Mathematics 9709 · Probability & Statistics 2

Confidence interval for a proportion

Same margin idea, but now you estimate a proportion (a vote, a defect rate). The sample proportion $\hat p=r/n$ is approximately normal for large $n$ , with standard error $\sqrt{\hat p\hat q/n}$ . The headline question: does the interval stay on one side of the boundary?

p̂ ± k·√(p̂q̂ / n)

Because $E(\hat P)=p$ and $\operatorname{Var}(\hat P)=p(1-p)/n$ , the proportion behaves like a mean with its own standard error. Use $\hat p=r/n$ , $\hat q=1-\hat p$ and the same $k$ table. It is a normal approximation (needs $np>5$ , $nq>5$ ) with no continuity correction, unlike the Chapter 5 probabilities.

Proportion CI = p̂ ± k·√(p̂q̂/n), p̂ = r/n. Same k table; no continuity correction. Read it against the boundary (0.5, a pass rate, …).

A poll of 200 with 108 for Sofia gives p̂ = 0.54, 95% CI (0.471, 0.609). The interval straddles 0.5, so we can't be confident she'll win — the classic poll caution.

Worked example

Start simple: opinion poll vs 0.5

200 voters polled, 108 favour Sofia (the winner needs over 50%). Find a 95% CI for her support.

$\hat p=\tfrac{108}{200}=0.54$ , $\hat q=0.46$ .
$0.54\pm1.96\sqrt{\tfrac{0.54\times0.46}{200}}=0.54\pm0.069$ .
CI $(0.471,\ 0.609)$ — it straddles 0.5.

Answer

(0.471,\ 0.609)

Despite $\hat p>0.5$ , her true share could be below half.

Worked example

Build up: 99% CI vs a classification boundary

A puzzle is “easy” if over 70% solve it in time, else “average”. Of 120 people, 87 solved it. (a) Find a 99% CI for the proportion and classify it. (b) Of 200 such samples, how many 99% CIs would contain the true proportion?

$\hat p=\tfrac{87}{120}=0.725$ , $k=2.576$ .
$0.725\pm2.576\sqrt{\tfrac{0.725\times0.275}{120}}=0.725\pm0.105$ , so CI $(0.62,\ 0.83)$ .
(a) The CI crosses 0.70, so it is “easy or average” — not decided.
(b) A 99% CI captures $p$ 99% of the time: $200\times0.99=198$ .

Answer

(0.62,\ 0.83)\text{ crosses }0.70;\quad 198\text{ of the 200 CIs.}

Reverse: what sample size hits a target margin?

The margin is $k\sqrt{\hat p\hat q/n}$ . Set it equal to the allowed error $E$ and solve for $n$ : $n=\Big(\tfrac{k}{E}\Big)^2\hat p\hat q$ .

Worked example

Sample size for a required accuracy

A college will switch days if 80% of apprentices agree. How many should be polled to be 90% certain of measuring that 80% support to within 5%?

$\hat p=0.8$ , $E=0.05$ , 90% so $k=1.645$ .
$n=\Big(\tfrac{1.645}{0.05}\Big)^2\times0.8\times0.2=173.2$ .
A sample size must round up — rounding down to 173 leaves the margin just above 5%, so take $n=174$ .

Answer

n=174

Examiner note

Use the same

k

(inverse normal for odd levels), keep the square root in

\sqrt{\hat p\hat q/n}

, and write the answer as an interval. For a sample size,

n=(k/E)^2\hat p\hat q

. (9709/62/M/J/23 Q1) (9709/62/M/J/24 Q3)

Common mistake

Dropping the square root in

\sqrt{\hat p\hat q/n}

; applying a continuity correction (not used here); confusing the proportion SE with the mean SE

\sigma/\sqrt n

; and not interpreting against the relevant boundary (0.5, a pass mark, …).

PreviousConfidence interval for a mean

Confidence interval for a proportion

p̂ ± k·√(p̂q̂ / n)

Proportion CI = p̂ ± k·√(p̂q̂/n), p̂ = r/n. Same k table; no continuity correction. Read it against the boundary (0.5, a pass rate, …).

A poll of 200 with 108 for Sofia gives p̂ = 0.54, 95% CI (0.471, 0.609). The interval straddles 0.5, so we can't be confident she'll win — the classic poll caution.

Worked example

Start simple: opinion poll vs 0.5

200 voters polled, 108 favour Sofia (the winner needs over 50%). Find a 95% CI for her support.

$\hat p=\tfrac{108}{200}=0.54$ , $\hat q=0.46$ .
$0.54\pm1.96\sqrt{\tfrac{0.54\times0.46}{200}}=0.54\pm0.069$ .
CI $(0.471,\ 0.609)$ — it straddles 0.5.

Answer

(0.471,\ 0.609)

Despite $\hat p>0.5$ , her true share could be below half.

Worked example

Build up: 99% CI vs a classification boundary

$\hat p=\tfrac{87}{120}=0.725$ , $k=2.576$ .
$0.725\pm2.576\sqrt{\tfrac{0.725\times0.275}{120}}=0.725\pm0.105$ , so CI $(0.62,\ 0.83)$ .
(a) The CI crosses 0.70, so it is “easy or average” — not decided.
(b) A 99% CI captures $p$ 99% of the time: $200\times0.99=198$ .

Answer

(0.62,\ 0.83)\text{ crosses }0.70;\quad 198\text{ of the 200 CIs.}

Reverse: what sample size hits a target margin?

The margin is $k\sqrt{\hat p\hat q/n}$ . Set it equal to the allowed error $E$ and solve for $n$ : $n=\Big(\tfrac{k}{E}\Big)^2\hat p\hat q$ .

Worked example

Sample size for a required accuracy

A college will switch days if 80% of apprentices agree. How many should be polled to be 90% certain of measuring that 80% support to within 5%?

$\hat p=0.8$ , $E=0.05$ , 90% so $k=1.645$ .
$n=\Big(\tfrac{1.645}{0.05}\Big)^2\times0.8\times0.2=173.2$ .
A sample size must round up — rounding down to 173 leaves the margin just above 5%, so take $n=174$ .

Answer

n=174

Examiner note

Use the same

k

(inverse normal for odd levels), keep the square root in

\sqrt{\hat p\hat q/n}

, and write the answer as an interval. For a sample size,

n=(k/E)^2\hat p\hat q

. (9709/62/M/J/23 Q1) (9709/62/M/J/24 Q3)

Common mistake

Dropping the square root in

\sqrt{\hat p\hat q/n}

; applying a continuity correction (not used here); confusing the proportion SE with the mean SE

\sigma/\sqrt n

; and not interpreting against the relevant boundary (0.5, a pass mark, …).