(commute)$conf.int              # just the interval

# ── z-interval from raw data (BSDA package) ────────────────────────
# library(BSDA);  z.test(commute, sigma.x = 6)$conf.int

# ── Sample size ────────────────────────────────────────────────────
n_for_mean <- function(sigma, E, conf = 0.95)
  ceiling((qnorm(1 - (1 - conf)/2) * sigma / E)^2)
n_for_mean(10, 2)      # 97
n_for_mean(10, 1)      # 385

# ── CI for a variance / SD ─────────────────────────────────────────
ci_var <- function(s, n, conf = 0.95) {
  a  <- 1 - conf
  lo <- (n - 1) * s^2 / qchisq(1 - a/2, n - 1)
  hi <- (n - 1) * s^2 / qchisq(a/2,     n - 1)
  c(var_lower = lo, var_upper = hi, sd_lower = sqrt(lo), sd_upper = sqrt(hi))
}
round(ci_var(6.2, 15), 3)

# ── What "95% confident" actually means — simulate it ──────────────
set.seed(7)
covered <- replicate(1000, {
  s  <- rnorm(20, mean = 100, sd = 15)
  ci <- t.test(s)$conf.int
  ci[1] <= 100 && 100 <= ci[2]
})
mean(covered)          # ≈ 0.95 — the intervals move, μ never does

Python¶

import numpy as np
from scipy import stats

# ── From summary statistics ────────────────────────────────────────
def ci_z(xbar, sigma, n, conf=0.95):
    z = stats.norm.ppf(1 - (1 - conf) / 2)
    e = z * sigma / np.sqrt(n)
    return xbar, e, xbar - e, xbar + e

ci_z(1250, 85, 50)      # (1250, 23.560, 1226.44, 1273.56)

def ci_t(xbar, s, n, conf=0.95):
    t = stats.t.ppf(1 - (1 - conf) / 2, df=n - 1)
    e = t * s / np.sqrt(n)
    return xbar, e, xbar - e, xbar + e

ci_t(27.4, 6.2, 15)     # (27.4, 3.4335, 23.9665, 30.8335)

stats.t.ppf(0.975, df=14)   # critical t -> 2.144787
stats.norm.ppf(0.975)       # critical z -> 1.959964

# ── From raw data ──────────────────────────────────────────────────
# (a different 15-commute sample, to show the raw-data route)
commute = np.array([22, 31, 28, 19, 35, 27, 24, 33, 29, 21, 30, 26, 32, 25, 29])

stats.t.interval(0.95,
                 df=commute.size - 1,
                 loc=commute.mean(),
                 scale=stats.sem(commute))          # scipy's SEM = s/√n

stats.norm.interval(0.95, loc=commute.mean(), scale=6 / np.sqrt(commute.size))

# statsmodels gives it in one call
# from statsmodels.stats.weightstats import DescrStatsW
# DescrStatsW(commute).tconfint_mean(alpha=0.05)

# ── Sample size ────────────────────────────────────────────────────
def n_for_mean(sigma, E, conf=0.95):
    z = stats.norm.ppf(1 - (1 - conf) / 2)
    return int(np.ceil((z * sigma / E) ** 2))

n_for_mean(10, 2)       # 97
n_for_mean(10, 1)       # 385

# ── CI for a variance / SD ─────────────────────────────────────────
def ci_var(s, n, conf=0.95):
    a = 1 - conf
    lo = (n - 1) * s**2 / stats.chi2.ppf(1 - a/2, n - 1)
    hi = (n - 1) * s**2 / stats.chi2.ppf(a/2,     n - 1)
    return lo, hi, np.sqrt(lo), np.sqrt(hi)

ci_var(6.2, 15)

# ── Simulating the meaning of 95% ──────────────────────────────────
rng = np.random.default_rng(7)
hits = 0
for _ in range(1000):
    s = rng.normal(100, 15, 20)
    lo, hi = stats.t.interval(0.95, s.size - 1, s.mean(), stats.sem(s))
    hits += lo <= 100 <= hi
hits / 1000             # ≈ 0.95

Quick Reference¶

Task	Excel	R	Python
Critical z	`NORM.S.INV(1-α/2)`	`qnorm(1-α/2)`	`norm.ppf(1-α/2)`
Critical t	`T.INV.2T(α, df)`	`qt(1-α/2, df)`	`t.ppf(1-α/2, df)`
Margin of error (z)	`CONFIDENCE.NORM(α,σ,n)`	`qnorm(...)*σ/sqrt(n)`	`norm.ppf(...)*σ/√n`
Margin of error (t)	`CONFIDENCE.T(α,s,n)`	`qt(...)*s/sqrt(n)`	`t.ppf(...)*s/√n`
Standard error	`STDEV.S/SQRT(COUNT)`	`sd(x)/sqrt(length(x))`	`stats.sem(x)`
CI from raw data	ToolPak ▸ Descriptive Stats	`t.test(x)$conf.int`	`stats.t.interval(...)`
Sample size	`ROUNDUP((z*σ/E)^2,0)`	`ceiling((qnorm*σ/E)^2)`	`ceil((zσ/E)*2)`
CI for `σ²`	`(n-1)s²/CHISQ.INV.RT`	`(n-1)*s^2/qchisq(...)`	`(n-1)s*2/chi2.ppf(...)`

Common Mistakes¶

Using z when σ is unknown. If you computed s from the data, use t.
Using df = n instead of n − 1.
Confusing Excel's T.INV (one tail) with T.INV.2T (two tails). For a 95% CI: T.INV.2T(0.05, df) or T.INV(0.975, df) — both give 2.145 at df = 14.
Saying "there is a 95% probability that μ is in this interval".
Rounding the sample size down. Always round up.
Building a CI from a convenience sample. The formula assumes random sampling (08-01).

Exercises: 09-01: Exercises — Confidence Interval for a Mean

⬅️ Previous: 08-02: Sampling Distributions and the Central Limit Theorem ➡️ Next: 09-02: Confidence Interval for a Proportion