6.4 - Confidence intervals for a proportion

6.4 - Confidence intervals for a proportion

A survey's sample proportion is an estimate, not the exact population proportion. In this final part of 6.4, you will calculate its estimated standard error, build a large-sample confidence interval and explain how confidence and sample size affect the precision of a reported percentage.

Estimating a population proportion

Two random surveys both report 40%40\% support, but one has 200200 responses and the other 800800. Should their uncertainty be the same? Their estimates agree, but the larger independent sample has a smaller standard error.

Call the population proportion pp. For nn independent observations with the same probability pp of the outcome being counted, the number RR of successes has distribution B(n,p)\mathrm B(n,p). The random sample proportion is P^=R/n\widehat P=R/n; after observing rr successes, its value is p^=r/n\hat p=r/n.

Because E(R)=np\mathrm E(R)=np and Var⁡(R)=np(1−p)\operatorname{Var}(R)=np(1-p), scaling by 1/n1/n gives

E(P^)=p,Var⁡(P^)=p(1−p)n.\mathrm E(\widehat P)=p,\qquad \operatorname{Var}(\widehat P)=\frac{p(1-p)}n.

Thus the sample proportion is an unbiased estimate of pp under the stated model.

Sample proportion and estimated standard error

For rr successes in nn independent observations,

p^=rn,estimated SE=p^(1−p^)n.\hat p=\frac rn,\qquad \text{estimated SE}=\sqrt{\frac{\hat p(1-\hat p)}n}.

For a large sample with a suitable normal approximation,

P^ approximately ∼N(p,p(1−p)n).\widehat P\ \text{approximately}\ \sim \mathrm N\left(p,\frac{p(1-p)}n\right).

The unknown pp in the standard error is estimated by p^\hat p.

The approximate distribution is in the formula booklet. The plug-in standard-error expression is not stated separately: obtain it by replacing the unknown proportion with its sample estimate.

The normal approximation is inherited from the binomial model. It requires a large sample and enough successes and failures for a bell-shaped approximation to be reasonable. The earlier checks are np>5np>5 and n(1−p)>5n(1-p)>5; with pp unknown, inspect the observed counts np^=rn\hat p=r and n(1−p^)=n−rn(1-\hat p)=n-r. Both should exceed 55 before using this procedure. These are practical suitability checks, not a guarantee of exact confidence coverage. A large nn alone is insufficient when almost every response is the same.

Worked example 1: estimating a proportion's standard error

Question. In 200200 independent random responses, 8080 support a proposal. Find the sample proportion and its estimated standard error.

Plan. Divide the success count by the sample size, then use the estimated binomial proportion variance.

p^=80200=0.4,SE⁡^=0.4(0.6)200=0.0346410….\hat p=\frac{80}{200}=0.4,\qquad \widehat{\operatorname{SE}}= \sqrt{\frac{0.4(0.6)}{200}} =0.0346410\ldots.

Answer. Estimated proportion 0.40.4, or 40%40\%; estimated standard error 0.03460.0346, about 3.463.46 percentage points.

Check. The observed success/failure counts are 8080 and 120120, both exceeding 55. Squaring the standard error and multiplying by 200200 recovers 0.24=p^(1−p^)0.24=\hat p(1-\hat p).

Quadruple the sample size while keeping the observed proportion at 0.40.4. The larger count supplies more independent information.

Worked example 2: the same proportion in a larger sample

Question. In 800800 independent random responses, 320320 support the proposal. Find the sample proportion and estimated standard error.

Plan. Recalculate the standard error using n=800n=800.

p^=320800=0.4,SE⁡^=0.4(0.6)800=0.0173205….\hat p=\frac{320}{800}=0.4,\qquad \widehat{\operatorname{SE}}= \sqrt{\frac{0.4(0.6)}{800}} =0.0173205\ldots.

Answer. Estimated proportion 0.40.4; estimated standard error 0.01730.0173, about 1.731.73 percentage points.

Check. The standard error is half its previous value because the sample size is four times as large. The success/failure counts 320320 and 480480 also satisfy the suitability checks.

Keep proportions as decimals throughout the formula. Convert to percentages at the end: a proportion difference 0.030.03 is 33 percentage points. Do not enter 4040 in place of 0.40.4, which would make 1−p^1-\hat p negative.

Watch out

The population proportion pp is fixed but unknown; p^\hat p changes between samples. The standard error above is estimated using p^\hat p, so its numerical value also changes between samples.

Calculate another estimate without rounding its proportion first.

Constructing the interval

A confidence interval for pp uses the same central-normal idea as an interval for a mean. Its centre is the observed proportion; its margin is a normal critical value multiplied by the proportion's estimated standard error.

Large-sample confidence interval for a proportion

For a sufficiently large independent random sample with suitable success and failure counts, an approximate central 100(1−α)%100(1-\alpha)\% confidence interval is

p^±zp^(1−p^)n,Φ(z)=1−α2.\hat p\pm z\sqrt{\frac{\hat p(1-\hat p)}n}, \qquad \Phi(z)=1-\frac\alpha2.

The interval estimates population proportion pp.

The confidence-interval formula is not in the formula booklet: construct it from the supplied sample-proportion approximation. Common normal critical values are supplied in the tables.

This is an approximation, because both the normal distribution and the substitution of p^\hat p for pp are used. If there are very few successes or failures, do not force this formula into use. An interval extending below 00 or above 11 is a warning that this approximation is unsuitable near a boundary, not evidence of an impossible population proportion. Simply cutting the endpoints off does not repair its confidence properties.

Worked example 3: a 95 percent proportion interval

Question. A random sample of 200200 independent responses contains 8080 supporters. Calculate an approximate 95%95\% confidence interval for the population proportion supporting the proposal.

Plan. Check both counts, then use z=1.960z=1.960 and the estimated standard error.

Here p^=0.4\hat p=0.4, with 8080 successes and 120120 failures. Both counts exceed 55.

0.4±1.9600.4(0.6)200=0.4±0.0678964….0.4\pm1.960\sqrt{\frac{0.4(0.6)}{200}} =0.4\pm0.0678964\ldots.

Answer. Approximately (0.332,0.468)(0.332,0.468), or 33.2%33.2\% to 46.8%46.8\%.

Check. The unrounded limits 0.3321036…0.3321036\ldots and 0.4678964…0.4678964\ldots average to 0.40.4. Their distance from the centre is 1.9601.960 times the estimated standard error.

Keep the observed proportion and confidence level fixed, but quadruple the sample size. Only the standard error changes.

Worked example 4: a narrower interval

Question. A random sample of 800800 independent responses contains 320320 supporters. Calculate an approximate 95%95\% confidence interval for the population proportion.

Plan. Use the same centre 0.40.4 and critical value, with the larger sample size.

0.4±1.9600.4(0.6)800=0.4±0.0339482….0.4\pm1.960\sqrt{\frac{0.4(0.6)}{800}} =0.4\pm0.0339482\ldots.

Answer. Approximately (0.366,0.434)(0.366,0.434), or 36.6%36.6\% to 43.4%43.4\%.

Check. The margin is half that in Example 3. The midpoint stays 0.40.4, and the success/failure counts 320,480320,480 remain suitable.

The interpretation concerns the fixed population proportion. Say that we are 95%95\% confident the population proportion lies between the limits; over repeated samples, the approximate procedure captures the true proportion about 95%95\% of the time when its assumptions and approximation are appropriate. It does not assert that 95%95\% of individuals have the characteristic.

Exam tip

State the sample proportion, substituted interval expression and both limits. If reporting percentages, convert both limits consistently and retain several significant figures until the end.

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