Confidence intervals — derivation for mean, proportion: Question bank

Before we start, three plain-word reminders so no symbol is unearned:

Keep straight which things are fixed ( $μ$ ) and which wobble ( $\overset{ˉ}{X}$ , the band). Nearly every trap below is a fixed-vs-random confusion in disguise.

True or false — justify

The probability that the true mean lies inside the computed interval

[501.02, 502.98]

is 0.95.

False. Once computed, both endpoints and

μ

are fixed numbers, so

μ

is either in or out — probability 0 or 1. The 95% describes the long-run success rate of the method, not this one band.

A 99% confidence interval is more likely to be "correct" for a single given sample than a 95% one.

True in the long-run sense: a higher confidence level means a larger fraction of intervals capture

μ

. The cost is that the 99% interval is wider (uses a bigger critical value), so it is less precise.

Widening the confidence level from 95% to 99% makes the interval narrower because we are "more sure."

False. More confidence demands a bigger critical value (

z_{0.005} = 2.576 > 1.96

), which widens the band. Certainty and precision trade off against each other.

For the same data, the

t

-interval is always at least as wide as the

z

-interval.

True. For any finite

n - 1

degrees of freedom,

t_{α /2, n - 1} > z_{α /2}

because the

t

-distribution has fatter tails, so its critical value is larger. They converge only as

n \to \infty

— see Student's t-distribution.

Doubling the sample size

n

halves the margin of error.

False. The margin scales like

1/ n

, so quadrupling

n

halves it; doubling

n

shrinks it only by a factor

1/ 2 \approx 0.71

. This is why big precision gains get expensive — see Sample Size Determination.

If a 95% CI for

μ

excludes the value 0, then a two-sided hypothesis test of

μ = 0

at level

α = 0.05

rejects.

True. This is the CI–test duality: the CI is exactly the set of null values that would not be rejected, so a value outside it is rejected — see Hypothesis Testing.

The confidence interval tells us the range in which 95% of individual data values fall.

False. It bounds the mean

μ

, whose spread is the standard error

σ / n

, not the raw data spread

σ

. A range for individual values (a prediction/tolerance interval) is much wider.

For a proportion, the standard error

\overset{p}{^} (1 - \overset{p}{^}) / n

is largest when

\overset{p}{^} = 0.5

True. The function

p (1 - p)

peaks at

p = 0.5

(value

0.25

) and drops toward the extremes, so a 50/50 split is the hardest to pin down — this drives worst-case sample-size planning.

The Central Limit Theorem guarantees

\overset{ˉ}{X}

is exactly normal for any sample size.

False. It gives an approximate normal shape for large

n

; for small

n

from a skewed population the approximation can be poor. See Central Limit Theorem.

Spot the error

"We're 95% confident, so we look up the 95th percentile

z = 1.645

Error: two-sided intervals split

α = 0.05

into two tails of

0.025

each, so we need

z_{0.025} = 1.96

. The value

1.645

is a one-sided 95% bound.

"Since we don't know

σ

, we just use

s

in place of it and keep the

z

critical value, even for

n = 12

Error: replacing

σ

by the random

s

adds uncertainty, so for small

n

we must use the wider

t_{n - 1}

critical value, not

z

. Only for large

n

do they roughly agree.

"Margin of error

= z_{α /2} σ

Error: the missing

n

makes the interval absurdly wide. The correct margin is

z_{α /2} σ / n

because we bound the mean, whose spread is the standard error — see Standard Error.

"With 3 successes in 20 trials,

\overset{p}{^} = 0.15

, we apply the Wald interval directly."

Error: the validity check

n \overset{p}{^} = 20 (0.15) = 3 < 5

fails, so the normal approximation is unreliable. With so few successes the Wald interval can even spill below 0; use an exact or adjusted method instead.

"The

t

-distribution has

n

degrees of freedom."

Error: it has

n - 1

. Computing

s

imposes the constraint

\sum (X_{i} - \overset{ˉ}{X}) = 0

, using up one degree of freedom and leaving

n - 1

free deviations.

"Because our 95% interval is

[65.7, 74.3]

, if we sampled again we'd get a mean in that range 95% of the time."

Error: the interval is about the fixed parameter

μ

, not about future sample means. A future

\overset{ˉ}{X}

could easily fall outside; the 95% refers to how often intervals capture

μ

"A wider interval means our estimate is better."

Error: a wider interval means less precision — we've pinned

μ

down over a larger range. Narrow intervals (from larger

n

or smaller

σ

) are the precise ones; width and quality are inverted here.

Why questions

Why do we split

α

into two equal tails instead of putting it all in one?

Because a symmetric two-sided interval brackets

μ

from both above and below, so we reserve

α /2

probability in each tail; the central band then holds

1 - α

. A one-tailed split gives a one-sided bound, a different question.

Why does the variance of

\overset{ˉ}{X}

shrink as

σ^{2} / n

rather than staying

σ^{2}

Averaging

n

independent values lets their random ups and downs partly cancel; the variance of the sum is

n σ^{2}

, and dividing by

n

scales variance by

1/ n^{2}

, leaving

σ^{2} / n

Why must a proportion be treated as a special case of a mean?

A 0/1 (Bernoulli) trial has mean

p

and variance

p (1 - p)

, so

\overset{p}{^}

is literally the sample mean of these indicators; the whole mean machinery transfers with

σ^{2}

replaced by

p (1 - p)

— see Bernoulli & Binomial Distributions.

Why do we plug

\overset{p}{^}

into the standard error when the true SE uses the unknown

p

We can't compute

p (1 - p) / n

without knowing

p

, so we substitute our best estimate

\overset{p}{^}

. This is the "Wald" approximation and is why the interval is only reliable when successes and failures are both plentiful.

Why does the

t

-interval get closer to the

z

-interval as

n

grows?

With more data,

s

estimates

σ

more accurately, so the extra uncertainty that fattened the

t

-tails vanishes and

t_{n - 1} \to N (0, 1)

Why does "95% confidence" refer to the procedure and not to one interval?

Randomness lives in the sampling, not in

μ

. Before sampling, the yet-to-be-built band has a 95% chance of catching the fixed

μ

; after we observe data, the band is fixed too, so no probability remains to speak of.

Edge cases

What happens to the CI as

n \to \infty

(with

σ

fixed)?

The standard error

σ / n \to 0

, so the margin shrinks to zero and the interval collapses onto the true

μ

— infinite data pins the mean exactly.

What if

\overset{p}{^} = 0

(zero successes observed)?

The Wald SE

\overset{p}{^} (1 - \overset{p}{^}) / n = 0

, giving the degenerate interval

[0, 0]

, which is clearly wrong —

p

isn't certainly zero. This is exactly where the approximation breaks and an adjusted (e.g. Wilson) interval is required.

What if the population is already exactly normal but

n = 2

The

z

-approximation via CLT isn't needed for shape (it's normal already), but

σ

is unknown, so we still use

t_{n - 1} = t_{1}

, whose tails are extremely fat — the interval is very wide, honestly reflecting how little two points tell us.

What happens to the margin when the confidence level is 100%?

The critical value

z_{α /2}

diverges to infinity (no finite band can be certain under a normal model), so a 100% interval is

(- \infty, \infty)

— technically always correct, practically useless.

What if

σ = 0

(no variability in the population)?

The standard error is

0

, so every sample gives

\overset{ˉ}{X} = μ

exactly and the interval collapses to the single point

{μ}

— with no randomness there is nothing to be uncertain about.

If two researchers each build a 95% CI from independent samples and the intervals don't overlap, is one of them definitely wrong?

Not necessarily — non-overlap is strong evidence the means differ but each interval individually still had a 5% miss chance, and non-overlap is a conservative (not exact) test for a difference.

Is $μ$ random or fixed? ::: Fixed and unknown; the interval is the random object (before data are seen). Does higher confidence give a narrower or wider interval? ::: Wider — a larger critical value is needed to catch $μ$ more often. When is the proportion standard error largest? ::: At $\overset{p}{^} = 0.5$ , since $p (1 - p)$ peaks there. Why $n - 1$ and not $n$ degrees of freedom? ::: Estimating $s$ imposes one constraint $\sum (X_{i} - \overset{ˉ}{X}) = 0$ , removing one free deviation. Non-overlapping 95% CIs — proof of difference? ::: Strong but not definitive evidence; it's a conservative check, and each interval still had its own 5% miss rate.

Connections

Confidence intervals — derivation for mean, proportion — the parent derivation these traps stress-test.
Central Limit Theorem — why the normal shape (and its limits for small skewed $n$ ).
Standard Error — the $σ / n$ that so many errors drop.
Student's t-distribution — fatter tails, $n - 1$ df, $z$ -vs- $t$ traps.
Hypothesis Testing — CI–test duality behind the "excludes 0" item.
Bernoulli & Binomial Distributions — proportion-as-mean and the $p (1 - p)$ peak.
Sample Size Determination — the $1/ n$ scaling behind the "doubling" trap.

Question bank — Confidence intervals — derivation for mean, proportion