Confidence interval for a proportion | CAIE 9709 Probability & Statistics 2 Notes

Probability & Statistics 2

Mathematics9709

Use the observed proportion in the standard error

\hat p\pm k\sqrt{\frac{\hat p(1-\hat p)}{n}}

Symbols

$\hat p=r/n$ = sample proportion (none)
$k$ = central normal critical value (none)
$n$ = sample size (none)

This is a large-sample normal approximation, so the sample must be random and both the observed successes and failures must be sufficiently numerous.
No continuity correction is used in this confidence interval.
The result estimates the population proportion p, not the number of successes in a future sample.

Read the interval against a claim

0.02250.227p̂ = 0.125claim 0.25

0.25 is outside [0.0225, 0.227]

The interval estimates the population proportion.

An interval estimates the population proportion, not the percentage of individual observations.

Construct the interval, then compare with a claim

Worked example

Current-paper phone ownership interval

In a random sample of 40 students, 5 own a particular phone.

\hat p=5/40=0.125

0.125\pm1.96\sqrt{\frac{0.125(0.875)}{40}}

Answer

0.0225<p<0.227

The manufacturer's claim 0.25 lies outside this interval. (9709/61/M/J/25 Q6(c))

Key idea

If a claim lies inside the interval, it remains plausible. If it lies outside, it is inconsistent with this interval at the corresponding confidence level.

A given width can produce two proportions

The factor $\hat p(1-\hat p)$ is unchanged when $\hat p$ is replaced by $1-\hat p$ . Therefore a width-only question can have two symmetric answers.

Worked example

Current-paper reverse-width problem

A 90% interval from 200 spins has full width 0.1066. Let the sample proportion be a.

m=\frac{0.1066}{2(1.645)}

a(1-a)=200m^2

a^2-a+0.20997=0

Answer

a=0.300\quad\text{or}\quad a=0.700

(9709/62/M/J/25 Q4)

Your turnIndependent proportion-interval check [4 marks]

In a random sample of 250 people, 102 support a proposal. Find an approximate 90% confidence interval for the population proportion.

Show worked answer

\hat p=102/250=0.408

SE=\sqrt{\frac{0.408(0.592)}{250}}

SE=0.0311

0.408\pm1.645(0.0311)

Answer

0.357<p<0.459

Common mistake

Keep the square root over the whole variance expression. Do not use a continuity correction, and do not discard the second root in a reverse-width question unless the context rules it out.

Use the observed proportion in the standard error

\hat p\pm k\sqrt{\frac{\hat p(1-\hat p)}{n}}

Symbols

$\hat p=r/n$ = sample proportion (none)
$k$ = central normal critical value (none)
$n$ = sample size (none)

This is a large-sample normal approximation, so the sample must be random and both the observed successes and failures must be sufficiently numerous.
No continuity correction is used in this confidence interval.
The result estimates the population proportion p, not the number of successes in a future sample.

Read the interval against a claim

0.02250.227p̂ = 0.125claim 0.25

0.25 is outside [0.0225, 0.227]

The interval estimates the population proportion.

An interval estimates the population proportion, not the percentage of individual observations.

Construct the interval, then compare with a claim

Worked example

Current-paper phone ownership interval

In a random sample of 40 students, 5 own a particular phone.

\hat p=5/40=0.125

0.125\pm1.96\sqrt{\frac{0.125(0.875)}{40}}

Answer

0.0225<p<0.227

The manufacturer's claim 0.25 lies outside this interval. (9709/61/M/J/25 Q6(c))

Key idea

If a claim lies inside the interval, it remains plausible. If it lies outside, it is inconsistent with this interval at the corresponding confidence level.

A given width can produce two proportions

The factor $\hat p(1-\hat p)$ is unchanged when $\hat p$ is replaced by $1-\hat p$ . Therefore a width-only question can have two symmetric answers.

Worked example

Current-paper reverse-width problem

A 90% interval from 200 spins has full width 0.1066. Let the sample proportion be a.

m=\frac{0.1066}{2(1.645)}

a(1-a)=200m^2

a^2-a+0.20997=0

Answer

a=0.300\quad\text{or}\quad a=0.700

(9709/62/M/J/25 Q4)

Your turnIndependent proportion-interval check [4 marks]

In a random sample of 250 people, 102 support a proposal. Find an approximate 90% confidence interval for the population proportion.

Show worked answer

\hat p=102/250=0.408

SE=\sqrt{\frac{0.408(0.592)}{250}}

SE=0.0311

0.408\pm1.645(0.0311)

Answer

0.357<p<0.459

Common mistake

Keep the square root over the whole variance expression. Do not use a continuity correction, and do not discard the second root in a reverse-width question unless the context rules it out.