Mathematics 9709 · Probability & Statistics 2

Confidence interval for a mean

A confidence interval is your estimate plus-or-minus a margin for sampling luck: $\bar x\pm k\cdot\tfrac{\sigma}{\sqrt n}$ . “95% confident” means that if you resampled many times, 95% of the intervals you'd build would capture $\mu$ — not that this one has a 95% chance.

Estimate ± k × standard error

Same standard error as the test, $\sigma/\sqrt n$ (or $s/\sqrt n$ for a large sample). Pick $k$ from the level: $90\%\to1.645,\ 95\%\to1.960,\ 98\%\to2.326,\ 99\%\to2.576$ . A higher level needs a bigger $k$ , so the interval gets wider (more confident, but less precise).

CI = x̄ ± k·SE. Higher level → bigger k → wider interval. The exam payoff is the reading: does the interval contain the claimed value?

The 95% confidence band x̄ ± 1.96·SE = (62.9, 82.1). It reaches past the threshold 80, so the true mean could exceed 80 — the interval drives the decision.

Worked example

Start simple: σ known, decide vs a threshold

Pond samples are normal with $\sigma=12$ . Six 100 ml samples give counts 56, 102, 48, 74, 88, 67. (a) Find a 95% CI for the mean count. (b) If a mean above 80 calls for action, what do you advise?

$\bar x=\tfrac{1}{6}(56+\dots+67)=72.5$ ; $\text{SE}=\tfrac{12}{\sqrt6}=4.90$ .
$72.5\pm1.96\cdot4.90=72.5\pm9.6$ , so the CI is $(62.9,\ 82.1)$ .
The upper end 82.1 is above 80, so the true mean could exceed 80.

Answer

(62.9,\ 82.1)

The interval reaches past 80, so advise action — and that more samples be taken.

Worked example

Build up: 99% CI vs a target

Sweet packets claim 100 g, with $\sigma=5$ . Ten packets give $\bar x=103.8$ . Find a 99% CI; does the mechanism need adjusting?

$103.8\pm2.576\cdot\tfrac{5}{\sqrt{10}}=103.8\pm4.07$ .
CI $(99.7,\ 107.9)$ — it dips below the stated 100 g.

Answer

(99.7,\ 107.9)

The interval dips below 100 g, so the true mean could be under target — the mechanism may need adjustment.

Why not just sample huge? The margin only shrinks like 1/√n

The width is $2k\cdot\sigma/\sqrt n$ , so to halve the margin you must quadruple $n$ . Sixteen times the sample buys only a quarter of the margin — drag $n$ below and watch the curve tighten fast at first, then crawl.

n = 1·X̄ ~ N(50, 10²/n)·SE = σ/√n = 10.00

Drag n: the sampling distribution keeps the same centre but narrows by 1/√n. Doubling n does not halve the margin — that takes 4× the sample.

Worked example

σ unknown, large sample, then reverse to the level

60 strawberries: $\sum x=972,\ \sum x^2=17304.78$ g. (a) Find a 90% CI for the mean mass. (b) A CI from this sample has width 3.65 g — what confidence level is that?

$\bar x=\tfrac{972}{60}=16.2$ ; $s^2=\tfrac{1}{59}\big(17304.78-\tfrac{972^2}{60}\big)=26.41$ , so $s=5.14$ .
(a) $16.2\pm1.645\cdot\tfrac{5.14}{\sqrt{60}}=16.2\pm1.09$ , giving $(15.1,\ 17.3)$ .
(b) Width $=2k\cdot\tfrac{s}{\sqrt n}$ : $2k\cdot\tfrac{5.14}{\sqrt{60}}=3.65\Rightarrow k=2.75$ .
$\Phi(2.75)=0.997$ , so the level is $100\,(2{\cdot}0.997-1)=99.4\%$ .

Answer

(15.1,\ 17.3)\quad\text{and}\quad\alpha\approx99.4\%

Level C%	k = Φ⁻¹((1 + C/100)/2)
90%	1.645
92%	1.751
95%	1.960
98%	2.326
99%	2.576

Examiner note

For a non-standard level (92%, 93%, …) read

k

from the inverse normal,

\Phi(k)=\tfrac{1+C/100}{2}

; write the answer as an interval

(a,b)

, not a single margin. Large

n

means the population need not be Normal (CLT). (9709/62/O/N/24 Q2) (9709/62/O/N/23 Q2)

Working backwards from a given interval

For a symmetric CI $(a,b)$ : centre $\bar x=\tfrac{a+b}{2}$ , half-width $\tfrac{b-a}{2}=k\cdot\text{SE}$ . Recover $k$ from the half-width, then the level is $100\,(2\Phi(k)-1)\%$ (e.g. $k=1.751\Rightarrow92\%$ ). The full width is $2k\cdot\text{SE}$ , so keep the 2.

Common mistake

The wrong

k

; dividing by

n

not

\sqrt n

; forgetting the factor 2 in the width; and misreading “95% confident” as “95% probability

\mu

is in this interval” (safe phrasing: 95% of such intervals contain

\mu

PreviousHypothesis test for a mean NextConfidence interval for a proportion

Confidence interval for a mean

Estimate ± k × standard error

CI = x̄ ± k·SE. Higher level → bigger k → wider interval. The exam payoff is the reading: does the interval contain the claimed value?

The 95% confidence band x̄ ± 1.96·SE = (62.9, 82.1). It reaches past the threshold 80, so the true mean could exceed 80 — the interval drives the decision.

Worked example

Start simple: σ known, decide vs a threshold

Pond samples are normal with $\sigma=12$ . Six 100 ml samples give counts 56, 102, 48, 74, 88, 67. (a) Find a 95% CI for the mean count. (b) If a mean above 80 calls for action, what do you advise?

$\bar x=\tfrac{1}{6}(56+\dots+67)=72.5$ ; $\text{SE}=\tfrac{12}{\sqrt6}=4.90$ .
$72.5\pm1.96\cdot4.90=72.5\pm9.6$ , so the CI is $(62.9,\ 82.1)$ .
The upper end 82.1 is above 80, so the true mean could exceed 80.

Answer

(62.9,\ 82.1)

The interval reaches past 80, so advise action — and that more samples be taken.

Worked example

Build up: 99% CI vs a target

Sweet packets claim 100 g, with $\sigma=5$ . Ten packets give $\bar x=103.8$ . Find a 99% CI; does the mechanism need adjusting?

$103.8\pm2.576\cdot\tfrac{5}{\sqrt{10}}=103.8\pm4.07$ .
CI $(99.7,\ 107.9)$ — it dips below the stated 100 g.

Answer

(99.7,\ 107.9)

The interval dips below 100 g, so the true mean could be under target — the mechanism may need adjustment.

Why not just sample huge? The margin only shrinks like 1/√n

n = 1·X̄ ~ N(50, 10²/n)·SE = σ/√n = 10.00

Drag n: the sampling distribution keeps the same centre but narrows by 1/√n. Doubling n does not halve the margin — that takes 4× the sample.

Worked example

σ unknown, large sample, then reverse to the level

60 strawberries: $\sum x=972,\ \sum x^2=17304.78$ g. (a) Find a 90% CI for the mean mass. (b) A CI from this sample has width 3.65 g — what confidence level is that?

$\bar x=\tfrac{972}{60}=16.2$ ; $s^2=\tfrac{1}{59}\big(17304.78-\tfrac{972^2}{60}\big)=26.41$ , so $s=5.14$ .
(a) $16.2\pm1.645\cdot\tfrac{5.14}{\sqrt{60}}=16.2\pm1.09$ , giving $(15.1,\ 17.3)$ .
(b) Width $=2k\cdot\tfrac{s}{\sqrt n}$ : $2k\cdot\tfrac{5.14}{\sqrt{60}}=3.65\Rightarrow k=2.75$ .
$\Phi(2.75)=0.997$ , so the level is $100\,(2{\cdot}0.997-1)=99.4\%$ .

Answer

(15.1,\ 17.3)\quad\text{and}\quad\alpha\approx99.4\%

Level C%	k = Φ⁻¹((1 + C/100)/2)
90%	1.645
92%	1.751
95%	1.960
98%	2.326
99%	2.576

Examiner note

For a non-standard level (92%, 93%, …) read

k

from the inverse normal,

\Phi(k)=\tfrac{1+C/100}{2}

; write the answer as an interval

(a,b)

, not a single margin. Large

n

means the population need not be Normal (CLT). (9709/62/O/N/24 Q2) (9709/62/O/N/23 Q2)

Working backwards from a given interval

Common mistake

The wrong

k

; dividing by

n

not

\sqrt n

; forgetting the factor 2 in the width; and misreading “95% confident” as “95% probability

\mu

is in this interval” (safe phrasing: 95% of such intervals contain

\mu